Classical Mechanics, Work, and Energy

Scalar and Vector Quantities are two important ways of describing physical quantities in science, especially in mechanics. To understand motion, forces, work, and energy, you must know whether a quantity needs only size or both size and direction.

A scalar quantity has magnitude only. Magnitude means the numerical size of the quantity, together with its unit. For example, if a car travels at 20 m/s, that value tells us how fast it is moving, but not the direction. So speed is a scalar.

A vector quantity has magnitude and direction. For example, if a car moves at 20 m/s east, we now know both how fast it moves and the direction of motion. This makes velocity a vector.

This difference matters because two quantities with the same magnitude can represent different physical situations if their directions are different. A force of 10 N upward is not the same as a force of 10 N downward, even though the magnitudes are equal.

Why this idea matters in mechanics

In classical mechanics, many important quantities are vectors. Motion happens in space, so direction often changes the result. If two people push a box with equal forces in opposite directions, the box may not move at all. This shows that direction must be included when combining vector quantities.

On the other hand, some quantities do not depend on direction. For example, mass, time, distance, speed, energy, and work are scalars. These quantities can be fully described by a number and a unit.

Common scalar quantities

• Mass — for example, 5 kg
• Time — for example, 12 s
• Temperature — for example, 25°C
• Distance — for example, 100 m
• Speed — for example, 15 m/s
• Work — for example, 40 J
• Energy — for example, 250 J
• Power — for example, 60 W

Common vector quantities

• Displacement — for example, 10 m north
• Velocity — for example, 8 m/s west
• Acceleration — for example, 3 m/s² downward
• Force — for example, 12 N east
• Momentum — for example, 18 kg·m/s south

How to tell whether a quantity is scalar or vector

1. Ask: Is direction needed to describe it completely?
2. If direction is not needed, it is a scalar.
3. If direction is needed, it is a vector.

For example, saying an object traveled 50 m gives only the total path length, so distance is a scalar. But saying an object moved 50 m east describes its change in position with direction, so displacement is a vector.

Distance vs displacement

Students often confuse these two quantities. Distance is the total length of the path traveled. It is a scalar. Displacement is the straight-line change in position from start to finish, including direction. It is a vector.

If a student walks 3 m east and then 3 m west, the total distance traveled is:

$$3 + 3 = 6 \text{ m}$$

But the student ends where they started, so the displacement is:

$$0 \text{ m}$$

This example shows that scalar and vector quantities can behave very differently.

Speed vs velocity

Speed tells how fast an object moves. It does not include direction, so it is a scalar. Velocity tells both speed and direction, so it is a vector.

If a bicycle moves at 10 m/s south, its speed is 10 m/s, while its velocity is 10 m/s south.

Representing vectors

Vectors are often shown using arrows. The length of the arrow represents magnitude, and the arrowhead shows direction.

In writing, a vector may be described with words such as north, south, east, west, up, or down. In diagrams, vectors may be drawn on axes. For example, a force to the right might be shown as a horizontal arrow.

Adding scalars

Scalars are added using ordinary arithmetic. For example, if one container has 2 kg of sand and another has 3 kg, the total mass is:

$$2 + 3 = 5 \text{ kg}$$

No direction is involved, so scalar addition is simple.

Adding vectors

Vectors cannot always be added by just adding their magnitudes, because direction matters. When vectors point in the same direction, you add their magnitudes. When they point in opposite directions, you subtract their magnitudes.

For example, if two forces act on a box:

• Force 1: 5 N east
• Force 2: 3 N east

Then the resultant force is:

$$5 + 3 = 8 \text{ N east}$$

But if the forces are:

• Force 1: 5 N east
• Force 2: 3 N west

Then the resultant force is:

$$5 - 3 = 2 \text{ N east}$$

The direction of the larger magnitude is kept.

Vector addition using components on a straight line

When vectors lie along the same straight line, it is helpful to choose one direction as positive and the opposite direction as negative.

For example, let east be positive. Then:

• 6 m east becomes \(+6\) m

• 4 m west becomes \(-4\) m

The total is:

$$+6 + (-4) = +2 \text{ m}$$

So the resultant displacement is 2 m east.

Vectors at right angles

Sometimes vectors point in different directions, such as one east and one north. In that case, they form a right angle. To find the magnitude of the resultant vector, we use the Pythagorean theorem.

If one displacement is 3 m east and another is 4 m north, the magnitude of the resultant displacement is:

$$R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m}$$

The final direction is between east and north. At this level, it is enough to say the object moved 5 m northeast if an exact angle is not required.

Worked Example 1: Classifying quantities

Question: Classify each of the following as scalar or vector:

• 12 kg
• 7 m/s north
• 50 J
• 9.8 m/s² downward

Solution:

• 12 kg is scalar because mass has magnitude only.
• 7 m/s north is vector because velocity includes direction.
• 50 J is scalar because energy has magnitude only.
• 9.8 m/s² downward is vector because acceleration includes direction.

Answer: scalar, vector, scalar, vector.

Worked Example 2: Distance and displacement

Question: A student walks 5 m east and then 2 m west. Find the distance and displacement.

Step 1: Find distance.

Distance is total path length:

$$5 + 2 = 7 \text{ m}$$

Step 2: Find displacement.

Take east as positive:

$$+5 + (-2) = +3 \text{ m}$$

So the displacement is 3 m east.

Answer: Distance = 7 m, Displacement = 3 m east.

Worked Example 3: Adding forces in opposite directions

Question: Two students push a cart. One pushes with 15 N east. The other pushes with 9 N west. What is the resultant force?

Solution:

The forces are in opposite directions, so subtract magnitudes:

$$15 - 9 = 6 \text{ N}$$

The larger force is east, so the resultant force is 6 N east.

Answer: 6 N east.

Worked Example 4: Perpendicular vectors

Question: A robot moves 6 m east and then 8 m north. What is the magnitude of its resultant displacement?

Solution:

The two displacements are at right angles, so use the Pythagorean theorem:

$$R = \sqrt{6^2 + 8^2}$$

$$R = \sqrt{36 + 64}$$

$$R = \sqrt{100} = 10 \text{ m}$$

The direction is toward the northeast.

Answer: 10 m northeast in magnitude and general direction.

Important mistakes to avoid

• Do not confuse speed and velocity. Speed is scalar; velocity is vector.
• Do not confuse distance and displacement. Distance is total path length; displacement is overall change in position with direction.
• Do not ignore direction when adding vectors. Opposite directions reduce the result.
• Do not treat every quantity in mechanics as a vector. For example, work and energy are scalars.

Quick comparison table

• Scalar: magnitude only
• Vector: magnitude and direction
• Scalar examples: mass, time, distance, speed, energy, work
• Vector examples: displacement, velocity, acceleration, force, momentum

Brief summary

Scalar quantities describe how much of something there is. Vector quantities describe how much and in what direction. This difference is essential in mechanics because motion and force often depend on direction.

When working with vectors, always pay attention to direction before adding or comparing values. If you can correctly tell whether a quantity is scalar or vector, and you can combine simple vectors, you have an important foundation for studying motion, forces, work, and energy.

One-Dimensional Kinematics is the study of motion along a straight line. In this topic, we describe how an object moves without worrying about the forces causing the motion. That means we focus on quantities such as position, displacement, velocity, speed, and acceleration.

This lesson is important because it builds the foundation for later topics in mechanics, including forces, energy, and momentum. If you can describe motion clearly in one dimension, you will be much more prepared for the rest of physics.

In one-dimensional motion, an object moves along a single axis, usually labeled the x-axis. We choose one direction to be positive and the opposite direction to be negative. For example, motion to the right may be positive, while motion to the left is negative.

1. Position and Reference Point

Position tells where an object is located relative to a chosen reference point. We usually use the symbol \(x\) for position.

For example, if a car is 30 meters to the right of a signpost, we could say its position is \(x = +30\text{ m}\). If it is 10 meters to the left, then its position could be \(x = -10\text{ m}\).

The sign of position matters because it tells direction relative to the origin. The origin is simply the zero point on the axis.

2. Distance vs. Displacement

These two ideas are related, but they are not the same.

• Distance is the total length of the path traveled. It does not include direction.
• Displacement is the change in position. It includes direction.

Displacement is calculated by:

Here, \(x_i\) is the initial position and \(x_f\) is the final position.

If a student walks from \(x = 2\text{ m}\) to \(x = 9\text{ m}\), the displacement is:

If that same student then walks back to \(x = 5\text{ m}\), the total distance traveled is \(7 + 4 = 11\text{ m}\), but the overall displacement is:

This shows that distance depends on the whole path, while displacement depends only on starting and ending positions.

3. Time and Motion

Motion describes how position changes over time. Time is usually represented by \(t\) and measured in seconds \((\text{s})\).

To describe motion, we often ask:

• Where is the object?
• How fast is it moving?
• Is it speeding up, slowing down, or staying at the same speed?

4. Speed and Velocity

Speed tells how fast an object moves. It does not include direction.

Average speed is:

Velocity tells both how fast and in what direction an object moves. Because it includes direction, velocity can be positive or negative.

Average velocity is:

where \(\Delta x\) is displacement and \(\Delta t\) is the time interval.

If an object has a positive velocity, it is moving in the positive direction. If it has a negative velocity, it is moving in the negative direction.

It is very important not to confuse speed and velocity:

• Speed uses distance.
• Velocity uses displacement.

5. Acceleration

Acceleration describes how velocity changes over time. It does not only mean “speeding up.” An object also accelerates when it slows down or changes direction.

Average acceleration is given by:

where:

• \(v_i\) = initial velocity
• \(v_f\) = final velocity
• \(\Delta t\) = time interval

The unit of acceleration is meters per second squared, written as \(\text{m/s}^2\).

A positive acceleration means velocity is changing in the positive direction. A negative acceleration means velocity is changing in the negative direction.

Be careful: the sign of acceleration does not automatically tell you whether an object is speeding up or slowing down. You must compare the direction of velocity and acceleration.

• If velocity and acceleration have the same sign, the object speeds up.
• If velocity and acceleration have opposite signs, the object slows down.

6. Constant Acceleration Equations

Many one-dimensional kinematics problems involve constant acceleration. This means the acceleration stays the same during the motion.

When acceleration is constant, we can use the following equations:

These equations are sometimes called the kinematic equations. They work only when acceleration is constant.

When solving problems, first identify what values you know and what quantity you need to find. Then choose the equation that connects those variables.

7. Common Variables in Kinematics

• \(x_i\): initial position
• \(x_f\): final position
• \(\Delta x\): displacement
• \(v_i\): initial velocity
• \(v_f\): final velocity
• \(a\): acceleration
• \(t\): time

Keeping track of signs is one of the most important skills in one-dimensional motion. Decide on a positive direction at the start and use it consistently.

8. Worked Example 1: Finding Displacement and Average Velocity

A runner moves from \(x_i = 4\text{ m}\) to \(x_f = 19\text{ m}\) in \(3\text{ s}\). Find the displacement and average velocity.

Step 1: Find displacement.

Step 2: Find average velocity.

Answer: The displacement is \(15\text{ m}\), and the average velocity is \(5\text{ m/s}\) in the positive direction.

9. Worked Example 2: Using Acceleration

A bicycle increases its velocity from \(2\text{ m/s}\) to \(8\text{ m/s}\) in \(3\text{ s}\). Find its average acceleration.

Use the acceleration formula:

Substitute the values:

Answer: The average acceleration is \(2\text{ m/s}^2\).

10. Worked Example 3: Motion with Constant Acceleration

A car starts from rest and accelerates at \(4\text{ m/s}^2\) for \(5\text{ s}\). Find:

1. its final velocity
2. its displacement

Given:

• \(v_i = 0\text{ m/s}\)
• \(a = 4\text{ m/s}^2\)
• \(t = 5\text{ s}\)

Part 1: Final velocity

Part 2: Displacement

Answer: The final velocity is \(20\text{ m/s}\), and the displacement is \(50\text{ m}\).

11. Worked Example 4: Negative Velocity and Negative Acceleration

A ball moving to the left has an initial velocity of \(-6\text{ m/s}\). It continues accelerating to the left at \(-2\text{ m/s}^2\) for \(4\text{ s}\). Find its final velocity.

Use:

Substitute:

Answer: The final velocity is \(-14\text{ m/s}\). Because both velocity and acceleration are negative, the ball is speeding up in the negative direction.

12. Interpreting Motion Graphs

Graphs are a major part of kinematics. In one-dimensional motion, the most common graphs are:

• position-time graphs
• velocity-time graphs
• acceleration-time graphs

Position-Time Graphs

A position-time graph shows how position changes over time.

• The slope of a position-time graph represents velocity.
• A positive slope means positive velocity.
• A negative slope means negative velocity.
• A zero slope means the object is at rest.
• A steeper slope means a greater speed.

If the graph is a straight line, the velocity is constant. If the graph curves, the velocity is changing.

Velocity-Time Graphs

A velocity-time graph shows how velocity changes over time.

• The slope of a velocity-time graph represents acceleration.
• A horizontal line means constant velocity and zero acceleration.
• An upward slope means positive acceleration.
• A downward slope means negative acceleration.

The area under a velocity-time graph gives displacement.

For example, if velocity is constant at \(3\text{ m/s}\) for \(4\text{ s}\), then displacement is the area of a rectangle:

Acceleration-Time Graphs

An acceleration-time graph shows how acceleration changes over time.

• A horizontal line means constant acceleration.
• If the line is above the time axis, acceleration is positive.
• If the line is below the time axis, acceleration is negative.

The area under an acceleration-time graph gives the change in velocity.

13. How to Tell What Motion Is Happening

You can often describe motion by looking at the signs of velocity and acceleration.

• Positive velocity, positive acceleration: moving in the positive direction and speeding up.
• Positive velocity, negative acceleration: moving in the positive direction and slowing down.
• Negative velocity, negative acceleration: moving in the negative direction and speeding up.
• Negative velocity, positive acceleration: moving in the negative direction and slowing down.

14. Free Fall as One-Dimensional Motion

Vertical motion near Earth is a common example of one-dimensional kinematics. If air resistance is ignored, the acceleration is due to gravity alone.

The acceleration due to gravity is approximately:

If upward is chosen as positive, then acceleration due to gravity is:

If downward is chosen as positive, then acceleration due to gravity is:

What matters most is being consistent with your sign convention.

15. Common Mistakes to Avoid

• Confusing distance with displacement.
• Confusing speed with velocity.
• Ignoring negative signs.
• Using a constant-acceleration equation when acceleration is not constant.
• Forgetting that an object can have zero velocity at an instant and still have nonzero acceleration.

16. Problem-Solving Tips

1. Draw a simple line diagram if helpful.
2. Choose a positive direction.
3. List known values with correct signs.
4. Identify the unknown quantity.
5. Select the equation that matches the knowns and unknown.
6. Check whether the answer makes sense physically.

17. Brief Summary

One-dimensional kinematics describes motion along a straight line using position, displacement, velocity, and acceleration. Displacement tells the change in position, velocity tells the rate of change of position, and acceleration tells the rate of change of velocity.

When acceleration is constant, the kinematic equations let us solve for unknown quantities such as final velocity, displacement, or time. Motion graphs also help us understand movement: slope and area on these graphs have clear physical meanings.

With careful attention to direction, signs, and units, you can solve a wide variety of motion problems in one dimension.

Two-Dimensional Kinematics and Projectiles

In many real-world situations, objects do not move in just one straight line. A soccer ball kicked through the air, a basketball shot toward a hoop, or water sprayed from a hose all move in two dimensions. This means their motion has both a horizontal part and a vertical part.

The study of this kind of motion is called two-dimensional kinematics. When an object moves through the air under the influence of gravity alone, we call it a projectile, and its motion is called projectile motion.

The key idea of projectile motion is very important: horizontal and vertical motions are independent of each other. Gravity affects the vertical motion, but it does not directly affect the horizontal motion if air resistance is ignored.

This lesson will show you how to break motion into horizontal and vertical parts, use equations to describe each part, and solve common projectile problems.

1. What is two-dimensional motion?

In one-dimensional motion, an object moves along a straight line, such as up and down or left and right. In two-dimensional motion, the object moves in a plane, so its position changes in both the horizontal direction and the vertical direction.

To analyze this motion, we treat it as two separate one-dimensional motions happening at the same time:

• Horizontal motion along the x-axis
• Vertical motion along the y-axis

This makes complicated motion much easier to understand.

2. Projectile motion assumptions

In basic 11th Grade physics, we usually make these assumptions for projectile motion:

• Air resistance is ignored.
• The only force acting after launch is gravity.
• Gravity acts downward with constant acceleration.

Near Earth’s surface, the acceleration due to gravity is

When we choose upward as positive, the vertical acceleration is

Since no horizontal force acts on the projectile, the horizontal acceleration is

3. The shape of a projectile’s path

A projectile usually follows a curved path called a parabola. This happens because:

• Horizontally, the object moves at constant velocity.
• Vertically, the object speeds up downward because of gravity.

Even though the path is curved, you can solve the motion by looking at the horizontal and vertical directions separately.

4. Breaking velocity into components

If a projectile is launched at an angle, its initial velocity must be split into horizontal and vertical components.

If the initial speed is \(v_0\) and the launch angle is \(\theta\), then:

Here:

• \(v_{0x}\) is the initial horizontal velocity
• \(v_{0y}\) is the initial vertical velocity

This step is one of the most important parts of solving projectile problems. Once the velocity is separated into components, each direction can be handled with the correct equations.

5. Horizontal motion equations

Because horizontal acceleration is zero, horizontal velocity stays constant:

The horizontal displacement is:

This means that if you know the horizontal velocity and the time, you can find how far the projectile travels horizontally.

6. Vertical motion equations

The vertical motion is affected by gravity, so it behaves like any other motion with constant acceleration.

The main vertical equations are:

If upward is positive, then \(a_y = -9.8\ \text{m/s}^2\).

7. Time connects both directions

The horizontal and vertical motions are independent, but they share one common quantity: time. The time the projectile spends in the air is the same for both directions.

This idea lets us connect horizontal and vertical information. Often, we find the time from the vertical motion first, then use that time in the horizontal equation.

8. Important projectile quantities

There are several common quantities you may be asked to find.

• Time of flight: total time the projectile is in the air
• Maximum height: greatest vertical position reached
• Range: total horizontal distance traveled

For a projectile launched and landing at the same height:

At the highest point, the vertical velocity becomes zero:

So the time to reach maximum height is:

The total time of flight is twice this value:

The maximum height is:

The horizontal range is:

These formulas are very useful, but they only work directly when the projectile starts and lands at the same vertical level.

9. Step-by-step strategy for solving projectile problems

1. Draw a simple sketch of the situation.
2. Choose positive directions, usually right and up.
3. Break the initial velocity into horizontal and vertical components.
4. List what is known and what must be found.
5. Use vertical motion equations to find time or height.
6. Use the same time in the horizontal motion equation.
7. Check whether your answer is reasonable.

10. Worked Example 1: Horizontal launch from a table

A ball rolls off a table with a horizontal speed of \(4.0\ \text{m/s}\). The table is \(1.2\ \text{m}\) high. How long is the ball in the air, and how far from the table does it land?

Step 1: Identify the motion

The ball is launched horizontally, so:

• \(v_{0x} = 4.0\ \text{m/s}\)
• \(v_{0y} = 0\)
• \(y = -1.2\ \text{m}\)
• \(a_y = -9.8\ \text{m/s}^2\)

Step 2: Find the time using vertical motion

Use

Substitute the values:

Step 3: Find the horizontal distance

Answer: The ball is in the air for about \(0.49\ \text{s}\) and lands about \(2.0\ \text{m}\) from the table.

11. Worked Example 2: Launch at an angle

A ball is kicked with a speed of \(20\ \text{m/s}\) at an angle of \(30^\circ\) above the horizontal. Find its initial velocity components.

Step 1: Use trigonometry

Using \(\cos 30^\circ \approx 0.866\) and \(\sin 30^\circ = 0.5\):

Answer:

• Initial horizontal velocity: \(17.3\ \text{m/s}\)
• Initial vertical velocity: \(10.0\ \text{m/s}\)

This example shows the first step in almost every angled projectile problem.

12. Worked Example 3: Time of flight, maximum height, and range

A projectile is launched at \(20\ \text{m/s}\) at an angle of \(30^\circ\) and lands at the same height from which it was launched. Find:

• the time of flight
• the maximum height
• the range

From the previous example:

Step 1: Find the time to reach maximum height

At the top, \(v_y = 0\), so:

Step 2: Find the total time of flight

Step 3: Find the maximum height

Use

Step 4: Find the range

Answer:

• Time of flight: \(2.04\ \text{s}\)
• Maximum height: \(5.1\ \text{m}\)
• Range: \(35.3\ \text{m}\)

13. Worked Example 4: Projectile launched from a height

A rock is thrown horizontally from a cliff \(45\ \text{m}\) high with a speed of \(12\ \text{m/s}\). How long does it take to hit the ground, and how far from the base of the cliff does it land?

Step 1: List known values

• \(v_{0x} = 12\ \text{m/s}\)
• \(v_{0y} = 0\)
• \(y = -45\ \text{m}\)
• \(a_y = -9.8\ \text{m/s}^2\)

Step 2: Solve for time from vertical motion

Step 3: Solve for horizontal distance

Answer: The rock hits the ground after about \(3.03\ \text{s}\) and lands about \(36.4\ \text{m}\) from the base of the cliff.

14. Common mistakes to avoid

• Mixing horizontal and vertical quantities. Use x-equations for horizontal motion and y-equations for vertical motion.
• Forgetting that gravity only affects vertical motion. Horizontal acceleration is zero when air resistance is ignored.
• Using the total initial velocity where a component is needed. For angled launches, always find \(v_{0x}\) and \(v_{0y}\) first.
• Using the same-height formulas when launch and landing heights are different. In that case, use the general kinematics equations.
• Sign errors. Be consistent with your choice of positive and negative directions.

15. How projectile motion connects to energy and mechanics

Projectile motion is part of classical mechanics because it combines motion and forces. Gravity is the force that changes the projectile’s vertical velocity.

This topic also connects to energy. As a projectile rises, some kinetic energy changes into gravitational potential energy. As it falls, gravitational potential energy changes back into kinetic energy. Even so, to predict the path, we usually use kinematics equations and vector components.

16. Key ideas to remember

• Projectile motion is two-dimensional motion under gravity.
• Horizontal and vertical motions are independent.
• Horizontal velocity stays constant if air resistance is ignored.
• Vertical motion has constant downward acceleration \((-9.8\ \text{m/s}^2)\).
• For angled launches, split the initial velocity into components using sine and cosine.
• Time is the link between horizontal and vertical motion.

Brief Summary

Two-dimensional kinematics helps us understand motion that happens both horizontally and vertically. In projectile motion, the horizontal part has constant velocity, while the vertical part changes because of gravity. By breaking velocity into components and solving each direction separately, you can find time in the air, maximum height, and horizontal range.

Galilean Relativity and Reference Frames

In mechanics, we often describe motion by answering questions like: How fast is the object moving? In what direction? How far did it travel? The important idea is that the answer can depend on who is observing the motion. This is where reference frames and Galilean relativity come in.

A reference frame is the point of view from which motion is measured. It includes a choice of position and time measurements. For example, a student standing on the sidewalk and a passenger sitting in a moving bus are using different reference frames.

Galilean relativity says that the laws of mechanics are the same in all inertial reference frames. An inertial reference frame is a frame that is either at rest or moving at a constant velocity in a straight line. In such frames, objects obey the usual laws of motion in the same way.

This means there is no special “best” inertial frame. If one person is in a train moving steadily and another person is standing on the ground, both can use Newtonian ideas to describe motion correctly. Their measurements of position and velocity may be different, but the basic laws remain the same.

To understand this topic well, we need to study three main ideas:

• Motion is relative
• Different observers can measure different velocities and displacements
• Galilean transformations connect measurements in different inertial frames

1. Motion is relative

Imagine you are sitting in a car moving smoothly along a straight road. A water bottle beside you appears to be at rest relative to you. But to a person standing on the sidewalk, the bottle is moving at the same speed as the car. So is the bottle moving or not?

The answer is: it depends on the reference frame. Relative to the passenger, the bottle has velocity 0. Relative to the ground, the bottle has the velocity of the car.

This shows that velocity is relative. You must always say relative to what? when describing motion.

2. Position and displacement depend on the reference frame

The position of an object tells where it is compared with an origin. If different observers choose different origins or are moving relative to each other, they may assign different positions to the same object.

Displacement is the change in position:

$$\Delta x = x_f - x_i$$

If two observers are in different reference frames, they may measure different values of position at a given time. However, if the frames move with constant velocity relative to each other, their measurements are related in a simple way.

3. Velocity in different reference frames

Suppose frame S is the ground frame, and frame S' is moving to the right with constant velocity \(v\) relative to S. If an object moves with velocity \(u'\) in S', then its velocity \(u\) in S is:

$$u = u' + v$$

This is called the Galilean velocity transformation.

If you know the velocity in the ground frame and want the velocity in the moving frame, you rearrange the equation:

$$u' = u - v$$

This is often described as relative velocity.

For example, if a person walks forward inside a train, their speed relative to the ground is the speed of the person relative to the train plus the speed of the train relative to the ground, as long as both are moving in the same direction.

4. Galilean transformation for position

If frame S' moves at constant velocity \(v\) relative to frame S, and the origins line up at time \(t = 0\), then the positions are related by:

$$x' = x - vt$$

Or equivalently:

$$x = x' + vt$$

These equations help us switch from one inertial frame to another.

In Galilean relativity, time is treated as the same for all observers:

$$t' = t$$

This matches everyday experience in classical mechanics.

5. Why inertial frames matter

Galilean relativity works for inertial frames, meaning frames that are not accelerating. If you are in a car moving at constant velocity, the car is approximately an inertial frame. But if the car suddenly speeds up, slows down, or turns, you feel pushed. That frame is no longer inertial.

In a non-inertial frame, motion can seem strange unless extra effects are considered. For this lesson, we focus on inertial frames only.

6. Everyday examples of reference frames

• Walking on a moving walkway: Your speed relative to the ground is different from your speed relative to the walkway.
• Throwing a ball on a bus: To a passenger, the ball may move straight up and down. To a person outside, the ball moves forward while going up and down.
• Boats in a river: The boat’s velocity relative to the water differs from its velocity relative to the shore.

These situations all show that motion is described differently in different frames, even though the physical event is the same.

Worked Example 1: Passenger walking in a train

A train moves to the right at \(20\,\text{m/s}\) relative to the ground. A passenger walks to the right inside the train at \(2\,\text{m/s}\) relative to the train. What is the passenger’s velocity relative to the ground?

Step 1: Identify the moving frame and the object’s velocity in that frame.

The train is moving at \(v = 20\,\text{m/s}\). The passenger’s velocity relative to the train is \(u' = 2\,\text{m/s}\).

Step 2: Use the Galilean velocity transformation.

$$u = u' + v$$

$$u = 2 + 20 = 22\,\text{m/s}$$

Answer: The passenger’s velocity relative to the ground is \(22\,\text{m/s}\) to the right.

Worked Example 2: Walking opposite the train’s motion

A train moves to the right at \(15\,\text{m/s}\) relative to the ground. A passenger walks to the left at \(3\,\text{m/s}\) relative to the train. Find the passenger’s velocity relative to the ground.

Step 1: Choose a positive direction.

Let right be positive. Then the train’s velocity is \(v = +15\,\text{m/s}\), and the passenger’s velocity relative to the train is \(u' = -3\,\text{m/s}\).

Step 2: Apply the formula.

$$u = u' + v$$

$$u = -3 + 15 = 12\,\text{m/s}$$

Answer: The passenger’s velocity relative to the ground is \(12\,\text{m/s}\) to the right.

Even though the passenger is walking left inside the train, the train is moving fast enough to carry the passenger to the right overall.

Worked Example 3: Ball thrown upward on a moving bus

A bus moves at a constant velocity of \(10\,\text{m/s}\) to the right. A student inside the bus throws a ball straight upward. Describe the motion of the ball as seen by:

1. the student on the bus
2. a person standing on the ground

Solution:

To the student on the bus, the ball goes straight up and straight down, because the student and the ball both already share the bus’s forward motion.

To the person on the ground, the ball has:

• an upward and downward motion
• a forward horizontal motion of \(10\,\text{m/s}\)

So the person on the ground sees the ball follow a curved path.

This example shows that different observers can describe the same event in different ways, but both descriptions are correct in their own reference frames.

Worked Example 4: Using the position transformation

A cart moves to the right at \(5\,\text{m/s}\) relative to the ground. At time \(t = 4\,\text{s}\), a box is at position \(x = 30\,\text{m}\) in the ground frame. What is the box’s position in the cart’s frame, assuming both origins matched at \(t=0\)?

Step 1: Use the Galilean position transformation.

$$x' = x - vt$$

Step 2: Substitute the values.

$$x' = 30 - (5)(4)$$

$$x' = 30 - 20 = 10\,\text{m}$$

Answer: In the cart’s frame, the box is at \(10\,\text{m}\).

7. Key ideas to remember about relative velocity

• If two motions are in the same direction, add the velocities.
• If they are in opposite directions, use signs carefully and subtract as needed.
• Always state which frame the velocity is measured in.
• A velocity by itself is incomplete unless you say relative to what.

8. What Galilean relativity tells us about mechanics

If you perform a mechanics experiment in a smoothly moving train, the results are the same as if you performed it in a room at rest. For example, if you drop a ball, it falls in the same way relative to you. This is because the laws of mechanics do not change from one inertial frame to another.

This idea was very important in the development of physics. It showed that constant motion cannot be detected just by doing ordinary mechanics experiments inside a closed system. Only changes in motion, like acceleration, can be directly felt.

9. Common mistakes

• Forgetting the reference frame: Saying “the object moves at \(10\,\text{m/s}\)” without saying relative to what can be misleading.
• Ignoring direction: Velocity includes direction, so signs matter.
• Confusing speed and velocity: Speed is just how fast; velocity includes direction.
• Using Galilean relativity in accelerating frames: The simple formulas here are for inertial frames only.

10. Final summary

Galilean relativity explains how motion is described from different inertial reference frames. Position, displacement, and velocity can have different values for different observers, but the laws of mechanics remain the same in all inertial frames.

The most important equations are:

$$u = u' + v$$

$$x' = x - vt$$

$$t' = t$$

These equations show how to connect measurements made by observers moving at constant velocity relative to one another. Once you understand reference frames, many motion problems become much clearer.

Newton's First Law and Inertia are foundational ideas in classical mechanics. They explain why objects do not change their motion unless something causes that change.

In everyday life, it may seem like moving objects naturally slow down and stop on their own. For example, a rolling ball eventually stops, and a sliding book comes to rest. However, this does not happen because motion “runs out.” It happens because forces such as friction and air resistance act on the object.

Newton's First Law states:

An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction, unless acted on by a net external force.

This law is sometimes called the law of inertia because inertia is the property that resists changes in motion.

To understand this law, we need to be clear about what motion means. In physics, motion includes both speed and direction. If either speed changes, direction changes, or both change, then the object's motion has changed.

That means the following are all changes in motion:

• starting to move
• stopping
• speeding up
• slowing down
• turning

According to Newton's First Law, any of these changes require a net external force.

Net force means the overall force on an object after adding all the forces together. Forces can cancel each other out or combine.

If the net force is zero, then the object's motion does not change. This means:

• if it is at rest, it stays at rest
• if it is moving, it keeps moving in a straight line at constant speed

If the net force is not zero, then the object's motion changes.

This idea connects directly to Newton's Second Law, which says

$$F_{\text{net}} = ma$$

If $$F_{\text{net}} = 0$$, then

$$a = 0$$

Zero acceleration means the velocity stays constant. Constant velocity can mean either rest or motion in a straight line at steady speed.

Inertia is the tendency of an object to resist changes in its state of motion. Every object has inertia.

An object with a lot of inertia is harder to start moving, harder to stop, and harder to change direction. An object with less inertia changes motion more easily.

The amount of inertia depends on the object's mass. More mass means more inertia.

This is why:

• a soccer ball is easier to kick than a bowling ball
• a shopping cart is easier to push when empty than when full
• a small car is usually easier to push by hand than a truck

Mass is a measure of how much matter is in an object, but in this topic it is especially important because it measures inertial mass, or how strongly an object resists acceleration.

So, when we say inertia is “driven by inertial mass,” we mean that the greater the mass, the greater the resistance to changing motion.

A common misunderstanding is that a force is needed to keep something moving. That is not correct. A force is needed to change motion, not to maintain constant motion.

For example, imagine a hockey puck gliding across smooth ice. If friction is very small, the puck can travel a long distance while keeping nearly the same velocity. On an ideal frictionless surface, it would continue moving indefinitely unless another force acted on it.

In real life, friction and air resistance usually make objects slow down. These forces can hide Newton's First Law because they are often present even when we do not notice them.

Balanced and unbalanced forces are very important here.

• Balanced forces: the net force is zero
• Unbalanced forces: the net force is not zero

Balanced forces do not change motion. Unbalanced forces do.

Consider a book resting on a table. Two main forces act on it:

• gravity pulls downward
• the table pushes upward with a normal force

These forces are equal in size and opposite in direction, so the net force is zero. The book stays at rest.

Now consider a car moving at constant speed on a straight road. The engine provides a forward force, while friction and air resistance act backward. If these forces balance, the net force is zero, so the car continues at constant velocity.

If the driver presses the gas pedal harder and the forward force becomes greater than the backward forces, the net force is forward. The car speeds up.

If the driver brakes, the net force is backward. The car slows down.

Seatbelt situations are classic examples of inertia.

When a car stops suddenly, the passengers' bodies tend to keep moving forward because of inertia. The seatbelt provides the external force that changes their motion and brings them to rest with the car.

When a car starts moving suddenly, passengers may seem to fall backward. In reality, their bodies tend to stay at rest while the car moves forward underneath them.

This is why seatbelts are so important: they apply the force needed to safely change the motion of your body.

Worked Example 1: Book on a table

A book is lying still on a table. Is Newton's First Law satisfied? What is the net force?

Step 1: Identify the motion. The book is at rest.

Step 2: Ask whether its motion is changing. It is not changing.

Step 3: Apply Newton's First Law. If motion is not changing, the net external force must be zero.

Step 4: Identify forces. Gravity pulls down, and the table pushes up.

Since these forces balance,

$$F_{\text{net}} = 0$$

Answer: Yes, Newton's First Law is satisfied. The book remains at rest because the net force is zero.

Worked Example 2: Shopping cart and inertia

An empty shopping cart and a full shopping cart are both at rest. You push each one with the same force. Which one changes motion more easily, and why?

Step 1: Compare the masses. The full cart has more mass than the empty cart.

Step 2: Connect mass to inertia. More mass means more inertia.

Step 3: Decide which object resists motion change more. The full cart resists more because it has greater inertia.

Answer: The empty cart changes motion more easily. The full cart has greater inertial mass, so it is harder to accelerate.

Worked Example 3: Constant speed car

A car moves in a straight line at a constant speed of \(20\,\text{m/s}\). The forward force from the engine is \(500\,\text{N}\), and friction plus air resistance total \(500\,\text{N}\) backward. What is the net force, and what happens to the car's motion?

Step 1: Add the forces with direction.

Forward force: \(+500\,\text{N}\)

Backward force: \(-500\,\text{N}\)

$$F_{\text{net}} = 500 + (-500) = 0\,\text{N}$$

Step 2: Use Newton's First Law. With zero net force, there is no change in motion.

Answer: The net force is \(0\,\text{N}\). The car continues moving at constant speed in the same direction.

Worked Example 4: Sudden stop and passenger motion

A bus moving forward suddenly stops. A standing passenger lurches forward. Why does this happen?

Step 1: Identify what changed motion first. The bus changed motion first because the brakes applied a force to the bus.

Step 2: Think about the passenger's body. The passenger's body was moving forward with the bus.

Step 3: Apply inertia. The passenger's body tends to keep its forward motion.

Step 4: Explain the result. Until another force acts on the passenger, the body keeps moving forward, so the passenger lurches forward.

Answer: The passenger moves forward because of inertia. The body resists the sudden change in motion when the bus stops.

Key ideas to remember

• Newton's First Law describes what happens when the net external force is zero.
• An object keeps its current state of motion unless a net external force changes it.
• Inertia is the tendency to resist changes in motion.
• Mass measures inertia: more mass means more inertia.
• Balanced forces do not change motion; unbalanced forces do.
• Constant motion does not require a net force.

Quick check questions

1. If the net force on an object is zero, can it still be moving? Yes. It can move at constant speed in a straight line.
2. Why is it harder to push a truck than a bicycle? The truck has more mass and therefore more inertia.
3. Why does a ball rolling on grass stop? Because friction and air resistance create a net force opposite the motion.
4. When a car turns, is its motion changing? Yes. A change in direction is a change in velocity.

Brief Summary

Newton's First Law says that objects resist changes in motion. If the net external force is zero, an object stays at rest or keeps moving with constant velocity. This resistance to change is called inertia, and the amount of inertia depends on mass. Understanding this law helps explain motion in everyday life, from seatbelts to moving carts to cars traveling at constant speed.

Net Force and Equilibrium

Have you ever played tug-of-war, pushed a toy car, or watched a book sit still on a table? In all of these situations, forces are at work.

A force is a push or a pull. Forces can make objects start moving, stop moving, speed up, slow down, or change direction.

Sometimes more than one force acts on an object at the same time. To understand what the object will do, we look at the net force.

Net force means the total force acting on an object after we combine all the pushes and pulls.

If forces work in the same direction, we add them. If forces work in opposite directions, we subtract them.

We can write that idea like this:

Same direction: \(3 + 2 = 5\)

Opposite directions: \(7 - 4 = 3\)

When the net force is 0, the forces are balanced. When the net force is not 0, the forces are unbalanced.

Balanced forces and unbalanced forces help us understand equilibrium.

Equilibrium means the forces are balanced so the net force is zero.

• Static equilibrium: the object is not moving, and the net force is 0.
• Dynamic equilibrium: the object is moving at a steady speed in a straight line, and the net force is 0.
• If the net force is not 0, the object is accelerating. That means its motion is changing. It might speed up, slow down, or turn.

For 4th Grade, you can think of accelerating as changing motion.

To help us show forces, we use a simple picture called a free-body diagram.

A free-body diagram is a drawing that shows an object and the forces acting on it with arrows.

Each arrow shows:

• the direction of the force
• the size of the force, often written with a number

For example, if two students pull a box in opposite directions, we can draw one arrow pointing left and one arrow pointing right.

Main Idea 1: Forces can act in opposite directions

Imagine a box on the floor. One person pushes it to the right with 6 units of force. Another person pushes it to the left with 2 units of force.

The forces are opposite, so we subtract:

$$6 - 2 = 4$$

The net force is 4 units to the right. The box will change its motion to the right.

Main Idea 2: Balanced forces make net force 0

Now imagine two people pushing the same box. One pushes right with 5 units. The other pushes left with 5 units.

We subtract because the forces are opposite:

$$5 - 5 = 0$$

The net force is 0. The forces are balanced.

If the box is standing still, it stays still. That is static equilibrium.

If the box is already moving straight ahead at a steady speed, it keeps moving that way. That is dynamic equilibrium.

Main Idea 3: Forces also act up and down

Forces do not only act left and right. They can also act up and down.

Think about a book resting on a table. Gravity pulls the book down. The table pushes up on the book.

If those forces are the same size, the book does not move up or down.

That means the up-and-down forces are balanced. The net force is 0, so the book is in static equilibrium.

Main Idea 4: Look at each direction carefully

When solving a net force problem, it helps to follow these steps:

1. Find all the forces acting on the object.
2. Draw arrows to show the forces.
3. Check which forces are in the same direction and which are opposite.
4. Add or subtract to find the net force.
5. Decide if the object is in equilibrium or accelerating.

You can ask yourself:

• Is the net force 0?
• If yes, is the object still or moving steadily?
• If no, which way is the stronger force?

Worked Example 1: A simple push

A toy car is pushed to the right with 4 units of force. There is no force pushing it left.

Free-body diagram idea:

• Right arrow: 4

Since there is only one force, the net force is 4 units to the right.

$$4 = 4$$

The forces are unbalanced. The toy car will change motion to the right.

Answer: Net force = 4 units right. The car is accelerating.

Worked Example 2: Tug-of-war

One team pulls a rope left with 8 units of force. The other team pulls right with 8 units of force.

Free-body diagram idea:

• Left arrow: 8
• Right arrow: 8

The forces are opposite, so we subtract:

$$8 - 8 = 0$$

The net force is 0. The forces are balanced.

If the rope is not moving, it is in static equilibrium.

Answer: Net force = 0. The rope is in static equilibrium if it stays still.

Worked Example 3: A stronger force wins

A wagon is pulled to the right with 10 units of force. Friction pushes to the left with 3 units of force.

Free-body diagram idea:

• Right arrow: 10
• Left arrow: 3

Subtract the opposite forces:

$$10 - 3 = 7$$

The net force is 7 units to the right.

Because the net force is not 0, the wagon is not in equilibrium.

Answer: Net force = 7 units right. The wagon is accelerating to the right.

Worked Example 4: Moving steadily

A scooter is moving straight forward. A push forward is 6 units. A force backward is also 6 units.

Free-body diagram idea:

• Forward arrow: 6
• Backward arrow: 6

Subtract the opposite forces:

$$6 - 6 = 0$$

The net force is 0. The forces are balanced.

The scooter is already moving, and it keeps moving steadily in a straight line.

Answer: Net force = 0. This is dynamic equilibrium.

How to read a free-body diagram

Here is a simple way to read one:

• If the arrows are the same size in opposite directions, the forces are balanced.
• If one arrow is bigger, that direction has the stronger force.
• The object changes motion in the direction of the net force.

Important things to remember

• Net force is the total force after adding and subtracting forces.
• Balanced forces mean net force is 0.
• Unbalanced forces mean net force is not 0.
• Static equilibrium means not moving, with net force 0.
• Dynamic equilibrium means moving steadily in a straight line, with net force 0.
• If net force is not 0, the object is accelerating, or changing motion.

Try thinking about these real-life examples

• A lamp sitting on a desk: static equilibrium
• A bus moving straight at the same speed: dynamic equilibrium
• A soccer ball being kicked: unbalanced force, so it accelerates

Brief Summary

Forces are pushes and pulls. The net force tells us the total force on an object. If the net force is 0, the object is in equilibrium. If it is standing still, that is static equilibrium. If it is moving steadily in a straight line, that is dynamic equilibrium. If the net force is not 0, the object changes motion.

Newton’s Second Law and Net Force explains how forces cause objects to speed up, slow down, or change direction. This law connects three important ideas: force, mass, and acceleration. If you understand how these three are related, you can predict how an object will move when forces act on it.

In everyday life, objects are often pushed or pulled by more than one force at a time. A soccer ball may be kicked while gravity pulls it downward. A box on the floor may be pushed forward while friction pushes backward. To understand motion correctly, we must look at the net force, which is the overall force after all individual forces are combined.

This lesson will show how to use Newton’s Second Law, how to find net force, and how to use free-body diagrams to organize forces clearly.

1. What is Newton’s Second Law?

Newton’s Second Law states that the acceleration of an object depends on two things:

• the net force acting on the object
• the mass of the object

The law is written mathematically as:

where:

• \(F_{\text{net}}\) = net force, measured in newtons (N)
• \(m\) = mass, measured in kilograms (kg)
• \(a\) = acceleration, measured in meters per second squared \((m/s^2)\)

This equation means:

• If net force increases, acceleration increases.
• If mass increases, acceleration decreases for the same net force.
• If the net force is zero, the acceleration is zero.

So, a light object accelerates more easily than a heavy object when the same force is applied. That is why an empty shopping cart is easier to speed up than a full one.

2. What is net force?

Net force is the total force acting on an object after all forces are added together, including their directions. Because force is a vector, direction matters.

If two forces act in the same direction, they add.

If two forces act in opposite directions, they subtract.

For example, if a person pushes a box with 20 N to the right and friction acts with 5 N to the left, then the net force is:

The box will accelerate to the right because the net force points to the right.

3. Balanced and unbalanced forces

When forces are balanced, the net force is zero.

If the net force is zero, then:

This does not always mean the object is at rest. It could be:

• staying at rest, or
• moving at a constant velocity

When forces are unbalanced, the net force is not zero. Then the object accelerates. Acceleration can mean:

• speeding up
• slowing down
• changing direction

4. Mass and acceleration

Mass measures how much matter an object has. In Newton’s Second Law, mass also tells us how difficult it is to change an object’s motion. A larger mass means more resistance to acceleration.

From the equation

we can see that acceleration is directly proportional to net force and inversely proportional to mass.

This means:

• double the net force  double the acceleration
• double the mass  half the acceleration

5. Units in Newton’s Second Law

The unit of force is the newton (N). One newton is the force needed to accelerate a 1 kg mass at \(1\,m/s^2\).

Always check that your units match:

• mass in kilograms, not grams
• acceleration in \(m/s^2\)
• force in newtons

If mass is given in grams, convert to kilograms before using the formula.

6. Free-body diagrams

A free-body diagram is a simple drawing that shows all the forces acting on one object. It helps you identify the net force and decide which direction the object will accelerate.

To draw a free-body diagram:

1. Draw the object as a box or dot.
2. Draw arrows for each force acting on the object.
3. Label each arrow with the force name and, if known, its size.
4. Use arrow direction to show the direction of each force.

Common forces you may see include:

• weight or gravitational force, downward
• normal force, upward from a surface
• friction, opposite motion or attempted motion
• applied force, a push or pull
• tension, along a rope or string

For an object resting on a flat surface, the upward normal force often balances the downward weight. In that case, the vertical net force is zero, and any acceleration must come from the horizontal forces.

7. Steps for solving force problems

When solving Newton’s Second Law problems, use a clear process:

1. Identify the object you are analyzing.
2. Draw a free-body diagram.
3. Choose a positive direction.
4. Find the net force by adding forces with signs.
5. Use \(F_{\text{net}} = ma\).
6. Solve for the missing value.
7. Include units and direction in your answer.

Worked Example 1: Finding acceleration from net force

A 4 kg cart experiences a net force of 12 N to the right. What is its acceleration?

Step 1: Write the formula

Step 2: Solve for acceleration

Step 3: Substitute values

Answer: The cart accelerates at \(3\,m/s^2\) to the right.

This example shows that acceleration points in the same direction as the net force.

Worked Example 2: Finding net force from two opposite forces

A box is pushed with 30 N to the right. Friction acts with 18 N to the left. The box has a mass of 6 kg. What is the box’s acceleration?

Step 1: Find the net force

Step 2: Use Newton’s Second Law

Step 3: Substitute values

Answer: The box accelerates at \(2\,m/s^2\) to the right.

If the friction had been equal to the push, the net force would have been zero and there would be no acceleration.

Worked Example 3: Solving for force

A 1.5 kg ball accelerates at \(8\,m/s^2\) upward after being hit. What net force acts on the ball?

Step 1: Use the formula

Step 2: Substitute values

Answer: The net force is 12 N upward.

This is the net force, not just one individual force. If more than one force acts on the ball, their combined effect must equal 12 N upward.

Worked Example 4: Using a free-body diagram idea

A 10 kg sled is pulled to the right with a force of 50 N. Friction acts to the left with 20 N. The sled moves across level snow. Find the acceleration.

Step 1: Identify forces

• pulling force: 50 N right
• friction: 20 N left
• weight downward
• normal force upward

Because the sled is on level ground, the normal force and weight balance each other vertically. So we only need the horizontal net force.

Step 2: Find horizontal net force

Step 3: Calculate acceleration

Answer: The sled accelerates at \(3\,m/s^2\) to the right.

8. Important ideas about direction

Acceleration always points in the direction of the net force, not necessarily in the direction of motion.

For example, a car moving to the right while braking may still be moving right, but if the net force is to the left, its acceleration is to the left. That leftward acceleration causes the car to slow down.

This is a very important idea: motion and acceleration do not have to point in the same direction.

9. Common mistakes to avoid

• Using total force instead of net force: Always combine all forces first.
• Ignoring direction: Forces in opposite directions must subtract.
• Mixing units: Convert grams to kilograms when needed.
• Forgetting balanced vertical forces: On a level surface, normal force and weight often cancel.
• Thinking zero net force means zero motion: Zero net force means zero acceleration, not necessarily zero velocity.

10. Quick check questions

• If the same force acts on two objects, which one has greater acceleration: the one with smaller mass or larger mass?
• If two forces of 15 N and 15 N act in opposite directions, what is the net force?
• If a 5 kg object has a net force of 25 N, what is its acceleration?
• If an object moves at constant velocity, what must be true about its net force?

Answers:

• The object with smaller mass.
• 0 N.
• \(a = 25/5 = 5\,m/s^2\).
• Its net force must be zero.

11. Summary

Newton’s Second Law shows that the motion of an object changes when a net force acts on it. The relationship is given by:

To use this law correctly, you must identify all the forces on the object, combine them to find the net force, and then solve for force, mass, or acceleration. Free-body diagrams are a useful tool because they help you organize forces and avoid mistakes. Once you know the net force, you can determine how the object will accelerate.

Work-Energy Theorem sounds like a big idea, but it can be understood in a simple way. It tells us how work and motion are connected.

When a force pushes or pulls an object and makes it move, work is done. The Work-Energy Theorem says that the net work done on an object changes its kinetic energy. Kinetic energy is the energy of motion.

In simple words: if you do work on something, you can make it move faster, slower, or stop. The amount of change in its motion energy depends on the total work done on it.

We can write the idea like this:

$$\text{Net Work} = \text{Change in Kinetic Energy}$$

Or in math symbols:

$$W_{\text{net}} = \Delta KE$$

Here, \(\Delta\) means change in.

Let’s learn the parts.

• Work: A force makes an object move.
• Net work: The total work from all the pushes and pulls together.
• Kinetic energy: Energy an object has because it is moving.

If net work is positive, the object usually speeds up. Its kinetic energy increases.

If net work is negative, the object usually slows down. Its kinetic energy decreases.

If net work is zero, the object’s kinetic energy does not change. It keeps moving at the same speed, or it stays still.

What is kinetic energy?

Kinetic energy depends on how fast something is moving and how much matter it has. For this lesson, the most important thing to remember is this: faster motion means more kinetic energy.

A rolling marble has kinetic energy. A running dog has kinetic energy. A moving bicycle has kinetic energy. If they move faster, they have more kinetic energy.

What is work?

Work happens when two things occur:

1. A force is used, like a push or a pull.
2. The object moves.

If you push hard on a wall and it does not move, no work is done on the wall in science terms.

If you push a wagon and it rolls forward, work is done.

What does “net work” mean?

Sometimes more than one force acts on an object. Some forces help the motion, and some oppose it. Net work means we look at the total effect of all those forces together.

For example, if you push a box forward but friction pulls backward, the box feels both. The net work depends on which effect is stronger.

How the theorem works

The Work-Energy Theorem connects cause and effect:

• Cause: Net work is done on an object.
• Effect: The object’s kinetic energy changes.

This means:

• More positive net work \(\rightarrow\) more motion energy
• More negative net work \(\rightarrow\) less motion energy

Think of a playground swing.

When you push a swing, you do work on it. The swing moves faster, so its kinetic energy increases.

If you grab the swing gently to slow it down, you do negative work. The swing loses kinetic energy.

Everyday examples

• Kicking a soccer ball makes it speed up.
• Brakes on a bike slow the bike down.
• Pushing a shopping cart makes it move faster.
• Grass and dirt can slow a rolling ball because they oppose its motion.

A helpful way to remember it

You can remember the theorem like this:

Work changes motion energy.

If the total work helps the motion, the object speeds up.

If the total work fights the motion, the object slows down.

Worked Example 1: Pushing a toy car

A toy car is sitting still. A child pushes it, and the car starts moving fast across the floor.

What happened?

The child did work on the toy car because the child used a force and the car moved. The net work was positive, so the toy car’s kinetic energy increased.

Answer: Positive net work made the car gain kinetic energy and speed up.

Worked Example 2: A rolling ball on grass

A ball rolls from the sidewalk onto grass. On the grass, it slows down and stops.

What happened?

The grass pushes against the motion of the ball. This does negative work on the ball. Because of that, the ball loses kinetic energy.

Answer: Negative net work made the ball’s kinetic energy decrease, so it slowed down and stopped.

Worked Example 3: Two forces on a sled

A child pulls a sled forward. Snow rubbing on the bottom of the sled pushes backward.

Imagine the forward pull does more work than the backward rubbing.

What happens to the sled?

Because the forward effect is bigger, the net work is positive. That means the sled’s kinetic energy increases.

Answer: The sled speeds up because the total work is positive.

Now imagine the backward rubbing is just as strong as the forward pull, so the effects balance.

What happens then?

If the net work is zero, the sled’s kinetic energy does not change.

Answer: The sled keeps the same speed if it is already moving, or stays still if it is not moving.

Worked Example 4: Braking a bicycle

A bicycle is moving quickly. The rider squeezes the brakes, and the bike slows down.

What happened?

The brakes create a force opposite the motion. That does negative work on the bicycle. So the kinetic energy decreases.

Answer: Negative net work from the brakes caused the bike to slow down.

Important ideas to notice

• If an object speeds up, its kinetic energy increases.
• If an object slows down, its kinetic energy decreases.
• If its speed stays the same, its kinetic energy stays the same.
• The change depends on the net work, not just one force.

Simple comparison chart

• Positive net work: object speeds up, kinetic energy increases
• Negative net work: object slows down, kinetic energy decreases
• Zero net work: no change in kinetic energy

Common mistakes to avoid

• Thinking every push does work. If the object does not move, no work is done.
• Looking at only one force instead of the total of all forces.
• Forgetting that slowing down means kinetic energy is being reduced.

Let’s check your understanding

1. A skateboarder pushes off and moves faster. Is the net work positive, negative, or zero?
   Answer: Positive.
2. A rolling book bag is slowed by carpet. Does its kinetic energy increase or decrease?
   Answer: Decrease.
3. A toy train moves at the same speed on a straight track. What is happening to its kinetic energy?
   Answer: It stays the same.

Brief Summary

The Work-Energy Theorem says that the total work done on an object changes its kinetic energy.

When net work is positive, the object gains motion energy and speeds up. When net work is negative, the object loses motion energy and slows down. When net work is zero, its kinetic energy stays the same.

This big idea helps explain many everyday motions, like pushing a cart, kicking a ball, or braking a bike.

Newton's Third Law and Interaction Pairs

When two objects interact, they push or pull on each other. Newton's Third Law explains what happens in every one of these interactions.

Newton's Third Law: If object A exerts a force on object B, then object B exerts a force on object A that is equal in magnitude and opposite in direction.

In symbols, this idea is often written as

$$\vec{F}_{A \to B} = -\vec{F}_{B \to A}$$

This means the two forces have the same size but point in opposite directions.

These two forces are called an interaction pair or an action-reaction pair. They always happen together. One force cannot exist without the other.

Important idea: The two forces in an interaction pair act on different objects. Because they act on different objects, they do not cancel each other out.

This is one of the most common mistakes students make. Equal and opposite forces only cancel if they act on the same object. Interaction pairs act on different objects, so each object must be analyzed separately.

Why Newton's Third Law matters

• It helps explain why objects move when they push or pull on each other.
• It shows why forces always come in pairs.
• It is important for understanding walking, jumping, rockets, collisions, and many everyday situations.

Main teaching point 1: Forces always come from interactions

A force does not appear by itself. A force happens because two objects interact. For example:

• A book and a table interact.
• A foot and the ground interact.
• A bat and a ball interact.
• The Earth and a falling object interact.

Whenever you see one object exerting a force, you should immediately ask: What is the matching force back on the first object?

Main teaching point 2: Equal in magnitude, opposite in direction

If object A pushes on object B with a force of 20 N to the right, then object B pushes on object A with a force of 20 N to the left.

The forces are equal because the interaction is mutual. They are opposite because each object pushes or pulls back against the other.

So if

$$\vec{F}_{A \to B} = 20\,\text{N to the right}$$

then

$$\vec{F}_{B \to A} = 20\,\text{N to the left}$$

Main teaching point 3: Interaction pairs act on different objects

This point is the key to understanding the law correctly.

Suppose a student pushes on a wall. The student exerts a force on the wall. At the same time, the wall exerts a force on the student.

• The force on the wall is caused by the student.
• The force on the student is caused by the wall.

These are an interaction pair. They are equal and opposite, but they do not cancel because one acts on the wall and the other acts on the student.

Main teaching point 4: How to identify an interaction pair

To decide whether two forces form an interaction pair, use this checklist:

1. Do the two forces involve the same two objects?
2. Are the forces the same type of force? For example, both contact forces or both gravitational forces.
3. Do they act on different objects?
4. Are they equal in magnitude and opposite in direction?

If all of these are true, the two forces are an interaction pair.

Main teaching point 5: Do not confuse interaction pairs with balanced forces

Balanced forces are different. Balanced forces act on the same object and add up to zero.

For example, a book resting on a table has:

• Weight downward from the Earth
• Normal force upward from the table

If the book is not moving up or down, these two forces are balanced.

But they are not an interaction pair, because they both act on the book, not on different objects.

The interaction pairs in that situation are:

• Table on book and book on table
• Earth on book and book on Earth

Main teaching point 6: Common examples of Newton's Third Law

• Walking: Your foot pushes backward on the ground, and the ground pushes forward on your foot.
• Swimming: A swimmer pushes water backward, and the water pushes the swimmer forward.
• Jumping: You push downward on the ground, and the ground pushes upward on you.
• Rocket motion: The rocket pushes gases downward, and the gases push the rocket upward.
• Collisions: In a crash, each object exerts an equal and opposite force on the other.

Worked Example 1: A hand pushes a box

A hand pushes a box with a force of 15 N to the right. Identify the interaction pair.

Step 1: Identify the two objects.
The two objects are the hand and the box.

Step 2: State the force given.
The hand exerts a force on the box of 15 N to the right.

Step 3: Use Newton's Third Law.
The box exerts a force on the hand of 15 N to the left.

Answer:

• Force of hand on box: 15 N to the right
• Force of box on hand: 15 N to the left

These two forces form an interaction pair.

Worked Example 2: A book resting on a table

A book sits still on a table. A student says, "The weight of the book and the normal force from the table are a Newton's Third Law pair." Is the student correct?

Step 1: Name the forces.

• Weight: Earth pulls the book downward.
• Normal force: Table pushes the book upward.

Step 2: Check where the forces act.
Both of these forces act on the book.

Step 3: Apply the rule.
Interaction pairs must act on different objects.

Conclusion: No, the student is not correct.

These two forces may be balanced if the book is at rest, but they are not a Third Law pair.

The correct interaction pairs are:

• Table on book and book on table
• Earth on book and book on Earth

Worked Example 3: Walking forward

A student walks forward across the floor. Explain the interaction pair that helps the student move.

Step 1: Identify the interacting objects.
The two objects are the student's foot and the floor.

Step 2: Describe the first force.
The foot pushes backward on the floor.

Step 3: Describe the matching force.
The floor pushes forward on the foot.

Step 4: Explain the motion.
The forward force from the floor acts on the student and helps the student move forward.

Answer: The interaction pair is the foot pushing backward on the floor and the floor pushing forward on the foot.

This example shows why friction can help motion. Without enough friction between foot and floor, the floor cannot provide the forward push needed for walking.

Worked Example 4: Two skaters push apart

Two skaters stand still on ice and push off each other. One skater has a mass of 50 kg and the other has a mass of 70 kg. Which skater experiences the larger force?

Step 1: Apply Newton's Third Law.
When the skaters push on each other, each exerts an equal and opposite force on the other.

Step 2: Compare the forces.
The force on the 50 kg skater and the force on the 70 kg skater are equal in magnitude.

Step 3: Explain why they may move differently.
Even though the forces are equal, the skaters can have different changes in motion because their masses are different.

Answer: Neither skater experiences a larger force. The forces are equal and opposite.

Important misconception: Equal forces do not always cause equal motion. Motion also depends on mass.

How Newton's Third Law connects to everyday motion

Many students wonder: if forces are always equal and opposite, why does anything move at all?

The answer is that the equal and opposite forces act on different objects. For example, when you jump, the ground pushes you upward, and you push the ground downward. You move because the upward force acts on you. The Earth also feels your downward force, but because the Earth is so massive, its motion is too small to notice.

Force-pair language you should practice

It is helpful to name forces clearly in the form:

"force of A on B"

Then the matching interaction pair is:

"force of B on A"

Examples:

• Force of bat on ball ↔ force of ball on bat
• Force of Earth on apple ↔ force of apple on Earth
• Force of tire on road ↔ force of road on tire

This naming method helps prevent confusion.

Quick check: Is it a Third Law pair?

Ask yourself these questions:

• Are there exactly two interacting objects?
• Does one force act on the first object and the other on the second?
• Are the forces the same kind of interaction?
• Are they equal and opposite?

If yes, then it is a Newton's Third Law pair.

Brief summary

Newton's Third Law says that forces always come in pairs. If one object exerts a force on a second object, the second object exerts an equal and opposite force on the first.

These interaction pairs act on different objects, so they do not cancel each other out. To identify them correctly, always name the two objects and match "force of A on B" with "force of B on A."

Frictional Forces are forces that oppose motion or the tendency of motion between two surfaces that are touching.

Friction is part of everyday life. It helps you walk without slipping, allows car tires to grip the road, and lets you hold objects in your hand. At the same time, friction can also make motion harder by slowing objects down and turning some mechanical energy into thermal energy.

In 11th Grade mechanics, friction is especially important because it affects how forces change motion. To solve problems involving friction, you need to understand the difference between static friction and kinetic friction, how to calculate each one, and how the normal force affects friction.

1. What is friction?

Friction is a contact force that acts parallel to the surfaces in contact. Its direction is always opposite the direction of motion, or opposite the direction the object is trying to move.

Even surfaces that look smooth have tiny bumps and irregularities. When two surfaces press against each other, these microscopic rough spots resist sliding. That resistance is friction.

Friction depends mainly on:

• the type of surfaces in contact
• how hard the surfaces are pressed together, measured by the normal force

Friction does not mainly depend on the visible area of contact in the simple models used in high school physics.

2. The normal force and why it matters

The normal force, written as \(F_N\), is the support force exerted by a surface on an object in contact with it. It acts perpendicular to the surface.

For an object resting on a flat horizontal surface with no other vertical forces, the normal force equals the object's weight:

$$F_N = mg$$

where:

• \(m\) = mass of the object
• \(g\) = gravitational field strength, about \(9.8\,\text{m/s}^2\) on Earth

If the normal force becomes larger, friction usually becomes larger too. This is why a heavy box is harder to slide than a light box made of the same material.

3. Static friction

Static friction acts when two surfaces are not sliding relative to each other. It prevents motion from starting.

If you push a box gently and it does not move, static friction is balancing your push. If you push harder, static friction also increases, up to a maximum value.

The maximum static friction is:

$$f_s^{\max} = \mu_s F_N$$

where:

• \(f_s^{\max}\) = maximum static friction
• \(\mu_s\) = coefficient of static friction
• \(F_N\) = normal force

The coefficient of static friction, \(\mu_s\), is a number that depends on the pair of surfaces. Rougher surfaces usually have a larger value.

An important idea is that static friction is not always equal to \(\mu_s F_N\). Instead, static friction adjusts as needed to prevent motion, up to its maximum value.

So for static friction:

$$f_s \leq \mu_s F_N$$

If the applied force is smaller than the maximum static friction, the object stays at rest.

4. Kinetic friction

Kinetic friction acts when two surfaces are sliding past each other. Once motion has started, the friction is usually a little smaller than the maximum static friction.

The kinetic friction force is calculated using:

$$f_k = \mu_k F_N$$

where:

• \(f_k\) = kinetic friction
• \(\mu_k\) = coefficient of kinetic friction
• \(F_N\) = normal force

Usually, \(\mu_k < \mu_s\). This means it often takes more force to start moving an object than to keep it moving.

5. Comparing static and kinetic friction

• Static friction: acts when there is no sliding
• Kinetic friction: acts when surfaces are sliding
• Static friction changes as needed, up to a maximum
• Kinetic friction is usually constant in simple problems
• Usually, \(\mu_s\) is greater than \(\mu_k\)

6. Direction of friction

Friction always acts to oppose relative motion or attempted motion between surfaces.

For example:

• If you push a box to the right and it does not move, static friction acts to the left.
• If the box slides to the right, kinetic friction acts to the left.
• If an object tends to slide down a ramp, friction acts up the ramp.

Always determine first what direction the object is moving, or trying to move, before choosing the direction of friction.

7. Friction on a horizontal surface

On a flat surface, if no other vertical forces are acting, the normal force is:

$$F_N = mg$$

Then friction can be found by:

$$f_s^{\max} = \mu_s mg$$

$$f_k = \mu_k mg$$

This makes many friction problems straightforward.

8. Friction and Newton's Second Law

Friction is often included in force balance and acceleration problems.

Newton's Second Law says:

$$\sum F = ma$$

If an object is moving along a horizontal surface, the net horizontal force is often:

$$F_{\text{applied}} - f_k = ma$$

If the object is not moving, then the horizontal forces balance:

$$F_{\text{applied}} = f_s$$

as long as the applied force does not exceed \(f_s^{\max}\).

9. Friction and energy

Friction is important in work and energy because it often transforms mechanical energy into thermal energy.

When friction acts over a distance, the work done by friction is negative because it opposes motion.

The work done by friction can be written as:

$$W_f = -fd$$

for a constant friction force acting opposite the motion, where:

• \(W_f\) = work done by friction
• \(f\) = friction force
• \(d\) = distance moved

This negative work helps explain why sliding objects slow down if no other force keeps pushing them.

10. Worked Example 1: Maximum static friction on a flat surface

Problem: A \(12\,\text{kg}\) box rests on a horizontal floor. The coefficient of static friction is \(\mu_s = 0.40\). Find the maximum static friction force.

Step 1: Find the normal force.

Because the box is on a flat surface:

$$F_N = mg = (12)(9.8) = 117.6\,\text{N}$$

Step 2: Use the static friction formula.

$$f_s^{\max} = \mu_s F_N = (0.40)(117.6) = 47.04\,\text{N}$$

Answer: The maximum static friction is approximately \(47\,\text{N}\).

Meaning: Any applied force less than or equal to \(47\,\text{N}\) will not make the box move. A greater force will start the box sliding.

11. Worked Example 2: Static friction adjusts to match the applied force

Problem: The same \(12\,\text{kg}\) box is pushed with a horizontal force of \(30\,\text{N}\). What is the friction force?

Step 1: Compare the applied force to the maximum static friction.

From Example 1:

$$f_s^{\max} = 47\,\text{N}$$

The applied force is only \(30\,\text{N}\), which is less than \(47\,\text{N}\).

Step 2: Decide what happens.

The box does not move, so static friction balances the push.

$$f_s = 30\,\text{N}$$

Answer: The friction force is \(30\,\text{N}\), acting opposite the push.

Important lesson: Static friction is not automatically equal to \(47\,\text{N}\). It only provides the amount needed, up to that maximum.

12. Worked Example 3: Kinetic friction and acceleration

Problem: A \(10\,\text{kg}\) crate is sliding across a horizontal floor. The coefficient of kinetic friction is \(\mu_k = 0.25\). A horizontal force of \(40\,\text{N}\) pulls the crate. Find the acceleration.

Step 1: Find the normal force.

$$F_N = mg = (10)(9.8) = 98\,\text{N}$$

Step 2: Find the kinetic friction.

$$f_k = \mu_k F_N = (0.25)(98) = 24.5\,\text{N}$$

Step 3: Find the net force.

The pull is to the right and friction is to the left:

$$F_{\text{net}} = 40 - 24.5 = 15.5\,\text{N}$$

Step 4: Use Newton's Second Law.

$$a = \frac{F_{\text{net}}}{m} = \frac{15.5}{10} = 1.55\,\text{m/s}^2$$

Answer: The crate accelerates at \(1.55\,\text{m/s}^2\) to the right.

13. Worked Example 4: Work done by friction

Problem: A \(5.0\,\text{kg}\) block slides \(3.0\,\text{m}\) across a floor with a coefficient of kinetic friction \(\mu_k = 0.20\). Find the work done by friction.

Step 1: Find the normal force.

$$F_N = mg = (5.0)(9.8) = 49\,\text{N}$$

Step 2: Find the kinetic friction.

$$f_k = \mu_k F_N = (0.20)(49) = 9.8\,\text{N}$$

Step 3: Calculate the work done by friction.

Because friction opposes the motion:

$$W_f = -fd = -(9.8)(3.0) = -29.4\,\text{J}$$

Answer: The work done by friction is \(-29.4\,\text{J}\).

Meaning: Friction removes \(29.4\,\text{J}\) of mechanical energy from the block-and-floor system, usually as thermal energy.

14. Common mistakes to avoid

• Using \(f_s = \mu_s F_N\) every time. This only gives the maximum static friction, not always the actual static friction.
• Forgetting to find the normal force first. Friction depends on \(F_N\), so you usually need it before finding friction.
• Using the wrong coefficient. Use \(\mu_s\) when the object is not sliding and \(\mu_k\) when it is sliding.
• Ignoring direction. Friction always opposes motion or attempted motion.
• Confusing weight and normal force. On a flat surface they may be equal, but they are different forces with different directions.

15. Problem-solving steps for friction questions

1. Identify whether the object is at rest or sliding.
2. Draw or imagine the forces acting on the object.
3. Find the normal force.
4. Use the correct friction formula:

• For static friction: $$f_s \leq \mu_s F_N$$
• Maximum static friction: $$f_s^{\max} = \mu_s F_N$$
• For kinetic friction: $$f_k = \mu_k F_N$$

5. Use Newton's laws or work-energy ideas to answer the question.
6. Check whether your answer makes sense in size and direction.

16. Brief summary

Friction is a force that opposes motion between surfaces in contact. Static friction prevents motion from starting and can vary up to a maximum value, while kinetic friction acts during sliding and is usually constant in simple problems.

Both types of friction depend on the coefficient of friction and the normal force. By identifying the kind of friction present and applying the formulas carefully, you can solve problems involving motion, acceleration, and energy loss due to friction.

Drag Forces and Terminal Velocity

When objects move through air or water, they do not move freely. The fluid pushes back on them. This backward force is called drag force or fluid resistance.

Drag is important in many everyday situations. A falling skydiver speeds up at first, then stops speeding up and falls at a steady rate. A bicyclist must pedal harder to go faster because air resistance increases. A boat moving through water also experiences drag.

This lesson explains what drag force is, how it changes with speed, and why a falling object can reach a constant speed called terminal velocity.

1. What is drag force?

Drag force is a force that opposes the motion of an object through a fluid. In science, a fluid means a substance that can flow, such as a liquid or a gas. So both air and water are fluids.

If an object moves downward through air, drag acts upward. If a car moves forward, drag acts backward. Drag always points in the direction opposite the object's motion relative to the fluid.

Several things affect the size of the drag force:

• Speed: faster motion usually means more drag
• Shape: wide or flat shapes usually have more drag than streamlined shapes
• Surface area: larger area facing the fluid gives more drag
• Type of fluid: water usually causes more drag than air because it is denser

For many 11th Grade problems, we focus most on one key idea: drag force increases as speed increases.

2. Forces on a falling object

Imagine dropping a ball from rest. At the moment it is released, gravity pulls it downward. The gravitational force is its weight:

where:

• \(m\) = mass of the object

• \(g\) = gravitational field strength, about \(9.8\,\text{m/s}^2\) on Earth

At first, the object is moving very slowly, so the drag force is very small. That means gravity is larger than drag, so there is a net force downward. A net force causes acceleration, so the object speeds up as it falls.

As the object gets faster, the drag force grows. Now the upward drag becomes more significant. The net downward force becomes smaller, so the acceleration becomes smaller too.

Eventually, the drag force becomes equal in size to the weight:

At that moment, the forces are balanced. The net force is zero:

When the net force is zero, acceleration is zero. The object does not keep speeding up. Instead, it continues falling at a constant speed. This constant speed is called terminal velocity.

3. What is terminal velocity?

Terminal velocity is the maximum constant speed reached by an object falling through a fluid when drag force balances the object's weight.

At terminal velocity:

• weight downward = drag upward
• net force = 0
• acceleration = 0
• velocity is constant

This does not mean the object has stopped moving. It means the object is still moving, but its speed is no longer changing.

A common misunderstanding is to think that if velocity is large, acceleration must also be large. That is not true. An object can move quickly with zero acceleration if its speed stays constant.

4. How drag changes with speed

In simple school-level models, drag is often described as increasing with speed. In some situations, drag is approximately proportional to speed:

In other situations, especially at higher speeds, drag is often closer to proportional to the square of speed:

For this lesson, the most important idea is not which formula is used, but that greater speed leads to greater drag. That is why a falling object cannot keep accelerating forever in a fluid like air.

5. Motion of a falling object in air

The motion of a falling object in air can be described in stages:

1. Just released: speed is zero or very small, so drag is almost zero.
2. Early fall: gravity is much larger than drag, so the object accelerates downward quickly.
3. Later fall: speed increases, so drag increases. The acceleration becomes smaller.
4. Terminal velocity: drag equals weight, so acceleration becomes zero and speed stays constant.

This explains why graphs of speed versus time for a falling object in air usually rise at first and then level off.

6. Factors that affect terminal velocity

Different objects have different terminal velocities. The terminal velocity depends on the balance between weight and drag.

An object with greater weight usually needs a greater drag force to balance it. Since drag increases with speed, that often means the object must fall faster before reaching terminal velocity.

Here are some general trends:

• Greater mass: often leads to a larger terminal velocity if shape and area stay similar
• Larger surface area: usually leads to a smaller terminal velocity because drag becomes large at lower speeds
• More streamlined shape: usually leads to a larger terminal velocity because drag is reduced
• Denser fluid: usually leads to a smaller terminal velocity because drag is greater

A parachute is a great example. Opening a parachute greatly increases surface area. This increases drag, so the skydiver reaches a new, much lower terminal velocity.

7. Free fall compared with falling with drag

In free fall, air resistance is ignored. Then gravity is the only force acting, so the object accelerates downward at about \(9.8\,\text{m/s}^2\).

But in real air, drag is often important. Then the acceleration is less than \(9.8\,\text{m/s}^2\), and it decreases as speed increases.

So:

• Free fall: only gravity acts, acceleration is constant downward
• Falling with drag: gravity and drag both act, acceleration changes and may become zero at terminal velocity

8. Force diagrams

Force diagrams help show what is happening.

At the start of a fall:

• Downward weight \(mg\): large
• Upward drag \(F_d\): very small
• Net force: downward

During the fall:

• Downward weight \(mg\): unchanged
• Upward drag \(F_d\): increasing
• Net force: still downward, but smaller

At terminal velocity:

• Downward weight \(mg\)
• Upward drag \(F_d\)
• Equal in size, opposite in direction
• Net force: zero

Notice that the weight does not change much during the fall, but the drag force does change because it depends on speed.

Worked Example 1: Identifying the net force

A falling object has a weight of \(20\,\text{N}\) downward and a drag force of \(5\,\text{N}\) upward. What is the net force, and what happens to the object's motion?

Step 1: Write the forces.

Weight: \(20\,\text{N}\) downward

Drag: \(5\,\text{N}\) upward

Step 2: Find the net force.

Step 3: Interpret the result.

Because the net force is downward, the object accelerates downward. Its speed increases as it falls.

Answer: The net force is \(15\,\text{N}\) downward, so the object speeds up downward.

Worked Example 2: Recognizing terminal velocity

A raindrop falls with a weight of \(0.12\,\text{N}\) downward and experiences \(0.12\,\text{N}\) of drag upward. Is it accelerating?

Step 1: Compare the forces.

The forces are equal in size and opposite in direction.

Step 2: Find the net force.

Step 3: Use Newton's second law idea.

If net force is zero, acceleration is zero.

Conclusion: The raindrop is not accelerating. If it is moving, it is moving at constant speed. That speed is its terminal velocity.

Worked Example 3: Finding weight at terminal velocity

A skydiver of mass \(70\,\text{kg}\) is falling at terminal velocity. What is the drag force acting on the skydiver? Use \(g = 9.8\,\text{m/s}^2\).

Step 1: Find the weight.

Step 2: Apply terminal velocity condition.

At terminal velocity, drag force equals weight in size.

Direction: Since the skydiver is falling downward, drag acts upward.

Answer: The drag force is \(686\,\text{N}\) upward.

Worked Example 4: Comparing two situations

A skydiver first falls without a parachute and later opens a parachute.

Explain what happens to:

• drag force
• acceleration
• terminal velocity

Step 1: Consider surface area.

Opening the parachute greatly increases the surface area facing the air.

Step 2: Effect on drag.

Because of the larger area, the drag force becomes much larger.

Step 3: Effect on motion right after opening.

Just after the parachute opens, drag may become larger than weight for a short time. Then the net force is upward, so the skydiver slows down.

Step 4: New terminal velocity.

As the skydiver slows, drag decreases until it again matches weight. Then the skydiver reaches a new terminal velocity.

Answer:

• Drag force increases a lot
• The skydiver slows down after the parachute opens
• The new terminal velocity is much smaller than before

9. Energy ideas connected to drag

As an object falls, gravitational potential energy decreases. Some of that energy becomes kinetic energy as the object speeds up.

But when drag is present, not all of the lost gravitational potential energy becomes kinetic energy. Some energy is transferred to the surrounding fluid, often as thermal energy and sound.

This helps explain why the speed stops increasing. Energy is still being transferred, but drag prevents all of it from going into increasing kinetic energy.

10. Common mistakes to avoid

• Thinking terminal velocity means the object stops: It does not stop; it moves at constant speed.
• Thinking weight gets smaller as the object falls: Weight stays about the same near Earth's surface.
• Thinking drag is always constant: Drag usually changes with speed.
• Thinking zero net force means zero velocity: Zero net force means zero acceleration, not zero velocity.
• Ignoring direction: Weight acts downward; drag acts opposite the motion.

11. Key ideas to remember

• Drag force is a resistive force from a fluid.
• Drag acts opposite the direction of motion.
• As speed increases, drag increases.
• A falling object accelerates at first because weight is larger than drag.
• Terminal velocity happens when drag equals weight.
• At terminal velocity, net force and acceleration are zero, but speed is not zero.

Brief Summary

When an object moves through air or water, it experiences drag force in the opposite direction of motion. For a falling object, gravity pulls downward while drag pushes upward. As the object speeds up, drag increases until it balances the object's weight. At that point, the net force is zero, acceleration becomes zero, and the object falls at a constant speed called terminal velocity.

Gravity, Mass, and Weight

Everything around us is pulled by gravity. Gravity is a force that pulls objects toward each other. On Earth, gravity pulls things down toward the ground.

This lesson will help you understand three important ideas: gravity, mass, and weight. These words are related, but they do not mean the same thing.

Mass is how much matter is in an object. Matter is the “stuff” that makes up everything. A bowling ball has more mass than a tennis ball because it has more matter.

Weight is the pull of gravity on an object. Weight can change depending on where you are, because gravity can be stronger or weaker in different places.

So here is the big idea:

• Mass tells how much matter an object has.
• Weight tells how hard gravity is pulling on that object.

For example, if you take a backpack from Earth to the Moon, the backpack’s mass stays the same because it is still made of the same amount of matter. But its weight changes because the Moon’s gravity is weaker than Earth’s gravity.

What is gravity?

Gravity is a force that pulls objects together. Earth is very large, so it has enough gravity to pull people, animals, water, and other objects toward it.

That is why:

• balls fall when you drop them,
• rain falls from clouds,
• you stay on the ground instead of floating away.

The Sun, Moon, and planets all have gravity too. Bigger objects usually have stronger gravity because they have more matter.

Mass stays the same

Mass does not change just because you move to a new place. If a rock has a mass of 2 kilograms on Earth, it still has a mass of 2 kilograms on the Moon.

You can think of mass as part of the object itself. It is one of the object’s own properties.

Weight can change

Weight depends on gravity. If gravity changes, weight changes too.

On Earth, gravity is stronger than on the Moon. So an object weighs more on Earth than it does on the Moon.

Here is a simple rule:

• Stronger gravity = more weight
• Weaker gravity = less weight

A simple math rule for weight

Scientists often show the relationship between mass and weight with this rule:

$$\text{weight} = \text{mass} \times \text{gravity}$$

We can also write it like this:

$$W = m \times g$$

In this rule:

• \(W\) means weight
• \(m\) means mass
• \(g\) means the pull of gravity

For 4th grade, you can think of \(g\) as a number that tells how strong gravity is in a place.

On Earth, we will use:

$$g = 10$$

On the Moon, gravity is much weaker. We will use:

$$g = 2$$

These numbers help us compare how weight changes in different places.

Worked Example 1: Finding weight on Earth

A toy has a mass of 3 units. What is its weight on Earth?

Use the rule:

$$W = m \times g$$

Substitute the numbers:

$$W = 3 \times 10$$

Multiply:

$$W = 30$$

Answer: The toy’s weight on Earth is 30 units.

Worked Example 2: Same mass, different weight on the Moon

The same toy has a mass of 3 units. What is its weight on the Moon?

Use the same rule:

$$W = m \times g$$

Now use Moon gravity:

$$W = 3 \times 2$$

Multiply:

$$W = 6$$

Answer: The toy’s weight on the Moon is 6 units.

Notice what changed:

• The mass stayed 3.
• The weight changed from 30 to 6 because gravity changed.

Worked Example 3: Comparing two objects

A book has a mass of 2 units. A box has a mass of 5 units. Which has more weight on Earth?

First, find the book’s weight:

$$W = 2 \times 10 = 20$$

Now, find the box’s weight:

$$W = 5 \times 10 = 50$$

Compare the answers:

• Book weight = 20 units
• Box weight = 50 units

Answer: The box has more weight on Earth because it has more mass.

Worked Example 4: Finding mass from weight

An object weighs 40 units on Earth. Its mass is unknown. What is its mass?

Start with the rule:

$$W = m \times g$$

We know:

• \(W = 40\)
• \(g = 10\)

So:

$$40 = m \times 10$$

To find \(m\), divide 40 by 10:

$$m = 4$$

Answer: The object’s mass is 4 units.

How gravity affects motion

Gravity is one reason objects move. If you drop a pencil, gravity pulls it downward. If you toss a ball up, gravity pulls it back down.

Without gravity, things would not fall the same way. Gravity helps control how objects move near Earth and other planets.

Earth, Moon, and other planets

Different worlds have different gravity. That means your weight would not be the same everywhere.

Here are some important ideas:

• On a place with stronger gravity, you would weigh more.
• On a place with weaker gravity, you would weigh less.
• Your mass would stay the same in both places.

This is why astronauts can bounce more easily on the Moon. Their mass is the same, but their weight is smaller because the Moon’s gravity is weaker.

Important differences to remember

• Gravity is a pulling force.
• Mass is how much matter is in an object.
• Weight is the force of gravity pulling on an object.

A quick comparison chart

• Mass: stays the same, tells how much matter
• Weight: can change, depends on gravity
• Gravity: the pull that causes weight

Common mistake to avoid

Sometimes people say mass and weight are the same thing. They are not.

If you say, “I weigh less on the Moon,” that is correct. If you say, “I have less mass on the Moon,” that is not correct. Your mass stays the same.

Let’s review with one more example

A ball has a mass of 4 units.

• On Earth: $$W = 4 \times 10 = 40$$
• On the Moon: $$W = 4 \times 2 = 8$$

The ball’s mass is still 4 units in both places. Only its weight changes.

Summary

Gravity is a force that pulls objects together. Earth’s gravity pulls things toward the ground.

Mass is the amount of matter in an object, and it stays the same wherever the object goes. Weight is the pull of gravity on that object, so weight can change from place to place.

Use the rule $$W = m \times g$$ to find weight. If gravity gets stronger, weight gets bigger. If gravity gets weaker, weight gets smaller.

Uniform Circular Motion is the motion of an object traveling in a circle at constant speed. Even though the speed stays the same, the object is still accelerating because its direction is constantly changing.

This idea is very important in mechanics because it shows that an object can accelerate without speeding up or slowing down. In uniform circular motion, the acceleration points toward the center of the circle. This is called centripetal acceleration.

Examples of uniform circular motion include a satellite orbiting Earth in a nearly circular path, a car moving around a circular track at constant speed, or a ball tied to a string and swung in a horizontal circle.

1. Why circular motion needs a force

According to Newton's first law, an object will keep moving in a straight line unless a force acts on it. So if an object is moving in a circle, something must be constantly changing its direction.

That force must act toward the center of the circle. A center-seeking force is called a centripetal force. Without it, the object would leave the circular path and move off in a straight line tangent to the circle.

The word centripetal means "center-seeking." It does not describe a new kind of force. Instead, it describes the role a force is playing. For example:

• Tension can provide centripetal force for a ball on a string.
• Friction can provide centripetal force for a car turning on a road.
• Gravity can provide centripetal force for planets and satellites.

2. Speed, velocity, and acceleration in a circle

It is important to distinguish between speed and velocity.

• Speed tells how fast an object moves.
• Velocity tells both speed and direction.

In uniform circular motion, the speed is constant, but the velocity changes because the direction changes at every point on the circle.

Since acceleration is the rate of change of velocity, a change in direction means there is acceleration. This acceleration always points toward the center of the circle.

3. Centripetal acceleration

The size of the centripetal acceleration is given by

where:

• \(a_c\) = centripetal acceleration in meters per second squared, \(\text{m/s}^2\)
• \(v\) = speed in meters per second, \(\text{m/s}\)
• \(r\) = radius of the circle in meters, \(\text{m}\)

This equation shows two important patterns:

• If speed increases, centripetal acceleration increases a lot because speed is squared.
• If the radius gets larger, centripetal acceleration gets smaller.

So a fast object moving in a small circle needs a very large inward acceleration.

4. Centripetal force

Newton's second law says that force equals mass times acceleration:

For circular motion, the inward force is the centripetal force:

Substituting the formula for centripetal acceleration gives

where:

• \(F_c\) = centripetal force in newtons, \(\text{N}\)
• \(m\) = mass in kilograms, \(\text{kg}\)
• \(v\) = speed in meters per second
• \(r\) = radius in meters

This force is not extra or separate from the real forces acting on the object. It is the net inward force that keeps the object moving in a circle.

5. Direction of velocity and force

At any point in circular motion:

• The velocity points tangent to the circle.
• The centripetal force points toward the center.
• The centripetal acceleration also points toward the center.

This means the force is perpendicular to the motion. A perpendicular force changes direction, not speed.

This is why an object in uniform circular motion can keep the same speed while still accelerating.

6. Period and frequency

Sometimes circular motion is described using period and frequency.

• Period, \(T\), is the time for one complete revolution.
• Frequency, \(f\), is the number of revolutions per second.

They are related by

In one full revolution, the object travels the circumference of the circle:

So the speed can also be written as

Since \(f = 1/T\), we can also write

These formulas are useful when the problem gives time per revolution or revolutions per second instead of speed.

7. Worked Example 1: Finding centripetal acceleration

A car moves at a constant speed of \(12\,\text{m/s}\) around a circular track of radius \(18\,\text{m}\). Find its centripetal acceleration.

Step 1: Write the formula

Step 2: Substitute the values

Step 3: Calculate

Answer: The centripetal acceleration is \(8\,\text{m/s}^2\), directed toward the center of the track.

Worked Example 2: Finding centripetal force

A \(0.50\,\text{kg}\) ball is whirled in a circle of radius \(1.2\,\text{m}\) at a speed of \(6.0\,\text{m/s}\). What centripetal force is needed?

Step 1: Use the formula

Step 2: Substitute

Step 3: Calculate

Answer: The required centripetal force is \(15\,\text{N}\), toward the center of the circle.

Worked Example 3: Using period to find speed and acceleration

A point on the edge of a rotating ride moves in a circle of radius \(4.0\,\text{m}\). It completes one revolution every \(2.0\,\text{s}\). Find:

1. the speed
2. the centripetal acceleration

Step 1: Find speed using period

Using \(\pi \approx 3.14\):

Step 2: Find centripetal acceleration

Answer:

• Speed \(\approx 12.6\,\text{m/s}\)
• Centripetal acceleration \(\approx 39.7\,\text{m/s}^2\)

Worked Example 4: Identifying the force that causes circular motion

A car turns around a flat circular road. What provides the centripetal force?

On a flat road, the force that points toward the center is friction between the tires and the road. That friction acts sideways on the car and keeps it turning.

If there is not enough friction, the car cannot get the needed centripetal force, so it will slide outward from the circular path.

8. Common misunderstandings

• Misunderstanding 1: "If speed is constant, acceleration is zero."
  Not true in circular motion. The direction changes, so velocity changes, so there is acceleration.
• Misunderstanding 2: "Centripetal force is a separate force."
  Centripetal force is the name for the net inward force. It can be tension, friction, gravity, or another force.
• Misunderstanding 3: "The force points in the direction of motion."
  In uniform circular motion, the centripetal force points toward the center, not forward.
• Misunderstanding 4: "An object wants to move outward."
  The object naturally continues in a straight-line path. The circular path happens only because an inward force keeps changing its direction.

9. How uniform circular motion connects to work and energy

In uniform circular motion, the centripetal force is perpendicular to the direction of motion. Because of this, the force changes the object's direction but not its speed.

That means the object's kinetic energy stays constant as long as the speed stays constant. So in uniform circular motion, the inward force does not increase or decrease the kinetic energy.

10. Key ideas to remember

• Uniform circular motion means moving in a circle at constant speed.
• Even at constant speed, the object accelerates because its direction changes.
• The acceleration is called centripetal acceleration and points toward the center.
• The required inward force is called centripetal force.
• The main formulas are:

Brief Summary

Uniform circular motion happens when an object moves around a circle at constant speed. Although the speed stays the same, the direction changes continuously, so the object has an acceleration toward the center of the circle. This inward acceleration is called centripetal acceleration, and it is caused by a centripetal force such as tension, friction, or gravity.

Universal Gravitation explains why objects with mass attract each other. It is the reason an apple falls to Earth, the Moon stays in orbit around Earth, and planets move around the Sun. In 11th Grade science, this idea connects forces, motion, and energy in one powerful rule.

Newton's law of universal gravitation says that every mass attracts every other mass. This attraction depends on two main things: the sizes of the masses and the distance between them.

The mathematical form of the law is:

In this equation:

• \(F\) is the gravitational force between two objects, measured in newtons (N).
• \(G\) is the universal gravitational constant: \(6.67 \times 10^{-11}\, \text{N·m}^2/\text{kg}^2\).
• \(m_1\) and \(m_2\) are the masses of the two objects, measured in kilograms (kg).
• \(r\) is the distance between the centers of the two objects, measured in meters (m).

This equation shows two important patterns:

• If either mass increases, the gravitational force increases.
• If the distance between the objects increases, the gravitational force decreases very quickly because distance is squared.

That squared distance matters a lot. If the distance doubles, the force becomes:

So doubling the distance makes the force one-fourth as strong. If the distance triples, the force becomes one-ninth as strong.

Gravity is always attractive. Unlike some other forces, gravity does not push masses apart. It always pulls them toward each other.

It is also important to understand that the force acts on both objects. If Earth pulls on the Moon, then the Moon pulls on Earth with the same size force in the opposite direction. This follows Newton's third law.

Even small objects attract each other gravitationally, but the force is usually too tiny to notice unless at least one of the masses is very large, like a planet, moon, or star.

Why do we feel weight? Your weight is the gravitational force between you and Earth. Near Earth's surface, this force is often written as:

Here, \(m\) is your mass and \(g\) is the gravitational field strength near Earth, about \(9.8\, \text{m/s}^2\) or \(9.8\, \text{N/kg}\).

This formula is actually a special case of universal gravitation. Near Earth's surface, the distance from you to Earth's center stays almost constant, so the gravitational force can be simplified to \(mg\).

If we use Newton's law for an object of mass \(m\) near Earth, we get:

Here, \(M_E\) is Earth's mass and \(r\) is the distance from Earth's center. Comparing this with \(F = mg\), we can write:

This tells us that gravity depends on the planet's mass and size. A more massive planet gives stronger gravity, but if its radius is also very large, that can reduce the gravity at its surface.

Planetary orbits happen because gravity provides the force that keeps planets and moons moving in curved paths. Without gravity, a planet would move in a straight line. Gravity continuously pulls it inward, changing its direction so it travels around the Sun.

For an object in a circular orbit, gravity acts as the centripetal force. That means:

In this equation:

• \(M\) is the mass of the large central object, such as the Sun or Earth.
• \(m\) is the mass of the orbiting object.
• \(r\) is the orbital radius.
• \(v\) is the orbital speed.

The orbiting object's mass cancels out, giving:

This means orbital speed depends on the mass of the central object and the distance from it. Objects closer to a planet or star must move faster to stay in orbit.

We can also find the orbital period, which is the time it takes to complete one orbit. For a circular orbit, the distance traveled in one orbit is the circumference \(2\pi r\), so:

If we substitute the orbital speed formula, we get:

This helps explain why planets farther from the Sun take longer to complete an orbit.

Key ideas to remember about universal gravitation are:

• Every mass attracts every other mass.
• The force is stronger for larger masses.
• The force is weaker when distance increases.
• Gravity is the force that keeps planets, moons, and satellites in orbit.
• Weight is the gravitational force on an object near a planet's surface.

Now let's work through some examples.

Worked Example 1: Comparing force when distance changes

Two objects are separated by a distance \(r\). If the distance becomes \(3r\), how does the gravitational force change?

Step 1: Write the original relationship.

Since gravitational force is inversely proportional to \(r^2\):

Step 2: Replace \(r\) with \(3r\).

Step 3: Compare the new force to the old force.

Answer: The gravitational force becomes one-ninth of the original force.

Worked Example 2: Calculating gravitational force between two masses

Find the gravitational force between two masses of \(10\, \text{kg}\) and \(20\, \text{kg}\) separated by \(2.0\, \text{m}\).

Step 1: List the known values.

• \(m_1 = 10\, \text{kg}\)
• \(m_2 = 20\, \text{kg}\)
• \(r = 2.0\, \text{m}\)
• \(G = 6.67 \times 10^{-11}\, \text{N·m}^2/\text{kg}^2\)

Step 2: Use Newton's law of gravitation.

Step 3: Substitute the values.

Answer: The gravitational force is approximately \(3.34 \times 10^{-9}\, \text{N}\).

This is an extremely small force, which is why gravity between everyday objects is not noticeable.

Worked Example 3: Finding gravitational field strength near a planet

A planet has mass \(6.0 \times 10^{24}\, \text{kg}\) and radius \(6.4 \times 10^6\, \text{m}\). Find the gravitational field strength at its surface.

Step 1: Use the formula for gravitational field strength.

Step 2: Substitute the values.

Step 3: Square the radius.

Step 4: Continue the calculation.

Answer: The gravitational field strength is about \(9.8\, \text{m/s}^2\).

This is very close to Earth's gravitational field strength, so this planet would have a surface gravity similar to Earth.

Worked Example 4: Orbital speed of a satellite

A satellite orbits Earth at a distance of \(7.0 \times 10^6\, \text{m}\) from Earth's center. Earth's mass is \(6.0 \times 10^{24}\, \text{kg}\). Find the satellite's orbital speed.

Step 1: Use the orbital speed formula.

Step 2: Substitute the values.

Step 3: Multiply the values in the numerator.

Step 4: Divide by \(r\).

Step 5: Take the square root.

Answer: The orbital speed is about \(7.6 \times 10^3\, \text{m/s}\), or 7.6 km/s.

This shows that satellites must move very fast to stay in orbit around Earth.

Common mistakes to avoid

• Do not forget to square the distance in the denominator.
• Use the distance between the centers of objects, not just the distance between their surfaces.
• Make sure all units are in SI units: kilograms, meters, and newtons.
• Do not confuse mass with weight. Mass is the amount of matter; weight is the gravitational force on that mass.
• When solving orbit problems, remember that gravity is what provides the centripetal force.

Brief Summary

Newton's law of universal gravitation states that every two masses attract each other with a force given by \(F = G\frac{m_1m_2}{r^2}\). This force becomes stronger with larger masses and weaker with greater distance. Near a planet's surface, gravity gives objects weight, and in space, gravity keeps moons, planets, and satellites in orbit.

Lesson: Work and Mechanical Power

In everyday language, the word work can mean any effort, such as studying or cleaning. In science, however, work has a very specific meaning. Work is done when a force causes an object to move through a distance. If there is no movement, then no mechanical work is done, even if you feel tired.

This idea is important in classical mechanics because work connects force, motion, and energy. When work is done on an object, energy is transferred to or from that object. Closely related to work is power, which tells us how quickly that work is done.

In this lesson, you will learn what work and mechanical power mean, how to calculate them, what units they use, and how to solve problems involving these ideas.

1. What is Work?

In physics, work happens when all of the following are true:

• A force acts on an object,
• The object moves, and
• At least part of the force is in the direction of the motion.

If a force acts but the object does not move, the work done by that force is zero.

The formula for work is:

where:

• W = work done
• F = force applied
• d = displacement of the object
• \theta = angle between the force and the direction of motion

When the force is in exactly the same direction as the motion, then \(\theta = 0^\circ\), and since \(\cos 0^\circ = 1\), the formula becomes:

This simpler form is often used in introductory problems.

2. Units of Work

The SI unit of work is the joule, abbreviated as J.

One joule is equal to one newton of force applied through one meter:

So if you push with a force of 10 N and the object moves 3 m in the direction of the force, the work done is 30 J.

3. Positive, Negative, and Zero Work

Work can be positive, negative, or zero.

• Positive work happens when the force and displacement are in the same direction.
• Negative work happens when the force acts opposite to the displacement.
• Zero work happens when there is no displacement or when the force is perpendicular to the motion.

For example, friction usually does negative work because it opposes motion. A person carrying a backpack while walking on level ground does no work on the backpack in the horizontal direction if the force they apply is upward and the motion is horizontal.

4. When is No Work Done?

Students often think that any force means work is being done, but that is not always true. Here are important cases where work is zero:

• The object does not move.
• The force is perpendicular to the displacement.
• The applied force has no component in the direction of motion.

For example, if you push hard on a wall and it does not move, then the displacement is zero, so:

5. Work and Energy

Work is a way of transferring energy. If you do positive work on an object, the object gains energy. If negative work is done on an object, the object loses energy.

This is why work is closely connected to mechanical energy. For example, when you lift an object, you do work against gravity, increasing the objects gravitational potential energy. When you push a cart and it speeds up, your work increases its kinetic energy.

6. What is Mechanical Power?

While work tells us how much energy is transferred, power tells us how fast that energy is transferred.

The formula for average power is:

where:

• P = power
• W = work done
• t = time taken

The SI unit of power is the watt, abbreviated as W.

One watt means one joule of work done each second:

A machine with greater power does not always do more total work. It may simply do the same amount of work in less time.

7. Another Useful Power Formula

When a force causes motion at a constant speed in the same direction, power can also be written as:

where:

• F = force
• v = velocity in the direction of the force

This formula is useful for moving objects like cars, elevators, and conveyor belts.

If the force and velocity are not in the same direction, only the component of force in the direction of motion contributes to power.

8. Worked Examples

Example 1: Simple Work Calculation

A student pushes a box with a force of 25 N across the floor for 4 m in the same direction as the force. How much work is done?

Step 1: Write the formula.

Step 2: Substitute the values.

Step 3: Calculate.

Answer: The work done is 100 J.

Example 2: Work with a Force Opposing Motion

A friction force of 8 N acts on a sliding book as it moves 5 m forward. How much work does friction do?

Since friction acts opposite to the direction of motion, the angle is \(180^\circ\). Because \(\cos 180^\circ = -1\):

Answer: Friction does -40 J of work. The negative sign shows that friction removes mechanical energy from the object.

Example 3: Average Power

A motor does 1200 J of work in 6 s. Find its average power.

Step 1: Use the power formula.

Step 2: Substitute the values.

Step 3: Calculate.

Answer: The average power is 200 W.

Example 4: Using \(P = Fv\)

A machine pulls a cart with a force of 300 N at a constant speed of 2.5 m/s in the same direction. What power does the machine produce?

Step 1: Use the formula.

Step 2: Substitute the values.

Step 3: Calculate.

Answer: The machine produces 750 W of power.

9. Common Mistakes to Avoid

• Confusing force with work: Force is a push or pull, but work requires both force and displacement.
• Forgetting direction: If force acts opposite to motion, the work is negative.
• Using the wrong unit: Work is measured in joules, while power is measured in watts.
• Ignoring time in power problems: Power depends on how quickly work is done.
• Assuming effort means work: You may feel tired holding a heavy object still, but if it does not move, no mechanical work is done on it.

10. Key Ideas to Remember

• Work is done when a force causes displacement.
• The formula for work is \(W = Fd\cos\theta\).
• Work is measured in joules \((\text{J})\).
• Positive work adds energy; negative work removes energy.
• Power measures the rate of doing work.
• The formula for average power is \(P = \frac{W}{t}\).
• Power is measured in watts \((\text{W})\).
• When force and motion are in the same direction, power can be found using \(P = Fv\).

Brief Summary

Work and mechanical power are key ideas in mechanics. Work measures how much energy is transferred when a force moves an object, and power measures how quickly that transfer happens. By understanding the formulas, units, and direction of force, you can solve many real-world motion problems more confidently.

Centripetal Force and Circular Motion

Have you ever swung a bucket of water in a circle, watched a toy car go around a track, or seen the Moon move around Earth? These are all examples of circular motion. Circular motion means something is moving around in a circle.

But here is the big idea: an object does not go in a circle all by itself. To keep moving in a circle, it needs a force pulling or pushing inward. This inward force is called centripetal force.

The word centripetal means center-seeking. That helps us remember what this force does. It pulls or pushes toward the center of the circle.

Introduction: Why objects move in circles

If you roll a ball on the ground, it usually moves straight. Things like to keep moving in a straight line unless a force changes their motion.

So if something is moving in a circle, its direction must keep changing every moment. That change in direction happens because of a force pointing inward, toward the center of the circle.

Without that inward force, the object would stop going in a circle and move off in a straight path.

Main Teaching Point 1: What is circular motion?

Circular motion is motion around a center point. The object may move fast or slow, but if it keeps turning around the center, it is moving in a circle.

• A ball on a string being swung around
• A rider on a merry-go-round
• A car turning around a curved road
• The Moon moving around Earth

In each case, the object is not just moving forward. It is also being pulled or pushed inward so it keeps curving.

Main Teaching Point 2: What is centripetal force?

Centripetal force is the inward force that keeps something moving in a circle.

This force is not a new kind of force all by itself. It can be caused by different things:

• Tension from a string
• Friction between tires and a road
• Gravity between Earth and the Moon
• A push from a track or wall

The important thing is that the force points toward the center.

Main Teaching Point 3: What happens if the inward force stops?

If the inward force suddenly stops, the object will no longer move in a circle. It will move in a straight line.

That straight line points in the direction the object was going at that moment. You can think of it like letting go of a string while swinging a ball. The ball does not keep going in a circle. It flies off straight.

Main Teaching Point 4: Speed and circular motion

An object can move in a circle slowly or quickly. If it moves faster, it usually needs more inward force to stay in the circle.

If there is not enough inward force, the object may slide outward from the curved path. For example, a car that turns too fast on a curve may skid.

We can say this idea simply with words and a small math pattern:

More speed means more centripetal force is needed.

Scientists often write this idea like this:

This means the needed inward force gets bigger as speed gets bigger. For 4th grade, the main thing to remember is: faster circle motion needs stronger inward force.

Main Teaching Point 5: Bigger circles and smaller circles

The size of the circle also matters. A smaller circle needs a stronger inward pull to make the object turn more sharply.

A larger circle has a gentler turn, so it may need less inward force if the speed stays the same.

Main Teaching Point 6: Real-world examples

1. Ball on a string

When you swing a ball on a string, the string pulls inward. That pull is the centripetal force.

2. Car on a curve

When a car turns, friction between the tires and the road helps pull the car inward around the curve.

3. Moon around Earth

The Moon moves around Earth because gravity pulls it inward. Gravity is the centripetal force in this case.

4. Roller coaster or curved track

The track pushes on the car and changes its direction, helping it move in a circle or loop.

Main Teaching Point 7: What are banked curves?

Some roads and tracks are tilted on curves. These are called banked curves.

A banked curve helps a car turn more safely. The tilted road gives part of the push needed to guide the car inward around the curve.

This is why race tracks and some highway curves are not flat. The tilt helps with circular motion.

Main Teaching Point 8: Rotational inertia, in a simple way

Rotational inertia is a fancy way to talk about how an object resists changes in turning motion.

For 4th grade, we can think of it this way: heavier things, or things with mass spread out farther, are harder to start turning, stop turning, or change their turning.

For example, a small toy wheel is easier to spin than a heavy merry-go-round with children on it. The bigger, heavier turning system is harder to change.

You do not need to memorize the big words. Just remember: some objects are harder to spin or turn than others.

Worked Example 1: Swinging a ball on a string

Question: Mia swings a ball on a string in a circle. What provides the centripetal force?

Step 1: Ask what is pulling the ball inward.

Step 2: The string is attached to the ball and pulls it toward the center.

Answer: The pull of the string provides the centripetal force.

Worked Example 2: Car turning on a road

Question: A car goes around a curve. What helps provide the centripetal force?

Step 1: Think about what keeps the car from sliding straight.

Step 2: The tires grip the road.

Step 3: That grip is called friction.

Answer: Friction between the tires and the road helps provide the centripetal force.

Worked Example 3: What if the force stops?

Question: Leo is swinging a rubber stopper on a string. The string breaks. What happens next?

Step 1: The string was providing the inward force.

Step 2: When the string breaks, the inward force is gone.

Step 3: Without the inward force, the stopper cannot keep moving in a circle.

Answer: The stopper flies off in a straight line.

Worked Example 4: Faster turning

Question: Two toy cars go around circles of the same size. One car moves slowly, and one car moves quickly. Which car needs more centripetal force?

Step 1: Compare the speeds.

Step 2: The faster car changes direction more quickly.

Step 3: Faster circular motion needs more inward force.

Answer: The faster car needs more centripetal force.

Helpful Math Idea

Sometimes we compare how long it takes to go around one circle. This is called the time for one lap.

If one object goes around in 2 seconds and another goes around in 4 seconds, the first object is moving around the circle faster.

We can write simple comparisons like this:

So the object taking 2 seconds for one lap is the faster one.

Things to Remember

• Circular motion means moving around in a circle.
• Centripetal force is a force that points inward, toward the center.
• Without centripetal force, an object moves in a straight line.
• Faster motion in a circle needs more inward force.
• Strings, gravity, friction, and tracks can all provide centripetal force.
• Banked curves help objects turn.
• Some objects are harder to start or stop turning because of rotational inertia.

Quick Check

1. What does centripetal mean?
2. What direction does centripetal force point?
3. What provides the centripetal force for the Moon moving around Earth?
4. If a car takes a curve faster, does it need more or less inward force?

Answers to Quick Check

1. Center-seeking.
2. Toward the center of the circle.
3. Gravity.
4. More inward force.

Brief Summary

Objects move in circles only when a force keeps pulling or pushing them inward. That inward force is called centripetal force. It can come from a string, friction, gravity, or a track. If the force stops, the object moves straight. Faster circular motion needs more inward force.

Momentum and Collisions

Have you ever noticed that a fast-moving soccer ball can be hard to stop, but a slow-rolling ball is easy to catch? Or that a heavy shopping cart is harder to stop than an empty one? This happens because moving objects have momentum.

Momentum is a way to describe how much motion an object has. An object with more mass or more speed has more momentum. Big, fast objects usually have a lot of momentum. Small, slow objects usually have less.

Momentum helps us understand what happens when objects collide, or bump into each other. In a collision, momentum can move from one object to another. By studying momentum, we can make good predictions about how objects will move after they crash, bounce, or stick together.

1. What is momentum?

Momentum depends on two things:

• Mass — how much matter is in an object
• Velocity — speed in a certain direction

The rule for momentum is:

$$\text{momentum} = \text{mass} \times \text{velocity}$$

We can write it like this:

$$p = m \times v$$

In this formula:

• \(p\) means momentum
• \(m\) means mass
• \(v\) means velocity

If mass gets bigger, momentum gets bigger. If velocity gets bigger, momentum also gets bigger.

For example, if two bicycles move at the same speed, the heavier one has more momentum. If two balls have the same mass, the faster one has more momentum.

2. Direction matters

Momentum includes direction because velocity includes direction. That means an object moving to the right has different momentum from an object moving to the left.

We can use positive and negative numbers to show direction:

• Right might be positive
• Left might be negative

So if a ball has momentum of \(+12\), it might be moving right. If it has momentum of \(-12\), it has the same amount of momentum but is moving left.

3. How to calculate momentum

To find momentum:

1. Find the mass of the object.
2. Find the velocity of the object.
3. Multiply them.

Worked Example 1

A toy car has a mass of 2 kg and moves to the right at 3 m/s. What is its momentum?

Use the formula:

$$p = m \times v$$

Substitute the numbers:

$$p = 2 \times 3 = 6$$

The momentum is 6 kg·m/s to the right.

Worked Example 2

A ball has a mass of 4 kg and rolls to the left at 2 m/s. What is its momentum?

If we choose left as negative, then \(v = -2\).

$$p = m \times v$$

$$p = 4 \times (-2) = -8$$

The momentum is -8 kg·m/s. The negative sign tells us the ball is moving left.

4. What is a collision?

A collision happens when two objects hit each other. They might:

• bounce apart, like pool balls
• slow down and move away
• stick together, like two pieces of clay

During a collision, pushes and pulls happen very quickly. The objects can change speed or change direction. But if no outside force changes the total motion, the total momentum stays the same.

This is called conservation of momentum.

5. Conservation of momentum

Conservation of momentum means:

$$\text{total momentum before} = \text{total momentum after}$$

This does not mean each object keeps the same momentum. It means the whole system together keeps the same total momentum.

For two objects, we can write:

$$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$$

You do not need to memorize all the symbols. Just remember this idea:

• Add the momentum of all objects before the collision.
• Add the momentum of all objects after the collision.
• The totals should match.

6. Two common kinds of collisions

Elastic collision

In an elastic collision, the objects bounce off each other. A good simple example is two marbles hitting and bouncing apart.

At this grade level, the most important idea is that in an elastic collision:

• the objects bounce
• the total momentum stays the same

Inelastic collision

In an inelastic collision, the objects do not bounce apart normally. They may stick together or move together after the crash.

A simple example is two lumps of clay that crash and stick.

In an inelastic collision:

• the objects may stick together
• the total momentum still stays the same

7. Predicting collisions

Momentum helps us predict what will happen after objects collide.

If one object has a lot of momentum and hits a lighter object that is not moving, the lighter object will usually move away quickly. If two equal objects collide, they may trade motion in a simple way.

When objects stick together, we add their momentum before the collision, then share that total momentum across the combined mass after the collision.

Worked Example 3: Inelastic collision

A 3 kg cart moves right at 4 m/s. It crashes into a 1 kg cart that is standing still. The carts stick together. What happens after the collision?

Step 1: Find momentum before the collision.

First cart:

$$p_1 = 3 \times 4 = 12$$

Second cart:

It is not moving, so its velocity is 0.

$$p_2 = 1 \times 0 = 0$$

Total momentum before:

$$12 + 0 = 12$$

Step 2: Find the total mass after the collision.

They stick together, so:

$$3 + 1 = 4\text{ kg}$$

Step 3: Find the new velocity.

The total momentum after must still be 12.

$$p = m \times v$$

$$12 = 4 \times v$$

$$v = 3\text{ m/s}$$

After the collision, the two carts move together at 3 m/s to the right.

Why did they slow down?

The moving cart had to share its momentum with the other cart. Because the total mass became larger, the new speed became smaller.

Worked Example 4: Elastic collision idea

A 2 kg ball moves right at 5 m/s and hits another 2 kg ball that is standing still. After they collide, the first ball stops. What is the momentum of the second ball?

Step 1: Find total momentum before.

First ball:

$$p_1 = 2 \times 5 = 10$$

Second ball:

$$p_2 = 2 \times 0 = 0$$

Total before:

$$10 + 0 = 10$$

Step 2: Use conservation of momentum.

After the collision, the first ball stops, so its momentum is 0.

That means the second ball must have momentum of 10 so the total stays the same.

Step 3: Find the second ball's velocity.

$$p = m \times v$$

$$10 = 2 \times v$$

$$v = 5\text{ m/s}$$

The second ball moves at 5 m/s to the right.

This is a simple example of an elastic collision, where the objects bounce and momentum is conserved.

8. Important ideas to remember

• Momentum is the amount of motion an object has.
• Momentum depends on mass and velocity.
• The formula is $$p = m \times v$$
• Direction matters. Right and left have different signs.
• In a collision, total momentum before and after stays the same.
• In an elastic collision, objects bounce.
• In an inelastic collision, objects may stick together.

9. Real-life examples

• A bowling ball has more momentum than a tennis ball moving at the same speed because it has more mass.
• A fast skateboard has more momentum than a slow skateboard with the same rider.
• Train cars that connect after bumping are like an inelastic collision.
• Billiard balls that bounce apart are like an elastic collision.

10. Quick check for yourself

• If a truck and a bicycle move at the same speed, which has more momentum?
• If the same ball rolls faster, does its momentum get bigger or smaller?
• When objects collide, what stays the same for the whole system?
• If two objects stick together after a collision, is that elastic or inelastic?

Summary

Momentum tells us how much motion an object has, and we calculate it with $$p = m \times v$$. More mass or more speed means more momentum. Because direction matters, momentum can be positive or negative.

When objects collide, the total momentum of the system stays the same. In an elastic collision, objects bounce apart. In an inelastic collision, objects may stick together. By using momentum, we can predict how objects move before and after a collision.

Kinetic and Potential Energy are two main forms of mechanical energy. They help us describe how objects move and how energy can be stored and transferred.

In everyday life, you see these ideas everywhere. A moving bicycle has kinetic energy. A book held above the floor has gravitational potential energy. A stretched rubber band has elastic potential energy.

This lesson explains what kinetic energy and potential energy are, how to calculate them, and how they change from one form into another.

1. What is Energy?

Energy is the ability to cause change or do work. In mechanics, we often focus on mechanical energy, which includes the energy of motion and stored energy due to position or shape.

The two important types in this lesson are:

• Kinetic energy: energy an object has because it is moving
• Potential energy: stored energy due to position or condition

2. Kinetic Energy

Kinetic energy depends on an object's mass and speed. A heavier object has more kinetic energy than a lighter one moving at the same speed. Also, if an object moves faster, its kinetic energy increases greatly.

The formula for kinetic energy is:

$$KE = \frac{1}{2}mv^2$$

where:

• \(KE\) = kinetic energy in joules (J)
• \(m\) = mass in kilograms (kg)
• \(v\) = speed in meters per second (m/s)

Important idea: speed is squared. This means if speed doubles, kinetic energy becomes four times as large.

3. Gravitational Potential Energy

Gravitational potential energy is the energy stored because of an object's height above a reference point, usually the ground.

The higher an object is, the more gravitational potential energy it has. A heavier object also has more gravitational potential energy at the same height.

The formula is:

$$GPE = mgh$$

where:

• \(GPE\) = gravitational potential energy in joules (J)
• \(m\) = mass in kilograms (kg)
• \(g\) = gravitational field strength, about \(9.8\,\text{m/s}^2\) on Earth
• \(h\) = height in meters (m)

If an object is lifted higher, work is done against gravity, and that energy is stored as gravitational potential energy.

4. Elastic Potential Energy

Elastic potential energy is stored when an object is stretched or compressed, such as a spring, bow, or rubber band.

For an ideal spring, the elastic potential energy is:

$$EPE = \frac{1}{2}kx^2$$

where:

• \(EPE\) = elastic potential energy in joules (J)
• \(k\) = spring constant in newtons per meter (N/m)
• \(x\) = stretch or compression distance in meters (m)

This formula shows that stretching a spring twice as far stores four times as much elastic potential energy.

5. Units of Energy

All forms of energy in this lesson are measured in joules, written as \(J\).

One joule is a small amount of energy, so larger situations may involve many joules.

6. How Energy Changes Form

One of the most important ideas in mechanics is that energy can be transferred and converted from one form to another.

For example:

• A ball held in the air has gravitational potential energy.
• When it falls, that potential energy changes into kinetic energy.
• Just before it hits the ground, most of the energy is kinetic.

Similarly:

• A stretched bow stores elastic potential energy.
• When released, that energy changes into the kinetic energy of the arrow.

7. Conservation of Mechanical Energy

If we ignore friction and air resistance, the total mechanical energy of a system stays constant. This is called conservation of mechanical energy.

That means:

$$\text{Total Mechanical Energy} = KE + PE$$

and in an ideal situation:

$$KE_i + PE_i = KE_f + PE_f$$

The subscripts \(i\) and \(f\) mean initial and final.

This does not mean energy disappears. It means energy changes form while the total stays the same.

8. Real-World Effects: Friction and Air Resistance

In real situations, friction and air resistance often cause some mechanical energy to change into heat and sound. Because of this, the kinetic energy and potential energy alone may not stay constant.

For example, when a ball rolls across the floor, friction slowly reduces its kinetic energy. The energy is not destroyed. It is transferred mainly as heat.

9. Key Relationships to Remember

• Kinetic energy increases with mass.
• Kinetic energy increases with the square of speed.
• Gravitational potential energy increases with mass.
• Gravitational potential energy increases with height.
• Elastic potential energy increases with the square of stretch or compression distance.

10. Worked Example 1: Finding Kinetic Energy

A \(2.0\,\text{kg}\) ball rolls at \(3.0\,\text{m/s}\). Find its kinetic energy.

Use the formula:

$$KE = \frac{1}{2}mv^2$$

Substitute the values:

$$KE = \frac{1}{2}(2.0)(3.0)^2$$

$$KE = 1.0 \times 9.0 = 9.0\,\text{J}$$

Answer: The ball has \(9.0\,\text{J}\) of kinetic energy.

11. Worked Example 2: Finding Gravitational Potential Energy

A \(5.0\,\text{kg}\) backpack is placed on a shelf \(2.0\,\text{m}\) high. Find its gravitational potential energy.

Use:

$$GPE = mgh$$

Substitute the values:

$$GPE = (5.0)(9.8)(2.0)$$

$$GPE = 98\,\text{J}$$

Answer: The backpack has \(98\,\text{J}\) of gravitational potential energy relative to the floor.

12. Worked Example 3: Using Conservation of Mechanical Energy

A \(1.5\,\text{kg}\) rock is dropped from a height of \(10\,\text{m}\). Ignore air resistance. What is its kinetic energy just before it hits the ground?

At the top, the rock is not moving yet, so its initial kinetic energy is \(0\). Its energy is all gravitational potential energy.

First find the initial gravitational potential energy:

$$GPE = mgh = (1.5)(9.8)(10) = 147\,\text{J}$$

Since mechanical energy is conserved, just before hitting the ground the gravitational potential energy has changed into kinetic energy:

$$KE = 147\,\text{J}$$

Answer: The rock's kinetic energy just before impact is \(147\,\text{J}\).

13. Worked Example 4: Elastic Potential Energy

A spring with spring constant \(k = 200\,\text{N/m}\) is compressed by \(0.10\,\text{m}\). Find the elastic potential energy stored.

Use:

$$EPE = \frac{1}{2}kx^2$$

Substitute the values:

$$EPE = \frac{1}{2}(200)(0.10)^2$$

$$EPE = 100(0.01) = 1.0\,\text{J}$$

Answer: The spring stores \(1.0\,\text{J}\) of elastic potential energy.

14. Comparing the Different Forms

It is useful to compare the three energy equations from this lesson:

• $$KE = \frac{1}{2}mv^2$$
• $$GPE = mgh$$
• $$EPE = \frac{1}{2}kx^2$$

These formulas show what each type of energy depends on:

• Kinetic energy depends on mass and speed.
• Gravitational potential energy depends on mass, gravity, and height.
• Elastic potential energy depends on spring stiffness and how far it is stretched or compressed.

15. Common Mistakes to Avoid

• Using mass in grams instead of kilograms. Always convert to kilograms first.
• Forgetting to square the speed in the kinetic energy formula.
• Using the wrong height. Height must be measured from the chosen reference point.
• Forgetting units. Energy should be given in joules.
• Mixing speed and velocity ideas. In kinetic energy, the formula uses speed, so direction does not matter.

16. Quick Check for Understanding

1. If the speed of a car doubles, by what factor does its kinetic energy change?
2. Which has more gravitational potential energy: a \(3\,\text{kg}\) object at \(4\,\text{m}\) or a \(6\,\text{kg}\) object at \(2\,\text{m}\)?
3. When a stretched spring is released, what form of energy is usually produced?
4. If friction is ignored, what happens to total mechanical energy?

Answers:

1. It becomes 4 times as large.
2. They are equal, because \(mgh\) is the same in both cases.
3. Usually kinetic energy.
4. It stays constant.

17. Summary

Kinetic energy is the energy of motion, given by \(KE = \frac{1}{2}mv^2\). Potential energy is stored energy. In this lesson, the two main kinds were gravitational potential energy, \(GPE = mgh\), and elastic potential energy, \(EPE = \frac{1}{2}kx^2\).

These forms of energy can change into one another. In ideal conditions without friction or air resistance, total mechanical energy is conserved. Understanding these ideas helps explain falling objects, moving vehicles, springs, and many other everyday physical systems.

Conservation of Momentum is a big idea about what happens when objects bump, hit, or stick together.

Momentum is the motion energy of a moving object. An object with more mass or more speed has more momentum.

We can find momentum with a simple math rule:

$$\text{momentum} = \text{mass} \times \text{speed}$$

We can also write it like this:

$$p = m \times v$$

Here, \(p\) means momentum, \(m\) means mass, and \(v\) means speed.

The word conservation means something stays the same. So, conservation of momentum means that in a closed situation, the total momentum stays the same before and after objects collide.

This does not mean each object keeps the same momentum. It means the total amount together stays the same.

Think of two toy carts on a smooth track. One cart rolls into another cart. After the hit, they may bounce apart or stick together. Even though their speeds change, the total momentum of both carts together stays the same.

Important idea: We use conservation of momentum best when outside pushes and pulls are very small. That means we are looking at a system that is almost isolated, or closed off from outside forces.

For 4th Grade, you can think of an isolated system like this: nothing from the outside is changing the motion very much. The objects mostly only push on each other.

Main Teaching Points

1. Momentum depends on mass and speed.

• A heavy object moving slowly can have a lot of momentum.
• A light object moving fast can also have momentum.
• If mass gets bigger, momentum gets bigger.
• If speed gets bigger, momentum gets bigger.

For example:

$$p = m \times v$$

If a cart has mass \(2\) and speed \(3\), then:

$$p = 2 \times 3 = 6$$

Its momentum is \(6\).

2. Total momentum means adding the momentum of all the objects.

If two objects are moving, we add their momentum together to get the total momentum.

Before a collision:

$$\text{total momentum before} = p_1 + p_2$$

After a collision:

$$\text{total momentum after} = p_1 + p_2$$

And conservation of momentum says:

$$\text{total momentum before} = \text{total momentum after}$$

3. Direction matters.

If one object moves to the right and another moves to the left, they are moving in opposite directions.

To keep track of direction, we can say:

• Right is positive
• Left is negative

So if an object has momentum \(8\) to the right, we can write \(+8\).

If an object has momentum \(8\) to the left, we can write \(-8\).

This helps us add momentum correctly.

4. Collisions can happen in different ways.

There are two common kinds of collisions we can talk about:

• Elastic collision: the objects bounce apart after they hit.
• Inelastic collision: the objects stick together, or do not bounce apart much.

In both kinds of collisions, total momentum stays the same.

So even if objects bounce or stick, conservation of momentum still works.

5. Momentum can move from one object to another.

When one object hits another, some or all of its momentum can be passed along.

That is why one moving billiard ball can make another ball move. It is also why one rolling toy car can make another toy car start rolling.

The first object may slow down, and the second object may speed up. But the total momentum stays the same.

Worked Examples

Example 1: Finding the momentum of one object

A toy cart has mass \(3\) and speed \(2\).

Use the rule:

$$p = m \times v$$

Substitute the numbers:

$$p = 3 \times 2 = 6$$

Answer: The cart has momentum \(6\).

This is the first step in solving momentum problems. We find each object's momentum first.

Example 2: One moving cart hits a cart at rest and they stick together

Cart A has mass \(2\) and speed \(4\) to the right.

Cart B has mass \(2\) and speed \(0\).

They collide and stick together.

First, find the total momentum before the collision.

Cart A momentum:

$$p_A = 2 \times 4 = 8$$

Cart B momentum:

$$p_B = 2 \times 0 = 0$$

Total before:

$$8 + 0 = 8$$

Since momentum is conserved, the total after the collision must also be \(8\).

Now the carts are stuck together, so their masses add:

$$2 + 2 = 4$$

We know:

$$\text{momentum after} = \text{mass together} \times \text{new speed}$$

So:

$$8 = 4 \times \text{new speed}$$

$$\text{new speed} = 2$$

Answer: The stuck-together carts move at speed \(2\) to the right.

Notice that the speed changed, but the total momentum stayed the same.

Example 3: Two carts move toward each other

Let right be positive and left be negative.

Cart A has mass \(3\) and speed \(3\) to the right.

Cart B has mass \(1\) and speed \(5\) to the left.

Find the total momentum before the collision.

Cart A momentum:

$$p_A = 3 \times 3 = +9$$

Cart B momentum:

$$p_B = 1 \times 5 = -5$$

Total before:

$$+9 + (-5) = 4$$

So the total momentum is \(4\) to the right.

If the carts stick together, their total mass is:

$$3 + 1 = 4$$

Now use conservation of momentum:

$$4 = 4 \times \text{new speed}$$

$$\text{new speed} = 1$$

Answer: After they stick together, they move at speed \(1\) to the right.

This example shows why direction matters. One cart's momentum worked against the other's because they were moving opposite ways.

Example 4: A bounce-apart collision

Cart A has mass \(2\) and speed \(3\) to the right.

Cart B has mass \(2\) and is standing still.

After they collide, Cart A stops, and Cart B moves away.

Find the total momentum before.

Cart A momentum:

$$p_A = 2 \times 3 = 6$$

Cart B momentum:

$$p_B = 2 \times 0 = 0$$

Total before:

$$6 + 0 = 6$$

Momentum is conserved, so total after is also \(6\).

After the collision, Cart A stops, so its momentum is \(0\).

That means Cart B must have momentum \(6\).

Use:

$$p = m \times v$$

$$6 = 2 \times v$$

$$v = 3$$

Answer: Cart B moves at speed \(3\) to the right.

This is a simple example of an elastic collision, where the objects bounce apart instead of sticking together.

How to Solve Conservation of Momentum Problems

1. Find each object's mass and speed.
2. Decide the direction. Choose one direction as positive.
3. Find each object's momentum using \(p = m \times v\).
4. Add the momenta to get the total before the collision.
5. Use conservation of momentum: total before = total after.
6. Solve for the missing speed or momentum.

This pattern works for many collision questions.

Common Mistakes to Watch For

• Forgetting to add all objects together. We care about the total momentum of the whole system.
• Forgetting direction. Momentum to the left and momentum to the right should not be treated the same.
• Mixing up speed and momentum. Speed can change a lot, but total momentum stays the same.
• Thinking momentum stays the same for each object. It is the total that stays the same.

Real-Life Connections

• A bowling ball hits pins and transfers momentum to them.
• A shopping cart bumps another cart and makes it move.
• Two skaters pushing off each other move in opposite directions.
• A ball bouncing off another ball shows momentum changing hands.

These examples help us see that momentum is all around us whenever moving objects interact.

Brief Summary

Momentum is found by multiplying mass and speed: \(p = m \times v\).

Conservation of momentum means that when objects collide in a closed system, the total momentum before the collision equals the total momentum after the collision.

This is true when objects bounce apart and when they stick together.

To solve problems, find each object's momentum, pay attention to direction, add them together, and set total before equal to total after.

Conservation of Mechanical Energy is one of the most useful ideas in classical mechanics. It helps us understand motion by tracking how energy changes form instead of tracking every force at every moment.

In many situations, an object’s energy changes back and forth between kinetic energy and potential energy. If no energy is lost to friction, air resistance, or other non-conservative forces, then the total mechanical energy stays constant.

This lesson explains what mechanical energy is, when it is conserved, how to solve problems using it, and how to recognize when the method can and cannot be used.

1. What is mechanical energy?

Mechanical energy is the energy an object has because of its motion and its position.

• Kinetic energy is energy of motion.
• Potential energy is stored energy due to position or shape.

The total mechanical energy is:

$$E_{mech} = K + U$$

where:

• (K) is kinetic energy
• (U) is potential energy

For motion near Earth’s surface, the most common forms are:

$$K = \frac{1}{2}mv^2$$

$$U_g = mgh$$

Here, (m) is mass, (v) is speed, (g) is gravitational field strength (about (9.8\,\text{m/s}^2)), and (h) is height.

If a spring is involved, elastic potential energy may also appear:

$$U_s = \frac{1}{2}kx^2$$

where (k) is the spring constant and (x) is the amount the spring is stretched or compressed.

2. What does “conservation” mean?

When a quantity is conserved, its total amount stays the same. For mechanical energy, this means:

$$K_i + U_i = K_f + U_f$$

The subscripts i and f mean initial and final.

This equation says that if energy changes from one form to another, the total remains constant.

For example:

• As an object falls, gravitational potential energy decreases.
• At the same time, kinetic energy increases.
• The total mechanical energy stays the same if no energy is lost.

3. When is mechanical energy conserved?

Mechanical energy is conserved when only conservative forces do work on the system.

At this level, the main conservative forces you should know are:

• Gravity
• Spring force

Mechanical energy is not conserved if significant non-conservative forces transfer energy out of the system, such as:

• Friction
• Air resistance

In those cases, some mechanical energy is transformed into other forms, such as thermal energy or sound.

Important idea: Energy itself is always conserved, but mechanical energy is only conserved when there are no non-conservative losses.

4. Understanding energy changes in motion

Imagine a ball thrown straight upward.

• At the bottom, its speed is large, so kinetic energy is large.
• As it rises, its speed decreases, so kinetic energy decreases.
• Its height increases, so gravitational potential energy increases.
• At the top, speed is momentarily zero, so kinetic energy is zero and potential energy is greatest.

On the way down, the opposite happens. Potential energy changes back into kinetic energy.

This constant conversion between (K) and (U) is the heart of conservation of mechanical energy.

5. Choosing a reference level for potential energy

Gravitational potential energy depends on height, but height must be measured from a chosen reference level. You may choose any point to be (h=0).

For example, you could choose:

• the ground as zero height,
• the bottom of a ramp as zero height, or
• a tabletop as zero height.

The choice does not change the final answer as long as you use it consistently throughout the problem.

6. A step-by-step method for solving problems

1. Identify the initial and final positions.
2. Decide whether mechanical energy is conserved.
3. Write the energy equation: (K_i + U_i = K_f + U_f).
4. Substitute the correct formulas, such as (K = \frac{1}{2}mv^2) and (U_g = mgh).
5. Cancel any quantities that appear on both sides if possible.
6. Solve for the unknown.
7. Check whether the answer makes physical sense.

7. Worked Example 1: Falling object from rest

A rock is dropped from a height of (20\,\text{m}). Ignore air resistance. What is its speed just before it hits the ground?

Step 1: Identify initial and final energies.

At the top:

• initial speed is zero, so (K_i = 0)
• gravitational potential energy is (U_i = mgh)

At the ground:

• take (h=0), so (U_f = 0)
• the rock has kinetic energy (K_f = \frac{1}{2}mv^2)

Step 2: Use conservation of mechanical energy.

$$K_i + U_i = K_f + U_f$$

$$0 + mgh = \frac{1}{2}mv^2 + 0$$

The mass cancels:

$$gh = \frac{1}{2}v^2$$

$$v^2 = 2gh$$

$$v = \sqrt{2gh}$$

Step 3: Substitute values.

$$v = \sqrt{2(9.8)(20)}$$

$$v = \sqrt{392} \approx 19.8\,\text{m/s}$$

Answer: The rock’s speed just before hitting the ground is about (19.8\,\text{m/s}).

Key idea: The mass did not matter. For objects falling without air resistance, speed depends only on the change in height.

8. Worked Example 2: Object moving up a ramp

A cart moves up a frictionless ramp with an initial speed of (12\,\text{m/s}). How high does it rise before stopping?

Step 1: Describe the motion.

At the bottom:

• kinetic energy is present
• take potential energy as zero

At the highest point:

• speed is zero, so kinetic energy is zero
• all the mechanical energy is gravitational potential energy

Step 2: Write the energy equation.

$$\frac{1}{2}mv^2 + 0 = 0 + mgh$$

Cancel mass:

$$\frac{1}{2}v^2 = gh$$

$$h = \frac{v^2}{2g}$$

Step 3: Substitute values.

$$h = \frac{12^2}{2(9.8)} = \frac{144}{19.6} \approx 7.35\,\text{m}$$

Answer: The cart rises to a height of about (7.35\,\text{m}).

Key idea: Kinetic energy was converted completely into gravitational potential energy.

9. Worked Example 3: Pendulum-like motion

A (2.0\,\text{kg}) object is released from rest at a height of (1.5\,\text{m}) above its lowest point. What is its speed at the lowest point? Ignore air resistance.

Step 1: Write known quantities.

• (m = 2.0\,\text{kg})
• (h = 1.5\,\text{m})
• initial speed = 0

Step 2: Apply conservation of mechanical energy.

$$mgh = \frac{1}{2}mv^2$$

Cancel mass:

$$gh = \frac{1}{2}v^2$$

$$v = \sqrt{2gh}$$

Step 3: Calculate.

$$v = \sqrt{2(9.8)(1.5)}$$

$$v = \sqrt{29.4} \approx 5.42\,\text{m/s}$$

Answer: The speed at the lowest point is about (5.42\,\text{m/s}).

Notice: Even though the mass was given, it canceled again. This often happens in gravity-only energy problems.

10. Worked Example 4: Spring and motion

A spring with spring constant (200\,\text{N/m}) is compressed by (0.10\,\text{m}) and used to launch a (0.50\,\text{kg}) cart on a frictionless surface. What is the cart’s speed when the spring returns to its natural length?

Step 1: Identify energy forms.

At the start:

• spring potential energy is present
• cart starts from rest, so kinetic energy is zero

At the natural length of the spring:

• spring potential energy is zero
• the cart has kinetic energy

Step 2: Write the equation.

$$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$$

The (\frac{1}{2}) cancels:

$$kx^2 = mv^2$$

$$v = \sqrt{\frac{kx^2}{m}}$$

Step 3: Substitute values.

$$v = \sqrt{\frac{200(0.10)^2}{0.50}}$$

$$v = \sqrt{\frac{200(0.01)}{0.50}}$$

$$v = \sqrt{\frac{2.0}{0.50}} = \sqrt{4.0} = 2.0\,\text{m/s}$$

Answer: The cart’s speed is (2.0\,\text{m/s}).

Key idea: Spring potential energy can also transform into kinetic energy, just like gravitational potential energy can.

11. How friction changes the situation

So far, we have assumed friction and air resistance are absent. In real life, these forces often matter.

When friction acts, some mechanical energy is transformed into thermal energy. Then:

$$K_i + U_i \neq K_f + U_f$$

for mechanical energy alone.

That means the final kinetic energy is less than it would be in an ideal frictionless situation.

For this lesson, the key point is simple:

• If the problem says frictionless or ignore air resistance, mechanical energy is conserved.
• If friction or air resistance is important, mechanical energy is not conserved by itself.

12. Common mistakes to avoid

• Mixing up speed and velocity: kinetic energy uses speed, so direction does not matter in (\frac{1}{2}mv^2).
• Forgetting to choose a zero height: define your reference level clearly.
• Using conservation of mechanical energy when friction is present: check the forces first.
• Dropping terms too early: write the full equation before simplifying.
• Forgetting units: energy is measured in joules (\text{J}), speed in (\text{m/s}), and height in meters.

13. Quick concept check

Use these questions to test your understanding:

1. If a skateboarder rolls down a frictionless hill, what happens to gravitational potential energy and kinetic energy?
2. At the highest point of a tossed ball’s motion, what is true about its kinetic energy?
3. Why can mass often cancel in gravity-only energy problems?
4. What type of force must be absent for mechanical energy to remain constant?

Answers:

1. Potential energy decreases while kinetic energy increases.
2. Its kinetic energy is zero if the ball is momentarily at rest.
3. Because mass appears in both (mgh) and (\frac{1}{2}mv^2).
4. Non-conservative forces such as friction and air resistance must be absent or negligible.

14. Final summary

Conservation of mechanical energy means that in a system with no friction or air resistance, the total of kinetic energy and potential energy stays constant.

As objects move, energy can change form:

• gravitational potential energy can become kinetic energy,
• kinetic energy can become gravitational potential energy,
• spring potential energy can become kinetic energy, and vice versa.

The main equation is:

$$K_i + U_i = K_f + U_f$$

If you can identify the energy forms at the start and end of motion, this equation often makes solving mechanics problems much easier than using forces and acceleration directly.

Linear Momentum and Impulse are two important ideas in mechanics that help us describe motion during collisions, kicks, catches, and other short interactions. They connect the motion of an object before and after a force acts on it.

In this lesson, you will learn what momentum is, how to calculate it, what impulse means, and how impulse changes momentum. You will also see how these ideas explain real-life events like airbags, sports, and car crashes.

1. What is linear momentum?

Linear momentum is the quantity of motion an object has. It depends on two things:

• the object’s mass
• the object’s velocity

The formula for linear momentum is:

$$p = mv$$

where:

• (p) = momentum
• (m) = mass
• (v) = velocity

Momentum is a vector, which means it has both size and direction. The direction of momentum is the same as the direction of velocity.

The SI unit for momentum is:

$$\text{kg}\cdot\text{m/s}$$

This means that a heavy object moving slowly can have the same momentum as a light object moving quickly.

Important ideas about momentum:

• If mass increases, momentum increases.
• If velocity increases, momentum increases.
• If velocity is negative, momentum is negative, showing direction.
• If an object is at rest, its momentum is zero.

2. Examples of momentum in everyday life

A truck moving at a moderate speed can have much more momentum than a bicycle moving fast because the truck has much greater mass.

A baseball thrown toward a batter has momentum because it has both mass and velocity. When the bat hits the ball, the ball’s momentum changes.

A parked car has no momentum because its velocity is zero.

3. What is impulse?

Impulse describes the effect of a force acting over a period of time. A force applied for a longer time produces a greater change in motion than the same force applied for a very short time.

The formula for impulse is:

$$J = F\Delta t$$

where:

• (J) = impulse
• (F) = force
• (\Delta t) = time interval

The unit of impulse is:

$$\text{N}\cdot\text{s}$$

This unit is equivalent to momentum units:

$$\text{N}\cdot\text{s} = \text{kg}\cdot\text{m/s}$$

4. The impulse-momentum theorem

The most important relationship in this topic is that impulse equals change in momentum.

$$J = \Delta p$$

Since momentum is (p = mv), the change in momentum is:

$$\Delta p = mv_f - mv_i$$

So the impulse-momentum theorem can be written as:

$$F\Delta t = mv_f - mv_i$$

This equation tells us that when a force acts on an object for some amount of time, the object’s momentum changes.

Why this matters:

• A larger force causes a larger change in momentum.
• A longer time of contact also causes a larger change in momentum.
• For the same change in momentum, increasing the time decreases the force.

5. Why increasing time can reduce force

This idea is very important for safety. If the momentum of an object must change, spreading that change over a longer time reduces the force.

From the equation

$$F = \frac{\Delta p}{\Delta t}$$

you can see that if (\Delta p) stays the same and (\Delta t) gets larger, then (F) gets smaller.

Examples:

• Airbags increase the time over which a passenger stops, reducing force.
• Seat belts help your body slow down over a slightly longer time.
• Gym mats increase stopping time when a person lands.
• Catching a ball by moving your hands backward increases the stopping time and reduces the force on your hands.

6. Sign and direction in momentum problems

Because momentum is a vector, direction matters. In one-dimensional problems, we often choose one direction as positive.

For example:

• right or forward = positive
• left or backward = negative

If an object changes direction, its momentum changes sign. This can create a large change in momentum.

For instance, if a ball is moving at (+8\,\text{m/s}) and then rebounds at (-6\,\text{m/s}), the momentum change includes both the slowing down and the reversal of direction.

7. Worked Example 1: Finding momentum

A (4.0\,\text{kg}) cart moves to the right at (3.0\,\text{m/s}). Find its momentum.

Step 1: Use the formula

$$p = mv$$

Step 2: Substitute values

$$p = (4.0)(3.0)$$

$$p = 12\,\text{kg}\cdot\text{m/s}$$

Answer: The momentum is (12\,\text{kg}\cdot\text{m/s}) to the right.

Worked Example 2: Finding impulse from force and time

A force of (15\,\text{N}) acts on a ball for (0.20\,\text{s}). Find the impulse.

Step 1: Use the impulse formula

$$J = F\Delta t$$

Step 2: Substitute values

$$J = (15)(0.20)$$

$$J = 3.0\,\text{N}\cdot\text{s}$$

Answer: The impulse is (3.0\,\text{N}\cdot\text{s}).

Because impulse equals change in momentum, the ball’s momentum changed by (3.0\,\text{kg}\cdot\text{m/s}) in the direction of the force.

Worked Example 3: Using the impulse-momentum theorem

A (0.50\,\text{kg}) soccer ball is initially at rest. A player kicks it, and it moves forward at (12\,\text{m/s}). What impulse was given to the ball?

Step 1: Write initial and final momentum

Initial velocity: (v_i = 0)

Final velocity: (v_f = 12\,\text{m/s})

$$J = \Delta p = mv_f - mv_i$$

Step 2: Substitute values

$$J = (0.50)(12) - (0.50)(0)$$

$$J = 6.0 - 0$$

$$J = 6.0\,\text{kg}\cdot\text{m/s}$$

Answer: The impulse on the ball is (6.0\,\text{N}\cdot\text{s}) forward.

Worked Example 4: Finding average force during a collision

A (0.15\,\text{kg}) baseball traveling at (20\,\text{m/s}) toward a bat is hit straight back at (30\,\text{m/s}). The bat is in contact with the ball for (0.010\,\text{s}). Find the average force on the ball.

Step 1: Choose a positive direction

Let the direction toward the bat be positive.

Then:

• (v_i = +20\,\text{m/s})
• (v_f = -30\,\text{m/s}) because the ball goes in the opposite direction after being hit

Step 2: Find the change in momentum

$$\Delta p = mv_f - mv_i$$

$$\Delta p = (0.15)(-30) - (0.15)(20)$$

$$\Delta p = -4.5 - 3.0$$

$$\Delta p = -7.5\,\text{kg}\cdot\text{m/s}$$

Step 3: Use (F = \frac{\Delta p}{\Delta t})

$$F = \frac{-7.5}{0.010}$$

$$F = -750\,\text{N}$$

Answer: The average force on the ball is (750\,\text{N}) in the direction opposite to its initial motion.

The negative sign shows direction. The large force happens because the ball’s momentum changes a lot in a very short time.

8. Common mistakes to avoid

• Confusing velocity and speed: Momentum uses velocity, so direction matters.
• Ignoring negative signs: If an object changes direction, use positive and negative values correctly.
• Using mass in grams: Convert to kilograms before calculating momentum or impulse.
• Forgetting units: Momentum is in (\text{kg}\cdot\text{m/s}), and impulse is in (\text{N}\cdot\text{s}).
• Mixing up force and impulse: Force is measured in newtons, while impulse includes both force and time.

9. Problem-solving steps

When solving momentum and impulse questions, it helps to follow a clear process:

1. Identify what is given: mass, velocity, force, or time.
2. Choose a positive direction if motion is along one line.
3. Write the correct formula:

• (p = mv)
• (J = F\Delta t)
• (J = \Delta p = mv_f - mv_i)

4. Substitute values with correct units.
5. Solve carefully and include direction when needed.
6. Check if the answer makes physical sense.

10. Real-world importance

Linear momentum and impulse help explain many real situations:

• why a punch follows through
• why helmets and padding reduce injury
• why a truck is harder to stop than a skateboard
• why rebound motion can create a large force

These ideas are also important in engineering, transportation safety, sports science, and accident analysis.

Brief Summary

Linear momentum is the product of mass and velocity: (p = mv). Impulse is the product of force and time: (J = F\Delta t).

The key idea is the impulse-momentum theorem:

$$F\Delta t = \Delta p$$

This means that impulse changes an object’s momentum. If the same momentum change happens over a longer time, the force is smaller. That is why safety devices like airbags, helmets, and mats are so effective.

Universal Gravitation is the idea that every object with mass pulls on every other object with mass.

This means Earth pulls on you, you pull on Earth, the Moon pulls on Earth, and even two small objects like books pull on each other. Usually, the pull between small objects is too tiny to notice, but the pull becomes very important when one or both objects are very massive, like planets, moons, and stars.

Gravity is an attractive force. That means it pulls objects together, not apart.

In this lesson, you will learn what affects the strength of gravity, how distance changes gravitational force, and how to use a simple formula to compare different situations.

1. What affects gravitational force?

The strength of gravity between two objects depends on two main things:

• The masses of the objects
• The distance between their centers

If the masses are larger, the gravitational force is stronger.

If the distance between the objects increases, the gravitational force becomes weaker.

Scientists describe this relationship with the law of universal gravitation:

$$F = G\frac{m_1m_2}{d^2}$$

In this formula:

• \(F\) is gravitational force
• \(G\) is a constant number used by scientists
• \(m_1\) and \(m_2\) are the masses of the two objects
• \(d\) is the distance between the centers of the objects

For 7th Grade, the most important part is to understand the pattern in the formula, not to memorize the value of \(G\).

2. Mass and gravity: direct relationship

The formula shows that mass is in the top part of the fraction. This means gravitational force changes directly with mass.

If one mass gets bigger, gravity gets stronger.

If both masses get bigger, gravity gets even stronger.

Here are some mass patterns:

• If one mass doubles, the gravitational force doubles.
• If both masses double, the gravitational force becomes 4 times as great, because \(2 \times 2 = 4\).
• If one mass is cut in half, the gravitational force is cut in half.

3. Distance and gravity: inverse-square relationship

The distance is squared in the bottom of the fraction: \(d^2\).

This is very important. It means gravity does not just weaken a little when distance increases. It weakens much faster.

This is called an inverse-square relationship.

Here are some distance patterns:

• If the distance doubles, the force becomes \(\frac{1}{2^2} = \frac{1}{4}\) as strong.
• If the distance triples, the force becomes \(\frac{1}{3^2} = \frac{1}{9}\) as strong.
• If the distance is cut in half, the force becomes \(\frac{1}{(1/2)^2} = 4\) times as strong.

This is why distance matters so much in gravity.

4. Why don’t we notice gravity between everyday objects?

Every object with mass pulls on every other object. So yes, a pencil pulls on a notebook, and a student pulls on a desk.

But these forces are extremely small because everyday objects have small masses compared with planets and stars.

Earth is so massive that its gravitational pull is strong enough to keep you on the ground, make objects fall, and keep the Moon in orbit.

5. Gravity on Earth, the Moon, and in space

Earth’s gravity pulls objects toward Earth’s center. That is why dropped objects fall downward.

The Moon also has gravity, but because the Moon has less mass than Earth, its pull is weaker.

Gravity is also what keeps planets moving around the Sun and moons moving around planets.

People sometimes say there is “no gravity” in space, but gravity is still there. In space, astronauts may seem to float because they are falling around Earth while moving forward very fast.

6. Weight and mass

It is important to know the difference between mass and weight.

• Mass is how much matter an object has.
• Weight is the force of gravity pulling on that mass.

Your mass stays the same wherever you go.

Your weight can change if gravity changes.

For example, a person would have the same mass on Earth and on the Moon, but would weigh less on the Moon because the Moon’s gravity is weaker.

7. Worked Example 1: What happens if one mass doubles?

Suppose two objects pull on each other with a force of 10 units.

If one object’s mass doubles and the other mass and distance stay the same, what happens to the force?

Step 1: Look at the formula.

$$F = G\frac{m_1m_2}{d^2}$$

Step 2: If one mass doubles, the top of the fraction doubles.

Step 3: So the force also doubles.

The new force is 20 units.

Answer: Doubling one mass doubles the gravitational force.

8. Worked Example 2: What happens if the distance doubles?

Now suppose the starting gravitational force is 12 units.

If the distance between the objects doubles, what is the new force?

Step 1: Use the inverse-square idea.

If distance doubles, force becomes \(\frac{1}{2^2} = \frac{1}{4}\) as strong.

Step 2: Find one-fourth of 12.

$$12 \div 4 = 3$$

Answer: The new force is 3 units.

9. Worked Example 3: Both masses double

Suppose the original force is 5 units.

If both masses double and the distance stays the same, what happens?

Step 1: One doubled mass makes the force 2 times as great.

Step 2: The second doubled mass also multiplies the force by 2.

Step 3: Multiply the changes.

$$2 \times 2 = 4$$

Step 4: Multiply the original force by 4.

$$5 \times 4 = 20$$

Answer: The new force is 20 units.

10. Worked Example 4: One mass doubles and distance triples

This example combines two changes.

Suppose the original force is 18 units. One mass doubles, and the distance triples. What is the new force?

Step 1: Doubling one mass multiplies the force by 2.

Step 2: Tripling the distance makes the force \(\frac{1}{9}\) as strong, because \(3^2 = 9\).

Step 3: Combine the changes.

$$2 \times \frac{1}{9} = \frac{2}{9}$$

Step 4: Find \(\frac{2}{9}\) of 18.

$$18 \times \frac{2}{9} = 4$$

Answer: The new force is 4 units.

11. Common mistakes to avoid

• Mistake: Thinking only large objects have gravity.
  Correct idea: All objects with mass have gravity.
• Mistake: Thinking distance and force change in the same way.
  Correct idea: Distance is squared, so it changes the force much more.
• Mistake: Confusing mass and weight.
  Correct idea: Mass is the amount of matter; weight is the pull of gravity on that matter.
• Mistake: Thinking there is no gravity in space.
  Correct idea: Gravity exists throughout space.

12. Real-world examples of universal gravitation

• Earth pulls falling objects toward the ground.
• The Moon stays in orbit around Earth because of gravity.
• Earth stays in orbit around the Sun because of gravity.
• Ocean tides are affected by the Moon’s gravitational pull.

13. Key ideas to remember

• Gravity is a force of attraction between any two objects with mass.
• More mass means stronger gravitational force.
• More distance means weaker gravitational force.
• The force changes with the square of distance: \(d^2\).
• Weight depends on gravity, but mass does not change from place to place.

Brief Summary

Universal gravitation explains that all objects with mass attract each other. The gravitational force becomes stronger when masses increase and weaker when distance increases. Because distance is squared in the formula, even small increases in distance can greatly reduce the force. This law helps explain falling objects, orbits, and why planets, moons, and stars affect one another.

Collisions and Conservation of Momentum

When objects crash into each other, bounce apart, or stick together, we call the event a collision. Collisions happen in everyday life: a bat hitting a ball, two shopping carts bumping, or a car crash. To understand what happens in a collision, one of the most important ideas in physics is momentum.

Momentum is the quantity of motion an object has. It depends on both the object’s mass and its velocity. A heavy truck moving slowly can have the same momentum as a small car moving quickly.

The formula for momentum is

$$p = mv$$

where:

• \(p\) is momentum
• \(m\) is mass
• \(v\) is velocity

Because velocity has direction, momentum also has direction. This means momentum is a vector. In one-dimensional problems, we usually show direction using positive and negative signs.

For example:

• Motion to the right might be positive.
• Motion to the left might be negative.

This sign choice matters a lot in collision problems.

The Law of Conservation of Momentum says that in an isolated system, the total momentum before a collision equals the total momentum after the collision.

An isolated system is a system with no significant external forces acting on it. During a short collision, we often assume outside forces are small enough to ignore.

Mathematically, conservation of momentum is written as

$$p_{\text{before}} = p_{\text{after}}$$

For two objects, this becomes

$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

where:

• \(m_1, m_2\) are the masses of the two objects
• \(v_{1i}, v_{2i}\) are their initial velocities
• \(v_{1f}, v_{2f}\) are their final velocities

This equation is the main tool for solving collision problems.

Why is momentum conserved? During a collision, the two objects push on each other with equal and opposite forces. These internal forces can change the momentum of each object, but the total momentum of the system stays constant as long as there is no net external force.

It is important to understand that individual momentum can change, but total momentum stays the same. One object may slow down while another speeds up, yet the overall momentum of the system is conserved.

Types of Collisions

There are two main types of collisions you will study:

• Elastic collisions
• Inelastic collisions

In all collisions, momentum is conserved if the system is isolated.

The difference between elastic and inelastic collisions involves kinetic energy.

Elastic Collision

An elastic collision is a collision in which both momentum and kinetic energy are conserved.

Kinetic energy is given by

$$KE = \frac{1}{2}mv^2$$

In a perfectly elastic collision, objects bounce off each other without losing kinetic energy to heat, sound, or permanent deformation.

Examples of nearly elastic collisions include:

• Billiard balls colliding
• Air track gliders bumping
• Gas particles colliding

Inelastic Collision

An inelastic collision is a collision in which momentum is conserved, but kinetic energy is not conserved.

Some kinetic energy is transformed into other forms of energy, such as:

• Heat
• Sound
• Deformation of the objects

A special type of inelastic collision is a perfectly inelastic collision. In this case, the objects stick together after the collision and move as one object.

For a perfectly inelastic collision, the conservation of momentum equation becomes

$$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$$

because both objects have the same final velocity after they stick together.

Important Idea: Momentum is always conserved in isolated collisions, but kinetic energy is only conserved in elastic collisions.

How to Solve Collision Problems

1. Choose a positive direction. For example, right is positive.
2. Write the known masses and velocities.
3. Include signs on velocities. Leftward motion should be negative if right is positive.
4. Apply conservation of momentum.
5. If needed, use kinetic energy information to decide whether the collision is elastic or inelastic.
6. Check your answer. Does the direction make sense? Are the units reasonable?

Units of Momentum

Momentum is measured in

$$\text{kg} \cdot \text{m/s}$$

These are the units you should expect in your final answer.

Worked Example 1: Finding Momentum

A 4.0 kg cart moves to the right at 3.0 m/s. What is its momentum?

Step 1: Use the formula.

$$p = mv$$

Step 2: Substitute values.

$$p = (4.0)(3.0) = 12.0$$

Answer:

$$p = 12.0\ \text{kg} \cdot \text{m/s}$$

Since the cart is moving to the right, we could also write the momentum as +12.0 kg·m/s.

Worked Example 2: Perfectly Inelastic Collision

A 2.0 kg cart moving right at 4.0 m/s collides with a 3.0 kg cart at rest. The carts stick together. What is their final velocity?

Step 1: Identify what kind of collision this is.

The carts stick together, so this is a perfectly inelastic collision.

Step 2: Write the momentum equation.

$$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$$

Step 3: Substitute values.

$$ (2.0)(4.0) + (3.0)(0) = (2.0 + 3.0)v_f $$

$$ 8.0 = 5.0v_f $$

Step 4: Solve.

$$ v_f = \frac{8.0}{5.0} = 1.6\ \text{m/s} $$

Answer: The combined carts move at

$$1.6\ \text{m/s to the right}$$

Check the idea: The final speed is less than 4.0 m/s because the moving cart now has to move both masses together.

Worked Example 3: Two Objects Bounce Apart

A 1.5 kg ball moves right at 6.0 m/s and hits a 2.0 kg ball at rest. After the collision, the 1.5 kg ball moves left at 2.0 m/s. What is the final velocity of the 2.0 kg ball?

Step 1: Choose a positive direction.

Take right as positive. Then the first ball’s final velocity is \(-2.0\ \text{m/s}\).

Step 2: Write conservation of momentum.

$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

Step 3: Substitute values.

$$ (1.5)(6.0) + (2.0)(0) = (1.5)(-2.0) + (2.0)v_{2f} $$

$$ 9.0 = -3.0 + 2.0v_{2f} $$

Step 4: Solve.

$$ 12.0 = 2.0v_{2f} $$

$$ v_{2f} = 6.0\ \text{m/s} $$

Answer:

$$6.0\ \text{m/s to the right}$$

What happened? The first ball bounced backward, so the second ball had to gain enough forward momentum to keep total momentum constant.

Worked Example 4: Opposite Directions

A 1000 kg car travels east at 8.0 m/s and hits a 1200 kg car traveling west at 5.0 m/s. The cars lock together. Find their final velocity.

Step 1: Define positive direction.

Let east be positive. Then:

• \(v_{1i} = +8.0\ \text{m/s}\)
• \(v_{2i} = -5.0\ \text{m/s}\)

Step 2: Use the perfectly inelastic formula.

$$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$$

Step 3: Substitute values.

$$ (1000)(8.0) + (1200)(-5.0) = (2200)v_f $$

$$ 8000 - 6000 = 2200v_f $$

$$ 2000 = 2200v_f $$

Step 4: Solve.

$$ v_f = \frac{2000}{2200} \approx 0.91\ \text{m/s} $$

Answer:

$$0.91\ \text{m/s east}$$

Why east? The total momentum before the collision was positive, so the combined cars must move east after the collision.

Momentum and Kinetic Energy Comparison

Students often confuse momentum and kinetic energy. They are related, but they are not the same.

• Momentum: \(p = mv\)
• Kinetic energy: \(KE = \frac{1}{2}mv^2\)

Notice that momentum depends on velocity to the first power, but kinetic energy depends on velocity squared. This is why doubling velocity doubles momentum but makes kinetic energy four times larger.

In collisions:

• Momentum is conserved in an isolated system.
• Kinetic energy is conserved only in elastic collisions.

Common Mistakes to Avoid

• Forgetting direction: Velocity and momentum can be positive or negative.
• Using speed instead of velocity: Speed has no direction, but momentum needs direction.
• Assuming kinetic energy is always conserved: This is only true for elastic collisions.
• Mixing up before and after values: Keep initial and final quantities clearly labeled.
• For perfectly inelastic collisions, forgetting that both objects share the same final velocity.

Real-World Applications

• Car safety: Engineers study momentum during crashes to design seat belts, airbags, and crumple zones.
• Sports: The motion of balls, bats, rackets, and players involves momentum transfer.
• Rocket motion: Momentum helps explain how rockets move when gases are pushed backward.
• Particle physics: Scientists analyze collision products to understand tiny particles.

Key Takeaways

• Momentum is given by \(p = mv\).
• Momentum has both magnitude and direction.
• In an isolated system, total momentum before a collision equals total momentum after the collision.
• Elastic collisions conserve both momentum and kinetic energy.
• Inelastic collisions conserve momentum, but not kinetic energy.
• In a perfectly inelastic collision, objects stick together and move with a common final velocity.

Brief Summary

Collisions are analyzed using the law of conservation of momentum. As long as the system is isolated, total momentum stays constant before and after the collision. Elastic collisions conserve both momentum and kinetic energy, while inelastic collisions conserve only momentum. By carefully using mass, velocity, and direction, you can predict how objects move after they collide.

Conservation of Momentum is a big idea in motion and collisions. It helps us understand what happens when objects hit each other, stick together, or bounce apart.

Momentum depends on both mass and velocity. A heavy object moving slowly can have a lot of momentum, and a lighter object moving quickly can also have a lot of momentum.

The formula for momentum is:

$$p = mv$$

In this formula, p means momentum, m means mass, and v means velocity.

Because velocity includes direction, momentum also has direction. This means we must pay attention to whether an object is moving to the right, left, forward, or backward.

For example, if we say right is positive, then an object moving left has a negative velocity. That means its momentum is negative too.

1. What does “conservation” mean?

In science, conservation means something stays the same overall. For momentum, this means the total momentum of a closed system remains constant.

A closed system is a group of objects that are not affected by outside forces very much during the event we are studying. In many collision problems, we treat the colliding objects as a closed system for a short amount of time.

This leads to the law of conservation of momentum:

$$\text{Total momentum before} = \text{Total momentum after}$$

Or, written with symbols:

$$p_{\text{before}} = p_{\text{after}}$$

For two objects, this becomes:

$$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$$

The velocities before the collision are written without primes, and the velocities after the collision are written with primes.

2. Why momentum is useful

Momentum helps us predict motion during collisions. Even when objects change speed after crashing into each other, the total momentum of the system stays the same if no important outside forces act on it.

This is especially useful when:

• objects collide and stick together,
• objects bounce apart,
• one object starts moving after being hit by another.

3. Direction matters

Since momentum depends on velocity, direction must be included in every calculation.

Suppose a 2 kg cart moves to the right at 3 m/s. Its momentum is:

$$p = mv = (2)(3) = 6\text{ kgm/s}$$

If the same cart moves to the left at 3 m/s, and we choose right as positive, then:

$$p = (2)(-3) = -6\text{ kgm/s}$$

The negative sign shows direction, not a different kind of momentum.

4. Types of collisions

There are two common types of collisions you may study at this level.

Elastic collision: the objects bounce off each other. Momentum is conserved.

Inelastic collision: the objects do not bounce apart normally. They may stick together or move off with less separation. Momentum is still conserved.

For 8th Grade science, the most important idea is this: in both elastic and inelastic collisions, total momentum stays the same in a closed system.

5. Steps for solving momentum problems

1. List the masses and velocities.
2. Choose a positive direction. For example, right is positive.
3. Find total momentum before the collision.
4. Use conservation of momentum. Set total momentum before equal to total momentum after.
5. Solve for the unknown.
6. Check the sign of your answer. A negative answer means the object moves in the opposite direction.

6. Worked Example 1: Finding momentum

A 4 kg scooter moves to the right at 2 m/s. What is its momentum?

Step 1: Use the formula.

$$p = mv$$

Step 2: Substitute the values.

$$p = (4)(2)$$

Step 3: Multiply.

$$p = 8\text{ kgm/s}$$

Answer: The scooter has a momentum of 8 kgm/s to the right.

7. Worked Example 2: Two objects stick together

A 3 kg cart moving right at 4 m/s crashes into a 2 kg cart at rest. The carts stick together. What is their velocity after the collision?

Step 1: Write the known values.

• m_1 = 3\text{ kg}
• v_1 = 4\text{ m/s}
• m_2 = 2\text{ kg}
• v_2 = 0\text{ m/s}

Because the carts stick together, they move with one final velocity, which we will call v_f.

Step 2: Find total momentum before.

$$p_{\text{before}} = m_1v_1 + m_2v_2$$

$$p_{\text{before}} = (3)(4) + (2)(0) = 12$$

So the total momentum before is:

$$12\text{ kgm/s}$$

Step 3: Write the momentum after.

After the collision, the total mass is:

$$3 + 2 = 5\text{ kg}$$

So:

$$p_{\text{after}} = (5)v_f$$

Step 4: Use conservation of momentum.

$$p_{\text{before}} = p_{\text{after}}$$

$$12 = 5v_f$$

Step 5: Solve.

$$v_f = \frac{12}{5} = 2.4\text{ m/s}$$

Answer: The stuck-together carts move at 2.4 m/s to the right.

8. Worked Example 3: Objects bounce apart

A 2 kg cart moves right at 5 m/s and hits a 1 kg cart at rest. After the collision, the 2 kg cart moves right at 2 m/s. What is the final velocity of the 1 kg cart?

Step 1: Write what is known.

• m_1 = 2\text{ kg}
• v_1 = 5\text{ m/s}
• m_2 = 1\text{ kg}
• v_2 = 0\text{ m/s}
• v_1' = 2\text{ m/s}
• v_2' = ?

Step 2: Find total momentum before.

$$p_{\text{before}} = (2)(5) + (1)(0) = 10$$

Step 3: Write total momentum after.

$$p_{\text{after}} = (2)(2) + (1)(v_2')$$

$$p_{\text{after}} = 4 + v_2'$$

Step 4: Set them equal.

$$10 = 4 + v_2'$$

Step 5: Solve.

$$v_2' = 6\text{ m/s}$$

Answer: The 1 kg cart moves at 6 m/s to the right.

9. Worked Example 4: Using negative velocity

A 1 kg ball moves right at 6 m/s and collides with a 1 kg ball moving left at 2 m/s. After the collision, they stick together. What is their final velocity?

Step 1: Choose right as positive.

• Ball 1: m_1 = 1\text{ kg}, \; v_1 = 6\text{ m/s}
• Ball 2: m_2 = 1\text{ kg}, \; v_2 = -2\text{ m/s}

Step 2: Find total momentum before.

$$p_{\text{before}} = m_1v_1 + m_2v_2$$

$$p_{\text{before}} = (1)(6) + (1)(-2) = 4$$

Step 3: Add the masses for after the collision.

$$m_{\text{total}} = 1 + 1 = 2\text{ kg}$$

Step 4: Use conservation of momentum.

$$4 = 2v_f$$

$$v_f = 2\text{ m/s}$$

Answer: The two balls move together at 2 m/s to the right.

10. Common mistakes to avoid

• Forgetting direction. Left and right should not both be positive.
• Mixing up mass and velocity. Momentum is mass times velocity, not mass plus velocity.
• Forgetting to add total momentum. You must include all objects in the system.
• Not noticing when objects stick together. If they stick together, they share one final velocity.
• Ignoring units. Momentum is measured in kgm/s.

11. Real-world connections

Conservation of momentum explains many everyday events.

• When one billiard ball hits another, momentum is transferred.
• When bumper cars crash, their motion changes based on mass and velocity.
• When a person jumps from a small boat, the boat moves backward.
• In sports, a moving ball can cause another object to move after impact.

In all of these examples, the total momentum of the system stays the same if outside forces are small during the collision.

12. Quick review

• Momentum is the product of mass and velocity: $$p = mv$$
• Momentum has direction.
• In a closed system, total momentum before a collision equals total momentum after.
• This is true for both elastic and inelastic collisions.
• If objects stick together, they move with the same final velocity.

Final idea to remember: During a collision, the motion of individual objects can change, but the total momentum of the whole system stays constant if no important outside force acts on it.

Rotational Kinematics and Dynamics is the study of how objects spin, what causes them to spin, and how to describe that motion using physics. It is closely connected to the motion of objects moving in straight lines, called linear motion. Many ideas in rotation have a matching idea in linear motion.

For example, in linear motion we talk about distance, speed, acceleration, force, mass, and momentum. In rotational motion, we talk about angular displacement, angular velocity, angular acceleration, torque, moment of inertia, and angular momentum. Learning these pairs makes rotational motion much easier to understand.

This lesson will explain the main quantities used in rotational motion, the equations that connect them, and how torque causes rotation. By the end, you should be able to describe spinning objects and solve basic rotational motion problems.

1. Angular quantities: describing rotation

When an object rotates, instead of measuring how far it moves in a straight line, we measure how far it turns. This turning is called angular displacement, and it is represented by the symbol \(\theta\).

Angular displacement is measured in radians. A full circle is \(2\pi\) radians, which is also \(360^\circ\). That means:

$$2\pi \text{ rad} = 360^\circ$$

To convert between degrees and radians:

• $$\theta(\text{rad}) = \theta(^\circ)\cdot \frac{\pi}{180}$$
• $$\theta(^\circ) = \theta(\text{rad})\cdot \frac{180}{\pi}$$

Angular velocity tells how quickly an angle changes. Its symbol is \(\omega\), and its unit is radians per second, or \(\text{rad/s}\).

$$\omega = \frac{\Delta \theta}{\Delta t}$$

If the angular velocity changes, then the object has angular acceleration, represented by \(\alpha\). Its unit is radians per second squared, or \(\text{rad/s}^2\).

$$\alpha = \frac{\Delta \omega}{\Delta t}$$

2. Rotation and linear motion: important connections

Rotational motion is easier to understand when you compare it to straight-line motion.

• Linear displacement \(x\) matches angular displacement \(\theta\)
• Linear velocity \(v\) matches angular velocity \(\omega\)
• Linear acceleration \(a\) matches angular acceleration \(\alpha\)
• Force \(F\) matches torque \(\tau\)
• Mass \(m\) matches moment of inertia \(I\)

If a point is on a rotating object at a distance \(r\) from the axis, its linear distance traveled along a circular path is related to the angle by:

$$s = r\theta$$

Its linear speed is related to angular speed by:

$$v = r\omega$$

Its tangential acceleration is related to angular acceleration by:

$$a_t = r\alpha$$

These formulas show that points farther from the center move faster and have greater tangential acceleration for the same rotational motion.

3. Rotational kinematics equations

When angular acceleration is constant, rotational motion follows equations that are almost the same as the constant-acceleration equations in linear motion.

$$\omega = \omega_0 + \alpha t$$

$$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$$

$$\omega^2 = \omega_0^2 + 2\alpha\theta$$

$$\theta = \frac{\omega_0 + \omega}{2}t$$

Here:

• \(\omega_0\) is the initial angular velocity
• \(\omega\) is the final angular velocity
• \(\alpha\) is angular acceleration
• \(t\) is time
• \(\theta\) is angular displacement

These equations are useful for wheels, fans, rotating platforms, gears, and any spinning object with constant angular acceleration.

4. Torque: the turning effect of a force

In linear motion, a force causes an object to accelerate. In rotational motion, a torque causes an object to gain angular acceleration.

Torque depends on three things:

• the size of the force
• the distance from the axis of rotation
• the angle at which the force is applied

The formula for torque is:

$$\tau = rF\sin\phi$$

where:

• \(\tau\) is torque in newton-meters \((\text{N}\cdot\text{m})\)
• \(r\) is the distance from the axis to the point where the force is applied
• \(F\) is the force
• \(\phi\) is the angle between the force and the lever arm

The torque is largest when the force is applied perpendicular to the lever arm, because then \(\sin 90^\circ = 1\).

This is why pushing a door at the handle works better than pushing near the hinges. The handle is farther from the axis, so the torque is larger.

5. Moment of inertia: rotational resistance to change

Mass measures how hard it is to change an object's linear motion. In rotation, the similar idea is called moment of inertia, represented by \(I\).

Moment of inertia tells how hard it is to change an object's rotational motion. It depends on:

• the object's mass
• how that mass is spread out from the axis of rotation

An object with more mass far from the axis has a larger moment of inertia, so it is harder to start or stop spinning.

For a single point mass:

$$I = mr^2$$

This formula shows that distance from the axis matters a lot, because it is squared.

So if two objects have the same mass but one has more of its mass farther from the center, that object will have the larger moment of inertia.

6. Newton's second law for rotation

In linear motion, Newton's second law is:

$$F = ma$$

In rotational motion, the matching law is:

$$\tau = I\alpha$$

This is one of the most important equations in rotational dynamics.

It tells us that angular acceleration increases when torque increases, and decreases when moment of inertia increases. So:

• more torque means faster spinning up
• larger moment of inertia means slower spinning up

7. Rotational kinetic energy

A moving object in a straight line has kinetic energy:

$$KE = \frac{1}{2}mv^2$$

A rotating object has rotational kinetic energy:

$$KE_{rot} = \frac{1}{2}I\omega^2$$

This means spinning objects store energy because of their motion. A bicycle wheel, ceiling fan, or record player all have rotational kinetic energy when spinning.

8. Angular momentum

Another important rotational quantity is angular momentum, represented by \(L\). For a rotating object:

$$L = I\omega$$

Angular momentum is important because in many situations it is conserved, meaning it stays constant if there is no external torque.

For example, when a figure skater pulls in their arms, their moment of inertia decreases. Since angular momentum stays the same, their angular velocity increases, so they spin faster.

9. Direction of rotation and signs

In rotational problems, direction matters. Usually:

• counterclockwise rotation is taken as positive
• clockwise rotation is taken as negative

This sign choice helps keep equations consistent. Always check the direction of torque, angular velocity, and angular acceleration.

Worked Example 1: Angular speed from angular displacement

A wheel turns through \(8\pi\) radians in \(4\) seconds. Find its angular velocity.

Step 1: Write the formula

$$\omega = \frac{\Delta \theta}{\Delta t}$$

Step 2: Substitute values

$$\omega = \frac{8\pi}{4}$$

Step 3: Simplify

$$\omega = 2\pi\ \text{rad/s}$$

Answer: The angular velocity is \(2\pi\ \text{rad/s}\).

Worked Example 2: Using constant angular acceleration

A fan starts from rest and accelerates at \(3\ \text{rad/s}^2\) for \(5\) s. Find:

• its final angular velocity
• its angular displacement

Step 1: Identify known values

• Initial angular velocity: \(\omega_0 = 0\)
• Angular acceleration: \(\alpha = 3\ \text{rad/s}^2\)
• Time: \(t = 5\ \text{s}\)

Step 2: Find final angular velocity

Use:

$$\omega = \omega_0 + \alpha t$$

Substitute:

$$\omega = 0 + (3)(5) = 15\ \text{rad/s}$$

Step 3: Find angular displacement

Use:

$$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$$

Substitute:

$$\theta = 0 + \frac{1}{2}(3)(5^2)$$

$$\theta = 1.5 \times 25 = 37.5\ \text{rad}$$

Answer:

• Final angular velocity: \(15\ \text{rad/s}\)
• Angular displacement: \(37.5\ \text{rad}\)

Worked Example 3: Finding torque

A student pushes on a wrench with a force of \(40\ \text{N}\). The force is applied perpendicular to the wrench at a distance of \(0.25\ \text{m}\) from the bolt. What torque is produced?

Step 1: Use the torque formula

$$\tau = rF\sin\phi$$

Because the force is perpendicular, \(\phi = 90^\circ\), so \(\sin 90^\circ = 1\).

Step 2: Substitute values

$$\tau = (0.25)(40)(1)$$

Step 3: Calculate

$$\tau = 10\ \text{N}\cdot\text{m}$$

Answer: The torque is \(10\ \text{N}\cdot\text{m}\).

Worked Example 4: Using \(\tau = I\alpha\)

A rotating object has a moment of inertia of \(2.0\ \text{kg}\cdot\text{m}^2\). A net torque of \(6.0\ \text{N}\cdot\text{m}\) acts on it. Find its angular acceleration.

Step 1: Write the equation

$$\tau = I\alpha$$

Step 2: Solve for \(\alpha\)

$$\alpha = \frac{\tau}{I}$$

Step 3: Substitute values

$$\alpha = \frac{6.0}{2.0}$$

Step 4: Calculate

$$\alpha = 3.0\ \text{rad/s}^2$$

Answer: The angular acceleration is \(3.0\ \text{rad/s}^2\).

10. Common mistakes to avoid

• Mixing degrees and radians: Most rotational formulas use radians, not degrees.
• Forgetting the radius: Linear speed and tangential acceleration depend on distance from the axis.
• Using force instead of torque: A force alone does not fully describe rotation; where it acts matters too.
• Ignoring direction: Clockwise and counterclockwise motion may need positive and negative signs.
• Confusing mass and moment of inertia: Moment of inertia depends not just on how much mass there is, but also where the mass is located.

11. Big-picture idea

Rotational motion follows the same logic as straight-line motion. In straight-line motion, forces change velocity. In rotational motion, torques change angular velocity. Mass resists changes in linear motion, while moment of inertia resists changes in rotational motion.

This parallel structure is very helpful:

• \(x \leftrightarrow \theta\)
• \(v \leftrightarrow \omega\)
• \(a \leftrightarrow \alpha\)
• \(F \leftrightarrow \tau\)
• \(m \leftrightarrow I\)

Once you see these connections, rotational kinematics and dynamics become much easier to organize and solve.

Summary

Rotational kinematics describes how objects spin using angular displacement \(\theta\), angular velocity \(\omega\), and angular acceleration \(\alpha\). These quantities are related to linear motion by formulas such as \(s=r\theta\), \(v=r\omega\), and \(a_t=r\alpha\).

Rotational dynamics explains what causes spinning motion. Torque is the turning effect of a force, and the equation $$\tau = I\alpha$$ shows that torque causes angular acceleration, while moment of inertia measures resistance to rotational change.

Rotating objects can also have rotational kinetic energy, $$KE_{rot}=\frac{1}{2}I\omega^2,$$ and angular momentum, $$L=I\omega.$$ These ideas help explain the motion of wheels, tools, machines, and many objects in everyday life.

Conservation of Angular Momentum is one of the most important ideas in rotational motion. It explains why spinning objects can speed up or slow down without any motor, why planets sweep around the Sun in a regular way, and why a figure skater spins faster when pulling in their arms.

To understand this idea, we first need to know what angular momentum means. Angular momentum is the rotational version of momentum. Regular momentum describes motion in a straight line. Angular momentum describes motion around an axis or center.

The key idea of this lesson is simple: if no net external torque acts on a system, its total angular momentum stays constant. This is called the law of conservation of angular momentum.

In symbols, angular momentum is often written as \(L\). For many 11th Grade problems involving a rigid object spinning about an axis, angular momentum is given by

where:

• \(L\) = angular momentum
• \(I\) = moment of inertia
• \(\omega\) = angular velocity

The moment of inertia tells us how mass is distributed relative to the axis of rotation. If more mass is farther from the axis, \(I\) is larger. If mass is closer to the axis, \(I\) is smaller.

The angular velocity \(\omega\) tells us how fast something is rotating. A larger \(\omega\) means faster spinning.

When no external torque acts, angular momentum before and after a change must be equal:

For many problems, this becomes

This equation is extremely useful. It shows that if the moment of inertia decreases, the angular velocity must increase to keep angular momentum constant. If the moment of inertia increases, the angular velocity must decrease.

Before going further, let us connect this to something you may already know. In straight-line motion, momentum is conserved when there is no net external force. In rotational motion, angular momentum is conserved when there is no net external torque.

Torque is the turning effect of a force. A force that causes an object to twist or rotate produces torque. If the net external torque is zero, then the total angular momentum does not change.

This gives us the main rule:

• No net external torque \(\rightarrow\) angular momentum stays constant.
• Net external torque present \(\rightarrow\) angular momentum can change.

It is important to notice that an object can change shape and still keep the same total angular momentum, as long as no external torque acts. This is why a skater, diver, or gymnast can spin faster simply by changing body position.

Let us look more closely at the relationship between \(I\) and \(\omega\):

• If mass moves closer to the axis, \(I\) decreases.
• If \(L\) stays constant, then \(\omega\) must increase.
• If mass moves farther from the axis, \(I\) increases.
• If \(L\) stays constant, then \(\omega\) must decrease.

This is the reason behind many real-life situations:

• A figure skater pulls in their arms and spins faster.
• A diver tucks into a ball to rotate faster in the air.
• Planets and comets move faster when they are closer to the Sun in their orbit.
• A person on a rotating stool spins faster when holding weights close to their body.

Why do planets move faster when closer to the Sun? The gravitational force from the Sun acts toward the Sun, so it does not create a turning effect that changes the planet's angular momentum about the Sun in the usual orbital picture. As a result, the planet's angular momentum stays nearly constant, and when the distance to the Sun decreases, the speed must increase.

This can be understood with a simple idea. For an object moving around a center, angular momentum depends on both how fast it moves and how far it is from the center. If it gets closer in, it must move faster so the total angular momentum remains the same.

Now let us work through some examples.

Worked Example 1: A figure skater pulls in her arms

A skater has an initial moment of inertia of \(4.0\,\text{kg·m}^2\) and an initial angular velocity of \(2.0\,\text{rad/s}\). She pulls in her arms, reducing her moment of inertia to \(2.0\,\text{kg·m}^2\). What is her new angular velocity?

Step 1: Write the conservation equation.

Step 2: Substitute the known values.

Step 3: Solve.

Answer: Her new angular velocity is \(4.0\,\text{rad/s}\).

What happened? Her moment of inertia was cut in half, so her spinning speed doubled.

Worked Example 2: A student on a rotating stool

A student spins on a stool while holding two masses. At first, the total moment of inertia is \(6.0\,\text{kg·m}^2\) and the angular velocity is \(3.0\,\text{rad/s}\). The student stretches their arms outward so the moment of inertia increases to \(9.0\,\text{kg·m}^2\). Find the new angular velocity.

Step 1: Use angular momentum conservation.

Step 2: Substitute values.

Step 3: Solve.

Answer: The new angular velocity is \(2.0\,\text{rad/s}\).

What happened? Since the mass moved farther from the axis, the moment of inertia increased, so the spinning speed decreased.

Worked Example 3: Finding final moment of inertia

A rotating platform has an initial angular velocity of \(5.0\,\text{rad/s}\) and an initial moment of inertia of \(2.4\,\text{kg·m}^2\). After a person changes position, the angular velocity becomes \(8.0\,\text{rad/s}\). Assuming no external torque, what is the final moment of inertia?

Step 1: Start with conservation of angular momentum.

Step 2: Substitute known values.

Step 3: Solve.

Answer: The final moment of inertia is \(1.5\,\text{kg·m}^2\).

What does this mean? The person moved mass closer to the axis, making the platform-person system easier to spin quickly.

Worked Example 4: Simple orbital idea

A planet moves around a star. At one point in its orbit it is \(4\) times farther from the star than at another point. If its angular momentum is conserved, how does its speed compare at the closer point?

For orbital motion, a simpler idea is enough: when the distance from the center becomes smaller, the speed must become larger to keep angular momentum constant.

If the planet is \(4\) times closer, its orbital speed becomes about \(4\) times larger, assuming the direction is appropriate for comparing angular momentum in this simplified way.

Answer: The planet moves faster when it is closer to the star, and in this simple comparison the speed is about 4 times greater.

Now let us discuss some common misunderstandings.

• Misunderstanding 1: “If something spins faster, it must have more angular momentum.”
  Not always. If the moment of inertia decreases at the same time, the angular momentum can stay the same.
• Misunderstanding 2: “Angular momentum is always conserved.”
  It is conserved only when there is no net external torque on the system.
• Misunderstanding 3: “Mass alone determines rotational motion.”
  Mass matters, but where the mass is located relative to the axis also matters. That is why moment of inertia is so important.
• Misunderstanding 4: “Pulling arms inward creates angular momentum.”
  No. Pulling inward changes the moment of inertia. The angular momentum stays the same if no external torque acts.

Here is a good step-by-step method for solving conservation of angular momentum problems:

1. Identify the system.
2. Check whether there is a net external torque.
3. If there is no net external torque, set initial angular momentum equal to final angular momentum.
4. Use \(L = I\omega\) when appropriate.
5. Solve for the missing quantity.
6. Check whether your answer makes physical sense. For example, if \(I\) decreases, \(\omega\) should increase.

Important idea: Conservation of angular momentum does not mean angular velocity must stay constant. It means the product of moment of inertia and angular velocity stays constant, as long as no external torque acts.

Also remember that angular momentum is a property of a system. When solving problems, be clear about what is included in the system. For example, if a skater and her arms are all part of one system, then moving her arms changes the moment of inertia of the whole system, but does not add external torque.

Let us connect this topic to the broader ideas of mechanics:

• Linear momentum is conserved when net external force is zero.
• Angular momentum is conserved when net external torque is zero.
• Mechanical energy may or may not stay constant depending on whether energy is lost or transferred.

That last point is important. In some real situations, angular momentum is conserved even when mechanical energy is not exactly conserved. For example, when a person pulls in their arms while spinning, they may do work with their muscles. So angular momentum can stay constant while rotational kinetic energy changes.

This means you should not assume that both energy and angular momentum are always conserved in every rotation problem. Always ask: Is there external torque? and Is energy being added or removed?

Brief Summary

Angular momentum is the rotational version of momentum and is given by \(L = I\omega\) for many spinning objects. If no net external torque acts on a system, angular momentum is conserved:

When the moment of inertia decreases, angular velocity increases. When the moment of inertia increases, angular velocity decreases. This principle explains spinning skaters, divers, rotating stools, and the motion of planets in orbit.