Classical Mechanics and Kinematics

Reference Frames and Coordinate Systems are basic tools for describing motion in physics. When we say that an object is moving, we are always comparing its position to something else. That “something else” is called a reference frame.

For example, a person sitting on a train may say that the seat beside them is at rest. But a person standing on the ground sees that same seat moving with the train. Both descriptions are correct because they are made from different reference frames.

A coordinate system is the set of axes we choose to measure position. In one dimension, we often use a single number line. In two dimensions, we usually use an x-axis and a y-axis. These axes help us describe where an object is and how it moves.

Understanding reference frames and coordinate systems is important because position, displacement, velocity, and even acceleration are described using them. If you do not know the reference frame or axis directions, the description of motion can be confusing or incomplete.

1. What is a Reference Frame?

A reference frame is the point of view from which motion is observed and measured. It includes:

• a location or observer,
• a way to measure position,
• and a clock to measure time.

In everyday life, common reference frames include:

• the ground,
• a moving car,
• a bicycle,
• an elevator,
• or even another object.

An object can be at rest in one frame and moving in another frame. This is one of the most important ideas in mechanics.

Suppose you place a backpack on the floor of a bus. To a passenger on the bus, the backpack is not moving. To a person on the roadside, the backpack moves forward along with the bus. So motion is relative, not absolute.

2. What is a Coordinate System?

A coordinate system gives a mathematical way to describe position. Before solving any motion problem, we choose:

• an origin, where position is zero,
• the direction of positive axes,
• and sometimes the scale of measurement.

In one-dimensional motion, the position of an object may be written as just one coordinate, such as \(x = 5\,\text{m}\). This means the object is 5 meters from the origin in the positive x-direction.

In two-dimensional motion, we write position as an ordered pair, such as \((x, y) = (3\,\text{m}, 4\,\text{m})\). This tells us how far the object is along the horizontal and vertical directions.

The choice of axis direction matters. If we choose east as positive, motion to the west is negative. If we choose upward as positive, downward motion is negative.

3. Position, Displacement, and Motion

Once a coordinate system is chosen, we can describe position clearly. The position of an object tells where it is relative to the origin.

The displacement is the change in position. In one dimension, displacement is:

$$\Delta x = x_f - x_i$$

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

Displacement depends on direction, so it can be positive, negative, or zero. This is different from distance, which is the total path length and is always positive.

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

$$\Delta x = 9 - 2 = 7\,\text{m}$$

If the student then walks back to \(x = 5\,\text{m}\), the displacement from the starting point becomes:

$$\Delta x = 5 - 2 = 3\,\text{m}$$

Even though the student walked more than 3 m in total, the displacement only depends on the start and end positions.

4. Velocity in Different Reference Frames

Velocity describes how position changes with time. In one dimension, average velocity is:

$$v_{avg} = \frac{\Delta x}{\Delta t}$$

The measured velocity depends on the reference frame.

Imagine a girl walking forward inside a train at \(2\,\text{m/s}\) relative to the train. If the train itself moves at \(15\,\text{m/s}\) relative to the ground, then the girl’s velocity relative to the ground is:

$$v_{girl,ground} = v_{girl,train} + v_{train,ground} = 2 + 15 = 17\,\text{m/s}$$

If she walks backward at \(2\,\text{m/s}\) relative to the train, then:

$$v_{girl,ground} = -2 + 15 = 13\,\text{m/s}$$

This shows that velocities can be combined when one frame moves relative to another. The sign depends on the axis direction you choose.

5. Rest and Motion Are Relative

In classical mechanics, there is no special “absolute rest” for everyday motion. Saying something is “at rest” only means it is not changing position in the chosen reference frame.

Examples:

• A book on a desk is at rest relative to the desk.
• The same book is moving relative to the Sun because Earth is moving.
• A passenger is at rest relative to a bus seat but moving relative to the road.

So, when describing motion, always ask: relative to what?

6. One-Dimensional Coordinate Systems

Many motion problems are one-dimensional, meaning motion happens along a straight line. In this case, we choose one axis, usually the x-axis.

For example, suppose a car moves along a road. We might choose:

• origin at a traffic light,
• positive direction to the east,
• negative direction to the west.

If the car is 50 m east of the light, then \(x = +50\,\text{m}\). If it is 20 m west, then \(x = -20\,\text{m}\).

This system helps us write motion clearly and avoid confusion about direction.

7. Two-Dimensional Coordinate Systems

In two dimensions, we use two perpendicular axes. Usually:

• the x-axis is horizontal,
• the y-axis is vertical.

A position might be written as \((x, y)\). For example, \((4, 3)\) means 4 units in the x-direction and 3 units in the y-direction.

This is useful for describing the motion of objects such as projectiles, boats crossing rivers, or people walking across a field.

In two dimensions, displacement can be written as changes in each direction:

$$\Delta x = x_f - x_i$$

$$\Delta y = y_f - y_i$$

If needed, the size of the displacement can be found using:

$$|\Delta \vec{r}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

This comes from the Pythagorean theorem.

8. Why Choosing Axes Carefully Matters

The physics does not change when you choose a different coordinate system, but the equations may become easier or harder to use.

For example:

• For motion on a straight road, choose the x-axis along the road.
• For vertical motion, choose the y-axis upward or downward.
• For motion on a slope, choose one axis along the slope if that helps simplify the problem.

A smart choice of axes makes signs and calculations easier to manage.

9. Worked Example 1: Position and Displacement in One Dimension

A student stands at \(x = -3\,\text{m}\) on a number line and walks to \(x = 8\,\text{m}\). Find the displacement.

Step 1: Write the formula.

$$\Delta x = x_f - x_i$$

Step 2: Substitute values.

$$\Delta x = 8 - (-3)$$

Step 3: Calculate.

$$\Delta x = 11\,\text{m}$$

Answer: The displacement is \(11\,\text{m}\) in the positive x-direction.

10. Worked Example 2: Velocity in Different Reference Frames

A boat moves at \(6\,\text{m/s}\) east relative to the water. The river flows at \(2\,\text{m/s}\) east relative to the riverbank. What is the boat’s velocity relative to the riverbank?

Step 1: Choose east as positive.

Boat relative to water: \(+6\,\text{m/s}\)

Water relative to bank: \(+2\,\text{m/s}\)

Step 2: Add the velocities.

$$v_{boat,bank} = v_{boat,water} + v_{water,bank}$$

$$v_{boat,bank} = 6 + 2 = 8\,\text{m/s}$$

Answer: The boat’s velocity relative to the riverbank is \(8\,\text{m/s}\) east.

11. Worked Example 3: Backward Motion Inside a Moving Vehicle

A boy walks at \(1.5\,\text{m/s}\) west relative to a bus. The bus moves at \(12\,\text{m/s}\) east relative to the ground. Find the boy’s velocity relative to the ground.

Step 1: Choose east as positive.

Boy relative to bus: \(-1.5\,\text{m/s}\)

Bus relative to ground: \(+12\,\text{m/s}\)

Step 2: Add the velocities.

$$v_{boy,ground} = v_{boy,bus} + v_{bus,ground}$$

$$v_{boy,ground} = -1.5 + 12 = 10.5\,\text{m/s}$$

Answer: The boy moves at \(10.5\,\text{m/s}\) east relative to the ground.

Even though he is walking west inside the bus, he is still moving east relative to the ground because the bus is moving faster eastward.

12. Worked Example 4: Position in Two Dimensions

A ball moves from point \((2, 1)\,\text{m}\) to point \((8, 5)\,\text{m}\). Find:

1. the displacement in x,
2. the displacement in y,
3. the size of the total displacement.

Step 1: Find changes in each direction.

$$\Delta x = 8 - 2 = 6\,\text{m}$$

$$\Delta y = 5 - 1 = 4\,\text{m}$$

Step 2: Find the size of displacement.

$$|\Delta \vec{r}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

$$|\Delta \vec{r}| = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}$$

$$|\Delta \vec{r}| \approx 7.2\,\text{m}$$

Answer:

• \(\Delta x = 6\,\text{m}\)
• \(\Delta y = 4\,\text{m}\)
• Total displacement \(\approx 7.2\,\text{m}\)

13. Common Mistakes to Avoid

• Forgetting to state the reference frame. Always say whether motion is measured relative to the ground, a vehicle, water, or something else.
• Ignoring sign conventions. Positive and negative directions must be chosen and used consistently.
• Confusing distance and displacement. Distance is total path length; displacement is change in position.
• Mixing frames incorrectly. Make sure each velocity is clearly labeled, such as “relative to the train” or “relative to the ground.”
• Using coordinates without defining the origin. Position numbers only make sense when the zero point is known.

14. Key Ideas to Remember

• Motion is described relative to a reference frame.
• A coordinate system uses an origin and axes to measure position.
• An object can be at rest in one frame and moving in another.
• Displacement is the change in position: \(\Delta x = x_f - x_i\).
• Velocity depends on the reference frame used.
• In two dimensions, position is written with coordinates such as \((x, y)\).

Brief Summary

Reference frames and coordinate systems allow us to describe motion clearly. A reference frame tells whose point of view we are using, while a coordinate system gives the axes and origin for measuring position. Because motion is relative, the same object can appear at rest or in motion depending on the observer. Careful choice of axes and correct use of signs help us calculate displacement and velocity accurately in one and two dimensions.

Scalar vs. Vector Quantities

In mechanics and kinematics, we describe motion using physical quantities such as distance, speed, displacement, and velocity. These quantities do not all behave in the same way. Some only need a size, while others need both a size and a direction.

This is the key difference between scalar and vector quantities. Understanding this difference helps you correctly describe motion and avoid common mistakes, especially when solving problems in one and two dimensions.

Definition of a Scalar

A scalar quantity is a quantity that has magnitude only. Magnitude means the numerical size of the quantity, along with its unit.

For example, if a car moves at 20 m/s, that tells us how fast it is moving, but not the direction. So speed is a scalar.

Common scalar quantities in mechanics include:

• Distance
• Speed
• Time
• Mass
• Energy

Definition of a Vector

A vector quantity is a quantity that has both magnitude and direction. To fully describe a vector, you must state how big it is and which way it points.

For example, saying an object moves at 20 m/s east gives both the magnitude, 20 m/s, and the direction, east. So velocity is a vector.

Common vector quantities in mechanics include:

• Displacement
• Velocity
• Acceleration
• Force
• Momentum

The Main Idea

If a quantity answers only how much?, it is usually a scalar. If it answers both how much? and which direction?, it is a vector.

This difference matters because scalars and vectors are combined differently. Scalars are usually added with ordinary arithmetic. Vectors must be added by taking direction into account.

Distance vs. Displacement

One of the most important scalar-vector pairs in kinematics is distance and displacement.

Distance is the total length of the path traveled. It does not care about direction, so it is a scalar.

Displacement is the straight-line change in position from the starting point to the ending point, including direction. It is a vector.

For example, imagine a student walks 3 m east and then 3 m west.

• Distance traveled = \(3 + 3 = 6\text{ m}\)
• Displacement = \(0\text{ m}\), because the student ends where they started

The distance is not zero because the student did move. The displacement is zero because there was no overall change in position.

Speed vs. Velocity

Another important pair is speed and velocity.

Speed tells how fast something moves. It is a scalar.

Velocity tells how fast something moves and in what direction. It is a vector.

Average speed is calculated using total distance:

Average velocity is calculated using displacement:

Because displacement includes direction, average velocity also includes direction.

How Vectors Are Represented

Vectors are often shown in one of two ways:

• With an arrow above the symbol, such as \(\vec{v}\)
• By writing magnitude and direction, such as 5 m/s north

In diagrams, vectors are drawn as arrows. The length of the arrow represents magnitude, and the arrowhead shows direction.

Adding Scalars and Vectors

Scalars are added normally. For example, if you walk 2 m and then 5 m more, the total distance is:

Vectors must be added with direction included.

If you move 4 m east and then 3 m east, the displacement is:

But if you move 4 m east and then 3 m west, the directions are opposite. We can treat one direction as positive and the other as negative:

So the net displacement is:

Why Direction Matters

Direction can completely change the result of a motion problem. Two objects can have the same speed but different velocities if they move in different directions.

For example, one car moving at 25 m/s north and another moving at 25 m/s south have the same speed, but different velocities.

Worked Example 1: Identifying Scalars and Vectors

Classify each quantity as a scalar or vector:

• 12 s
• 9 m/s west
• 50 m
• 3 m/s² downward

Solution

1. 12 s: This is time. Time has magnitude only, so it is a scalar.
2. 9 m/s west: This is a speed with direction, so it is velocity. It is a vector.
3. 50 m: This gives only size, not direction. It is a scalar.
4. 3 m/s² downward: Acceleration with a direction is a vector.

Worked Example 2: Distance and Displacement

A runner moves 100 m east, then 40 m west.

Find:

• Total distance
• Displacement

Solution

Step 1: Find distance

Distance is the total path traveled:

Step 2: Find displacement

Take east as positive and west as negative:

So the displacement is:

Answer:

• Distance = 140 m
• Displacement = 60 m east

Worked Example 3: Average Speed and Average Velocity

A student walks 30 m north in 20 s, then walks 10 m south in 10 s.

Find:

• Average speed
• Average velocity

Solution

Step 1: Total distance

Step 2: Total time

Step 3: Average speed

Step 4: Displacement

Take north as positive and south as negative:

So displacement is \(20\text{ m north}\).

Step 5: Average velocity

Answer:

• Average speed = 1.33 m/s
• Average velocity = 0.67 m/s north

Worked Example 4: Vector Addition in One Dimension

A toy car moves 8 m to the right, then 12 m to the left.

What is the car's final displacement?

Solution

Take right as positive and left as negative:

The negative sign means the final direction is left.

Final displacement:

This example shows that the final displacement follows the direction of the larger motion.

Common Mistakes to Avoid

• Mixing up distance and displacement: Distance is total path; displacement is overall change in position.
• Mixing up speed and velocity: Speed has no direction; velocity does.
• Ignoring direction in vector problems: A number alone is not enough for a vector.
• Adding opposite directions as if they were the same: Opposite directions should subtract, not simply add.

Quick Comparison Table

• Scalar: magnitude only
• Vector: magnitude and direction
• Distance: scalar
• Displacement: vector
• Speed: scalar
• Velocity: vector

How to Decide if a Quantity Is Scalar or Vector

1. Ask: Does it need a direction to be complete?
2. If no, it is a scalar.
3. If yes, it is a vector.

For example:

• 5 kg: no direction needed, so scalar
• 10 N upward: direction needed, so vector

Brief Summary

A scalar quantity has magnitude only, while a vector quantity has both magnitude and direction. In kinematics, distance and speed are scalars, while displacement and velocity are vectors.

When solving motion problems, always check whether direction matters. If it does, you are working with a vector, and direction must be included in your answer.

Vector Addition and Resolution are essential tools in mechanics. Many physical quantities, such as displacement, velocity, acceleration, and force, are vectors. A vector has both magnitude and direction.

In contrast, a scalar has magnitude only. Examples of scalars include mass, time, distance, and speed. Understanding the difference matters because vectors cannot be added in the same way as ordinary numbers unless their directions are taken into account.

This lesson explains how to add vectors and how to resolve a vector into perpendicular components. These skills are widely used in one-dimensional and two-dimensional motion problems.

1. What is a vector?

A vector is usually represented by an arrow. The length of the arrow shows the magnitude, and the arrowhead shows the direction.

• A force of 10 N to the right is a vector.
• A velocity of 20 m/s at 30° above the horizontal is a vector.
• A displacement of 5 m north is a vector.

In two dimensions, we often describe vectors using x- and y-axes. The horizontal direction is usually the x-axis, and the vertical direction is the y-axis.

2. Vector addition

When two or more vectors act together, we often need to find their resultant. The resultant is the single vector that has the same effect as all the given vectors combined.

There are two main situations:

• Vectors along the same straight line
• Vectors at different angles

2.1 Adding vectors in the same direction

If two vectors point in the same direction, their magnitudes are added directly.

For example, if a box is pulled by 4 N east and also by 6 N east, the resultant force is:

2.2 Adding vectors in opposite directions

If two vectors point in opposite directions, subtract the smaller magnitude from the larger one. The direction of the resultant is the direction of the larger vector.

For example, if one force is 9 N east and another is 5 N west, the resultant is:

2.3 Adding vectors at an angle

When vectors are not along the same line, simple addition is not enough. We must account for direction. Two common methods are:

• Graphical method using a scale diagram
• Analytical method using components and trigonometry

In Grade 12 mechanics, the analytical method is usually more accurate and more useful, especially in exam questions.

3. Resolution of a vector into components

To resolve a vector means to break it into parts along perpendicular directions, usually horizontal and vertical components.

Suppose a vector of magnitude \(A\) makes an angle \(\theta\) with the positive x-axis. Its components are:

Here:

• \(A_x\) is the horizontal component
• \(A_y\) is the vertical component

This comes from right triangle trigonometry. In a right triangle:

• \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}\)
• \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\)

So if the vector is the hypotenuse, the horizontal side is \(A\cos\theta\) and the vertical side is \(A\sin\theta\).

Important: these formulas are correct when the angle is measured from the horizontal. If the angle is measured from the vertical, the roles of sine and cosine switch relative to the axes.

4. Signs of components

Components can be positive or negative depending on direction.

• Right: positive x
• Left: negative x
• Up: positive y
• Down: negative y

For example:

• A vector pointing up and right has positive \(x\) and positive \(y\).
• A vector pointing down and left has negative \(x\) and negative \(y\).

Always think carefully about the direction before assigning signs.

5. Adding vectors using components

To add several vectors in two dimensions, follow these steps:

1. Resolve each vector into x- and y-components.
2. Add all x-components to get the total horizontal component, \(R_x\).
3. Add all y-components to get the total vertical component, \(R_y\).
4. Find the magnitude of the resultant using Pythagoras.
5. Find the direction using trigonometry.

If \(R_x\) and \(R_y\) are the resultant components, then the magnitude of the resultant \(R\) is:

The direction angle \(\theta\), measured from the positive x-axis, is found using:

So:

After calculating the angle, check the signs of \(R_x\) and \(R_y\) to make sure the vector lies in the correct quadrant.

6. Worked Example 1: Resolving one vector

A force of 20 N acts at an angle of 30° above the horizontal. Find its horizontal and vertical components.

Step 1: Write the known values

Magnitude: \(A = 20\text{ N}\)

Angle: \(\theta = 30^\circ\)

Step 2: Use the component formulas

Step 3: Substitute values

Answer:

• Horizontal component = \(17.32\text{ N}\) to the right
• Vertical component = \(10.0\text{ N}\) upward

7. Worked Example 2: Adding two perpendicular vectors

A student walks 3 m east and then 4 m north. Find the resultant displacement.

This forms a right triangle.

Step 1: Identify components

Step 2: Find magnitude

Step 3: Find direction

Answer: The resultant displacement is \(5\text{ m}\) at \(53.1^\circ\) north of east.

8. Worked Example 3: Adding two vectors at different angles

A boat is pulled by two ropes. One force is 50 N east. Another force is 40 N at 60° north of east. Find the resultant force.

Step 1: Resolve each vector into components

First force:

Second force:

Step 2: Add components

Step 3: Find magnitude

Step 4: Find direction

Answer: The resultant force is approximately \(78.1\text{ N}\) at \(26.4^\circ\) north of east.

9. Worked Example 4: Resultant with a negative component

A particle moves with one displacement of 8 m east and another displacement of 6 m at 45° south of west. Find the resultant displacement.

Step 1: Write the first vector in components

Step 2: Resolve the second vector

Since it is 45° south of west, both components are negative: left and down.

Step 3: Add components

Step 4: Find magnitude

Step 5: Find direction

The resultant has positive x and negative y, so it points south of east.

Answer: The resultant displacement is approximately \(5.67\text{ m}\) at \(48.4^\circ\) south of east.

10. Common mistakes to avoid

• Ignoring direction: Magnitude alone is not enough when working with vectors.
• Using sine and cosine incorrectly: Check whether the angle is measured from the horizontal or vertical.
• Forgetting negative signs: Leftward and downward components are negative if right and up are chosen as positive.
• Adding magnitudes directly: This only works when vectors are along the same line.
• Wrong direction for the final answer: Use the signs of the resultant components to decide the correct quadrant.

11. Why vector resolution is useful in mechanics

Many mechanics problems become much easier when vectors are split into horizontal and vertical parts. For example:

• Projectile motion uses horizontal and vertical velocity components.
• Forces on slopes are resolved into components parallel and perpendicular to the slope.
• Net force problems often require adding several force vectors.

By working with components, you turn a complicated angled vector problem into simpler horizontal and vertical calculations.

12. Key formulas

• Horizontal component: \(A_x = A\cos\theta\)
• Vertical component: \(A_y = A\sin\theta\)
• Resultant magnitude: \(R = \sqrt{R_x^2 + R_y^2}\)
• Resultant direction: \(\theta = \tan^{-1}(R_y/R_x)\)

13. Brief summary

Vector addition combines two or more vectors to find a single resultant vector. Vector resolution breaks one vector into perpendicular components, usually horizontal and vertical.

To solve two-dimensional vector problems, resolve each vector into components, add the x-components and y-components separately, then use Pythagoras and trigonometry to find the magnitude and direction of the resultant.

With practice, these methods become a powerful way to solve motion and force problems accurately.

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. We focus on quantities such as position, displacement, velocity, and acceleration.

This lesson is important because many real situations can be modeled as motion in one dimension, such as a car moving on a straight road, an elevator going up or down, or an object falling vertically. Once you understand one-dimensional motion, you will be better prepared for more advanced topics in mechanics.

1. Position and Reference Point

To describe motion, we must first choose a reference point and a coordinate system. In one-dimensional motion, we usually use an x-axis or a vertical y-axis. A position tells us where an object is located relative to the origin.

For example, if a car is at \(x = 5\text{ m}\), that means it is 5 meters to the right of the origin. If it is at \(x = -3\text{ m}\), it is 3 meters to the left of the origin. The sign matters because it tells us direction.

2. Distance vs. Displacement

These two ideas are related but not the same.

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

Displacement is calculated using:

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

If a student walks from \(2\text{ m}\) to \(10\text{ m}\), the displacement is:

If the student then walks back to \(6\text{ m}\), the total distance traveled is \(8 + 4 = 12\text{ m}\), but the overall displacement is:

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

3. Time and Motion

Motion happens over time. To describe how position changes, we compare displacement to the time interval. If an object changes position quickly, it has a larger velocity. If it changes slowly, its velocity is smaller.

The time interval is written as:

Time intervals are usually positive, because we measure how much time passes between two events.

4. Average Velocity

Average velocity tells us how quickly displacement happens over a time interval. It is defined as:

Average velocity includes direction because displacement includes direction. Its unit is meters per second, written as \(\text{m/s}\).

If a runner moves from \(x = 4\text{ m}\) to \(x = 20\text{ m}\) in \(4\text{ s}\), then:

A positive velocity means motion in the positive direction. A negative velocity means motion in the negative direction.

5. Instantaneous Velocity

In real life, an object's speed can change from moment to moment. Instantaneous velocity is the velocity at a specific instant in time. It tells us how fast and in what direction the object is moving at that exact moment.

For constant velocity motion, the average velocity and instantaneous velocity are the same. For changing motion, they can be different.

In 12th Grade physics, we often find instantaneous velocity from motion equations or from graphs. On a position-time graph, the instantaneous velocity is the slope of the tangent line at a point. If the graph is a straight line, the slope is constant, so the velocity is constant.

6. Acceleration

Acceleration measures how quickly velocity changes. If an object speeds up, slows down, or changes direction, it is accelerating.

Average acceleration is defined as:

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

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

Be careful: negative acceleration does not always mean slowing down. Whether an object speeds up or slows down depends on the signs of both 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.

7. Constant Acceleration Equations

When acceleration is constant, we can use the standard kinematics equations. These are very important in one-dimensional motion.

The main equations are:

These equations work only when acceleration is constant.

Here:

• \(v_i\) = initial velocity
• \(v_f\) = final velocity
• \(a\) = acceleration
• \(t\) = time interval
• \(\Delta x\) = displacement

A good strategy is to list the known values, identify the unknown value, and then choose the equation that connects them.

8. Free Fall as One-Dimensional Motion

Vertical motion under gravity is a common one-dimensional case. If air resistance is ignored, all falling objects near Earth have the same constant acceleration due to gravity.

If we choose upward as positive, then the acceleration is:

If we choose downward as positive, then \(a = +9.8\text{ m/s}^2\). The choice of sign depends on the coordinate system, but you must stay consistent.

An object thrown upward slows down as it rises because its velocity is upward while acceleration is downward. At the highest point, its instantaneous velocity is zero for an instant, but its acceleration is still \(-9.8\text{ m/s}^2\).

9. Graphs in One-Dimensional Kinematics

Graphs are powerful tools for understanding motion.

Position-Time Graph

• The slope gives velocity.
• A straight line means constant velocity.
• A curve means changing velocity.

Velocity-Time Graph

• The slope gives acceleration.
• The area under the graph gives displacement.
• A horizontal line means constant velocity.

Acceleration-Time Graph

• A horizontal line means constant acceleration.
• The area under the graph gives change in velocity.

When reading graphs, pay close attention to whether the values are positive or negative. That tells you the direction of motion or acceleration.

10. Solving Strategy for Kinematics Problems

1. Choose a positive direction.
2. Write down all known quantities with signs.
3. Identify what the question asks for.
4. Select the equation that matches the known and unknown variables.
5. Substitute carefully with units.
6. Check whether the sign and size of the answer make sense.

This organized method helps avoid mistakes, especially with negative signs.

Worked Example 1: Finding Displacement and Average Velocity

A cyclist moves from \(x_i = -2\text{ m}\) to \(x_f = 18\text{ m}\) in \(5\text{ s}\). Find the displacement and average velocity.

Step 1: Find displacement

Step 2: Find average velocity

Answer: The displacement is \(20\text{ m}\), and the average velocity is \(4\text{ m/s}\) in the positive direction.

Worked Example 2: Constant Acceleration from Rest

A car starts from rest and accelerates at \(3.0\text{ m/s}^2\) for \(6.0\text{ s}\). Find its final velocity and displacement.

Given:

• \(v_i = 0\text{ m/s}\)
• \(a = 3.0\text{ m/s}^2\)
• \(t = 6.0\text{ s}\)

Step 1: Final velocity

Step 2: Displacement

Answer: The final velocity is \(18\text{ m/s}\), and the car travels \(54\text{ m}\).

Worked Example 3: Slowing Down

A train is moving at \(25\text{ m/s}\) and slows down uniformly at \(-2.5\text{ m/s}^2\). How long does it take to stop, and how far does it travel before stopping?

Given:

• \(v_i = 25\text{ m/s}\)
• \(v_f = 0\text{ m/s}\)
• \(a = -2.5\text{ m/s}^2\)

Step 1: Time to stop

Step 2: Displacement

Use:

Answer: The train takes \(10\text{ s}\) to stop and travels \(125\text{ m}\) before stopping.

Worked Example 4: Vertical Motion Under Gravity

A ball is thrown straight upward with an initial velocity of \(20\text{ m/s}\). Ignore air resistance. Find:

• the time to reach the highest point
• the maximum height

Choose upward as positive. Then:

• \(v_i = 20\text{ m/s}\)
• \(v_f = 0\text{ m/s}\) at the top
• \(a = -9.8\text{ m/s}^2\)

Step 1: Time to highest point

Step 2: Maximum height

Answer: The ball takes about \(2.04\text{ s}\) to reach the top and rises about \(20.4\text{ m}\).

11. Common Mistakes to Avoid

• Confusing distance with displacement.
• Forgetting that velocity and acceleration can be negative.
• Using a constant-acceleration equation when acceleration is not constant.
• Mixing up initial and final values.
• Ignoring units.
• Using the wrong sign for gravity.

12. Key Ideas to Remember

• One-dimensional kinematics describes motion along a straight line.
• Displacement is change in position: \(\Delta x = x_f - x_i\).
• Average velocity is displacement divided by time.
• Instantaneous velocity is velocity at a specific moment.
• Acceleration is the rate of change of velocity.
• For constant acceleration, use the standard kinematics equations.
• Always choose a sign convention and use it consistently.

Brief Summary

One-dimensional kinematics explains how objects move in a straight line using position, displacement, velocity, and acceleration. Displacement tells how far and in what direction position changes, velocity describes how fast displacement changes, and acceleration describes how fast velocity changes. When acceleration is constant, we can use kinematics equations to solve for unknown quantities such as final velocity, time, and displacement. Careful attention to signs, units, and the meaning of each quantity is the key to solving these problems correctly.

Kinematic Graph Analysis is the skill of reading motion directly from graphs. In kinematics, graphs help us see how an object's position, velocity, and sometimes acceleration change with time. If you can interpret these graphs, you can understand motion without needing a long word problem.

This lesson focuses on two key ideas:

• How to find velocity from the slope of a position-time graph
• How to find displacement from the area under a velocity-time graph

These ideas connect algebra, geometry, and the basic calculus idea that rate of change is related to slope, while accumulated change is related to area.

1. Position-Time Graphs

A position-time graph shows where an object is at different times. Time is placed on the horizontal axis, and position is placed on the vertical axis.

For a position-time graph, the most important feature is the slope. The slope tells you how quickly position is changing with time, which is exactly the definition of velocity.

Mathematically, slope is:

$$\text{slope} = \frac{\Delta x}{\Delta t}$$

Since velocity is change in position divided by change in time, we have:

$$v = \frac{\Delta x}{\Delta t}$$

This means:

• 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 slope is constant, so the velocity is constant. If the graph curves, then the velocity is changing over time.

Average velocity and instantaneous velocity

When you calculate slope between two points on a position-time graph, you find the average velocity over that time interval:

$$v_{\text{avg}} = \frac{x_2 - x_1}{t_2 - t_1}$$

If you want the velocity at one exact moment on a curved graph, you need the slope of the graph at that point. In basic calculus language, this is the slope of the tangent line, which gives the instantaneous velocity.

For many 12th Grade problems, you may only need to:

• Find slope on a straight segment
• Estimate the slope of a tangent on a curved graph

2. Velocity-Time Graphs

A velocity-time graph shows how velocity changes with time. Time is on the horizontal axis, and velocity is on the vertical axis.

For a velocity-time graph, the key idea is area under the graph. The area between the graph and the time axis gives the object's displacement.

Displacement means change in position:

$$\Delta x = x_f - x_i$$

From a velocity-time graph:

$$\Delta x = \text{area under the } v\text{-}t \text{ graph}$$

If velocity is constant, the graph is a horizontal line. The area under it is a rectangle:

$$\Delta x = v \Delta t$$

If velocity changes linearly, the area may be a triangle or a trapezoid.

Common area formulas:

• Rectangle: $$A = bh$$
• Triangle: $$A = \frac{1}{2}bh$$
• Trapezoid: $$A = \frac{1}{2}(b_1 + b_2)h$$

In velocity-time graphs, be careful with sign:

• Area above the time axis gives positive displacement
• Area below the time axis gives negative displacement

This is important because displacement is different from distance.

• Displacement includes direction
• Distance is total ground covered and is always positive

To find total distance from a velocity-time graph, add the absolute values of all areas:

$$\text{distance} = \sum |\text{area}|$$

3. Connecting the Graphs

Position, velocity, and acceleration are connected. In this lesson, the main connections are:

• The slope of a position-time graph gives velocity
• The area under a velocity-time graph gives displacement

Using basic calculus notation, these ideas can be written as:

$$v = \frac{dx}{dt}$$

and

$$\Delta x = \int v\,dt$$

You do not need advanced calculus to use these ideas. On graphs, the derivative appears as slope, and the integral appears as area.

4. How to Analyze a Position-Time Graph

1. Read the axes carefully and check units.
2. Pick two clear points if the graph segment is straight.
3. Use $$v = \frac{\Delta x}{\Delta t}$$ to find slope.
4. Decide whether the object is moving forward, backward, or not moving.
5. Compare steepness to compare speeds.

If the graph changes direction, the slope changes sign. That means the velocity changes sign, so the object reverses direction.

5. How to Analyze a Velocity-Time Graph

1. Read the axes and units.
2. Break the graph into simple shapes like rectangles and triangles.
3. Find the area of each section.
4. Keep track of positive and negative areas.
5. Add the signed areas to find displacement.
6. Add the absolute values if you need total distance.

Worked Example 1: Velocity from a straight position-time graph

An object moves from position \(2\text{ m}\) at \(t = 1\text{ s}\) to position \(14\text{ m}\) at \(t = 5\text{ s}\). Find its velocity.

Step 1: Use the slope formula.

$$v = \frac{\Delta x}{\Delta t} = \frac{14 - 2}{5 - 1}$$

$$v = \frac{12}{4} = 3\text{ m/s}$$

Answer: The velocity is \(3\text{ m/s}\).

What this means: The position increases by 3 meters every second, so the object is moving in the positive direction at constant velocity.

Worked Example 2: Interpreting different slopes on a position-time graph

Suppose a position-time graph has three straight segments:

• From \(0\) s to \(2\) s, position goes from \(0\) m to \(6\) m
• From \(2\) s to \(4\) s, position stays at \(6\) m
• From \(4\) s to \(6\) s, position goes from \(6\) m to \(2\) m

Find the velocity in each interval.

First interval:

$$v = \frac{6 - 0}{2 - 0} = 3\text{ m/s}$$

The object moves forward.

Second interval:

$$v = \frac{6 - 6}{4 - 2} = 0\text{ m/s}$$

The object is stopped.

Third interval:

$$v = \frac{2 - 6}{6 - 4} = \frac{-4}{2} = -2\text{ m/s}$$

The object moves in the negative direction.

Answer:

• \(0\) to \(2\) s: \(3\text{ m/s}\)
• \(2\) to \(4\) s: \(0\text{ m/s}\)
• \(4\) to \(6\) s: \(-2\text{ m/s}\)

What this shows: A positive slope means forward motion, a flat line means rest, and a negative slope means backward motion.

Worked Example 3: Displacement from a velocity-time graph with positive velocity

An object moves with constant velocity \(4\text{ m/s}\) for \(5\text{ s}\). Use the velocity-time graph to find displacement.

The graph is a horizontal line at \(4\text{ m/s}\) from \(0\) to \(5\) s. The area under the graph is a rectangle.

$$\Delta x = bh = (5)(4) = 20\text{ m}$$

Answer: The displacement is \(20\text{ m}\).

Check with the formula:

$$\Delta x = v\Delta t = (4)(5) = 20\text{ m}$$

Worked Example 4: Displacement and distance from a velocity-time graph with direction change

An object has the following velocity-time graph:

• From \(0\) to \(3\) s, velocity increases linearly from \(0\) to \(6\text{ m/s}\)
• From \(3\) to \(5\) s, velocity is constant at \(6\text{ m/s}\)
• From \(5\) to \(7\) s, velocity is constant at \(-2\text{ m/s}\)

Find the total displacement and total distance.

Step 1: First section, triangle from 0 to 3 s

$$A_1 = \frac{1}{2}bh = \frac{1}{2}(3)(6) = 9\text{ m}$$

This area is above the axis, so displacement is \(+9\text{ m}\).

Step 2: Second section, rectangle from 3 to 5 s

$$A_2 = bh = (2)(6) = 12\text{ m}$$

This is also above the axis, so displacement is \(+12\text{ m}\).

Step 3: Third section, rectangle from 5 to 7 s

$$A_3 = bh = (2)(-2) = -4\text{ m}$$

Because the graph is below the axis, this contributes negative displacement.

Step 4: Add signed areas for displacement

$$\Delta x = 9 + 12 - 4 = 17\text{ m}$$

Step 5: Add absolute values for distance

$$\text{distance} = |9| + |12| + |-4| = 25\text{ m}$$

Answer:

• Total displacement = \(17\text{ m}\)
• Total distance = \(25\text{ m}\)

6. Curved Position-Time Graphs

Sometimes a position-time graph is curved instead of straight. A curved graph means the velocity is changing.

If the graph becomes steeper over time, the object is speeding up in the positive direction. If the graph becomes less steep, the object is slowing down. If the slope changes from positive to negative, the object turns around.

To estimate velocity at one instant on a curved graph, draw a tangent line at that point and find its slope:

$$v_{\text{instant}} \approx \frac{\Delta x}{\Delta t} \text{ for the tangent line}$$

This is the graph-based meaning of instantaneous velocity.

7. Common Mistakes to Avoid

• Mixing up slope and area. On a position-time graph, use slope for velocity. On a velocity-time graph, use area for displacement.
• Forgetting units. Slope on an \(x\)-\(t\) graph has units of \(\text{m/s}\). Area on a \(v\)-\(t\) graph has units of meters.
• Ignoring negative values. Negative slope means negative velocity. Negative area means negative displacement.
• Confusing displacement with distance. Displacement can be negative; distance cannot.
• Reading only the height of a velocity-time graph. The height gives velocity, but the area gives displacement.

8. Quick Comparison Table

• Position-time graph: slope gives velocity
• Velocity-time graph: area gives displacement
• Flat position-time graph: object is at rest
• Flat velocity-time graph: constant velocity
• Negative slope on position-time graph: moving in the negative direction
• Area below axis on velocity-time graph: negative displacement

9. Final Summary

Kinematic graph analysis lets you read motion directly from graphs. On a position-time graph, the slope tells you the velocity. On a velocity-time graph, the area under the curve tells you the displacement.

Always pay attention to sign, units, and whether the question asks for velocity, displacement, or distance. If you remember slope from position-time and area from velocity-time, you will be able to solve most graph-based kinematics problems with confidence.

Free Fall and Terminal Velocity are two important ideas in mechanics that explain how objects move when gravity pulls them downward. In this lesson, you will learn what happens when an object falls, why falling objects speed up at first, and why some objects eventually stop getting faster when moving through air.

These ideas help explain everyday events, such as why a coin falls faster than a feather in air, why raindrops do not keep speeding up forever, and why parachutes make falling safer. Understanding free fall and terminal velocity also connects directly to forces, acceleration, and Newton's laws of motion.

1. What is free fall?

Free fall is the motion of an object when the only force acting on it is gravity. This means we ignore air resistance and any other forces. Near Earth's surface, gravity pulls objects downward with a nearly constant acceleration.

This acceleration is called the acceleration due to gravity, written as \(g\). Its value is about:

$$g \approx 9.8\,\text{m/s}^2$$

This means that for every second an object is in free fall, its downward velocity increases by about \(9.8\,\text{m/s}\).

If we choose downward as positive, then for free fall:

$$a = g = 9.8\,\text{m/s}^2$$

If we choose upward as positive, then gravity acts downward, so:

$$a = -g = -9.8\,\text{m/s}^2$$

Both choices are correct. The most important thing is to stay consistent with your sign convention throughout the problem.

2. Equations for free fall

Because gravity provides nearly constant acceleration near Earth's surface, free-fall motion can be described using the same kinematics equations used for any motion with constant acceleration.

The main equations are:

$$v = u + at$$

$$s = ut + \frac{1}{2}at^2$$

$$v^2 = u^2 + 2as$$

Here:

• \(u\) = initial velocity
• \(v\) = final velocity
• \(a\) = acceleration
• \(t\) = time
• \(s\) = displacement

In free fall, \(a\) is replaced by \(g\) or \(-g\), depending on your sign convention.

3. Key ideas about free fall

• All objects in free fall have the same acceleration if air resistance is ignored.
• A heavy object and a light object fall with the same acceleration in a vacuum.
• Mass affects the gravitational force on an object, but not the free-fall acceleration near Earth when air resistance is absent.
• An object thrown upward is still in free fall after it leaves your hand, because gravity is the only force acting on it if air resistance is ignored.

This is why, in ideal conditions, a dropped hammer and feather would fall together on the Moon, where there is essentially no air resistance.

4. What changes in real life? Air resistance

In the real world, falling objects usually move through air. Air pushes against the motion of the object. This force is called air resistance or drag.

Drag acts in the direction opposite to motion. So if an object is falling downward, air resistance acts upward.

At the moment an object is released from rest, its speed is zero or very small, so air resistance is also very small. At that point, gravity is the main force, so the object starts accelerating downward at about \(g\).

As the object speeds up, air resistance increases. This upward force begins to reduce the net downward force. Because acceleration depends on net force, the object's downward acceleration becomes smaller.

5. Net force while falling through air

When an object falls through air, two main forces act on it:

• Weight downward: \(W = mg\)
• Air resistance upward

The net force is:

$$F_{\text{net}} = mg - F_{\text{drag}}$$

So the acceleration is:

$$a = \frac{F_{\text{net}}}{m} = \frac{mg - F_{\text{drag}}}{m}$$

At first, \(F_{\text{drag}}\) is small, so \(a\) is close to \(g\). Later, as \(F_{\text{drag}}\) grows, \(a\) becomes smaller.

6. What is terminal velocity?

Terminal velocity is the maximum constant speed reached by a falling object when air resistance becomes equal to the object's weight.

At terminal velocity:

$$F_{\text{drag}} = mg$$

So the net force becomes zero:

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

And therefore the acceleration becomes zero:

$$a = 0$$

This does not mean the object stops moving. It means the object continues falling at a constant velocity.

This is an important idea: zero acceleration does not mean zero velocity. It means velocity is no longer changing.

7. How speed changes during a fall

1. The object is released.
2. Gravity pulls it downward.
3. It speeds up, so air resistance increases.
4. The upward drag force gets closer to the downward weight.
5. Acceleration decreases.
6. Eventually drag equals weight.
7. The object falls at terminal velocity.

If there were no air, step 6 would never happen. The object would continue accelerating downward as long as gravity acted alone.

8. What affects terminal velocity?

Different objects have different terminal velocities. Terminal velocity depends on several factors, including:

• Mass: more massive objects often have larger weights and may reach higher terminal velocities.
• Surface area: objects with larger area facing the air experience more drag and usually have lower terminal velocities.
• Shape: streamlined shapes move through air more easily and often have higher terminal velocities.
• Air density: denser air creates more drag.

For example, a skydiver with arms and legs spread out has more air resistance and a lower terminal velocity than a skydiver in a tucked position. A parachute greatly increases surface area, causing drag to rise sharply and terminal velocity to decrease.

9. Free fall compared with terminal velocity

• Free fall: only gravity acts; acceleration is constant at \(g\).
• Falling with air resistance: gravity and drag both act; acceleration changes during the fall.
• Terminal velocity: drag equals weight; acceleration is zero and speed is constant.

10. Graph ideas

It is useful to imagine how graphs look during falling motion.

Velocity-time graph in ideal free fall: velocity changes at a constant rate, so the graph is a straight line with slope \(g\) or \(-g\), depending on the sign convention.

Velocity-time graph with air resistance: velocity increases quickly at first, then more slowly, and finally levels off at terminal velocity.

Acceleration-time graph in ideal free fall: acceleration is constant at \(g\) or \(-g\).

Acceleration-time graph with air resistance: acceleration starts near \(g\), then decreases toward zero as the object approaches terminal velocity.

11. Worked Example 1: Basic free fall from rest

A ball is dropped from rest. What is its velocity after \(3.0\) s, ignoring air resistance?

Step 1: Identify known values.

• Initial velocity: \(u = 0\)
• Time: \(t = 3.0\,\text{s}\)
• Acceleration: \(a = 9.8\,\text{m/s}^2\) downward

Step 2: Use the equation \(v = u + at\).

$$v = 0 + (9.8)(3.0)$$

$$v = 29.4\,\text{m/s}$$

Answer: The ball's velocity is \(29.4\,\text{m/s}\) downward after \(3.0\) s.

12. Worked Example 2: Distance fallen in free fall

A rock is dropped from rest. How far does it fall in \(2.5\) s, ignoring air resistance?

Step 1: Identify known values.

• \(u = 0\)
• \(t = 2.5\,\text{s}\)
• \(a = 9.8\,\text{m/s}^2\)

Step 2: Use \(s = ut + \frac{1}{2}at^2\).

$$s = 0 + \frac{1}{2}(9.8)(2.5)^2$$

$$s = 4.9(6.25)$$

$$s = 30.625\,\text{m}$$

Answer: The rock falls about \(30.6\,\text{m}\).

13. Worked Example 3: Object thrown upward

A ball is thrown straight upward with an initial velocity of \(20\,\text{m/s}\). How long does it take to reach its highest point? Ignore air resistance.

Idea: At the highest point, the velocity becomes zero for an instant.

Step 1: Choose upward as positive.

• \(u = 20\,\text{m/s}\)
• \(v = 0\)
• \(a = -9.8\,\text{m/s}^2\)

Step 2: Use \(v = u + at\).

$$0 = 20 - 9.8t$$

$$9.8t = 20$$

$$t = \frac{20}{9.8} \approx 2.04\,\text{s}$$

Answer: It takes about \(2.04\) s to reach the highest point.

This example shows that even while the ball moves upward, gravity still provides a downward acceleration of \(9.8\,\text{m/s}^2\).

14. Worked Example 4: Understanding terminal velocity with forces

A falling raindrop has a weight of \(0.020\,\text{N}\). At one moment, the air resistance on it is \(0.015\,\text{N}\) upward.

(a) Is it at terminal velocity?

No. Terminal velocity happens only when drag equals weight. Here:

$$0.015\,\text{N} \ne 0.020\,\text{N}$$

(b) What is the net force?

Taking downward as positive:

$$F_{\text{net}} = 0.020 - 0.015 = 0.005\,\text{N}$$

The net force is \(0.005\,\text{N}\) downward.

(c) What does this tell us about its motion?

Because the net force is downward, the raindrop is still accelerating downward. It has not yet reached terminal velocity.

If later the drag increases to \(0.020\,\text{N}\), then the net force becomes zero and the raindrop falls at constant speed.

15. Common mistakes to avoid

• Confusing velocity and acceleration: an object can move downward with large velocity but have zero acceleration if it is at terminal velocity.
• Forgetting sign convention: decide which direction is positive before using equations.
• Assuming heavier objects always fall faster: in free fall without air resistance, all objects accelerate equally.
• Thinking terminal velocity means stopping: terminal velocity means constant speed, not zero speed.
• Using free-fall equations when drag is important: the constant-acceleration equations only apply directly when acceleration is constant.

16. Real-life applications

• Skydiving: a skydiver speeds up after jumping, then reaches terminal velocity. Opening a parachute increases drag and lowers terminal velocity.
• Raindrops: raindrops do not keep accelerating forever because air resistance balances their weight.
• Sports: the motion of balls, especially over long distances, is affected by air resistance.
• Engineering and safety: understanding drag is important in designing parachutes and protective equipment.

17. Quick comparison table

• Only gravity acts: free fall
• Acceleration in free fall: \(9.8\,\text{m/s}^2\) downward
• Gravity + air resistance act: real falling motion in air
• At terminal velocity: drag = weight
• Net force at terminal velocity: zero
• Acceleration at terminal velocity: zero
• Velocity at terminal velocity: constant, not zero

18. Brief summary

Free fall is motion under the force of gravity alone. In ideal free fall near Earth, all objects accelerate downward at about \(9.8\,\text{m/s}^2\), regardless of mass.

In real air, falling objects also experience drag. As speed increases, drag increases, reducing the acceleration. When drag becomes equal to weight, the net force is zero, acceleration becomes zero, and the object falls at a constant speed called terminal velocity.

If you remember that gravity causes the fall, air resistance opposes the motion, and terminal velocity happens when forces balance, you will understand the main physics of falling objects.

Two-Dimensional Projectile Motion is the motion of an object that is launched into the air and then moves under the influence of gravity alone, assuming air resistance is negligible.

Examples include a kicked soccer ball, a basketball shot, or a rock thrown off a cliff. These objects move in a curved path called a trajectory.

The most important idea in projectile motion is this: horizontal motion and vertical motion can be analyzed separately. Even though the object follows one curved path, its motion can be broken into two independent parts.

In the horizontal direction, the projectile moves with constant velocity if air resistance is ignored. In the vertical direction, the projectile experiences constant acceleration due to gravity, which acts downward.

This combination of constant horizontal motion and accelerated vertical motion creates the familiar parabolic path.

1. Breaking motion into components

When an object 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
• \(\theta\) is measured from the horizontal

These two components are found using basic trigonometry. The horizontal component tells how fast the object moves across. The vertical component tells how fast it starts moving upward.

2. Horizontal motion

In projectile motion, horizontal acceleration is usually zero:

Because there is no horizontal acceleration, horizontal velocity stays constant:

The horizontal displacement after time \(t\) is:

This means the object covers equal horizontal distances in equal time intervals.

3. Vertical motion

In the vertical direction, gravity causes a constant downward acceleration:

Near Earth, \(g \approx 9.8\,\text{m/s}^2\). The negative sign means the acceleration is downward if upward is taken as positive.

The vertical velocity at time \(t\) is:

The vertical position after time \(t\) is:

If the projectile starts from a height \(y_0\) instead of the ground, then:

4. Why the path is a parabola

The horizontal position depends linearly on time, but the vertical position depends on time squared. Because of this, when time is eliminated, the path becomes a parabola.

You do not always need to derive the parabola equation, but it is important to understand that the curved path comes from combining constant horizontal motion with vertically accelerated motion.

5. Key quantities in projectile motion

• Time of flight: how long the projectile stays in the air
• Maximum height: the highest vertical point reached
• Range: the total horizontal distance traveled

These quantities are often found using vertical motion first, because gravity controls how long the projectile remains in the air.

6. Maximum height

At the highest point, the vertical velocity is zero:

Using

we get the time to reach maximum height:

The maximum height above the launch point is then:

7. Time of flight for launch and landing at the same height

If a projectile lands at the same height from which it was launched, the upward and downward parts of the motion are symmetric.

The total time of flight is:

This only works when launch height and landing height are the same.

8. Horizontal range for launch and landing at the same height

The range is horizontal velocity multiplied by time of flight:

Substituting component formulas gives:

This formula is useful only when the projectile lands at the same height from which it was launched.

9. Important ideas to remember

• The horizontal and vertical motions are independent.
• Gravity affects only the vertical motion, not the horizontal motion.
• At the highest point, the vertical velocity is zero, but the horizontal velocity is still not zero.
• A projectile can have zero vertical velocity at one instant and still be moving overall.
• The acceleration is always downward at \(9.8\,\text{m/s}^2\), even at the highest point.

Worked Example 1: Finding velocity components

A ball is launched at \(20\,\text{m/s}\) at an angle of \(30^\circ\) above the horizontal. Find the initial horizontal and vertical velocity components.

Step 1: Use the component formulas

Step 2: Substitute values

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

Answer: The horizontal component is \(17.32\,\text{m/s}\), and the vertical component is \(10.0\,\text{m/s}\).

Worked Example 2: Time of flight, maximum height, and range

A projectile is launched from level ground with speed \(20\,\text{m/s}\) at \(30^\circ\). Find:

• time of flight
• maximum height
• horizontal range

From Example 1, we already know:

Step 1: Time of flight

Step 2: Maximum height

Step 3: Range

Answer:

• Time of flight: \(2.04\,\text{s}\)
• Maximum height: \(5.10\,\text{m}\)
• Range: \(35.3\,\text{m}\)

Worked Example 3: Projectile launched horizontally

A ball rolls off a table with a horizontal speed of \(5.0\,\text{m/s}\). The table is \(1.2\,\text{m}\) high. How far from the table does the ball land?

This is still projectile motion. The initial vertical velocity is zero because the ball is launched horizontally.

Step 1: Analyze vertical motion to find time

Use:

Let the floor be \(y=0\), so \(y_0 = 1.2\,\text{m}\), \(v_{0y}=0\), and \(g=9.8\,\text{m/s}^2\):

Step 2: Use horizontal motion

Answer: The ball lands about \(2.48\,\text{m}\) from the table.

Worked Example 4: Projectile launched from a height

A rock is thrown from the top of a \(20\,\text{m}\) cliff with an initial speed of \(15\,\text{m/s}\) at \(40^\circ\) above the horizontal. Find the time it takes to hit the ground.

Step 1: Find the vertical component of initial velocity

Using \(\sin 40^\circ \approx 0.643\):

Step 2: Write the vertical position equation

Take the launch point as \(y_0 = 20\,\text{m}\) and the ground as \(y=0\):

Rearrange:

Step 3: Solve using the quadratic formula

This gives two solutions:

The negative time is not physically meaningful here, so we reject it.

Answer: The rock hits the ground after about \(3.23\,\text{s}\).

10. Common mistakes

• Mixing horizontal and vertical quantities: Use horizontal equations for horizontal motion and vertical equations for vertical motion.
• Forgetting signs: If upward is positive, then gravity is negative.
• Assuming velocity is zero at the top: Only the vertical component is zero. The projectile still has horizontal velocity.
• Using range formulas when heights are different: The shortcut formulas for time of flight and range only work when launch and landing heights are the same.
• Not resolving velocity into components: If launched at an angle, always find \(v_{0x}\) and \(v_{0y}\) first.

11. Problem-solving strategy

1. Draw a simple diagram of the motion.
2. Choose positive directions, usually rightward and upward.
3. Break the initial velocity into horizontal and vertical components.
4. List what is known and what must be found.
5. Use vertical motion to find time when needed.
6. Use horizontal motion to find range or horizontal position.
7. Check whether your answer is reasonable.

12. Final summary

Two-dimensional projectile motion is easiest to understand when you separate it into horizontal motion with constant velocity and vertical motion with constant acceleration due to gravity.

The horizontal acceleration is zero, while the vertical acceleration is always \(-9.8\,\text{m/s}^2\). By splitting the initial velocity into components, you can find time of flight, maximum height, and range.

If you remember that the two directions are independent, most projectile motion problems become much easier to solve.

Newton's First Law and Inertia

Introduction

In everyday life, objects seem to stop, start, or change direction all the time. A soccer ball slows down after being kicked, a car speeds up when the driver presses the gas pedal, and a book stays still on a table. Newton's First Law helps explain all of these situations.

Newton's First Law of Motion states that an object will remain at rest, or continue moving at a constant velocity in a straight line, unless acted on by a net external force.

This law is sometimes called the law of inertia because it describes the tendency of objects to resist changes in their motion.

Main Idea

There are two parts to Newton's First Law:

• If an object is at rest, it stays at rest unless a net external force acts on it.
• If an object is moving, it keeps moving with the same speed and in the same direction unless a net external force acts on it.

This means that motion does not require a constant force to keep going. Instead, a force is needed only to change motion, such as starting, stopping, speeding up, slowing down, or turning.

What is a Net Force?

A net force is the overall force acting on an object after adding all the individual forces together, including direction. If the forces balance, then the net force is zero.

When the net force is zero, the object's motion does not change. In symbols, we write:

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

If an object is already at rest and $$F_{\text{net}} = 0$$, it stays at rest. If it is already moving and $$F_{\text{net}} = 0$$, it keeps moving with constant velocity.

Constant velocity means both speed and direction stay the same. So if direction changes, velocity changes, which means there must be a net force.

Understanding Inertia

Inertia is the tendency of an object to resist changes in its state of motion. In simple terms, objects "want" to keep doing what they are already doing:

• A resting object resists being moved.
• A moving object resists being stopped or redirected.

The amount of inertia depends on mass. An object with more mass has more inertia, so it is harder to start moving, harder to stop, and harder to change direction.

For example, pushing a shopping cart is easier when it is empty than when it is full. The full cart has more mass, so it has more inertia.

Balanced and Unbalanced Forces

Newton's First Law is closely connected to the idea of balanced and unbalanced forces.

• Balanced forces: The forces cancel out, so $$F_{\text{net}} = 0$$.
• Unbalanced forces: The forces do not cancel out, so $$F_{\text{net}} \ne 0$$.

If forces are balanced, there is no change in motion. If forces are unbalanced, the object's motion changes.

Examples from Everyday Life

• A book on a desk stays still because the downward force of gravity is balanced by the upward support force from the desk.
• A hockey puck slides farther on ice than on rough ground because there is less friction opposing its motion.
• When a car suddenly stops, passengers lurch forward because their bodies continue moving due to inertia.
• Seat belts protect passengers by applying the force needed to stop their motion safely.

The Role of Friction

Many people think objects naturally stop moving on their own. But according to Newton's First Law, an object in motion would keep moving forever in a straight line if no net external force acted on it.

In real life, moving objects often slow down because of friction or air resistance. These forces act opposite the motion, creating a net force that changes the object's velocity.

That is why a rolling ball stops on the ground, but a spacecraft in outer space can keep moving for a very long time without needing continuous thrust.

Worked Example 1: Book Resting on a Table

A book is lying still on a table. Gravity pulls downward with a force of 12 N. The table pushes upward with a force of 12 N. What is the net force, and what happens to the book?

Step 1: Identify the forces.

• Downward force: 12 N
• Upward force: 12 N

Step 2: Find the net force.

Because the forces are equal and opposite, they cancel:

$$F_{\text{net}} = 12 - 12 = 0\text{ N}$$

Step 3: Apply Newton's First Law.

Since the net force is zero, the book's motion does not change. It remains at rest.

Answer: The net force is 0 N, so the book stays still.

Worked Example 2: Sliding Puck on Ice

A puck is sliding across smooth ice at a constant speed in a straight line. What can you say about the net force on the puck?

Step 1: Look at the motion.

The puck has constant speed and constant direction, so its velocity is constant.

Step 2: Use Newton's First Law.

If velocity is not changing, then there is no net external force changing the motion.

Answer: The net force is 0 N or very close to 0 N if friction is very small. The puck continues moving because of inertia.

Worked Example 3: Tug-of-War

Two students pull on a box in opposite directions. One pulls with 50 N to the right. The other pulls with 35 N to the left. What is the net force, and what happens to the box?

Step 1: Choose a positive direction.

Let right be positive.

Step 2: Add the forces with signs.

$$F_{\text{net}} = +50 + (-35) = +15\text{ N}$$

Step 3: Interpret the result.

The net force is 15 N to the right. Since the net force is not zero, the forces are unbalanced.

Step 4: Apply Newton's First Law.

The box will not keep its current motion unchanged. Its motion will change in the direction of the net force, which is to the right.

Answer: The net force is 15 N to the right, so the box's motion changes to the right.

Worked Example 4: Car and Passenger

A car moving forward suddenly brakes. Why does a passenger seem to move forward?

Step 1: Think about the motion before braking.

Before the brakes are applied, both the car and passenger are moving forward.

Step 2: What happens when the car brakes?

The brakes apply a force that slows the car. But the passenger's body tends to keep moving forward due to inertia.

Step 3: What stops the passenger?

The seat belt provides the external force needed to change the passenger's motion and bring them to rest with the car.

Answer: The passenger moves forward relative to the car because inertia resists the change in motion.

Common Misunderstandings

• Misunderstanding 1: Moving objects need a constant force to keep moving.
  Actually, an object will keep moving at constant velocity if the net force is zero.
• Misunderstanding 2: If an object is moving, a force must be acting in the direction of motion.
  Not always. If motion is constant, the net force can be zero.
• Misunderstanding 3: Inertia is a force.
  Inertia is not a force. It is a property of matter related to mass.
• Misunderstanding 4: Heavier objects fall faster because they have more inertia.
  Inertia affects resistance to changes in motion, but falling depends on gravity and other forces like air resistance.

Key Points to Remember

• Newton's First Law describes what happens when the net external force on an object is zero.
• An object at rest stays at rest, and an object in motion stays in motion with constant velocity, unless acted on by a net external force.
• Inertia is the tendency to resist changes in motion.
• More mass means more inertia.
• Balanced forces do not change motion; unbalanced forces do.
• Friction often explains why moving objects in daily life slow down.

Brief Summary

Newton's First Law explains that objects do not change their motion unless a net external force acts on them. If the net force is zero, an object either stays at rest or keeps moving with constant velocity. This behavior is caused by inertia, which is stronger in objects with greater mass. Understanding this law helps explain everyday events such as seat belt safety, sliding objects, and balanced forces.

Newton's Second Law and Net Force explains how forces change an object's motion. It connects three important ideas: force, mass, and acceleration. If you know the total force acting on an object and its mass, you can predict how its motion will change.

This lesson will help you understand what net force means, how to use Newton's Second Law, and how to solve problems in one and two dimensions. These ideas are central to classical mechanics and are used in many real-world situations, from pushing a cart to analyzing the motion of a car or rocket.

Newton's Second Law states that the acceleration of an object depends on the net force acting on it and its mass. In equation form,

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

Here, \(F_{\text{net}}\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.

This equation means two key things:

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

So, a small object speeds up more easily than a large object when the same force is applied.

Net force is the vector sum of all the forces acting on an object. A vector has both size and direction, so when adding forces, direction matters.

If forces act in the same direction, you add them. If they act in opposite directions, you subtract them. If forces act at angles, you usually break them into components and add the components separately.

For example, if a box is pushed to the right with \(12\,\text{N}\) and friction acts to the left with \(5\,\text{N}\), the net force is

$$F_{\text{net}} = 12 - 5 = 7\,\text{N} \text{ to the right}$$

The acceleration will also be to the right, because acceleration points in the same direction as the net force.

Important idea: Newton's Second Law uses the net force, not just one individual force. An object can have many forces acting on it at once, but only the total matters for its acceleration.

Common forces you may see in problems include:

• Weight: the force of gravity, usually downward
• Normal force: the support force from a surface
• Friction: a force that opposes motion or attempted motion
• Tension: a pulling force through a rope or cable
• Applied force: a push or pull from a person or object

Before using \(F_{\text{net}} = ma\), it helps to identify all the forces and their directions. A force diagram, often called a free-body diagram, can make this much easier.

In many situations, forces in one direction may cancel. For example, on a flat surface, the normal force upward and the weight downward are often equal in size, so the vertical net force is zero. That means there is no vertical acceleration, even if the object is moving horizontally.

This is a common point of confusion: motion and force are not the same thing. An object can move at constant velocity and still have forces acting on it, as long as those forces balance so that the net force is zero.

If the net force is zero, then

$$a = 0$$

This means the object either stays at rest or keeps moving at constant velocity.

If the net force is not zero, then the object accelerates. Acceleration can mean:

• speeding up,
• slowing down, or
• changing direction.

Units are important in Newton's Second Law:

• Force is measured in newtons \((\text{N})\)
• Mass is measured in kilograms \((\text{kg})\)
• Acceleration is measured in \(\text{m/s}^2\)

One newton is defined as:

$$1\,\text{N} = 1\,\text{kg}\cdot \text{m/s}^2$$

That matches the equation \(F=ma\).

How to solve Newton's Second Law problems

1. Identify the object or system you are analyzing.
2. List all forces acting on it.
3. Choose positive direction(s), such as right and up.
4. Find the net force by adding forces as vectors.
5. Use $$F_{\text{net}} = ma$$ in each direction.
6. Solve for the unknown, including the direction.

In one dimension, the equation is often written as

$$\sum F = ma$$

The symbol \(\sum F\) means the sum of all forces.

In two dimensions, you handle each direction separately:

$$\sum F_x = ma_x$$

$$\sum F_y = ma_y$$

This is useful when forces act at angles.

Worked Example 1: One force causing acceleration

A \(4.0\,\text{kg}\) cart is pushed with a net force of \(20\,\text{N}\) to the right. What is its acceleration?

Step 1: Write the known values.

• \(m = 4.0\,\text{kg}\)
• \(F_{\text{net}} = 20\,\text{N}\)

Step 2: Use Newton's Second Law.

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

$$a = \frac{F_{\text{net}}}{m}$$

$$a = \frac{20}{4.0} = 5.0\,\text{m/s}^2$$

Answer: The cart accelerates at \(5.0\,\text{m/s}^2\) to the right.

This example shows that acceleration is found by dividing net force by mass.

Worked Example 2: Opposing forces

A \(10\,\text{kg}\) box is pushed to the right with \(35\,\text{N}\). Friction acts to the left with \(15\,\text{N}\). Find the acceleration.

Step 1: Find the net force.

Take right as positive.

$$F_{\text{net}} = 35 - 15 = 20\,\text{N}$$

Step 2: Apply Newton's Second Law.

$$a = \frac{F_{\text{net}}}{m}$$

$$a = \frac{20}{10} = 2.0\,\text{m/s}^2$$

Answer: The box accelerates at \(2.0\,\text{m/s}^2\) to the right.

Notice that we did not use the \(35\,\text{N}\) force by itself. We had to include friction and find the net force first.

Worked Example 3: Solving for force

A car of mass \(1200\,\text{kg}\) accelerates at \(3.0\,\text{m/s}^2\) forward. What net force acts on the car?

Step 1: Write the equation.

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

Step 2: Substitute values.

$$F_{\text{net}} = (1200)(3.0) = 3600\,\text{N}$$

Answer: The net force is \(3600\,\text{N}\) forward.

This example shows that Newton's Second Law can be used to find force if mass and acceleration are known.

Worked Example 4: Two-dimensional force components

A \(2.0\,\text{kg}\) object has two horizontal forces acting on it. One force is \(8.0\,\text{N}\) east, and another force is \(6.0\,\text{N}\) north. Find the magnitude of the acceleration.

Step 1: Find the net force in each direction.

• \(F_x = 8.0\,\text{N}\)
• \(F_y = 6.0\,\text{N}\)

Step 2: Find the magnitude of the net force.

Since the forces are perpendicular, use the Pythagorean theorem:

$$F_{\text{net}} = \sqrt{8.0^2 + 6.0^2}$$

$$F_{\text{net}} = \sqrt{64 + 36} = \sqrt{100} = 10\,\text{N}$$

Step 3: Use Newton's Second Law.

$$a = \frac{F_{\text{net}}}{m} = \frac{10}{2.0} = 5.0\,\text{m/s}^2$$

Answer: The acceleration has magnitude \(5.0\,\text{m/s}^2\).

If needed, the direction of acceleration would match the direction of the net force.

Mass and acceleration relationship

Suppose you apply the same net force to two different objects. The object with smaller mass will have the larger acceleration. Mathematically,

$$a = \frac{F_{\text{net}}}{m}$$

This shows that acceleration is inversely related to mass. Doubling the mass cuts the acceleration in half if the net force stays the same.

Force and acceleration relationship

If the mass stays the same, acceleration is directly proportional to net force. Doubling the net force doubles the acceleration.

This is why an empty shopping cart is easier to speed up than a full one, and why pushing harder makes it speed up more quickly.

Balanced and unbalanced forces

• Balanced forces: net force is zero, so acceleration is zero.
• Unbalanced forces: net force is not zero, so acceleration is not zero.

Balanced forces do not always mean the object is at rest. They can also mean constant velocity.

Common mistakes to avoid

• Using one force instead of the net force.
• Ignoring direction when adding or subtracting forces.
• Confusing mass with weight. Mass is measured in \(\text{kg}\); weight is a force measured in newtons.
• Thinking that if an object is moving, a net force must be acting on it. Motion can continue with zero net force.
• Forgetting that acceleration points in the direction of the net force.

Tips for success

• Always draw or imagine the forces first.
• Choose a positive direction and stay consistent.
• Check units before solving.
• Ask yourself whether forces cancel in any direction.
• If the answer is acceleration, include both magnitude and direction when possible.

Brief Summary

Newton's Second Law states that the net force on an object equals its mass times its acceleration: $$F_{\text{net}} = ma$$. The key idea is that net force is the total of all forces, including their directions. If net force is zero, acceleration is zero; if net force is not zero, the object accelerates in the direction of the net force. By identifying forces carefully and adding them correctly, you can use this law to solve many motion problems in one and two dimensions.

Newton's Third Law explains how forces always come in pairs. It helps us understand why objects push, pull, recoil, accelerate, and interact with each other in every physical situation.

The law states: For every action force, there is an equal and opposite reaction force. A more precise way to say this is: If object A exerts a force on object B, then object B exerts a force of equal magnitude and opposite direction on object A.

In symbols, this can be 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.

Newton's Third Law is important because students often think a force can exist by itself. In reality, forces are always interactions between two objects. If one object pushes, the other object pushes back.

Introduction: What is an action-reaction pair?

An action-reaction pair is a pair of forces described by Newton's Third Law. These two forces are:

• equal in magnitude,
• opposite in direction,
• the same type of force, and
• acting on different objects.

That last point is the most important. Because the two forces act on different objects, they do not cancel each other out.

Forces only cancel when they act on the same object. This is a common source of confusion.

Main Teaching Point 1: Forces always involve two objects

Whenever you describe a force, you should ask two questions:

1. Which object is exerting the force?
2. Which object is receiving the force?

For example, if a hand pushes a wall, the hand exerts a force on the wall. At the same time, the wall exerts a force on the hand.

These two forces form an action-reaction pair:

• hand on wall
• wall on hand

They are equal and opposite, but one acts on the wall and the other acts on the hand.

Main Teaching Point 2: Equal and opposite does not mean balanced

Students sometimes think that if forces are equal and opposite, nothing can move. That is not always true.

Action-reaction forces do not cancel because they act on different objects. A force balance only matters when looking at all the forces on one single object.

For example, when you stand on the ground:

• Earth pulls you downward with gravity.
• The ground pushes you upward with a normal force.

Those two forces act on you, so they can balance.

But the Third Law pair to Earth's gravity on you is your gravity on Earth, not the normal force.

This distinction is very important:

• Balanced forces: multiple forces on the same object that may add to zero.
• Action-reaction pair: two forces on different objects.

Main Teaching Point 3: How to identify an action-reaction pair

To find a Third Law pair, use this checklist:

1. Find one force.
2. Ask: what two objects are interacting?
3. Reverse the order of the objects.
4. Keep the same force type.
5. Make the direction opposite and the magnitude equal.

Example: The bat hits the ball.

• Force 1: bat on ball
• Force 2: ball on bat

These are the Third Law pair.

Notice that the force of the bat on the ball is not paired with the ball's motion, the ball's weight, or air resistance. The pair must involve the same two objects interacting with each other.

Main Teaching Point 4: Why objects can respond differently

If action and reaction forces are equal, you may wonder why one object sometimes moves a lot while the other barely moves.

The reason is that the two forces act on different objects, and those objects may have different masses.

From Newton's Second Law,

$$\vec{F}_{\text{net}} = m\vec{a}$$

If the same size force acts on two different objects, the object with smaller mass usually has greater acceleration.

For example, when a truck collides with a small car, the truck and the car exert equal and opposite forces on each other. However, the car usually has a much larger acceleration because its mass is much smaller.

Main Teaching Point 5: Common examples of Newton's Third Law

• Walking: Your foot pushes backward on the ground, and the ground pushes forward on your foot.
• Swimming: You push water backward, and the water pushes you forward.
• Rocket motion: The rocket pushes gases downward, and the gases push the rocket upward.
• Book on a table: The book pushes down on the table, and the table pushes up on the book.
• Jumping: You push down on the ground, and the ground pushes you up.

In each case, motion happens because of interaction between objects.

Worked Example 1: Book resting on a table

A book sits at rest on a table. A student says, “The weight of the book and the normal force from the table are an action-reaction pair.” Is that correct?

Step 1: Identify the forces.

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

Both of these forces act on the book.

Step 2: Decide if they are a Third Law pair.

No. Since both forces act on the same object, they are not an action-reaction pair.

Step 3: Find the actual Third Law pairs.

• Earth on book ↔ book on Earth
• Table on book ↔ book on table

Conclusion: The weight and normal force may balance, but they are not a Third Law pair.

Worked Example 2: Walking forward

A student walks forward across the floor. Explain how Newton's Third Law makes this possible.

Step 1: Identify the interaction.

The student's foot pushes backward on the floor.

Step 2: Apply Newton's Third Law.

The floor pushes forward on the student's foot with an equal and opposite force.

Step 3: Explain the motion.

The forward force from the floor acts on the student and helps the student accelerate forward.

Conclusion: You move forward because you push backward on the ground, and the ground pushes you forward.

Worked Example 3: Recoil of a gun

When a gun fires a bullet, the bullet moves forward and the gun recoils backward. Explain using Newton's Third Law.

Step 1: Identify the interaction.

The gun exerts a forward force on the bullet.

Step 2: Find the reaction force.

The bullet exerts an equal and opposite backward force on the gun.

Step 3: Explain why the motions are different.

The force magnitudes are equal, but the bullet has a much smaller mass than the gun. So the bullet gets a much greater acceleration.

Using Newton's Second Law:

$$a = \frac{F_{\text{net}}}{m}$$

Smaller mass means larger acceleration for the same force.

Conclusion: The bullet shoots forward and the gun recoils backward because they exert equal and opposite forces on each other.

Worked Example 4: Collision between a truck and a car

A heavy truck collides with a small car. A student says, “The truck exerts a larger force because it is bigger.” Is the student correct?

Step 1: Apply Newton's Third Law.

No. During the collision, the truck exerts a force on the car, and the car exerts a force on the truck.

These forces are equal in magnitude and opposite in direction:

$$\vec{F}_{\text{truck on car}} = -\vec{F}_{\text{car on truck}}$$

Step 2: Explain why the damage or motion may differ.

The car and truck may experience different accelerations because their masses are different.

If the force magnitude is the same, then:

$$a = \frac{F}{m}$$

The smaller-mass car usually has the larger acceleration, so it may change speed more quickly or be damaged more severely.

Conclusion: The forces are equal, even though the effects on the two vehicles may be very different.

Common Mistakes to Avoid

• Mistake 1: Thinking action and reaction cancel each other.
  They do not cancel because they act on different objects.
• Mistake 2: Pairing forces that act on the same object.
  Those may be balanced forces, not Third Law pairs.
• Mistake 3: Thinking the bigger object exerts the bigger force in an interaction.
  The forces are equal in magnitude.
• Mistake 4: Forgetting to name both objects in the interaction.
  Always say “A on B” and “B on A.”

Quick Check Questions

1. When you push on a door, what force does the door exert?
2. Why do action-reaction forces not cancel?
3. What is the reaction force to Earth pulling on the Moon?
4. When a swimmer moves forward, what is being pushed backward?

Answers

1. The door pushes back on you with an equal and opposite force.
2. Because they act on different objects.
3. The Moon pulling on Earth.
4. The swimmer pushes the water backward.

Brief Summary

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

These action-reaction forces act on different objects, so they do not cancel each other. To identify the correct pair, always name the two interacting objects and reverse them: A on B and B on A.

Understanding this idea helps explain walking, collisions, recoil, rockets, and many other everyday examples of motion.

Free-Body Diagrams are one of the most useful tools in mechanics. They help you identify all the forces acting on one object so that you can apply Newton's laws correctly. If you can draw a good free-body diagram, you can solve many motion and force problems more easily and with fewer mistakes.

A free-body diagram, often called an FBD, is a simple sketch of a single object with arrows showing every force acting on it. The object is “freed” from its surroundings, which means you do not draw the other objects in detail. Instead, you show only the forces those objects exert on the chosen object.

This lesson will show you how to draw free-body diagrams, how to choose forces correctly, and how to use them in calculations involving equilibrium and acceleration.

Why free-body diagrams matter

In force problems, students often get confused because they mix up the motion of the object with the forces on it. A free-body diagram helps separate these ideas. Motion tells you what the acceleration is doing, while the free-body diagram tells you what forces are causing that acceleration.

For example, an object moving to the right does not necessarily have a force to the right. It could be moving right while slowing down, which would mean the net force is actually to the left. This is why drawing forces carefully matters.

What a free-body diagram includes

• One object, represented by a dot, box, or simple shape
• Arrows starting on the object and pointing in the direction of each force
• Labels for each force, such as weight, normal force, friction, or tension
• Optional coordinate axes, especially when working on slopes or in two dimensions

What a free-body diagram does not include

• Forces exerted by the object on other objects
• Extra background details that are not needed
• Velocity arrows unless the problem specifically asks for motion separately
• “Net force” as a separate force arrow; net force is the result of all the real forces combined

The common forces you should know

In 12th Grade mechanics, the most common forces in free-body diagrams are the following.

• Weight: the gravitational force acting on an object. It points straight downward toward Earth. Its size is usually given by \(W = mg\).
• Normal force: the support force from a surface. It acts perpendicular to the surface.
• Friction: a force parallel to a surface that opposes sliding or attempted sliding.
• Tension: the pulling force exerted by a string, rope, or cable. It acts along the string.
• Applied force: a push or pull from a person or another object.
• Air resistance: a force that opposes motion through air.

Important idea: only draw forces that are actually acting on the object. For example, if a box is resting on a table, you draw the weight downward and the normal force upward. You do not draw a “force of motion” just because the box may have moved earlier.

Step-by-step method for drawing a free-body diagram

1. Choose the object you are analyzing.
2. Draw the object alone as a simple box or dot.
3. Identify every interaction between the object and its surroundings.
4. Draw one arrow for each force acting on the object.
5. Label each force clearly.
6. Choose axes if you will write equations. Horizontal and vertical axes are common, but on an incline it is often easier to choose axes parallel and perpendicular to the slope.

How to connect a free-body diagram to Newton's second law

Once the forces are drawn, you add them by direction. Newton's second law says:

This means the net force on an object equals its mass times its acceleration. In most problems, you apply this separately in each direction:

If the object is not accelerating in one direction, then acceleration in that direction is zero, so the net force in that direction is also zero.

Balanced and unbalanced forces

If the forces on an object cancel out, the net force is zero. These are called balanced forces. The object may be at rest, or it may move with constant velocity.

If the forces do not cancel, the net force is not zero. These are unbalanced forces, and the object accelerates in the direction of the net force.

Example 1: A book resting on a table

A book of mass \(2.0\,\text{kg}\) rests on a horizontal table. Draw the free-body diagram and find the forces.

Step 1: Identify the object

The object is the book.

Step 2: Identify forces on the book

• Weight downward: \(W = mg\)
• Normal force upward from the table: \(N\)

Step 3: Calculate the weight

So the weight is \(19.6\,\text{N}\) downward.

Because the book is at rest and not accelerating vertically, the net vertical force is zero:

Free-body diagram description: one arrow downward labeled \(W = 19.6\,\text{N}\), and one arrow upward labeled \(N = 19.6\,\text{N}\).

Key lesson from this example: the normal force is not always equal to weight in every situation, but in this simple case it is, because the book is on a horizontal surface with no vertical acceleration.

Example 2: A box pushed across a floor

A \(5.0\,\text{kg}\) box is pushed to the right with an applied force of \(30\,\text{N}\). Friction acts to the left with magnitude \(10\,\text{N}\). Find the acceleration.

Step 1: Identify forces

• Applied force to the right: \(F_{\text{app}} = 30\,\text{N}\)
• Friction to the left: \(f = 10\,\text{N}\)
• Weight downward: \(W = mg\)
• Normal force upward: \(N\)

Step 2: Vertical direction

The box does not accelerate upward or downward, so:

Step 3: Horizontal direction

The box accelerates at \(4.0\,\text{m/s}^2\) to the right.

Free-body diagram description: rightward arrow for applied force, leftward arrow for friction, downward arrow for weight, upward arrow for normal force.

Key lesson from this example: include all real forces, then combine them by direction. The acceleration comes from the net force, not from one force by itself.

Example 3: A hanging mass on a rope

A \(3.0\,\text{kg}\) object hangs motionless from a rope. Draw the free-body diagram and find the tension.

Forces on the object

• Weight downward: \(W = mg\)
• Tension upward: \(T\)

Calculate the weight:

Because the object is motionless, the acceleration is zero:

So the tension is \(29.4\,\text{N}\) upward.

Key lesson from this example: tension pulls along the rope. For a hanging object at rest, tension equals weight.

Example 4: A block on an inclined plane

A \(4.0\,\text{kg}\) block rests on a frictionless incline that makes an angle of \(30^\circ\) with the horizontal. Draw the free-body diagram and find the component of weight parallel to the incline.

Step 1: Identify the forces

• Weight \(mg\) straight downward
• Normal force \(N\) perpendicular to the incline

There is no friction in this problem.

Step 2: Choose axes

For incline problems, it is often easiest to choose:

• the \(x\)-axis parallel to the incline
• the \(y\)-axis perpendicular to the incline

Step 3: Find the weight

The weight points straight down, but we split it into components relative to the incline:

With \(\theta = 30^\circ\):

The component of weight pulling the block down the incline is \(19.6\,\text{N}\).

If the plane is frictionless, this parallel component is the unbalanced force that causes the block to accelerate down the slope.

Key lesson from this example: on a slope, do not tilt the weight force. Weight always points vertically downward. It is the axes that are tilted, and then the weight is resolved into components.

Common mistakes to avoid

• Drawing forces that do not exist. There is no such thing as a “force of motion.”
• Forgetting the normal force when an object touches a surface.
• Drawing weight in the wrong direction. Weight always points straight downward.
• Confusing mass and weight. Mass is measured in kilograms; weight is a force measured in newtons.
• Using the net force as if it were a separate force. Net force is the sum of all the real forces.
• Assuming normal force always equals weight. This is only true in certain situations.
• For inclines, drawing the normal force straight up. The normal force must be perpendicular to the surface.

Helpful strategy for any problem

When you see a mechanics question, pause before doing calculations and ask:

1. What is the object I am studying?
2. What other things are touching it or pulling on it?
3. What forces come from those interactions?
4. In what directions do those forces act?
5. Is the object accelerating? If so, in what direction?

This habit makes your work clearer and helps you avoid sign mistakes and missing-force errors.

Brief summary

A free-body diagram is a simple drawing that shows all the forces acting on one object. To draw one, isolate the object, identify every real force, draw arrows in the correct directions, and label them clearly.

Once the diagram is complete, use Newton's second law, \(\sum F = ma\), to analyze the motion. Good free-body diagrams are the foundation for solving problems with weight, normal force, friction, tension, and forces on slopes.

Static and Kinetic Friction Models

Friction is a force that acts between surfaces that are touching. It resists motion, or the tendency of motion, between those surfaces. In classical mechanics, friction is important because it changes how objects move and how much force is needed to start or keep them moving.

There are two main friction models you need to know: static friction and kinetic friction. Static friction acts when surfaces are not sliding past each other. Kinetic friction acts when surfaces are sliding.

This lesson explains what these forces mean, how to calculate them, how they appear in free-body diagrams, and how to solve common problems involving friction.

1. What is friction?

Friction is a contact force. That means it only exists when two surfaces touch. It acts parallel to the surfaces and in the direction that opposes relative motion or attempted motion.

For example, if you push a box to the right across a floor, friction on the box acts to the left. If the box is not moving yet, static friction opposes the push. If the box is already sliding, kinetic friction opposes the sliding motion.

2. The normal force and why it matters

To calculate friction, you usually need the normal force, written as \(F_N\). The normal force is the support force a surface exerts on an object. It acts perpendicular to the surface.

On a flat horizontal surface, if there are no other vertical forces besides weight and the normal force, then

where:

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

But the normal force is not always equal to weight. If the surface is tilted, or if extra vertical forces act, then you must calculate the normal force from the situation.

3. Static friction

Static friction acts when two surfaces are in contact and not sliding relative to each other. It adjusts its size as needed to prevent motion, up to a maximum value.

This is very important: static friction is not always equal to \(\mu_s F_N\). Instead, it can take any value from zero up to a maximum:

The largest possible static friction is

where:

• \(f_s\) = static friction force
• \(\mu_s\) = coefficient of static friction
• \(F_N\) = normal force

If the force trying to move the object is smaller than \(f_{s,\max}\), static friction matches that force and the object stays at rest.

If the applied force becomes greater than \(f_{s,\max}\), static friction can no longer hold the object in place, and the object starts to move.

4. Kinetic friction

Kinetic friction acts when surfaces are sliding relative to each other. Its magnitude is modeled by

where:

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

Unlike static friction, kinetic friction is usually treated as a single fixed value for a given situation. Once the object is sliding, this friction force opposes the motion.

In most cases,

This means it is usually harder to start moving an object than to keep it moving.

5. Coefficient of friction

The coefficient of friction, written as \(\mu\), describes how strongly two surfaces interact. It depends on the materials in contact, such as rubber on concrete or wood on metal.

It has no units because it is a ratio. Larger values of \(\mu\) mean more friction. Smaller values mean less friction.

Commonly:

• \(\mu_s\) is used for static friction
• \(\mu_k\) is used for kinetic friction

6. Direction of friction

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

• If an object is sliding right, kinetic friction acts left.
• If you push right but the object stays still, static friction acts left.
• On an incline, friction acts along the slope and opposes the direction the object would move or is moving.

A common mistake is to think friction always acts opposite the applied force. That is not always true. Friction acts opposite the relative motion between surfaces.

7. Friction in free-body diagrams

When solving friction problems, begin with a free-body diagram. Show all forces acting on the object.

For an object on a horizontal surface, the forces often include:

• Weight downward: \(mg\)
• Normal force upward: \(F_N\)
• Applied force horizontally
• Friction horizontally opposite motion or attempted motion

After drawing the diagram, apply Newton's second law:

This allows you to connect friction to acceleration or equilibrium.

8. Static friction model in detail

Suppose you push a heavy box gently. If the box does not move, then the horizontal forces must balance. That means static friction has adjusted to exactly match your push.

For example, if you push with \(20\,\text{N}\) and the box does not move, then static friction is \(20\,\text{N}\) in the opposite direction.

If you increase your push to \(45\,\text{N}\) and the box still does not move, then static friction becomes \(45\,\text{N}\), as long as this is still less than or equal to the maximum static friction.

Only when your push exceeds \(f_{s,\max}\) does motion begin.

9. Kinetic friction model in detail

Once the box starts sliding, the friction changes from static to kinetic. At that point, the friction force is modeled by \(f_k = \mu_k F_N\).

If the applied force is greater than kinetic friction, the object accelerates. If the applied force equals kinetic friction, the object moves at constant velocity. If the applied force is less than kinetic friction, the object slows down.

10. Steps for solving friction problems

1. Identify whether the object is at rest or sliding.
2. Draw a free-body diagram.
3. Find the normal force.
4. Use the correct friction model:
   • Static: \(f_s \leq \mu_s F_N\)
   • Kinetic: \(f_k = \mu_k F_N\)
5. Apply Newton's laws to solve for the unknown quantity.
6. Check if your answer makes physical sense.

Worked Example 1: Finding kinetic friction on a flat surface

A \(10\,\text{kg}\) box slides across a floor. The coefficient of kinetic friction is \(\mu_k = 0.30\). Find the kinetic friction force.

Step 1: Find the normal force

Because the surface is horizontal and there are no extra vertical forces,

Step 2: Use the kinetic friction formula

Answer: The kinetic friction force is \(29.4\,\text{N}\), opposite the direction of motion.

Worked Example 2: Static friction before motion starts

A \(15\,\text{kg}\) crate sits on a floor with \(\mu_s = 0.40\). A student pushes it horizontally with a force of \(40\,\text{N}\). Will the crate move, and what is the static friction force?

Step 1: Find the normal force

Step 2: Find the maximum static friction

Step 3: Compare the push to the maximum static friction

The applied force is \(40\,\text{N}\), which is less than \(58.8\,\text{N}\). So static friction can balance it.

Therefore, the crate does not move, and the actual static friction force is

Answer: The crate stays at rest. Static friction is \(40\,\text{N}\), not \(58.8\,\text{N}\).

Worked Example 3: Force needed to start motion

A \(25\,\text{kg}\) object rests on a horizontal surface. The coefficient of static friction is \(\mu_s = 0.50\). What minimum horizontal force is needed to start moving it?

Step 1: Find the normal force

Step 2: Find the maximum static friction

To start motion, the applied force must be just greater than this value. So the minimum force needed is about

Answer: A horizontal force of about \(123\,\text{N}\) is needed to start motion.

Worked Example 4: Motion with kinetic friction

A \(8.0\,\text{kg}\) sled is pulled along level snow by a horizontal force of \(25\,\text{N}\). The coefficient of kinetic friction is \(\mu_k = 0.20\). Find the acceleration.

Step 1: Find the normal force

Step 2: Find kinetic friction

Step 3: Find the net force

Take the direction of pulling as positive:

Step 4: Use Newton's second law

Answer: The acceleration is about \(1.17\,\text{m/s}^2\) in the direction of the pull.

11. Friction on an incline

On an inclined surface, the normal force is smaller than the full weight because the surface only supports part of the object's weight.

For an incline at angle \(\theta\), the normal force is

Then friction is based on this normal force:

Friction acts up or down the slope depending on which way the object tends to move.

For example, if an object tends to slide down the incline, friction acts up the incline.

12. Important differences between static and kinetic friction

• Static friction: acts when there is no sliding
• Kinetic friction: acts when there is sliding
• Static friction: changes as needed up to a maximum
• Kinetic friction: usually modeled as \(\mu_k F_N\)
• Static friction: often larger than kinetic friction for the same surfaces

13. Common mistakes to avoid

• Using \(f_s = \mu_s F_N\) every time. This only gives the maximum static friction.
• Forgetting to calculate the normal force before finding friction.
• Assuming friction always equals the applied force. This is only true in some static cases.
• Using \(\mu_s\) when the object is sliding, or \(\mu_k\) when it is not sliding.
• Giving friction the wrong direction.

14. Quick strategy check

Ask yourself these questions:

• Is the object moving or not?
• What is the normal force?
• Do I need maximum static friction or actual static friction?
• What direction should friction point?
• Does the answer fit the motion described?

15. Brief summary

Friction is a force that opposes relative motion between surfaces in contact. Static friction prevents motion and can vary from zero up to \(\mu_s F_N\). Kinetic friction acts during sliding and is modeled by \(f_k = \mu_k F_N\).

To solve friction problems, first draw a free-body diagram, then find the normal force, choose the correct friction model, and apply Newton's laws. Remember that static friction is flexible up to a maximum, while kinetic friction is used once motion has started.

Hooke’s Law and Elastic Forces

When you stretch a spring or compress it, the spring pushes or pulls back. This restoring force is an example of an elastic force. In many everyday situations, springs, rubber bands, and other elastic materials resist changes to their shape.

Hooke’s Law describes how this restoring force behaves for many springs when the stretch or compression is not too large. It helps us predict how much force is needed to stretch a spring and how strongly the spring pulls back.

This idea is important in mechanics because elastic forces appear in many systems, such as car suspensions, trampolines, measuring scales, and oscillating objects. Understanding Hooke’s Law also connects force, motion, and energy.

1. What is an elastic force?

An elastic force is a force exerted by an object that has been stretched, compressed, or bent and is trying to return to its original shape. This is why it is called a restoring force.

For example:

• A stretched spring pulls inward.
• A compressed spring pushes outward.
• A stretched rubber band pulls back toward its original length.

The key idea is that the force acts in the direction that restores the object to its normal shape or length.

2. Hooke’s Law

For many springs, the restoring force is directly proportional to how far the spring is stretched or compressed from its natural length. This relationship is called Hooke’s Law.

Mathematically, Hooke’s Law is written as

where:

• \(F\) is the restoring force of the spring, measured in newtons (N)
• \(k\) is the spring constant, measured in newtons per meter (N/m)
• \(x\) is the displacement from the spring’s natural length, measured in meters (m)

The negative sign is very important. It shows that the spring force is always opposite to the displacement.

• If the spring is stretched to the right, the force is to the left.
• If the spring is compressed to the left, the force is to the right.

3. Meaning of the spring constant \(k\)

The spring constant tells us how stiff the spring is.

• A large \(k\) means a stiff spring. It takes more force to stretch or compress it.
• A small \(k\) means a softer spring. It is easier to stretch or compress.

For example, a spring with \(k=500\,\text{N/m}\) is much stiffer than one with \(k=50\,\text{N/m}\).

4. Natural length and displacement

The natural length of a spring is its length when no external force is stretching or compressing it. The quantity \(x\) in Hooke’s Law is not the total length of the spring. It is only the amount of stretch or compression compared with the natural length.

So if a spring has a natural length of \(20\,\text{cm}\) and is stretched to \(26\,\text{cm}\), then the displacement is

Always convert centimeters to meters before using Hooke’s Law in calculations.

5. Direction of the elastic force

The force from a spring depends on direction, so it is often treated as a vector. The sign of \(x\) helps us keep track of direction.

If we choose the right as positive:

• Stretching the spring to the right means \(x>0\), so \(F=-kx\) is negative, which means the force points left.
• Compressing the spring so that the end moves left means \(x<0\), so \(F=-kx\) is positive, which means the force points right.

This matches the idea that the spring always tries to return to equilibrium.

6. Equilibrium position

The equilibrium position is the position where the spring is at its natural length and the elastic force is zero.

At equilibrium,

If the spring is moved away from equilibrium, the restoring force appears and acts back toward equilibrium.

7. Graph of force vs displacement

Hooke’s Law shows a linear relationship between force and displacement. If you graph \(F\) against \(x\), you get a straight line through the origin with slope \(-k\).

This means:

• Doubling the displacement doubles the force.
• Tripling the displacement triples the force.
• The negative slope shows the force acts opposite to the displacement.

Because the relationship is linear, Hooke’s Law is often described as a linear restoring force law.

8. Limits of Hooke’s Law

Hooke’s Law does not work perfectly for every elastic material in every situation. It is usually valid only when the deformation is small enough.

If a spring is stretched too far, it may stop behaving linearly. In that case:

• the force may no longer be proportional to displacement, or
• the spring may become permanently deformed and not return to its original length.

This is why Hooke’s Law is best used within the elastic limit of the material.

9. Elastic potential energy

When a spring is stretched or compressed, work is done on it. That work is stored as elastic potential energy.

The formula is

where:

• \(U\) is elastic potential energy in joules (J)
• \(k\) is the spring constant in N/m
• \(x\) is the displacement from equilibrium in m

This formula applies for both stretching and compression because \(x^2\) is always positive.

10. Hooke’s Law and Newton’s Second Law

A spring force can cause acceleration. If a mass is attached to a spring and pulled away from equilibrium, the spring force acts toward equilibrium. By Newton’s Second Law,

If the spring is the main horizontal force, then

This means the acceleration also depends on displacement and points toward equilibrium. This is why spring-mass systems can move back and forth in oscillation.

11. Worked Example 1: Finding spring force

A spring has spring constant \(k=200\,\text{N/m}\). It is stretched by \(0.05\,\text{m}\). Find the restoring force.

Step 1: Write Hooke’s Law

Step 2: Substitute values

Answer: The restoring force is \(-10\,\text{N}\). The negative sign means the force is opposite the direction of stretch.

Interpretation: If the spring was stretched to the right, the spring pulls left with a force of \(10\,\text{N}\).

12. Worked Example 2: Finding the spring constant

A force of \(15\,\text{N}\) stretches a spring by \(0.30\,\text{m}\). Find the spring constant.

Here we use the size of the force:

Step 1: Rearrange for \(k\)

Step 2: Substitute values

Answer: The spring constant is \(50\,\text{N/m}\).

This tells us the spring needs \(50\,\text{N}\) of force for every meter of stretch.

13. Worked Example 3: Compression and direction

A spring has \(k=120\,\text{N/m}\). It is compressed by \(0.08\,\text{m}\). Take right as the positive direction. Find the spring force.

If the spring is compressed, the displacement is in the negative direction, so

Now apply Hooke’s Law:

Answer: The spring force is \(+9.6\,\text{N}\).

The positive sign means the force points to the right. This makes sense because the spring was compressed and pushes outward toward equilibrium.

14. Worked Example 4: Elastic potential energy

A spring with \(k=300\,\text{N/m}\) is stretched by \(0.10\,\text{m}\). Find the elastic potential energy stored.

Step 1: Use the energy formula

Step 2: Substitute values

Answer: The spring stores \(1.5\,\text{J}\) of elastic potential energy.

15. Common mistakes to avoid

• Using total length instead of displacement: Remember that \(x\) is the change in length, not the full length of the spring.
• Forgetting unit conversion: Convert centimeters to meters before substituting into formulas.
• Ignoring the negative sign: In \(F=-kx\), the negative sign shows direction.
• Assuming Hooke’s Law always works: It only applies when the spring is within its elastic limit.
• Mixing up force and energy formulas: Force uses \(F=-kx\), while energy uses \(U=\frac{1}{2}kx^2\).

16. Quick comparison of the two key formulas

• Spring force: $$F=-kx$$
• Elastic potential energy: $$U=\frac{1}{2}kx^2$$

The first formula tells how strongly the spring pushes or pulls. The second tells how much energy is stored in the deformation.

17. Real-life applications

• Vehicle suspension systems: Springs help absorb bumps and restore the vehicle to a smoother position.
• Spring scales: The amount of stretch is used to measure force or weight.
• Mattresses and seats: Elastic materials compress and push back.
• Mechanical devices: Many machines use springs to store and release energy.

18. Brief summary

Hooke’s Law describes the restoring force of a spring as directly proportional to displacement from equilibrium:

The spring constant \(k\) measures stiffness, and the negative sign shows that the force acts opposite the displacement. When a spring is stretched or compressed, it also stores elastic potential energy:

These ideas help us understand how elastic materials resist deformation and return toward their original shape.

Uniform Circular Motion and Centripetal Acceleration

When an object moves in a circle, its motion is different from straight-line motion. Even if the object’s speed stays constant, its velocity is changing because velocity includes both speed and direction.

This is the key idea behind uniform circular motion: an object moves around a circular path at a constant speed, but because its direction changes at every moment, it is still accelerating.

That acceleration points toward the center of the circle and is called centripetal acceleration. The word centripetal means “center-seeking.”

1. What is uniform circular motion?

Uniform circular motion happens when an object travels in a circle of radius \(r\) with constant speed \(v\). Examples include:

• a satellite orbiting Earth in a nearly circular path,
• a car moving around a circular track at steady speed,
• a stone tied to a string and swung in a horizontal circle,
• a point on the edge of a turning fan blade.

In all of these cases, the object keeps changing direction. Since acceleration is the rate of change of velocity, a change in direction means there must be acceleration, even if the speed does not change.

2. Why is there acceleration if the speed is constant?

Many students think acceleration only happens when something speeds up or slows down. But acceleration also happens when the direction of motion changes.

Imagine an object moving around a circle. At one point, its velocity might point east. A moment later, its velocity points northeast. The speed may be the same, but the direction is different, so the velocity has changed. Therefore, the object is accelerating.

For circular motion, this acceleration always points toward the center of the circle. It is perpendicular to the object’s instantaneous velocity.

3. Centripetal acceleration

The magnitude of centripetal acceleration is:

where:

• \(a_c\) = centripetal acceleration in \(\text{m/s}^2\),
• \(v\) = speed in \(\text{m/s}\),
• \(r\) = radius of the circular path in meters.

This formula shows two important ideas:

• If speed increases, centripetal acceleration increases a lot because speed is squared.
• If radius increases, centripetal acceleration decreases.

So a fast-moving object in a tight circle needs a large inward acceleration.

4. Centripetal force

If there is centripetal acceleration, there must also be a net force causing it. By Newton’s second law,

Substituting \(a_c = \frac{v^2}{r}\), we get:

where:

• \(F_c\) = centripetal force in newtons,
• \(m\) = mass in kilograms.

Centripetal force is not a new kind of force. It is the name for the net inward force that keeps the object moving in a circle.

Different physical forces can act as centripetal force, such as:

• tension in a string,
• gravity on a planet or satellite,
• friction between tires and the road,
• normal force in some curved paths.

5. Direction of velocity, acceleration, and force

In uniform circular motion:

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

This means velocity and centripetal acceleration are at right angles to each other.

6. Period and frequency

Sometimes circular motion is described using period and frequency.

• The period \(T\) is the time for one complete revolution.
• The 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 is:

Since \(f = \frac{1}{T}\), speed can also be written as:

These equations are useful when speed is not given directly.

7. Worked Example 1: Finding centripetal acceleration

A car moves around a circular track of radius \(50\,\text{m}\) at a constant speed of \(10\,\text{m/s}\). Find its centripetal acceleration.

Step 1: Write the formula.

Step 2: Substitute values.

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

8. Worked Example 2: Finding centripetal force

A \(0.50\,\text{kg}\) ball is swung in a horizontal circle of radius \(0.80\,\text{m}\) at a speed of \(4.0\,\text{m/s}\). Find the centripetal force on the ball.

Step 1: Use the formula.

Step 2: Substitute values.

Answer: The centripetal force is \(10\,\text{N}\), directed inward toward the center of the circle.

In this case, the force may be provided by the tension in the string.

9. Worked Example 3: Using period to find speed and acceleration

A point on the edge of a merry-go-round is \(3.0\,\text{m}\) from the center. It completes one revolution every \(6.0\,\text{s}\). Find:

1. the speed,
2. the centripetal acceleration.

Step 1: Find speed using \(v = \frac{2\pi r}{T}\).

Step 2: Find centripetal acceleration.

Answer:

• Speed \(\approx 3.14\,\text{m/s}\)
• Centripetal acceleration \(\approx 3.29\,\text{m/s}^2\)

10. Worked Example 4: Comparing two circular motions

Object A and Object B both move in circles of the same radius. Object B moves at twice the speed of Object A. How does the centripetal acceleration of B compare with that of A?

Use the formula:

If B has twice the speed, then \(v_B = 2v_A\).

So:

Answer: Object B has four times the centripetal acceleration of Object A.

This example shows why speed has such a strong effect in circular motion.

11. Common misconceptions

• “If speed is constant, acceleration is zero.”
  Not always. In circular motion, direction changes, so acceleration exists.
• “Centripetal force is an extra force added to the situation.”
  No. It is the name for the net inward force caused by ordinary forces like tension, friction, or gravity.
• “The force is in the direction of motion.”
  In uniform circular motion, the centripetal force points inward, while the velocity points tangent to the circle.
• “A larger radius always means a larger centripetal acceleration.”
  If speed stays the same, a larger radius actually gives a smaller centripetal acceleration because \(a_c = \frac{v^2}{r}\).

12. Real-life applications

• Cars turning corners: friction between tires and road provides the centripetal force.
• Planets and satellites: gravity provides the centripetal force that keeps them in orbit.
• Roller coasters: the track and seat provide forces needed for curved motion.
• Washing machines and centrifuges: spinning motion separates materials because of circular motion effects.

13. Problem-solving tips

• Check whether the motion is truly uniform circular motion: circular path and constant speed.
• Identify what provides the inward force: tension, friction, gravity, or another force.
• Use the correct formula based on what is given:

14. Brief summary

Uniform circular motion is motion in a circle at constant speed. Even though the speed stays the same, the object accelerates because its direction changes continuously.

This acceleration is called centripetal acceleration and always points toward the center of the circle. Its magnitude is given by \(a_c = \frac{v^2}{r}\), and the required inward net force is \(F_c = \frac{mv^2}{r}\).

Understanding circular motion means remembering that changing direction is a form of acceleration. That idea explains why an inward force is always needed to keep an object moving in a circle.

Universal Law of Gravitation

Gravity is the force that pulls objects toward one another. It is the reason apples fall, planets orbit the Sun, and the Moon stays in orbit around Earth. In 12th Grade science, the Universal Law of Gravitation helps us describe this attraction mathematically.

Newton proposed that every pair of masses in the universe attracts each other. This means gravity is not only something that happens near Earth. Any two objects with mass pull on each other, no matter where they are.

The strength of this gravitational force depends on two main things:

• the masses of the two objects, and
• the distance between their centers.

The mathematical form of Newton’s Universal Law of Gravitation is:

Here:

• \(F\) = gravitational force between the two objects, measured in newtons (N)
• \(G\) = universal gravitational constant
• \(m_1\) and \(m_2\) = masses of the two objects, measured in kilograms (kg)
• \(r\) = distance between the centers of the two masses, measured in meters (m)

The value of the gravitational constant is:

This number is very small, which tells us that gravity is a relatively weak force unless at least one of the objects has a very large mass, like a planet or star.

Important ideas from the formula

1. Force is directly proportional to mass. If one mass increases, the gravitational force increases. If both masses double, the force becomes four times larger because the masses are multiplied together.

2. Force is inversely proportional to the square of the distance. This is called the inverse-square law. If the distance between two objects doubles, the force becomes:

So doubling the distance makes the gravitational force one-fourth as large.

If the distance becomes three times larger, the force becomes:

This shows that distance has a very strong effect on gravitational force.

Direction of gravitational force

Gravitational force is always attractive. It acts along the straight line joining the centers of the two masses. If object A pulls on object B, then object B also pulls on object A with an equal force in the opposite direction. This follows Newton’s Third Law.

Point masses and spherical objects

The law is often written for point masses, which means objects small enough that all their mass can be treated as concentrated at one point. For large spherical bodies such as Earth, the same formula works if we measure the distance from center to center.

That is why we can calculate the gravitational attraction between Earth and the Moon using the distance between their centers.

Why objects fall to Earth

When you drop an object, Earth pulls it downward because Earth has a huge mass. The object also pulls Earth upward, but Earth’s mass is so large that its motion is too small to notice.

Near Earth’s surface, this gravitational pull gives objects weight. Weight is the force with which Earth attracts an object. It is written as:

Here \(m\) is the object’s mass and \(g\) is the gravitational field strength near Earth, about \(9.8\; \text{m/s}^2\).

This weight formula is related to the Universal Law of Gravitation. In fact, \(g\) comes from Earth’s mass and radius:

where \(M_E\) is Earth’s mass and \(R_E\) is Earth’s radius.

Common patterns to remember

• If one mass doubles, gravitational force doubles.
• If both masses double, gravitational force becomes four times larger.
• If distance doubles, gravitational force becomes one-fourth.
• If distance triples, gravitational force becomes one-ninth.
• If distance is halved, gravitational force becomes four times larger.

Worked Example 1: Finding the gravitational force between two small masses

Two objects have masses \(5\; \text{kg}\) and \(8\; \text{kg}\). Their centers are \(2\; \text{m}\) apart. Find the gravitational force between them.

Step 1: Write the formula.

Step 2: Substitute the values.

Step 3: Simplify.

Answer: The gravitational force is \(6.67 \times 10^{-10}\; \text{N}\).

This force is extremely small, which is why we do not notice gravitational attraction between everyday objects.

Worked Example 2: How changing distance affects force

Two masses are kept at a distance \(r\), and the gravitational force between them is \(36\; \text{N}\). What will the force be if the distance is doubled?

Because gravitational force follows an inverse-square law:

Answer: The new force is \(9\; \text{N}\).

This example shows that increasing distance reduces force very quickly.

Worked Example 3: Comparing two situations

Suppose the gravitational force between two objects is \(F\). What happens to the force if:

• one mass is tripled, and
• the distance between them is doubled?

Start with:

In the new situation, the force becomes:

Answer: The new force is \(\frac{3}{4}\) of the original force.

Even though the mass increased, the larger distance reduced the force.

Worked Example 4: Gravitational force between Earth and an object

A student has a mass of \(60\; \text{kg}\). Find the student’s weight near Earth’s surface using \(g = 9.8\; \text{m/s}^2\).

Use:

Substitute the values:

Answer: The student’s weight is \(588\; \text{N}\).

This is the gravitational force between Earth and the student near Earth’s surface.

Common mistakes to avoid

• Confusing mass and weight: mass is measured in kg, while weight is a force measured in N.
• Using diameter instead of center-to-center distance: in the formula, \(r\) is the distance between centers.
• Forgetting the square on distance: the formula uses \(r^2\), not just \(r\).
• Using wrong units: use kilograms for mass and meters for distance.
• Thinking gravity acts only on falling objects: gravity acts between all masses.

How this law explains motion in space

The Universal Law of Gravitation helps explain why planets orbit the Sun and why moons orbit planets. The gravitational force provides the pull needed to keep these objects moving in curved paths instead of flying off in straight lines.

It also explains why satellites remain in orbit around Earth. They are constantly being pulled by Earth’s gravity while moving forward.

Key takeaways

• Every object with mass attracts every other object with mass.
• The gravitational force is given by $$F = G\frac{m_1m_2}{r^2}$$
• The force increases with mass.
• The force decreases with the square of distance.
• Gravity is always an attractive force.
• Near Earth, weight is the gravitational force on an object.

Brief Summary

The Universal Law of Gravitation states that any two masses attract each other with a force that depends on their masses and the distance between them. The force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers. This law helps us understand falling objects, weight, planetary motion, and orbits in space.

Kepler's Laws of Planetary Motion explain how planets move around the Sun. These laws were developed by Johannes Kepler in the early 1600s after carefully studying astronomical data. They are called empirical laws because Kepler discovered them by observing patterns in data, not by starting with a theory.

Kepler's three laws describe the shape of planetary orbits, how a planet's speed changes as it moves, and how the time for an orbit depends on distance from the Sun. These ideas are important in classical mechanics because they connect motion in space to measurable patterns.

In this lesson, you will learn:

• why planetary orbits are not perfect circles,
• what it means for a planet to sweep out equal areas in equal times,
• how orbital period depends on orbital size,
• and how to solve simple problems using Kepler's laws.

Before Kepler, many people believed planets moved in perfect circles. Kepler showed that this was not correct. His work gave a much more accurate description of planetary motion and later helped Isaac Newton develop the law of gravitation.

Important orbit vocabulary:

• Orbit: the path of one object around another due to gravity.
• Ellipse: an oval-shaped curve.
• Focus: a special point inside an ellipse. An ellipse has two foci.
• Semi-major axis \\(a\\): half the longest width of an ellipse.
• Orbital period \\(T\\): the time needed to complete one orbit.

Now let us study each law one by one.

Kepler's First Law: The Law of Ellipses

Kepler's First Law states:

Every planet moves in an elliptical orbit, with the Sun at one focus of the ellipse.

This means the Sun is not at the center of the planet's orbit. Instead, it is located at one of the two foci of the ellipse. For many planets, the ellipse is close to a circle, so the difference is not always obvious. But the orbit is still technically an ellipse.

If a planet is closer to the Sun at one part of its orbit and farther away at another part, then its distance from the Sun changes during the year. This changing distance helps explain why the planet's speed also changes.

Two useful terms for this law are:

• Perihelion: the point where a planet is closest to the Sun.
• Aphelion: the point where a planet is farthest from the Sun.

Because the orbit is elliptical, the planet is not always the same distance from the Sun. This is different from uniform circular motion, where the radius stays constant.

Kepler's Second Law: The Law of Equal Areas

Kepler's Second Law states:

A line joining a planet and the Sun sweeps out equal areas in equal intervals of time.

Imagine drawing a line from the Sun to the planet. As the planet moves, that line sweeps out a region of space. Kepler found that if you compare two equal time intervals, the areas swept out are the same, even if the shapes are different.

This law tells us something very important about speed:

• When the planet is closer to the Sun, it moves faster.
• When the planet is farther from the Sun, it moves slower.

Near perihelion, the planet must move quickly so that the swept area in a given time stays equal to the area swept out elsewhere. Near aphelion, where the planet is farther away, it moves more slowly.

This does not mean the planet speeds up and slows down randomly. Its changing speed follows a precise pattern related to its changing distance from the Sun.

Kepler's Third Law: The Law of Periods

Kepler's Third Law states:

The square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit.

In symbols, this is written as:

$$T^2 \propto a^3$$

For planets orbiting the same star, such as the planets in our solar system orbiting the Sun, we can write this as:

$$\frac{T^2}{a^3}=\text{constant}$$

This law connects how long a planet takes to orbit with how far it is from the Sun on average. A planet with a larger orbit takes much longer to complete one revolution.

If two planets orbit the Sun, then:

$$\frac{T_1^2}{a_1^3}=\frac{T_2^2}{a_2^3}$$

This formula is especially useful for solving problems.

In our solar system, if \\(T\\) is measured in years and \\(a\\) is measured in astronomical units (AU), then the constant becomes 1, so the law becomes:

$$T^2=a^3$$

Here, \\(1\\) AU is the average distance from Earth to the Sun, and Earth has \\(T=1\\) year.

Why Kepler's Laws Matter

• They describe real planetary motion much better than old circular models.
• They help scientists predict where planets and other objects will be.
• They apply not only to planets, but also to comets, asteroids, and many satellites.
• They were a major step toward Newton's explanation of gravity.

Connecting the Three Laws

• The First Law tells the shape of the orbit: an ellipse.
• The Second Law tells how the speed changes: faster when closer, slower when farther.
• The Third Law tells how orbital time depends on orbit size.

Together, these laws give a complete description of planetary motion at a 12th Grade level.

Worked Example 1: Identifying the First Law

A student says, "Planets move in circles with the Sun at the center." Explain what is wrong with this statement using Kepler's First Law.

Step 1: State the correct law.

Kepler's First Law says planets move in elliptical orbits, not perfect circles.

Step 2: Describe the Sun's position.

The Sun is located at one focus of the ellipse, not at the center of the orbit.

Answer: The statement is wrong because planetary orbits are ellipses, and the Sun is at one focus rather than at the center.

Worked Example 2: Using the Second Law

A planet is moving around the Sun in an elliptical orbit. Compare its speed at perihelion and aphelion.

Step 1: Recall the Second Law.

Equal areas are swept out in equal times.

Step 2: Think about distance.

At perihelion, the planet is closer to the Sun. To sweep out the same area in the same time, it must move faster.

At aphelion, the planet is farther from the Sun. It can sweep the same area while moving more slowly.

Answer: The planet moves fastest at perihelion and slowest at aphelion.

Worked Example 3: Finding Orbital Period from Distance

A planet orbits the Sun at an average distance of \\(4\\) AU. Find its orbital period in years.

Step 1: Use Kepler's Third Law.

Since the planet orbits the Sun and units are AU and years:

$$T^2=a^3$$

Step 2: Substitute \\(a=4\\).

$$T^2=4^3=64$$

Step 3: Solve for \\(T\\).

$$T=\sqrt{64}=8$$

Answer: The orbital period is 8 years.

Worked Example 4: Comparing Two Planets

Planet A has an orbital period of \\(1\\) year and a semi-major axis of \\(1\\) AU. Planet B has a semi-major axis of \\(9\\) AU. Find the orbital period of Planet B.

Step 1: Write the comparison formula.

$$\frac{T_1^2}{a_1^3}=\frac{T_2^2}{a_2^3}$$

Step 2: Substitute the known values.

$$\frac{1^2}{1^3}=\frac{T_2^2}{9^3}$$

$$1=\frac{T_2^2}{729}$$

Step 3: Solve for \\(T_2\\).

$$T_2^2=729$$

$$T_2=\sqrt{729}=27$$

Answer: Planet B has an orbital period of 27 years.

Common Mistakes to Avoid

• Mistake 1: Thinking all orbits are circles. Kepler's First Law says they are ellipses.
• Mistake 2: Thinking a planet moves at constant speed. Kepler's Second Law shows speed changes.
• Mistake 3: Forgetting that the Third Law compares \\(T^2\\) with \\(a^3\\), not just \\(T\\) with \\(a\\).
• Mistake 4: Mixing units. For the simple form \\(T^2=a^3\\), use years for time and AU for distance in the solar system.

Quick Check for Understanding

1. According to Kepler's First Law, what is the shape of a planet's orbit?
2. Where is the Sun located in that orbit?
3. According to the Second Law, when does a planet move faster?
4. If a planet is farther from the Sun, would you expect its period to be shorter or longer?
5. What does the equation \\(T^2 \propto a^3\\) tell us?

Answers to the Quick Check

1. An ellipse.
2. At one focus of the ellipse.
3. When it is closer to the Sun.
4. Longer.
5. The square of the orbital period is proportional to the cube of the semi-major axis.

Brief Summary

Kepler's Laws describe how planets move around the Sun. The First Law says orbits are ellipses with the Sun at one focus. The Second Law says planets sweep out equal areas in equal times, so they move faster when closer to the Sun and slower when farther away. The Third Law says \\(T^2 \propto a^3\\), meaning planets with larger orbits take longer to complete one revolution.

Linear Momentum and Impulse are two closely connected ideas in mechanics. They help us understand how motion changes when forces act on objects, especially during collisions, catches, kicks, and crashes.

In everyday life, you can see these ideas in action when a soccer player kicks a ball, when airbags protect passengers, or when a bat hits a baseball. In each case, a force acts over a short time and changes how an object moves.

This lesson will explain what linear momentum is, what impulse is, how they are related, and how to solve common problems step by step.

1. What is Linear Momentum?

Linear momentum is the quantity of motion an object has. It depends on two things: the object's mass and its velocity.

The formula for linear momentum is

where:

• \(p\) = momentum

• \(m\) = mass

• \(v\) = velocity

Because velocity has direction, momentum also has direction. That means momentum is a vector quantity. If an object moves to the right, we may call its momentum positive. If it moves to the left, we may call it negative.

The SI unit of momentum is

A more massive object has more momentum if it moves at the same speed as a lighter one. Also, a faster object has more momentum if the masses are the same.

2. Understanding Momentum as a Vector

Since momentum depends on velocity, direction matters. For example, a car moving east and a car moving west at the same speed do not have the same momentum. Their momenta have the same size but opposite directions.

This becomes very important when solving problems involving collisions. A negative sign in momentum usually means the object is moving in the opposite direction from the one chosen as positive.

3. What is Impulse?

Impulse describes the effect of a force acting over a time interval. A small force acting for a long time can have the same effect as a large force acting for a short time.

The formula for impulse is

where:

• \(J\) = impulse

• \(F\) = average net force

• \(\Delta t\) = time interval

The unit of impulse is

This unit is equivalent to

So impulse and momentum have equivalent units.

4. The Impulse-Momentum Theorem

The most important relationship in this topic is that impulse equals the change in momentum.

This means

For an object with constant mass, this can also be written as

where:

• \(v_i\) = initial velocity

• \(v_f\) = final velocity

This equation tells us that changing an object's momentum requires either:

• a larger force,

• a longer time of action,

• or both.

5. Why Increasing Time Can Reduce Force

In many safety situations, the change in momentum is fixed. For example, if a passenger goes from moving forward to stopping, the momentum change is the same whether the stop is sudden or gradual.

Using

we see that if the stopping time increases, the force decreases. This is why airbags, padded helmets, and crumple zones are useful. They increase the time over which the momentum changes, reducing the force on the body.

6. Force-Time Graph and Impulse

If force changes during an interaction, impulse can still be found. On a force-time graph, the impulse is equal to the area under the graph.

For a constant force, the graph is a rectangle, so the area is simply

If the graph is triangular, the area is

This area still represents the change in momentum.

7. Steps for Solving Momentum and Impulse Problems

1. Identify what is given: mass, velocity, force, or time.

2. Choose a positive direction.

3. Use signs carefully for velocities and momentum.

4. Write the correct formula:

   • \(p = mv\)

   • \(J = F\Delta t\)

   • \(J = \Delta p = mv_f - mv_i\)

5. Substitute values with units.

6. Check whether the answer should include direction.

Worked Example 1: Finding Momentum

A 2.0 kg ball moves at 6.0 m/s to the right. Find its momentum.

Step 1: Use the formula

Step 2: Substitute values

Answer: The momentum is 12 kg·m/s to the right.

This example shows that momentum increases when mass or velocity increases.

Worked Example 2: Finding Impulse from Force and Time

A force of 15 N acts on an object for 0.40 s. Find the impulse.

Step 1: Use the formula

Step 2: Substitute values

Since \(1\,\text{N·s} = 1\,\text{kg·m/s}\), we can also write

Answer: The impulse is 6.0 N·s or 6.0 kg·m/s.

Worked Example 3: Using Impulse to Find Final Velocity

A 0.50 kg soccer ball is initially at rest. A player kicks it with an average force of 20 N for 0.10 s. What is the ball's final velocity?

Step 1: Find the impulse

Step 2: Use the impulse-momentum theorem

Because the ball starts at rest, \(v_i = 0\). So

Step 3: Solve for \(v_f\)

Answer: The ball's final velocity is 4.0 m/s in the direction of the force.

This example shows how force acting over time changes an object's motion.

Worked Example 4: Object Changes Direction

A 0.20 kg tennis ball is moving to the right at 8.0 m/s. It is hit and leaves to the left at 12 m/s. Find:

• the initial momentum,

• the final momentum,

• the change in momentum.

Step 1: Choose right as positive

Then:

• \(v_i = +8.0\,\text{m/s}\)

• \(v_f = -12\,\text{m/s}\)

Step 2: Find initial momentum

Step 3: Find final momentum

Step 4: Find change in momentum

Answer:

• Initial momentum: +1.6 kg·m/s

• Final momentum: -2.4 kg·m/s

• Change in momentum: -4.0 kg·m/s

The negative change in momentum means the impulse was toward the left.

8. Common Mistakes to Avoid

• Forgetting direction: Momentum must include direction because velocity has direction.

• Using speed instead of velocity: Speed has no direction, but momentum needs velocity.

• Ignoring negative signs: A negative momentum or impulse often means the direction is opposite to the positive direction chosen.

• Confusing force and impulse: Force is measured in newtons, while impulse is force multiplied by time.

• Using the wrong time: In impulse problems, use the time interval during which the force acts.

9. Real-World Applications

Linear momentum and impulse are useful in many real situations:

• Sports: A batter, golfer, or soccer player applies force over a short time to change a ball's momentum.

• Vehicle safety: Seat belts and airbags reduce force by increasing stopping time.

• Packaging: Bubble wrap and foam protect objects by lengthening the time of impact.

• Martial arts and catching: Pulling the hands back while catching increases stopping time and lowers force.

10. Key Equations to Remember

• Momentum:

  $$p = mv$$

• Impulse:

  $$J = F\Delta t$$

• Impulse-momentum theorem:

  $$J = \Delta p$$

• Combined form:

  $$F\Delta t = mv_f - mv_i$$

Brief Summary

Linear momentum measures an object's motion and is found using \(p = mv\). Since velocity has direction, momentum also has direction.

Impulse is the effect of a force acting over time and is found using \(J = F\Delta t\). Impulse equals the change in momentum, so

This idea helps explain collisions, kicks, catches, and safety features like airbags. If the time of impact increases, the force can decrease for the same change in momentum.

Conservation of Linear Momentum is one of the most important ideas in classical mechanics. It helps us understand what happens when objects collide, stick together, bounce apart, or even explode into pieces.

In this lesson, you will learn what linear momentum is, why it is conserved in a closed system, and how to solve problems involving perfectly elastic collisions, inelastic collisions, and explosions.

By the end, you should be able to identify a system, write a momentum equation correctly, and use it to find unknown speeds or directions after an interaction.

1. What is linear momentum?

Linear momentum is the quantity of motion an object has. It depends on two things: the object's mass and its velocity.

The formula for linear momentum is

$$p = mv$$

where:

• \(p\) = momentum
• \(m\) = mass
• \(v\) = velocity

Because velocity has direction, momentum also has direction. That means momentum is a vector quantity.

In one-dimensional problems, we usually choose one direction to be positive. For example, motion to the right may be positive, and motion to the left may be negative.

2. The principle of conservation of linear momentum

The law of conservation of linear momentum says:

In a closed system with no external net force, the total linear momentum remains constant.

This means the total momentum before an interaction equals the total momentum after the interaction.

Mathematically,

$$p_{\text{before}} = p_{\text{after}}$$

For two objects, this becomes

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

where:

• \(u_1, u_2\) are the initial velocities
• \(v_1, v_2\) are the final velocities

A closed system means we are only looking at the objects involved, and the external forces acting during the interaction are negligible or cancel out.

For many collision problems, the collision happens in a very short time. During that short time, external forces such as friction can often be ignored, so momentum is approximately conserved.

3. Why momentum is conserved in collisions

When two objects collide, they exert forces on each other. According to Newton's third law, these forces are equal in size and opposite in direction.

So if object 1 pushes on object 2, object 2 pushes back on object 1 with an equal and opposite force. These internal forces change the individual momenta of the objects, but the total momentum of the system stays the same.

One object may lose momentum while the other gains exactly the same amount.

4. Momentum and kinetic energy are not the same

Students often confuse conservation of momentum with conservation of kinetic energy. They are different ideas.

• Momentum is conserved in all collisions in a closed system.
• Kinetic energy is conserved only in perfectly elastic collisions.

Kinetic energy is given by

$$KE = \frac{1}{2}mv^2$$

Since velocity is squared, kinetic energy does not include direction. It is a scalar quantity.

5. Types of collisions

(a) Perfectly elastic collision

In a perfectly elastic collision:

• momentum is conserved
• kinetic energy is also conserved

Objects bounce off each other without losing total kinetic energy.

The equations are:

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

and

$$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$

(b) Inelastic collision

In an inelastic collision:

• momentum is conserved
• kinetic energy is not conserved

Some kinetic energy is transformed into other forms such as sound, heat, or deformation.

(c) Perfectly inelastic collision

This is a special type of inelastic collision in which the objects stick together after colliding.

Then both objects move with the same final velocity.

The momentum equation becomes

$$m_1u_1 + m_2u_2 = (m_1 + m_2)v$$

(d) Explosion or separation

In an explosion, a single object or system breaks into parts that move apart.

If the system starts at rest, its initial momentum is zero. Therefore, the total momentum after the explosion must also be zero.

That means the momenta of the pieces must balance each other.

6. A step-by-step method for solving momentum problems

1. Choose the system. Decide which objects are included.
2. Choose a positive direction. This helps you assign correct signs to velocities.
3. Write the total initial momentum.
4. Write the total final momentum.
5. Set them equal. Use conservation of momentum.
6. If needed, use kinetic energy information. This is only for perfectly elastic collisions.
7. Check signs and units. Momentum is usually measured in \(\text{kg m/s}\).

7. Important sign convention in one dimension

Suppose right is positive.

• An object moving right has positive velocity.
• An object moving left has negative velocity.

This sign convention is essential. If you ignore direction, you may get the correct magnitude but the wrong physical answer.

8. Worked Example 1: Basic momentum conservation

A \(2.0\,\text{kg}\) cart moves to the right at \(4.0\,\text{m/s}\). It collides with a \(1.0\,\text{kg}\) cart at rest. After the collision, the \(2.0\,\text{kg}\) cart moves at \(1.0\,\text{m/s}\) to the right. Find the final velocity of the \(1.0\,\text{kg}\) cart.

Step 1: Write known values

• \(m_1 = 2.0\,\text{kg}\)
• \(u_1 = +4.0\,\text{m/s}\)
• \(m_2 = 1.0\,\text{kg}\)
• \(u_2 = 0\)
• \(v_1 = +1.0\,\text{m/s}\)
• \(v_2 = ?\)

Step 2: Apply conservation of momentum

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

Substitute values:

$$2.0(4.0) + 1.0(0) = 2.0(1.0) + 1.0(v_2)$$

$$8.0 = 2.0 + v_2$$

$$v_2 = 6.0\,\text{m/s}$$

Answer: The \(1.0\,\text{kg}\) cart moves to the right at \(6.0\,\text{m/s}\).

9. Worked Example 2: Perfectly inelastic collision

A \(1200\,\text{kg}\) car moving at \(20\,\text{m/s}\) collides with a \(800\,\text{kg}\) car moving at \(10\,\text{m/s}\) in the same direction. The cars stick together. Find their common final velocity.

Step 1: Write the formula for a perfectly inelastic collision

$$m_1u_1 + m_2u_2 = (m_1+m_2)v$$

Step 2: Substitute values

$$1200(20) + 800(10) = (1200+800)v$$

$$24000 + 8000 = 2000v$$

$$32000 = 2000v$$

$$v = 16\,\text{m/s}$$

Answer: The combined cars move at \(16\,\text{m/s}\) in the original direction.

What happened to kinetic energy?

Momentum was conserved, but kinetic energy decreased because the cars stuck together and some energy was transformed into sound, heat, and deformation.

10. Worked Example 3: Perfectly elastic collision

A \(1.0\,\text{kg}\) ball moves to the right at \(5.0\,\text{m/s}\). It collides head-on with a \(1.0\,\text{kg}\) ball moving to the left at \(2.0\,\text{m/s}\). The collision is perfectly elastic. Find the final velocities.

Because the masses are equal in a perfectly elastic head-on collision, the two objects exchange velocities.

So:

• the first ball takes the second ball's initial velocity
• the second ball takes the first ball's initial velocity

Therefore,

• \(v_1 = -2.0\,\text{m/s}\)
• \(v_2 = +5.0\,\text{m/s}\)

Check with momentum:

Initial momentum:

$$p_i = 1.0(5.0) + 1.0(-2.0) = 3.0\,\text{kg m/s}$$

Final momentum:

$$p_f = 1.0(-2.0) + 1.0(5.0) = 3.0\,\text{kg m/s}$$

Check with kinetic energy:

Initial kinetic energy:

$$KE_i = \frac{1}{2}(1.0)(5.0)^2 + \frac{1}{2}(1.0)(2.0)^2 = 12.5 + 2.0 = 14.5\,\text{J}$$

Final kinetic energy:

$$KE_f = \frac{1}{2}(1.0)(2.0)^2 + \frac{1}{2}(1.0)(5.0)^2 = 2.0 + 12.5 = 14.5\,\text{J}$$

Answer: The first ball moves left at \(2.0\,\text{m/s}\), and the second ball moves right at \(5.0\,\text{m/s}\).

11. Worked Example 4: Explosion from rest

A firework of mass \(6.0\,\text{kg}\) explodes at rest into two pieces. One piece has mass \(2.0\,\text{kg}\) and moves to the right at \(9.0\,\text{m/s}\). Find the velocity of the other piece.

Step 1: Initial momentum

The firework is initially at rest, so

$$p_{\text{initial}} = 0$$

Step 2: Use conservation of momentum

The second piece has mass

$$m_2 = 6.0 - 2.0 = 4.0\,\text{kg}$$

Now write the momentum equation:

$$0 = m_1v_1 + m_2v_2$$

$$0 = 2.0(9.0) + 4.0v_2$$

$$0 = 18 + 4.0v_2$$

$$4.0v_2 = -18$$

$$v_2 = -4.5\,\text{m/s}$$

Answer: The second piece moves to the left at \(4.5\,\text{m/s}\).

The negative sign shows that it moves in the direction opposite to the first piece.

12. Comparing the three main situations

• Perfectly elastic collision: momentum conserved, kinetic energy conserved.
• Inelastic collision: momentum conserved, kinetic energy not conserved.
• Explosion: momentum conserved, pieces move apart so total momentum after still equals total momentum before.

13. Common mistakes to avoid

• Ignoring direction. Velocities must include positive or negative signs.
• Using kinetic energy conservation for every collision. Only perfectly elastic collisions conserve kinetic energy.
• Forgetting that stuck objects share one final velocity. In a perfectly inelastic collision, both objects move together after impact.
• Mixing up mass and weight. Momentum uses mass, not weight.
• Not defining the system clearly. Momentum is conserved for the whole closed system, not necessarily for each object alone.

14. How to tell which equation to use

Ask yourself these questions:

• Do the objects stick together? If yes, use the perfectly inelastic form.
• Does the problem say perfectly elastic? If yes, use both momentum and kinetic energy.
• Does the object break apart or explode? If yes, total momentum before equals total momentum after.
• Is one object initially at rest? Then one of the initial momentum terms is zero.

15. Key ideas to remember

• Momentum is given by \(p = mv\).
• Momentum is a vector, so direction matters.
• In a closed system, total momentum is conserved.
• All collisions conserve momentum if external forces are negligible.
• Only perfectly elastic collisions also conserve kinetic energy.
• Explosions follow the same momentum principle.

Brief Summary

Conservation of linear momentum means that in a closed system, the total momentum before an interaction is equal to the total momentum after it. This applies to collisions and explosions as long as external forces are negligible. In perfectly elastic collisions, both momentum and kinetic energy are conserved. In inelastic collisions, only momentum is conserved, and in perfectly inelastic collisions the objects stick together and move with one common velocity.

Center of Mass Dynamics is the study of how the average position of mass in an object or system moves, especially when forces act on it. This idea is very useful because even if a system has many parts moving in complicated ways, its center of mass often follows a much simpler motion.

In classical mechanics, the center of mass helps us connect the motion of a whole system to Newton’s laws. Instead of tracking every particle separately, we can often treat the entire system as if all its mass were concentrated at one special point.

This lesson will show you how to calculate the center of mass, how to understand its motion, and how external and internal forces affect it.

1. What is the center of mass?

The center of mass is the weighted average position of all the mass in a system. “Weighted” means that larger masses count more than smaller masses when finding the average position.

For two particles on a line, the center of mass position is

For more than two particles, the same pattern continues:

In two dimensions, we calculate the coordinates separately:

This means the center of mass is like the balance point of the system.

2. Physical meaning of the center of mass

If you hold an object exactly at its center of mass, it can balance. For a uniform ruler, the center of mass is at the middle. But for an object with uneven mass distribution, the center of mass may not be at the geometric center.

For a system of particles, the center of mass may even be located in empty space. For example, in a ring, the center of mass is at the center of the ring, even though there is no material there.

3. Center of mass and motion

The most important idea in center of mass dynamics is this: the motion of the center of mass depends only on external forces.

Internal forces are forces that parts of the system exert on each other. External forces come from outside the system.

According to Newton’s second law for a system,

where:

• \(\vec{F}_{\text{ext}}\) is the net external force on the system,
• \(M\) is the total mass of the system,
• \(\vec{a}_{\text{cm}}\) is the acceleration of the center of mass.

This equation is extremely powerful. It says that no matter how complicated the internal motion is, the center of mass accelerates as if the whole mass were a single particle acted on by the external force.

4. Why internal forces do not affect center of mass motion

Internal forces always come in pairs because of Newton’s third law. If one part of the system pushes on another, the second part pushes back with equal magnitude and opposite direction.

When we add all internal forces for the whole system, they cancel out. That is why only external forces determine the motion of the center of mass.

For example, when a firework explodes in the air, the pieces fly apart because of internal forces. But the center of mass of all the pieces still follows the same path it would have followed if the firework had not exploded, as long as the only external force is gravity.

5. Velocity and momentum of the center of mass

The velocity of the center of mass is the rate of change of its position:

The total momentum of a system is related to the center of mass velocity by

This means the system’s total momentum behaves as though all the mass were concentrated at the center of mass and moving with velocity \(\vec{v}_{\text{cm}}\).

If the net external force is zero, then

So the center of mass either remains at rest or moves with constant velocity. This is another way of stating conservation of momentum.

6. Important cases

• No external force: the center of mass moves with constant velocity.
• Constant external force: the center of mass has constant acceleration.
• Only gravity acts: the center of mass falls as if the whole object were a single particle in free fall.

This is why the path of the center of mass of a spinning or breaking object in the air can still be predicted using projectile motion.

7. Calculating center of mass in one dimension

Suppose two masses are on the \(x\)-axis. The larger mass pulls the center of mass closer to itself because it contributes more to the weighted average.

If two equal masses are placed at positions \(x_1\) and \(x_2\), then the center of mass lies exactly halfway between them:

If one mass is larger, the center of mass shifts toward the larger mass.

Worked Example 1: Two masses on a line

A \(2\,\text{kg}\) mass is at \(x = 1\,\text{m}\), and a \(4\,\text{kg}\) mass is at \(x = 7\,\text{m}\). Find the center of mass.

Step 1: Write the formula.

Step 2: Substitute values.

Answer: The center of mass is at \(x = 5\,\text{m}\).

This result makes sense because the \(4\,\text{kg}\) mass is heavier, so the center of mass is closer to \(7\,\text{m}\) than to \(1\,\text{m}\).

8. Calculating center of mass in two dimensions

In two dimensions, treat the \(x\)- and \(y\)-coordinates separately. Find the weighted average of the \(x\)-positions and the weighted average of the \(y\)-positions.

Worked Example 2: Three particles in a plane

Three masses are located at these points:

• \(2\,\text{kg}\) at \((0,0)\)
• \(1\,\text{kg}\) at \((6,0)\)
• \(3\,\text{kg}\) at \((0,4)\)

Find the center of mass.

Step 1: Find total mass.

Step 2: Calculate \(x_{\text{cm}}\).

Step 3: Calculate \(y_{\text{cm}}\).

Answer: The center of mass is at \((1\,\text{m}, 2\,\text{m})\).

9. Motion of the center of mass under external force

Once you know the total mass and the net external force, you can find the acceleration of the center of mass using

This is just Newton’s second law applied to the entire system.

Worked Example 3: Finding center of mass acceleration

A system consists of two blocks, one of mass \(3\,\text{kg}\) and the other of mass \(5\,\text{kg}\). A net external horizontal force of \(16\,\text{N}\) acts on the system. What is the acceleration of the center of mass?

Step 1: Find total mass.

Step 2: Apply the equation.

Answer: The center of mass accelerates at \(2\,\text{m/s}^2\).

Notice that we did not need to know how the two blocks push or pull on each other internally. Only the external force matters for center of mass motion.

10. Center of mass in explosions and separations

One of the most important applications of center of mass dynamics is in explosions, collisions, and objects breaking apart.

When a system breaks into pieces, the internal forces may cause the pieces to move in very different directions. However, the center of mass still obeys

If the net external force is zero, the center of mass continues with constant velocity before, during, and after the explosion.

Worked Example 4: Explosion in space

A stationary object in space explodes into two pieces: a \(2\,\text{kg}\) piece and a \(3\,\text{kg}\) piece. After the explosion, the \(2\,\text{kg}\) piece moves to the right at \(6\,\text{m/s}\). What is the velocity of the \(3\,\text{kg}\) piece?

Since the object was initially at rest and there is no net external force, total momentum is conserved.

Step 1: Initial total momentum.

Step 2: Set final total momentum equal to zero.

Answer: The \(3\,\text{kg}\) piece moves at \(4\,\text{m/s}\) to the left.

The center of mass remains at rest because the total momentum stays zero.

11. Center of mass and projectile motion

If an object is thrown into the air, its center of mass follows the same kind of parabolic path as any projectile, provided gravity is the only external force and air resistance is neglected.

This remains true even if the object rotates while moving. For example, a spinning baton or a gymnast in the air may turn and twist, but the center of mass still follows a smooth projectile path.

12. Common mistakes to avoid

• Using a simple average instead of a weighted average. The center of mass depends on mass values.
• Forgetting total mass in the denominator. Always divide by the sum of all masses.
• Mixing internal and external forces. Internal forces do not change center of mass motion.
• Assuming the center of mass must lie inside the object. It can lie outside or in empty space.
• Ignoring direction in momentum problems. Velocity and momentum can be positive or negative depending on direction.

13. Problem-solving strategy

1. Identify all masses and their positions.
2. Use weighted average formulas to find the center of mass location.
3. Find the total mass of the system.
4. Determine the net external force.
5. Use $$\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}$$ to find center of mass acceleration.
6. If no external force acts, use conservation of momentum.

14. Key formulas

• One-dimensional center of mass: $$x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}$$
• Two-dimensional center of mass: $$x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}, \quad y_{\text{cm}} = \frac{\sum m_i y_i}{\sum m_i}$$
• System form of Newton’s second law: $$\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}$$
• Total momentum: $$\vec{p}_{\text{total}} = M\vec{v}_{\text{cm}}$$

15. Brief summary

The center of mass is the weighted average position of mass in a system. It tells us where the system can be treated as if all its mass were concentrated.

The motion of the center of mass depends only on the net external force, not on internal forces between parts of the system. This makes it much easier to analyze complicated motion, collisions, explosions, and projectile motion.

If you remember the weighted average formulas and the equation $$\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}},$$ you can solve many center of mass problems clearly and efficiently.

Torque and Static Equilibrium are two important ideas in mechanics. They help us understand why objects rotate, why some objects stay balanced, and how engineers design structures that do not tip or collapse.

In earlier studies of motion, you learned that forces can change an object’s motion in a straight line. But forces can also cause rotation. For example, pushing on a door near its handle makes it swing easily, while pushing near the hinges does not. This turning effect of a force is called torque.

Static equilibrium happens when an object is completely at rest and remains at rest. For that to happen, the object must not move in a straight line and must not start rotating. So, in static equilibrium, both the net force and the net torque must be zero.

This lesson explains what torque is, how to calculate it, how to choose a pivot point, and how to test whether an object is in static equilibrium.

1. What is torque?

Torque is the turning effect of a force about a point or axis. It depends on three things:

• the size of the force,
• how far from the pivot the force acts,
• the angle between the force and the object.

The basic formula for torque is

where:

• \(\tau\) = torque,
• \(r\) = distance from the pivot to the point where the force is applied,
• \(F\) = force,
• \(\theta\) = angle between the direction of the force and the line from the pivot to the point of application.

The SI unit of torque is the newton-meter \((\text{N}\cdot\text{m})\).

Torque is largest when the force is applied perpendicular to the object. That is because \(\sin 90^\circ = 1\), so the formula becomes

If the force acts along the same line as the object, then \(\theta = 0^\circ\) and \(\sin 0^\circ = 0\), so the torque is zero. This is why pulling directly along a rod does not make it rotate.

2. Pivot point and lever arm

The pivot point, also called the axis of rotation, is the point about which the object could rotate. For a door, the hinges act as the pivot. For a seesaw, the support in the middle is the pivot.

Sometimes it is useful to think in terms of the lever arm, which is the perpendicular distance from the pivot to the line of action of the force. Then torque can also be written as

where \(d\) is the perpendicular distance.

This idea helps explain why a longer wrench makes it easier to loosen a bolt. A larger distance from the pivot produces a larger torque for the same force.

3. Direction of torque

Torque can cause rotation in either of two directions:

• counterclockwise
• clockwise

When solving problems, we usually choose one direction to be positive and the other to be negative. A common choice is:

• counterclockwise torque = positive
• clockwise torque = negative

The important thing is to stay consistent throughout the problem.

4. Conditions for static equilibrium

An object is in static equilibrium when it is at rest and remains at rest. That requires two separate conditions:

1. The net force must be zero.
2. The net torque must be zero.

Mathematically, this means

The force equations prevent straight-line motion. The torque equation prevents rotation.

If either condition is not met, the object will not remain in static equilibrium.

5. How to solve static equilibrium problems

These problems become much easier if you follow a clear method.

1. Draw a diagram. Show the object, the pivot, and all forces.
2. Label distances. Mark where each force acts relative to the pivot.
3. Choose a sign convention. Decide which torque direction is positive.
4. Write the torque equation. Add positive and negative torques and set the sum equal to zero.
5. Write the force equations if needed. Use \(\sum F_x = 0\) and \(\sum F_y = 0\).
6. Solve for the unknown.

A useful strategy is to choose the pivot at a point where unknown forces act. That makes their torque zero, which simplifies the equation.

6. Common sources of torque in equilibrium problems

• Weight: Acts downward through the object’s center of mass.
• Support forces: Normal forces, tensions, or reaction forces from supports.
• Applied forces: Pushes or pulls from a person or another object.

Remember that not every force produces torque about a chosen pivot. If a force acts directly through the pivot, its lever arm is zero, so its torque is zero.

Worked Example 1: Simple torque on a wrench

A mechanic applies a force of \(40\,\text{N}\) perpendicular to the end of a wrench that is \(0.25\,\text{m}\) long. Find the torque about the bolt.

Step 1: Identify values.

• \(F = 40\,\text{N}\)
• \(r = 0.25\,\text{m}\)
• \(\theta = 90^\circ\)

Step 2: Use the torque formula.

Answer: The torque is \(10\,\text{N}\cdot\text{m}\).

This example shows that a force applied farther from the pivot creates a larger turning effect.

Worked Example 2: Balanced seesaw

A child of weight \(300\,\text{N}\) sits \(2.0\,\text{m}\) from the pivot of a seesaw. How far from the pivot must a second child of weight \(200\,\text{N}\) sit to balance it?

For balance, the clockwise and counterclockwise torques must be equal in size.

Step 1: Write the torque balance equation.

Step 2: Substitute the known values.

Answer: The second child must sit \(3.0\,\text{m}\) from the pivot.

The lighter child must sit farther away to produce the same torque.

Worked Example 3: Uniform beam supported at one end

A uniform horizontal beam is \(4.0\,\text{m}\) long and weighs \(200\,\text{N}\). It is hinged at the wall on the left end. A downward force of \(100\,\text{N}\) is applied at the right end. What upward support force must act at the right end to keep the beam in static equilibrium?

Assume the beam is horizontal and the upward support force acts vertically.

Step 1: Identify where forces act.

• The beam’s weight acts at its center, \(2.0\,\text{m}\) from the hinge.
• The applied downward force acts at \(4.0\,\text{m}\).
• The unknown upward support force also acts at \(4.0\,\text{m}\).

Step 2: Choose the hinge as the pivot.

This is helpful because forces at the hinge produce zero torque.

Take counterclockwise as positive.

Step 3: Write the torque equation.

Step 4: Solve.

Answer: The upward support force must be \(200\,\text{N}\).

If needed, you could then use \(\sum F_y = 0\) to find the hinge’s vertical force.

Worked Example 4: Sign of torque and angle

A force of \(50\,\text{N}\) is applied to a rod \(0.60\,\text{m}\) from the pivot at an angle of \(30^\circ\) to the rod. Find the torque magnitude.

Step 1: Use the full torque formula.

Step 2: Calculate.

Answer: The torque magnitude is \(15\,\text{N}\cdot\text{m}\).

This example shows that if the force is not perpendicular, only the perpendicular part of the force creates torque.

7. Why both force and torque matter

It is possible for the net force on an object to be zero but the object still rotates. For example, two equal and opposite forces on opposite sides of a wheel can create rotation without causing forward motion.

It is also possible for the net torque to be zero while the net force is not zero. In that case, the object may move in a straight line without rotating.

That is why static equilibrium requires both conditions at the same time.

8. Real-life examples

• Opening a door: You push far from the hinges to create larger torque.
• Using a wrench: A longer wrench gives more torque with the same force.
• Seesaws: Balance depends on equal clockwise and counterclockwise torques.
• Bridges and shelves: Engineers design supports so net force and net torque are zero.

9. Common mistakes to avoid

• Using the full distance instead of the perpendicular distance. Torque depends on the perpendicular lever arm.
• Ignoring the object’s own weight. A beam’s weight often acts at its center.
• Forgetting the sign of torque. Clockwise and counterclockwise torques must be treated carefully.
• Checking only forces or only torques. Static equilibrium needs both.
• Using the wrong angle. In \(\tau = rF\sin\theta\), \(\theta\) is the angle between \(\vec{r}\) and \(\vec{F}\).

10. Key ideas to remember

• Torque measures the turning effect of a force.
• Torque is calculated using \(\tau = rF\sin\theta\).
• Greater force or greater distance from the pivot gives greater torque.
• An object in static equilibrium has no linear acceleration and no rotational acceleration.
• For static equilibrium: \(\sum F_x = 0\), \(\sum F_y = 0\), and \(\sum \tau = 0\).

Brief Summary

Torque is the rotational effect of a force, and it depends on force, distance from the pivot, and angle. Static equilibrium occurs when an object stays completely at rest, which means the net force is zero and the net torque is zero. To solve equilibrium problems, draw a force diagram, choose a pivot, calculate clockwise and counterclockwise torques, and make sure all forces balance as well.