Force, energy and momentum (3.4.1)

Solve mechanics problems using vectors, kinematics graphs, Newton’s laws, momentum and energy methods, including work, power and efficiency.

57 min

To subtract one vector from another, treat it as adding the negative: . This means reversing the direction of to get , then adding it to end-to-end.
The resultant vector runs from the tail of the first vector to the tip of the reversed vector . It represents the difference between the two original vectors, both in magnitude and direction

A graph with a grid background showing two vectors. The first vector, represented by an arrow labeled →v1, is green and moves horizontally to the right. The second vector, represented by an arrow labeled →vr, is red and moves diagonally downward to the left. There is also an arrow labeled −→v2, which is green and moves vertically downward.

The resultant of any two coplanar vectors can be determined by a scale drawing.

The addition of two coplanar displacement vectors is drawn in the example below:

For a scale, the magnitude of the resultant vector can be measured by a ruler as
The angle of the resultant vector to the horizontal is measured as

A graph showing vectors with labels: v1, v2, and vr. The angle θ is indicated at the origin where the vectors originate. The grid background is marked with horizontal and vertical lines.

The magnitude of the resultant vector, can be found from Pythagoras’ theorem:

Where and are the magnitudes of the two perpendicular vectors.

The diagram below shows the vector addition of two perpendicular displacement vectors.

A graph showing vectors in a coordinate system. The red vector is labeled with v_r, and it points diagonally upwards. The green vector is labeled with v_1 and v_2, pointing horizontally and vertically, respectively. An angle θ is indicated at the origin.

In the diagram above, and so:

Trigonometric relationships can be used to determine the direction of a resultant vector that is formed by two vectors and acting perpendicularly to one another.

For the diagram below, the angle, of the resultant vector to the horizontal can be found from:

A graph showing two vectors. The red vector labeled v_r points diagonally, with v_2 = 5 m at the top right and v_1 = 3 m at the bottom left. An angle θ is indicated at the origin.

In the diagram above is opposite the angle while is adjacent. Therefore, the angle can be calculated as follows:

It is useful to note that the magnitude and direction of any resultant vector can also be calculated for any coplanar vector using the cosine rule and the sine rule.

A vector can be resolved into its perpendicular components.

A force acting in the plane may be resolved into its and components. For a force with magnitude pointing at an angle to the X axis:

The horizontal component magnitude is
The vertical component magnitude is

A graph showing a vector in a Cartesian coordinate system. The vector is represented by a red arrow labeled F with its magnitude |F| = F. The angle θ is indicated at the base of the vector. The horizontal component of the vector is labeled Fx = F cos θ, and the vertical component is labeled Fy = F sin θ. The y-axis is vertical and the x-axis is horizontal.

There are many contexts where resolving a vector into its perpendicular components is useful, often when an object is constrained to move in one direction or when only one direction of the vector is relevant:

An example of this is in foot races, the wind velocity component parallel to the track must be calculated to determine the headwind or tailwind during a race.

Another example of this is projectile motion, in which an object is acted on by gravity so that its horizontal velocity component remains the same (ignoring air resistance) but its vertical velocity component varies.

When an object is in equilibrium, the forces acting on it are balanced and cancel each other out. When only two forces are acting on an object, it is in equilibrium if those forces are equal in magnitude and opposite in direction.

If instead three forces are acting on an object, then the triangle of forces can be used to solve equilibrium problems.

Forces acting on an object in equilibrium create a closed loop when you represent them to scale and arrange them tip-to-tail. In the case of three coplanar forces, this will result in a triangle of forces. Once the triangle of forces is set, we can determine the magnitude of any missing force using basic trigonometry.

How to set up a triangle of force is shown in the illustration below, where a box rests in equilibrium on a slope.

Start by drawing the free-body diagram of the forces acting on the body.
Select any vector and draw it separately.
From the tip of that vector, start the tail of a second vector.
From the tip of the second vector, start the tail of the third vector.
Write down all the known angles formed between the vectors.
In most cases, the triangle of forces will be a right triangle.
The most common forces are weight, tension, normal reaction, and friction.

A diagram showing a triangle with a blue shaded area. On the left side, there are vectors labeled N, f, and W, with an angle θ indicated. An arrow points to the right, leading to two configurations on the right side, showing the vectors W, N, and f arranged in two different ways, both including the angle θ.

The free-body diagram of a body resting on an inclined plane is shown below.

A diagram showing two triangular force diagrams. On the left, the vectors are labeled: W with an arrow, θ, N with an arrow, and f with an arrow. On the right, the vectors are labeled: W with an arrow, θ, N with an arrow, and f with an arrow, with the word 'or' in between the two diagrams.

Ensure that the direction of the vectors forms a closed loop.
Keep the vectors to scale.

A triangle diagram showing the forces acting on an object. The weight vector is labeled as W with an arrow pointing downwards, the normal force is labeled as N with an arrow pointing diagonally, and the friction force is labeled as f with an arrow pointing horizontally. The angle between the normal force and the horizontal axis is labeled as θ.

Don't

Change the direction of any vector.
Change the size of any vector.

Question walkthrough

Triangle of Forces for a Box in Equilibrium

Finds the tension in two strings holding a box in equilibrium, one horizontal and one at 45° below horizontal, by constructing a triangle of forces.

Imagine you’re using a wrench to tighten a bolt. To maximise the tightening effect, where on the wrench should you apply force?

The answer is at the end of the wrench. The reason behind this observation lies in what is called the moment of a force (torque).

An illustration showing the concepts of Torque, Force, Clamping force, and Tension in bolt. The image depicts a bolt being tightened with a wrench, with arrows indicating the direction of Torque and Force, and labels for Clamping force and Tension in bolt.

By definition, the moment of a force (torque) is the turning effect of a force about some axis or pivot. The SI unit of the moment of a force is expressed in newton metres

An axis of rotation is simply an imaginary line about which a body rotates. In our example, the axis of rotation passes vertically through the centre of the bolt.

The moment of a force can be expressed as:

Where:

is the magnitude of the force in newtons , and
is the perpendicular distance from the line of action of the force to the axis of rotation (pivot) in metres .

The SI unit of the moment of a force is expressed in newton metres

The line of action of a force, as shown in the figure below, is an imaginary line that extends infinitely in both directions along the direction of the force vector.

An illustration showing a pivot point labeled 'Pivot', a force indicated by an arrow, a line of action represented by a dashed line, and a perpendicular distance labeled 'd⊥'.

When dealing with moments of force, we have three cases:

Perpendicular line of action
In this case (diagram A), the line of action of the force is perpendicular to the axis of rotation. The moment of a force is easily calculated by multiplying the magnitude of the force by the line connecting the axis of rotation and the point of application of the force.
Non-perpendicular line of action
Here, the force’s line of action is non-perpendicular (diagram B). Given the angle that the force makes with the horizontal, the perpendicular distance between the line of action and the axis of rotation should be calculated using trigonometric relations.
Line of action through the axis of rotation
In the third case, the line of action of the force passes directly through the axis of rotation (diagram C). Consequently, the moment of the force is equal to zero, as there is no distance to create a turning effect.

Three diagrams labeled A, B, and C illustrating the concept of pivot and line of action. Diagram A shows a pivot with a force F acting vertically and a distance d_⊥. Diagram B depicts a pivot with a force F at an angle and a distance d_⊥ represented. Diagram C illustrates a pivot with a horizontal line of action and a force F acting in the opposite direction.

It is useful to note that drawing all the forces acting on the object in question can help you see which forces are perpendicular to the distance from the pivot. Not all the forces will provide a turning effect, and it is not unusual for a question to provide more forces than required to mislead you.

Question walkthrough

Moment of an Angled Force on a Rod

Finds the perpendicular distance from a hinge to the line of action of a force applied at an angle, then uses it to calculate the moment of the force.

How can you make any free object spin without causing any translational motion? The trick is to apply a pair of opposite but equal forces to the object, as shown in the figure below. This pair of forces is called a couple. The couple should always be parallel or coplanar, have different lines of action, and be perpendicular to the distance between the pair of forces.

A diagram of a steering wheel showing the forces F acting vertically upwards and downwards, and the distance d_⊥ marked horizontally across the wheel.

The moment of the couple is known as a torque. The magnitude of a torque can be calculated using the expression:

where:

is the magnitude of one of the forces, and
is the perpendicular distance between the two forces.

Some tips to keep in mind while dealing with a force couple:

The resultant force of a couple is zero.
Unlike moments of single forces, a couple does not need a pivot.

Question walkthrough

Identifying a Couple from Force Pairs

Identifies which pair of forces acting on a circular object forms a couple, by checking for equal magnitude, opposite direction, and different lines of action.

The principle of moments states that for a body in equilibrium, the sum of clockwise moments about any point must equal the sum of counterclockwise moments about that same point.

For a body to remain in equilibrium, the total moment generated by all the forces causing clockwise rotation must balance the total moment generated by all the forces causing counterclockwise rotation:

A diagram showing forces acting on a horizontal line with a point labeled O in the center. There are four forces: F1 pointing upwards with a clockwise (CW) circular arrow above it, F2 pointing downwards, F3 pointing downwards, and F4 pointing upwards with a counterclockwise (CCW) circular arrow above it.

Consider a rod that can rotate around a pivot , as illustrated in the figure above.

Force produces a CW rotation.
Force produces a CCW rotation.
Force produces a CW rotation.
Force produces a CCW rotation.

The rod is in equilibrium as the total moment around the pivot is balanced.

Question walkthrough

Balancing Moments on a Pivoted Beam

Finds the moments of the weight and an applied force about a pivot on an off-centre beam, then calculates the force needed for rotational equilibrium.

It is hard to study the motion of an irregular rigid body when a force is applied to it. In physics, to simplify calculations, we use a concept called the centre of mass (COM) to analyse the translational motion of an object. The centre of mass of an object is the point at which the mass of the object can be considered to be concentrated.

An illustration showing two irregular shapes with a red point labeled 'Centre of mass' between them. A force 'F' is indicated acting on the left shape, with a dashed line connecting the two shapes.

The centre of mass of an object:

does not depend on the orientation of the body
can be located outside or inside the physical boundaries of a body
does not depend on the gravitational field strength.

The centre of mass of simple geometric shapes is located at their geometric centres and can be found through symmetry. The centres of mass of a circle, a box, a triangle, and a doughnut are represented below by a red dot.

Four geometric shapes are displayed: a circle, an annulus (ring), a square, and a triangle. Each shape is filled with a light blue color and has a red dot at the center. Dashed lines indicate the axes of symmetry for each shape.

To determine the centre of mass of any regular shape:

Start by dividing it into two equal parts using a dashed line.
Next, divide the same shape into equal parts using a different dashed line.
The intersection of the two dashed lines indicates the centre of mass.

The centre of gravity (COG) is the point through which an object’s entire weight is considered to act, the resultant of the small weight forces distributed throughout the object. Because moments depend on the location of a force’s line of action, this is the point used when calculating the turning effect of weight in equilibrium and rotational problems.

On a free-body diagram, an object’s weight is drawn as a single vector acting downward from its centre of gravity, rather than as multiple forces spread across the object.

A purple toroidal shape with the words 'Centre of gravity' displayed inside an oval area, accompanied by a small red dot.

The centre of gravity of an object has the following properties:

It can be located outside or inside the physical boundaries of a body.
It can change in a non-uniform gravitational field.

In cases where the gravitational field strength can be considered uniform, such as a small object near the surface of Earth, a body’s centre of mass coincides with its centre of gravity.

A diagram showing two parts: A and B. Part A illustrates a pyramid on the surface of the earth with labels 'Centre of mass', 'Centre of gravity', and 'Surface of earth'. Part B features a globe labeled 'Earth' and a moon labeled 'Moon' with lines pointing to 'Centre of gravity' and 'Centre of mass'.

However, in cases involving a non-uniform gravitational field, such as a large celestial object in orbit of a massive body, the centre of mass (COM) is not in the same location as the centre of gravity (COG):

Image A: the triangle’s COM is identical to its COG because the gravitational field acting on it is uniform.

Image B: the Moon’s COM is at its geometric centre. However, the COG is closer to the Earth because the Moon’s gravitational field is significantly stronger on the side facing the Earth.

The location of the centre of mass of a body affects its stability. An object is:

Stable if its centre of mass lies vertically above its base (image A).
Unstable and will topple if its centre of mass lies vertically outside its base (image B).

Widening an object’s base or lowering its centre of mass increases its stability.

A) Stable: A vertical rectangular shape with a dashed line indicating the centre of mass located above the base. B) Unstable: A tilted rectangular shape with a dashed line indicating the centre of mass located above the base.

To determine the centre of gravity of an irregular-shaped object, complete the following steps:

Hang the body from a pivot near its edge and allow it to settle.
Use a market to draw a vertical line passing through the pivot.
Hang the body from a different pivot and allow it to settle.
Draw another vertical line passing through the new pivot.
The intersection of the two lines is the centre of gravity.

Since you are applying the procedure to a small object on the surface of the Earth, its centre of gravity is the same as its centre of mass.

An illustration showing two shapes. On the left, labeled 'Pivot 1' is a point on the shape with 'Line 1' extending vertically. On the right, 'Pivot 2' is marked on the second shape, with 'Line 1' and 'Line 2' extending from it, and a red dot indicating the 'Centre of mass'.

An object is said to be in equilibrium when both of the following conditions are met:

1. The object is in translational equilibrium: The sum of all forces acting on the object is equal to zero:

2. The object is in rotational equilibrium: The sum of all moments acting on the object about any point is zero:

When an object is in equilibrium, it can be at rest or moving at a constant velocity.

Question walkthrough

Proving a Beam is in Equilibrium

Proves a beam with three forces is in equilibrium by checking both translational equilibrium and rotational equilibrium using the principle of moments.

Displacement is a vector quantity characterised by its magnitude and direction. The net displacement of an object depends solely on its initial and final positions. Any intermediary stops are irrelevant. The magnitude of a displacement vector is equal to the length of the segment joining the initial and final positions.

An illustration comparing displacement and distance in motion. On the left, a runner moves from an initial position to a final position, with an intermediary position marked and the path labeled 'Displacement.' On the right, the same runner moves from an initial position to a final position, with an intermediary position marked and the path labeled 'Distance.'

Distance is a scalar quantity that is fully described by its magnitude. It represents the length of the path an object follows. Distance and displacement are not identical quantities. A person who walks on a circular path and returns to their initial position has a displacement of zero, but has covered a distance equal to the circumference of the circle.

It is important to note that distance travelled and displacement are equal only when an object is moving in a straight line in the positive direction.

Average velocity is a vector quantity defined as the change in the displacement of an object over time:

The direction of the average velocity vector is always the same as the direction of the displacement vector.

Average speed is a scalar quantity defined as the distance covered over time:

Average speed and average velocity are expressed in the SI system of units by metres per second

It is important to note that the magnitudes of average velocity and average speed are not the same. They are equal only when an object is moving in a straight line and in the positive direction.

Instantaneous speed, often simply referred to as speed, is how fast an object is moving at a specific moment. By looking at the speedometer of a moving car, you can read its speed. Speed is a scalar quantity with magnitude but no direction.

Instantaneous velocity, or simply velocity, is used when speed is specified in a given direction. For example, a car moving east at has a speed of and a velocity of due east. The magnitude of instantaneous velocity of a moving object is always equal to its speed.

Speed and velocity are expressed in the SI system of units by metres per second

Acceleration is a vector quantity defined as the change in velocity over time. It indicates how quickly an object’s velocity changes, whether in magnitude, direction, or both:

Acceleration is expressed in the SI system of units by metres per second squared

A common mistake is interpreting the sign of acceleration. Many assume that a negative acceleration indicates that an object is slowing down, while a positive acceleration indicates that the object is speeding up. However, this interpretation depends on the chosen positive direction. For example, a car could be speeding up in the negative direction.

To resolve this, multiply the velocity of the moving object by its acceleration. If you get a negative number, the object is slowing down; if you get a positive number, the object is speeding up. The car shown below is speeding up because the product of its velocity and acceleration is positive.

A diagram showing a car with the following labels: a = -3 m/s² (acceleration in red), v = -7 m/s⁻¹ (velocity in green), and a note stating 'Do multiply the acceleration and velocity quantities.' The positive direction is indicated with an arrow pointing to the right.

This car is speeding up as has a positive value.

A gray car is shown with the following labels: 'a = -3 m/s²' in red, 'v = -7 ms⁻¹' in green, and 'Positive direction' in black. Below the car, there is a warning that reads, 'Do not assume that this car is slowing down.'

Don't

This car is not slowing down as does not have a negative value.

If the initial velocity of the car was in the positive direction then the car would indeed be slowing down.

Question walkthrough

Distance vs Displacement on a Track

Compares distance and displacement, then average speed and average velocity, for an athlete running three sides of a rectangular track.

A displacement–time graph (x–t graph), illustrates how an object’s displacement varies with time. Displacement is plotted on the vertical axis, with time along the horizontal axis.

If the line on the x–t graph is moving away from the t-axis with respect to time, then the object is moving away from the origin. Similarly, if the line on the x–t graph is moving towards the t-axis with respect to time, then the object is moving towards the origin.

(A) Stationary, (B) Slowing down, (C) Speeding up, (D) Constant speed with graphs showing the relationship between x and t.

When an object is moving in the positive direction and away from the origin, a curve that is concave down, as in graph (B), indicates the object is slowing down, and a curve that is concave up, as in (C), indicates that the object is speeding up. A straight oblique line indicates that the object is moving at a constant speed, as in graph (D), or at rest, as shown in graph (A).

A velocity–time graph (v–t graph) shows how the velocity of an object varies with time. Velocity is plotted on the vertical axis with time along the horizontal axis.

If the line on the v–t graph is moving away from the t-axis with respect to time, then the object is speeding up. Similarly, if the line on the v–t graph is moving towards the t-axis with respect to time, then the object is slowing down. This is shown in graphs (B) and (C), respectively.

A horizontal line on the v–t graph, where indicates that the object is moving at a constant speed, as in graph (D); if the horizontal line is on the t-axis, then the object is stationary, as in graph (A).

(A) v vs. t graph showing Stationary, (B) v vs. t graph showing Slowing down, (C) v vs. t graph showing Speeding up, (D) v vs. t graph showing Constant speed

When the line on the v–t graph is above the t-axis, the object is moving in the positive direction and when the line is below the t-axis, this indicates that the object is moving in the negative direction.

A speed–time graph (v–t graph) is similar to a velocity-time graph in all its aspects except that the former does not indicate the direction of motion. Speed is plotted on the vertical axis and time along the horizontal axis. While a velocity–time graph can show lines that are below the t-axis, a speed–time graph does not.

Consider the following scenario:

A car has an initial velocity of
The car starts to slow down at a constant rate until it comes to a momentary stop.
The car then starts to speed up in the negative direction until it reaches a velocity of

A graph showing Velocity (ms^-1) versus Time (s) labeled (A) on the left, with a downward sloping line crossing the x-axis at 2 seconds and extending to -2 ms^-1 at 4 seconds. On the right, a graph showing Speed (ms^-1) versus Time (s) labeled (B), with a horizontal line at 0 ms^-1 across the time interval from 0 to 4 seconds.

The velocity–time graph of the car is shown in graph (A), and the speed–time graph of the same car is shown in graph (B). The negative values in the velocity–time graph are reflected as positive values in the speed–time graph.

An acceleration–time graph (a–t graph) shows the rate of change of the velocity with time. Acceleration is plotted on the vertical axis, while time is along the horizontal axis:

A horizontal line on the a–t graph that aligns with the t-axis indicates that the object is not accelerating, as shown in graph (A).

A horizontal line above or below the t-axis indicates that the object is moving at a constant acceleration, as shown in graph (B).

(A) a vs. t graph showing Zero acceleration; (B) a vs. t graph showing Constant acceleration

Data loggers are powerful instruments that are used to measure and analyse the motion of objects. A data logger is an electronic device that records data over time. It consists of sensors that can be used to record:

position,
speed,
acceleration

Using data from data loggers, we can draw the following graphs:

displacement–time
velocity–time
acceleration–time

Then analyse motion and get valuable insights.

Data loggers enhance motion analysis by offering precise, reliable, real-time measurements and immediate feedback, enabling quick adjustments. Their software also allows for real-time visualisation of motion graphs.

The gradient of the tangent to any point on the displacement–time graph is equal to the instantaneous velocity at that point:

Graph showing displacement (m) over time (s). On the left, a curve with a tangent labeled 'Tangent with a positive gradient'. On the right, a curve with a tangent labeled 'Tangent with a negative gradient'.

Gradient / velocity	Meaning
Positive gradient on displacement–time graph	Positive velocity
Negative gradient on displacement–time graph	Negative velocity
Positive velocity	Object moving in the positive direction
Negative velocity	Object moving in the negative direction

The numerical value of the gradient can be determined by drawing a right triangle using any two points on the tangent. The vertical side of the triangle is called the rise, and the horizontal side is called the run.

The gradient can be calculated using:

A graph showing displacement in meters (m) on the vertical axis and time in seconds (s) on the horizontal axis. A curved line represents the relationship between displacement and time. A red triangle is drawn on the graph, labeled 'Rise' and 'Run'.

The variation of the tangent’s gradient on a displacement–time graph can indicate whether an object is speeding up, slowing down, moving at a constant speed, or at rest.

(A) Tangent becoming steeper, (B) Tangent becoming less steep, (C) Horizontal tangent, (D) Unvarying tangent. Displacement (m) vs. Time (s) graphs.

Graph (A): A tangent that gets steeper with time indicates that the object is speeding up.
Graph (B): If the tangent gets less steep with time, the object is slowing down.
Graph (C): A horizontal tangent whose gradient is equal to zero, which indicates that the object is at rest.
Graph (D): A tangent whose gradient does not vary with time (i.e. constant) signifies that the object is moving at a constant speed.

The gradient of the tangent to any point on the velocity–time graph is equal to the acceleration at that point:

Two graphs showing displacement (m) over time (s). The left graph has a tangent with a positive gradient, while the right graph has a tangent with a negative gradient.

Gradient / acceleration	Meaning
Positive gradient on velocity–time graph	Positive acceleration
Negative gradient on velocity–time graph	Negative acceleration
Positive acceleration	Object’s velocity is increasing with time
Negative acceleration	Object’s velocity is decreasing with time

The variation of the tangent’s gradient on a velocity–time graph can indicate whether the acceleration of an object is uniform, non-uniform, or equal to zero:

(A) Varying tangent, (C) Horizontal tangent, (D) Unvarying tangent. Velocity ms⁻¹ plotted against Time (s) in three different graphs.

Graph (A): A gradient that changes with time indicates that the acceleration is non-uniform.
Graph (B): A horizontal tangent indicates that the object’s acceleration is zero.
Graph (C): Showcases a tangent where the gradient does not vary with time, signifying that the object is accelerating uniformly.

The area under the velocity–time graph is equal to the displacement of the object. When the curve on the velocity time graph is above the t-axis, then the area under it is positive. If the curve is below the t-axis, the area under the curve is negative.

A graph showing velocity in meters per second (ms^-1) on the vertical axis and time in seconds (s) on the horizontal axis. The graph features a green area labeled 'Positive area' and a blue area labeled 'Velocity ms^-1'.

The best way to determine the area under the graph is to divide it into regular geometric shapes of known area, such as rectangles, triangles, and trapezoids, and then sum the individual areas.

In a nonlinear velocity–time graph, you can get a rough estimate of the area under the curve by dividing the area into small trapezoids, triangles, and rectangles. The smaller the divisions, the more accurate the estimation will be.

Using trapezoids is particularly effective because it accounts for the curvature of the graph better than rectangles. Do not forget that areas under the t-axis have negative values.

A graph showing velocity in meters per second (ms^-1) on the vertical axis and time in seconds (s) on the horizontal axis. The graph features a wave-like curve with labeled points from 1 to 10 along the time axis.

The area under the graph is the sum of the individual areas of the shapes. For the graph above:

Question walkthrough

Finding Displacement from a Velocity-Time Graph

Splits a velocity-time graph into a triangle, trapezoid, and triangle to calculate the total displacement as the sum of the three areas.

A projectile is any object that is launched with an initial velocity and then moves solely under the influence of gravity in a uniform gravitational field. In the context of projectile motion, no other forces act on a projectile, such as thrust or air resistance.

A diagram illustrating projectile motion with a curved path. The horizontal axis is labeled 'Range'. An angle θ is shown at the starting point of the projectile, with an initial velocity 'u' indicated by an arrow. The vertical component of velocity is labeled 'Vy = 0 m s⁻¹'.

In projectile motion, horizontal and vertical velocity components can be considered independently:

The horizontal velocity of a projectile remains constant throughout its flight, as there is no horizontal acceleration (assuming air resistance is negligible).
The vertical velocity changes over time due to the acceleration caused by gravity.

The independent components combine to form a parabolic trajectory characteristic of projectile motion.

For a projectile launched at an angle, it is useful to resolve its velocity into horizontal and vertical components using trigonometry.

A diagram showing a right triangle with the vertical axis labeled 'R sin θ' and the horizontal axis labeled 'R cos θ'. An angle θ is indicated at the bottom left corner, and a vector u is shown pointing diagonally upwards to the right.

For a projectile launched with an initial velocity at an angle to the horizontal:

the horizontal component of velocity is
the vertical component of velocity is .

The initial vertical velocity of a projectile – equal to – determines the maximum height reached and the time of flight.

The initial horizontal velocity – equal to – affects the horizontal range (how far the projectile travels).

Two graphs showing the effect of changing vertical velocity and the effect of changing horizontal velocity. The top graph has a vertical axis labeled 'y' and a horizontal axis labeled 'x', with three curves in green, blue, and red representing different vertical velocities. The bottom graph also has a vertical axis labeled 'y' and a horizontal axis labeled 'x', with three curves in green, blue, and red representing different horizontal velocities.

Question walkthrough

Resolving Velocity into Horizontal and Vertical Components

Resolve a projectile's initial velocity into horizontal and vertical components using trigonometry.

Key terms in projectile motion:

Time of flight : This is the total time a projectile remains in the air, from the moment it’s launched until it hits the ground. The time of flight is determined solely by the initial vertical velocity.

Maximum height : The maximum height is the highest point the projectile reaches, where its vertical velocity momentarily becomes zero before it starts descending. The maximum height is determined by the initial vertical velocity.

Range : The range is the horizontal distance the projectile travels from the launch point until it lands. The range depends on the initial horizontal velocity and the time of flight.

A diagram illustrating projectile motion with labeled components: 'u' representing initial velocity, 'usin θ' indicating the vertical component, 'ucos θ' indicating the horizontal component, 'max height, H' showing the peak height, and 'Range, R' denoting the horizontal distance traveled.

Three common scenarios in projectile motion:

Vertical projection: The projectile is launched straight up, so there is no horizontal motion. Gravity directly opposes the vertical motion, slowing the projectile until it reaches its highest point, then pulling it back down.

Horizontal projection: The projectile is launched horizontally with no initial vertical velocity from an initial height above the ground. Its vertical motion begins only because gravity pulls it downward.

Projection at an angle: The most common scenario, where the initial velocity has both horizontal and vertical components. In this case, the velocity can be resolved into separate horizontal and vertical components, with gravity affecting only the vertical component.

Useful equations for a projectile with an initial velocity , at an angle , include:

A table displaying three physics formulas: 'Time of flight' with the formula T = 2u sin(θ) / g, 'Maximum height' with the formula H = (u sin(θ))² / 2g, and 'Horizontal range' with the formula R = 2u² sin(2θ) / g. © Medify

These equations are valid only for a projectile that returns to its initial launch height upon landing.

Question walkthrough

Time of Flight for Same-Height Projectile Launch

Calculate the total time of flight for a projectile launched and landing at the same height.

Question walkthrough

Maximum Height of a Projectile at an Angle

Calculate the maximum height reached by a projectile launched at an angle, using its vertical component of velocity.

Question walkthrough

Time of Flight from Projectile Range

Work backward from a projectile's range and launch angle to find its initial speed and total time of flight.

Learn how the vertical and horizontal components of velocity change (or don’t change) in projectile motion.

A graph showing a curved line with arrows indicating direction. Red arrows point upwards and downwards at various points along the curve, while blue arrows point horizontally to the left and right.

Remember that the vertical velocity of an object decreases with time at a constant rate during projectile motion.

The vertical velocity decreases until it reaches zero at the highest point before reversing direction.

The vertical velocity then continues to decrease at a constant rate until it returns to the surface.

The horizontal component of velocity remains constant throughout the entire motion.

A graph of a curve with black points along it. There are red arrows pointing upwards and downwards, and blue arrows pointing left and right, indicating directions along the curve.

Don't

Assume the projectile maintains the same speed throughout its flight: gravity continuously affects vertical motion.

Assume that the horizontal component of velocity decreases over time during projectile motion.

Projectile motion always ignores any effects due to drag.

Apply trigonometry correctly to resolve the horizontal and vertical components of an object’s trajectory in projectile motion.

SOH CAH TOA is a helpful mnemonic to remember the trigonometric ratios for right triangles:

A diagram showing a right triangle with a hypotenuse labeled 'u' in red. The vertical side is labeled 'usin θ' in blue, and the horizontal side is labeled 'ucos θ' in blue. There is a dashed line indicating the right angle.

Use cos(θ) for the horizontal component and sin(θ) for the vertical component when resolving velocity, where θ is measured from the horizontal.

A diagram illustrating a right triangle with a hypotenuse labeled 'u' in red. The vertical side is labeled 'ucos θ' in blue, and the horizontal side is labeled 'usin θ' in blue. A dashed line outlines the triangle.

Don't

Mix up sine and cosine: this will lead to incorrect calculations for range, height, and time of flight.

Question walkthrough

Time, Height and Range for an Elevated Launch

Calculate the time of flight, maximum height, and horizontal range for a projectile launched at an angle from an elevated platform.

Your respective data booklet contains four kinematic equations of motion for objects moving with constant acceleration. It is useful to note that a fifth equation (5) can be memorised to save time in your exams:

The variables in these equations are:

time ,
initial velocity ,
final velocity ,
acceleration , and
displacement .

All quantities apart from time are vectors, meaning they can be positive or negative depending on direction.

The choice of equation depends on the given variables for a problem. For instance, if the final velocity is not given or needed, equation (4) is often the most useful.

If an object moves right with a positive velocity, a negative velocity indicates motion to the left. You can choose which direction to take as positive as long as you remain consistent.

Similarly, positive displacement means movement in the chosen positive direction, while negative displacement means movement in the opposite direction.

Acceleration is:

positive when it causes the magnitude of velocity to increase in the positive direction or decrease in the negative direction.
negative when it causes the magnitude of velocity to decrease in the positive direction or increase in the negative direction.

When an object falls in a uniform gravitational field, the constant acceleration is determined by gravity. Near Earth’s surface, this acceleration is commonly denoted as This means that, neglecting air resistance, the velocity of a freely falling object increases by about every second.

A diagram illustrating a uniform gravitational field with arrows pointing downward. The text reads 'Uniform gravitational field' and 'g = 0.81 ms²' above the arrows, and 'Surface of earth' is labeled at the bottom.

The kinematic equations are only valid when air resistance is negligible. In this ideal case, all objects fall toward Earth at the same rate. In reality, factors such as an object’s mass and surface area influence air resistance, which in turn affects its motion.

The true value of can vary slightly depending on factors such as altitude and geographical location.

Sometimes, it may seem like a problem requiring the equations of motion to solve does not give you enough information to answer correctly. However, you often have to interpret some phrases in the question to get all the information. Here are some common phrases to watch out for:

“Starts from rest” – usually means that
“At maximum height” – the velocity at maximum height is zero.
“Falling due to gravity” – acceleration is
“Speed” – if the question asks to calculate speed, then you should ignore the sign of the velocity.

To analyse motion and collisions experimentally, various techniques and apparatus can be used to collect precise data on velocity, acceleration, and momentum.

Common Apparatus:

Trolleys & Air-Track Gliders – Minimise friction for accurate motion studies.
Ticker Timers – Produce dot traces on tape to measure speed and acceleration.
Light Gates & Data Loggers – Record time and velocity electronically for high accuracy.
Video Analysis – Captures motion for frame-by-frame study of velocity and collisions.

Choosing the appropriate method depends on the required precision and the type of motion being investigated.

Question walkthrough

Maximum Height of a Ball Thrown Upwards

Finds the maximum height reached by a ball thrown vertically upwards, using v² = u² + 2as with the final velocity equal to zero.

Question walkthrough

Time of Flight for a Ball Thrown Upwards

Finds the total time a ball is in the air after being thrown upwards and caught at the same height, using a SUVAT equation with s = 0.

Question walkthrough

Finding Acceleration Graphically from s and t²

Describes an experimental method for finding a trolley's acceleration down a ramp by plotting displacement against time squared and using the gradient.

An experiment to measure involving a steel ball-bearing, an electromagnet, a trap-door and a timer is shown below.

Diagram showing an Electromagnet, a height measurement scale with values from 0 to 70, the Height of fall = h, a Trapdoor, and an arrow pointing to To switch and timer.

An electromagnet releases the steel ball, which triggers a timer to start. When the ball lands on the trapdoor, a trigger stops the timer and records the time taken.

The distance the ball falls can be measured with a ruler. The relevant equation of motion is:

We know that the ball starts from rest so and Therefore:

If the ball bearing is small, there will be some air resistance, but it should be small. The influence of air resistance can be minimised by ensuring the final velocity is not too high or by using a vacuum chamber.

Question walkthrough

Finding g Graphically from a Pendulum

Describes an experimental method for finding g by measuring the time period of a pendulum for different lengths and plotting T² against length.

When an object moves through a fluid (liquid or gas), such as the air or water, it experiences a drag force due to the fluid.

A diagram showing a blue circle labeled 'Particle' in the center, with curved lines representing flow around it. An arrow indicates the 'Direction of motion' to the right.

Drag is a frictional force that opposes the motion of an object moving through the fluid, slowing the object down. Therefore, drag forces always act in the opposite direction to the object’s motion.

The magnitude of the drag force is dependent on multiple factors. These include:

the speed of the object
the shape of the object
the surface characteristics of the object, e.g. roughness
the density of the fluid
the cross-sectional area of the object.

The following equation of drag force is not required knowledge for your A-level study. However, the formula for drag, for a smooth object in a fluid of uniform density is given below for your understanding:

Where:

is the drag coefficient (dimensionless, depends on the shape of the object)
is the density of the fluid
is the cross-sectional area of the object
is the object’s speed.

The two factors that have the most significant impact on the drag force are the speed and cross-sectional area of the object.

A diagram showing a cylinder filled with fluid, labeled with 'Direction of motion' pointing downwards, 'Fluid' on the right side, and 'Cross-sectional area' in the middle of the cylinder.

The cross-sectional area of an object is the area of the shape formed when the object is cut perpendicular to a specific axis.

In the cylinder example above, the cross-sectional area would be the area of a circle, i.e.

When discussing drag force, the relevant cross-sectional area is the area of the shape formed perpendicular to the direction of motion. This represents the ‘face’ presented to the fluid, which will create the drag force on the object.

The two most important factors affecting the magnitude of the drag force are the object’s speed and the cross-sectional area: a larger cross-sectional area results in a greater drag force.

For most objects moving through a fluid, the drag force is directly proportional to the speed of the object squared, i.e.

A graph showing Drag (N) on the vertical axis and Speed (m s⁻¹) on the horizontal axis. The curve indicates that Drag is proportional to the square of speed, labeled as 'Drag ∝ speed²'. The horizontal lines represent forces 2F and F, with vertical dashed lines at speeds v and 2v.

As the graph shows above, if we double the speed, the drag force quadruples.

When objects move through the air, they experience a drag force known as air resistance.

For example, cars and aeroplanes typically have smooth and streamlined shapes to reduce the amount of air resistance they experience. This allows the vehicle to travel at higher speeds while also reducing the amount of fuel consumed.

An illustration of an airplane with airflow lines indicating the direction of air movement around it.

Air resistance significantly affects the motion of a projectile through the air.

The sketch graph shows the difference in the height and range of a projectile without and with the presence of air resistance.

A graph showing two trajectories: one in red labeled 'Without air resistance' and another in blue labeled 'With air resistance'. The y-axis is labeled 'y' and the x-axis is labeled 'Range'.

The factors affected by air resistance are as follows.

Height: without air resistance, a projectile follows a symmetric parabolic path. However, with air resistance, the vertical velocity component decreases at a greater rate as the projectile rises, reducing the maximum height it can reach.

Range: air resistance slows the projectile throughout its flight, reducing its horizontal velocity. As a result, the projectile covers less horizontal distance before landing, shortening the range compared to motion in a vacuum.

When an object is in free fall through a fluid, its weight remains constant throughout the fall. However, the drag force increases as the object’s speed increases.

At the instant an object begins to fall, the drag force is zero, and the total force acting on the object is due to its weight. The object accelerates at a rate equal to the acceleration due to free fall.

A diagram showing two scenarios of drag and weight. On the left, 'Drag < weight' with 't = 1 s' below the weight. On the right, 'Drag = weight' with 't = 10 s' below the weight. Both scenarios have arrows indicating drag upwards and weight downwards.

The image above shows that as the object falls, the speed increases, and so does the magnitude of the opposing drag force. The resultant force (net force) on the object decreases, and the instantaneous acceleration of the object decreases to less than

Eventually, the object will achieve constant speed due to the force of the weight and the force of the drag becoming equal. This is known as terminal velocity.

When an object is in free fall, its weight and drag force eventually become equal, and the object falls at a constant velocity known as terminal velocity.

The sketch graph below shows a velocity–time graph for an object in free fall through the air.

A graph showing velocity over time. The vertical axis is labeled 'Velocity' and the horizontal axis is labeled 'time'. A dashed line indicates 'terminal velocity'. Points are marked as t0, t1, and t2 along the time axis.

The object’s weight is equal to and the drag force is equal to The instantaneous acceleration of the object is The resultant force changes at each of the three times:

At the only force acting on the object is the weight, so the resultant force is equal to Therefore,

At the resultant force equals the difference between the weight and the drag force. Therefore:

At terminal velocity has been reached, i.e. Therefore, the resultant force is zero and

To investigate the motion of an object falling through a fluid under the influence of a drag force, you can use a motion sensor connected to a data logger or laptop.

In the setup below, a thin string passed over a pulley attaches the falling object to a light polystyrene ball. As the object falls through a liquid cylinder, such as water or glycerol, it pulls the polystyrene ball upwards.

The motion of the polystyrene ball in the air is identical to that of the object falling through the fluid, allowing you to analyse velocity–time and acceleration–time graphs without directly measuring the object’s motion in the liquid.

An illustration showing a setup with a pulley, string, polystyrene ball, motion sensor, object, and fluid. The polystyrene ball is hanging from the pulley via the string, while the object is submerged in the fluid.

Pointing the motion sensor directly at the falling object in the fluid is not practical due to several limitations.

The liquid can distort or scatter the sensor’s signal, resulting in inaccurate measurements.
Small objects may be difficult for the sensor to track reliably, especially in viscous fluids where turbulence or bubbles can interfere.

Monitoring the polystyrene ball in the air eliminates these issues, as the sensor operates more effectively in the air and provides clean, accurate data while still reflecting the object’s motion in the fluid.

An experiment involving paper cones falling through the air also demonstrates how air resistance affects falling objects and helps determine terminal velocity in air.

This can be done by dropping some paper cones from a height above the ground sufficient for the cone to reach terminal velocity (e.g. a few metres). Start a stopwatch at a certain reference point where the cone has reached terminal velocity and stop the timer once the cone hits the ground.

Repeat the process multiple times to calculate an average time to improve the accuracy. The terminal velocity, can then be calculated using the average time, and the distance, the cone fell:

The experiment can be repeated for cones of different surface areas, masses, and shapes to determine how these factors affect terminal velocity.

Factors that affect the terminal velocity of falling objects can be investigated.

Radius/surface area: Larger objects experience a greater drag force due to their increased surface area, which reduces their terminal velocity.

Viscosity: A more viscous fluid exerts a greater drag force, reducing terminal velocity. For example, honey is more viscous than water, so objects fall more slowly in honey than in water.

Mass/density: Heavier or denser objects require greater drag to balance their weight, leading to higher terminal velocities.

Shape/streamlining: Streamlined objects reduce drag, increasing terminal velocity. Non-streamlined objects experience greater drag and lower terminal velocity.

The graph below shows the velocity of a skydiver who deploys a parachute over time.

A graph showing velocity over time. The vertical axis is labeled 'Velocity' and the horizontal axis is labeled 'time'. The curve indicates two terminal velocities: 'Terminal velocity 1 parachute not open' before the point where 'Skydiver opens parachute', and 'Terminal velocity 2 parachute open' after the parachute is deployed. © Medify

The skydiver initially accelerates during free fall and eventually reaches terminal velocity as the force of air resistance balances the skydiver’s weight.
When the parachute is deployed, the skydiver immediately begins to decelerate, and their velocity decreases due to the much greater surface area and a large increase in air resistance.
With the parachute deployed, the skydiver continues decelerating until they reach terminal velocity again. The increased air resistance caused by the parachute enables the skydiver to reach a significantly lower terminal velocity, allowing them to land safely.

Question walkthrough

Finding a Mouse's Terminal Velocity

Derives and calculates terminal velocity from the balance of weight and a velocity-squared drag force (D=0.1v²), then compares the result to a human's much higher terminal velocity.

Newton’s First Law of Motion is the law of inertia. It states that an object at rest or moving with constant velocity will remain so unless acted upon by a net external force.

With no outside forces, a stationary object will not move. v = 0 ms⁻¹. With no outside forces, a moving object will not stop. v = +2 ms⁻¹. Surface.

The key idea is that objects resist change unless a net external force acts upon the object. Resisting changes in motion is called inertia.

An object moving at constant velocity has no overall resultant force.

A diagram showing a block on a surface with the words 'Velocity is constant' pointing to the right, 'Friction' labeled on the left, 'Surface' labeled below, and an arrow labeled 'Pull' at a 30° angle above the block.

In the diagrams above, the overall net force is zero. Let the pull force be and friction be

Note the angular dependence on the pull force. Acting at an angle, only the horizontal pull force component acts in the same direction as velocity and against friction.

The vertical component of the pull force acts in the same direction as the normal force from the surface. These combined forces act equally and opposite to the force due to gravity, resulting in a vertical force of zero.

Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

where:

is the force in ,
is mass in , and
is acceleration in .

The key idea is that greater forces acting on an object produce a greater acceleration. Conversely, a more massive object requires a greater force to achieve the same acceleration as an object with less mass.

Newton’s Third Law of Motion is the law of action and reaction. It states that any action produces an equal and opposite reaction.

The key idea is that forces occur in pairs: the action of exerting a force produces an equal and opposite force. Equal and opposite are used to describe forces that have the same magnitude but are acting in opposite directions:

Forces always come in pairs: you push on a wall, the wall pushes back. F1 = F2 as the wall does not move.

If only considering the magnitude of forces in a pair, and not direction, then:

The net force acting on an object is a single force that acts as the sum of all forces acting on an object. In this way, it describes the net effect of all the forces acting on an object. As each force acting on an object has a magnitude and direction, the net force is the vector addition of all of the forces present.

An illustration showing two sections. The top section is labeled 'All forces' with arrows indicating forces of 50 N to the left and 20 N to the left, resulting in a net force of 25 N to the right. The bottom section is labeled 'Net force' with an arrow indicating a force of 25 N to the left.

The direction and magnitude of the net force determine whether the object will accelerate:

If the forces on an object are balanced (net force is zero), the object will not accelerate – it will either remain stationary or move at a constant velocity.

If the forces are unbalanced (there is a nonzero net force), the object will accelerate in the direction of the net force.

Newton’s second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration :

Where:

is the force measured in newtons (N),
is the mass measured in kilograms (kg), and
is the acceleration measured in metres per second squared (.

The acceleration of an object is directly proportional to the net force acting on it. That means that the greater the force applied, the greater the acceleration for a constant mass.

Question walkthrough

Finding Net Force and Acceleration

Uses vector subtraction to find the net force on a box pulled against friction, then applies F=ma to calculate its acceleration.

The newton is the standard unit for measuring force in physics. One newton is defined as the force required to accelerate of mass at a rate of

If a toy car is to accelerate by , it will require a force of:

The SI unit for force is expressed as . This is because of how force relates to mass and acceleration through Newton’s second law of motion:

Where:

is force measured in newtons (N),
is mass measured in kilograms (kg), and
is acceleration measured in metres per second squared ().

Thus, when you multiply kilograms (kg) by metres per second squared (), the result is the unit of force, the newton (N):

Weight is the result of a gravitational field acting on a mass. As it is a force exerted on an object by gravity, it is given in newtons (N) and is considered a vector quantity. Weight always acts vertically downward, towards the centre of mass of a body.

The weight of an object can be calculated based on Newton’s second law:

Where:

is the weight force of an object (N),
is the mass of the object (kg), and
is gravitational acceleration.

The value of represents the acceleration of free fall or the strength of the gravitational field. This value is (or at sea level on Earth.

An object in free fall is one that is falling solely under the influence of gravity. In the absence of air resistance, all objects experience the same acceleration due to gravity regardless of their mass.

An illustration showing two panels labeled 'In air' and 'In a vacuum' with a hammer and a feather in each. Below, there is a panel labeled 'Experiment' featuring an astronaut standing with a hammer and a feather.

David Scott famously proved this during the Apollo 15 mission to the moon, where a dropped hammer and feather reached the ground at the same time.

Mass vs. weight. In everyday language, someone might say, “I weigh ." However, this is technically incorrect:

Mass is a scalar quantity measured in kilograms. It represents the quantity of matter that an object is made up of.
Weight is a vector quantity. It is a force, measured in newtons, that an object experiences due to its location in a gravitational field.

An illustration showing two figures with the same mass of 80 kg standing on two different celestial bodies. On the left, a figure stands on a gray planet with a weight of 128 N, and on the right, the same figure stands on Earth with a weight of 800 N. The background is blue.

An object’s mass is constant, but its weight varies depending on the strength of the gravitational field in which it is. For instance, the gravitational field strength on the Moon’s surface is , which means an object’s weight is approximately one-sixth what it would be on Earth.

Question walkthrough

Finding g from an Elevator's Scale

Uses the apparent weight reading on a scale in an accelerating elevator to determine the local gravitational acceleration via W = m(g+a).

A free-body diagram is a simplified representation of an object and the forces acting on it. It helps analyse the object’s dynamics by clearly showing all external forces acting on it, their directions and magnitudes.

An illustration showing a cube on a surface with red arrows indicating forces acting on it. The first image shows the cube upright with vertical arrows. The second image shows the cube tilted with diagonal arrows. The third image shows the cube at a different angle with arrows in various directions.

Free-body diagrams are very useful for solving problems and understanding how different forces interact. Only the forces acting on the object are shown, not the forces the object exerts on other objects.

Forces are represented as arrows originating from the object, with:

length indicating the relative magnitude of the force (longer arrow = greater force)
direction indicating the direction in which the force acts.

Each force is labeled with its type and its magnitude if known.

A diagram showing a cube on a surface with arrows indicating forces acting on it. The first image shows the cube resting on a flat surface with vertical arrows indicating forces. The second image shows the cube tilted with arrows indicating forces in different directions. The third image shows the cube at a different angle with arrows indicating forces in various directions.

Free body diagrams help in setting up equations using Newton’s laws of motion to solve for unknown forces or accelerations. They are crucial in problems involving objects on inclined planes, where forces must be resolved into components parallel and perpendicular to the incline.

When adding forces at an angle, always remember to label both the angle of application and the direction of the force with an arrow.

In the example above, the weight of a block on an inclined plane has been resolved into the weight force parallel to the slope, , and the weight perpendicular to the slope, .

If a constant force is applied to an object, the object will undergo a resulting acceleration, which will induce it to move. This motion can be studied in either one or two dimensions, such as along a flat surface or on an inclined plane.

In one-dimensional motion, movement occurs either vertically (up and down) or horizontally (left and right).

In two-dimensional motion, such as on a slope, both vertical and horizontal directions are involved. When dealing with slopes, it is often easier to resolve forces into parallel and perpendicular components rather than horizontal and vertical components.

A diagram showing a block on an inclined plane with forces labeled. The forces include N (normal force), W (weight), T (tension), and arrows indicating 'Perpendicular to slope' and 'Parallel to slope'.

A force acting at an angle to a surface can be split into two components.

Parallel component :

This is the component of the force acting along the surface (in the direction of motion on an incline).

Perpendicular component :

This is the component of the force acting perpendicular to the surface (typically balanced by the normal force in an inclined plane problem).

A diagram showing a block on an inclined plane with labels. The normal force is labeled 'N' in blue, the weight of the block is labeled 'W' in red, and the components of the weight are labeled 'Wcosθ' and 'Wsinθ' in red. The angle of inclination is labeled 'θ' in green.

In the diagram above, the weight of a block on an inclined slope may be resolved into parallel and perpendicular components to the slope.

The normal force and act perpendicular to the slope, while acts parallel to the slope.

Question walkthrough

Finding Normal Force on an Incline

Draws a free body diagram and resolves the weight of a block on a 30° incline into components to calculate the normal force, N = mg cos θ.

Momentum describes how much motion an object has and its resistance to change in velocity:

Where:

is the momentum in ,
is mass in , and
is velocity in .

Momentum is proportional to velocity. Velocity is a vector with a magnitude component (speed) and a direction. Hence, momentum is also a vector with both magnitude and direction.

A diagram comparing Scalar and Vector. Scalar is defined as having magnitude only, while Vector is defined as having magnitude and direction.

Scalar quantities only have magnitude, such as mass and speed, and therefore do not change with direction.

The laws of classical mechanics dictate that momentum is always conserved. This means that the total momentum of all interacting bodies before and after a collision is the same.

The conservation of momentum can be used to calculate the velocity of objects before and after a collision. For instance, two objects colliding as illustrated below:

CONSERVATION OF MOMENTUM. u1 ← m1 P_i = m1u1 m2 P_i = m2u2 u2 → v1 ← m1 P_f = m1v1 m2 P_f = m2v2 v2 → m1u1 + m2u2 = m1v1 + m2v2 © Medify

All forces are vectors. Finding a resultant force can be done by determining the horizontal and vertical components of all forces acting on an object.

Forces that act only in the vertical have no horizontal component, and the same applies to forces that only act along the horizontal; there is no vertical component.

An illustration of a buoy with various forces acting on it. The buoyancy, F_b, is 231 N and acts upwards. The drag, F_r, is 20 N and acts to the left. The pull, F_p, is 40 N and acts at a 20° angle to the right. The weight, F_w, is 245 N and acts downwards. The buoy has a mass of 25 kg.

In the example above, a buoy on a lake is being pulled through the water. The forces with vertical components are weight, buoyancy and the pull force. Overall vertical forces:

This means that forces are balanced in the vertical direction, with no overall vertical force.

Forces acting horizontally are the drag and pull.

Therefore, the overall force in the horizontal direction is .

The net force on an object is defined as the change in momentum over a period of time:

Where:

is the net force in ,
is the change in momentum in and
is the change in time in .

Newton’s Second Law is a specific case of net force where the object’s mass remains constant. However, what if the mass of an object with a net force applied to it changes with time? Then, only the general form is applicable:

Rockets ejecting fuel as they accelerate are a good example of this. The rocket loses mass as it accelerates.

ROCKET AT t with velocity v and mass m on the left, and ROCKET AT t + Δt with velocity v + Δv and mass m - Δm on the right.

At time the rocket has mass , but after a period of time , fuel of mass has been ejected from the rocket, making it less massive. The net force on the rocket has remained constant during this time:

Since the change in momentum remains constant over time, the rocket’s velocity must increase with time as its mass decreases.

Question walkthrough

Finding Mass Using Force and Momentum Change

Calculates a car's mass from driving force and change in velocity over time using F = dp/dt, without first finding acceleration.

Impulse is the effect of a force acting on an object over a given time period :

Where:

is the impulse in ,
is the force in and
is the change in time in .

Impulse can be considered the change in momentum of an object.

A change in momentum is a result of a change in velocity at constant mass:

Where:

is the impulse in ,
is change in momentum in ,
is mass in
and is the change in velocity in .

An excellent illustration of impulse is rain versus hail when standing under an umbrella. Raindrops are smaller and have less mass, and when they hit an umbrella, they have a relatively small change in momentum.

A man in a blue outfit is holding a blue umbrella. On the left side, there are blue raindrops falling from the umbrella, while on the right side, there are blue dots falling from the umbrella.

On the other hand, hail has a larger mass, and when it hits an umbrella, there is a greater change in momentum and, therefore, impulse. Standing in hail with an umbrella, a person will feel significantly more force per second than in rain.

Question walkthrough

Impulse Delivered to a Returning Tennis Ball

Calculate the impulse and direction of the force delivered by a racket to a tennis ball that rebounds at a lower speed.

Impulse can be determined graphically. In a force-time graph, the area under the graph equals the impulse. This is particularly important in instances where force is not constant but varies over time.

A graph showing Force (N) on the vertical axis and Time (s) on the horizontal axis. The area under the curve represents Impulse (Ns).

Calculating the area under the graph can be achieved through geometry or integration.

Geometry: If, for example, a force-time graph is linear, then the area under the graph is a triangle and can be calculated from:

where:

is the area in ,
is the base in , and
is the height in .

A graph showing Force (N) on the vertical axis and Time (s) on the horizontal axis. The left graph depicts a triangular shape representing Impulse (Ns) with a straight vertical line. The right graph shows a bell-shaped curve also representing Impulse (Ns).

Integration: For more complicated graphs, integration can be used to find the area. For this, the equation of the force-time relationship is needed. To calculate the area, the limits of the integral are needed – these are the times and where the graph crosses the x-axis. Then, construct the integral:

which gives the area under the curve.

It is important to note that you you do not require the explicit use of derivatives or integrals to solve problems in your A-level exams. The above has been provided for your holistic understanding.

Force-time graphs can be used in everyday physical scenarios. For instance, they can highlight the importance of seat belts.

A graph showing Force (kN) on the y-axis and Time (ms) on the x-axis. The blue line represents 'Without seat belt' and peaks at a greater force, while the red line represents 'With seat belt' and peaks at a lower force. The graph indicates 'Identical impulse as area under graph is the same' and shows time intervals marked as Δt.

The peak force is significantly larger without the seatbelt, which is more hazardous. The seat belt increases the time over which the force is spread, reducing the peak. However, both graphs have the same area under them, so the impulse is the same.

Question walkthrough

Finding impulse via integration

Finds the roots of a quadratic force–time function f(t)=−2.45t²+12t, then integrates between them to find the impulse delivered during a small collision.

The principle of conservation of momentum states that the total momentum of a closed system remains constant. A closed system is one in which no external forces act upon it.

A system can consist of many objects that interact with each other:

The objects can interact through contact forces, like snooker balls bouncing off each other.
Objects can also interact through non-contact forces, such as the electrostatic repulsion between two electrons or the gravitational attraction between planets.

In order for the principle of conservation of momentum to be obeyed, the internal forces of a closed system must not change the total momentum. This is a consequence of Newton’s laws of motion.

Newton’s third law states that if object A exerts a force on object B, object B will exert an equal and opposite force on object A. For example, if you push against a wall, the wall will push back against you with a force of equal magnitude, so that it remains stationary:

A stick figure pushing against a vertical gray wall, with arrows indicating forces F1 and F2. The equation F1 = F2 is shown above.

Newton’s second law states that the net force is equal to the change in momentum over a time period

Therefore, if the net force equals zero, then the change in momentum equals zero, and momentum is conserved.

The principle of conservation of momentum in a closed system can be stated mathematically as:

Momentum is a vector quantity. Therefore, both the magnitude and direction of the momentum vector are conserved for a closed system.

In one dimension, the direction of the momentum vector does not change after an interaction, since the objects move along one direction. For example, consider the head-on collision of Ball A and Ball B of equal mass :

Before: A (blue circle) moving at 5 m s⁻¹ to the right, B (green circle) moving at 3 m s⁻¹ to the left. After: A (blue circle) moving at velocity v to the left, B (green circle) moving at 5 m s⁻¹ to the right.

The momentum of an object with mass moving at speed is equal to:

Therefore, the principle of conservation of momentum for this collision leads to the expression:

Where is the final speed of Ball A. Cancelling the masses and rearranging gives:

If a collision between two objects occurs at an angle in two-dimensions, their final velocities will be in different directions.

Ball A moves at to the right, and collides at an angle with a stationary Ball B of equal mass; they move off in directions to each other.

Before: A 5 m s⁻¹ A → B. After: A 3 m s⁻¹, B, v.

Since momentum is conserved, the momentum vectors of the balls after the collision must sum to the momentum vector of Ball A before the collision.

The masses are the same, so the velocity vector triangle is the same shape as the momentum vector triangle. The vector triangle is a right-angle triangle, so the final velocity of Ball B can be found from Pythagoras’ theorem:

Therefore:

Conservation of momentum requires that momentum is conserved in any direction.

Before: A, v0, B. After: A, v1, θ1, B, v2, θ2. x, y.

The diagram illustrates a collision between Ball A of mass and initial velocity along the direction, colliding with Ball B of mass The collision results in final velocities and at angles of and to the horizontal.

Conservation of momentum in the direction leads to:

Since there is no initial momentum in the direction, the expression for the direction is:

In a collision, both the total momentum and the total energy are conserved. However, the kinetic energy before the collision can be converted to other forms of energy, such as heat or sound.

Perfectly elastic collisions: both the total momentum and the total kinetic energy are conserved.
- None of the kinetic energy is converted to other forms.

Inelastic collisions: the total momentum is conserved, but the total kinetic energy is not.
- Some of the kinetic energy is converted to other forms.

Question walkthrough

Proving a Collision is Elastic

Elastic collision

Question walkthrough

Classifying a Collision as Elastic or Inelastic

Determining collision type

Work done is defined as the energy transferred when a force moves an object through a distance:

Where:

is the displacement in metres , and
is the average force in the direction of the displacement in newtons .

A diagram showing a force of 1N acting over a distance of 1m, with a green filled rectangle representing work done. The text 'Work done = 1J' is displayed below the rectangles.

Work done is measured in joules, One joule is the work done when a force of one newton moves an object one metre in the direction of the force.

For an object to move at a constant velocity, the net force acting on it must be zero. If an object has an opposing force, such as friction or gravity, it will require a continuous applied force to move at a constant velocity.

Work done = (5 N)(2 m) = 10 J on the left side with a block being lifted with an applied force (tension) of 5 N and a weight of 5 N, and 2 m height. On the right side, Work done = (3 N)(2 m) = 6 J with a block being pushed with an applied force of 3 N and friction of 3 N over a distance of 2 m.

For an object moving at constant velocity, the work done by the applied force is equal to the work done against opposing forces like friction or gravity.

Question walkthrough

Calculating Work Done Lifting an Object

Uses W = mgh to find the work done lifting a 300kg object at constant velocity through a height of 10m.

Question walkthrough

Verifying Work Done via Two Methods

Calculates the work done by a 5N force pushing an object against 3N of friction, first directly via W=Fx, then by summing the separate contributions from acceleration and friction.

The direction of the force and the displacement of an object may not be the same. Generally, work done is given by:

Where:

is displacement,
is the average force, and
is the angle between and

Work done = F x cos Θ. Force pushing box into ground (no work done). F sin Θ, F cos Θ, Force in the direction of motion, Angle between F and x, x.

is the component of the force acting in the direction of motion.

Carefully note the direction of the force when calculating the work done,

The equation W = F x cos Θ is displayed above a diagram. The diagram shows a green rectangle representing an object, with an arrow labeled F indicating a force applied at an angle Θ. The horizontal distance x is marked between the object and another rectangle, with an additional label F cos Θ shown below the force arrow.

Multiply the displacement by the component of the force in the direction of the displacement to get work done.

An illustration showing the equation W = F x, where W represents work, F represents force, and x represents distance. A red arrow labeled F indicates the direction of the force applied at an angle θ, with a green rectangle representing an object being moved along a horizontal line.

Don't

Simply multiply force and displacement without considering their directions.

Question walkthrough

Calculating Work Done at an Angle

Uses W = Fd cos θ to find the work done lifting a box at constant velocity along a displacement inclined at 30° to the horizontal.

Doing work on an object transfers energy to it. For example, lifting a weight increases its gravitational potential energy, which equals the work done in lifting it.

The greater the magnitude of a force or the distance over which it is applied, the more energy is transferred:

Question walkthrough

Finding Height Gained from Energy Transfer

Uses the work done by a piston force to split transferred energy between heating water and lifting a platform and scooter, then finds the height gained using W = mgh.

Power is defined as the rate of doing work. In other words, power is the amount of work done per unit of time:

Where:

is power in watts ,
is work in joules , and
is time in seconds .

Another common unit of power is the joules per second

Differentiating between energy, work, and power can be challenging. The table below compares the three quantities and provides an example of each.

A table comparing Quantity, Energy, Work, and Power. Under Energy: Definition - An object or system’s capacity to do work; Nature - Fundamental quantity that exists in various forms. It can be transferred or transformed, but not created or destroyed; Formula - Various; Unit - Joule (J); Example - A bird has a kinetic energy of 50J. Under Work: Definition - Energy transferred due to an applied force; Nature - Depends on the applied force on an object over a distance; Formula - W = Fd; Unit - Joule (J); Example - 2000 J of work is done on the crate across the floor. Under Power: Definition - Rate of energy transfer or work done; Nature - Measure of how fast energy is transferred or work is done; Formula - P = W/t; Unit - Watt (W); Example - A 20 W lightbulb converts energy at a rate of 20 joules per second.

Question walkthrough

Calculating a Lift's Work and Power

Uses a lift's constant driving force and floor height to find distance travelled, then applies W = Fd and P = W/t to calculate work done and power output.

Power is the rate of doing work :

When a constant force moves an object at velocity in the same direction as the force, this simplifies to:

Where:

is the force in newtons , and
is velocity in metres per second .

This is one of the most widely used mechanical equations for engines, vehicles, and motors.

Question walkthrough

Finding Engine Power from Resistive Forces

Applies Newton's first law at constant velocity to equate driving force with resistive force, then uses P = Fv to calculate a car engine's power output in watts and kilowatts.

Efficiency is a term used to describe how well energy is converted in a mechanical system:

A system with high efficiency converts most of the input energy into useful output energy.
A system with low efficiency converts most of the input energy into wasted output energy.

High efficiency system: Total energy in leads to Useful energy out and Wasted energy out. Low efficiency system: Total energy in leads to Wasted energy out and Useful energy out.

Efficiency can be expressed as a decimal between 0 and 1 or as a percentage between 0% and 100%. A system with an efficiency of zero wastes all the input energy, and an efficiency of one (100%) is considered ideal. It converts all the input energy into useful energy. In reality, a system can never be 100% efficient.

Mathematically, the efficiency of a mechanical system is defined as the ratio of the useful output energy to the input energy:

To get the efficiency in percentage form, we apply the ratio below:

It is important to note that effiency can be anything between zero and one or 0% and 100%.

The efficiency can also be described in terms of power. Efficiency can also be defined as the ratio of the output power to the input power. In decimal form, the efficiency is:

And in percentage form, the efficiency is:

Question walkthrough

Calculating Elevator Motor Power and Efficiency

Calculates an elevator motor's output and input energy, its input and output power, and verifies the result matches the stated efficiency.

The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred or transformed from one form to another.

A diagram showing a rigid support at the top with three red circles labeled A, B, and C. Circle A is the centre point with K.E. = max. and P.E. = 0. Circles B and C are extreme points with K.E. = 0 and P.E. = max.

For example, for an ideal pendulum, gravitational potential energy is converted to kinetic energy and vice versa during each swing, but the total energy stays the same.

The principle of conservation of energy is best illustrated in an isolated system.

Isolated system: energy or matter cannot be exchanged with the surroundings. In an isolated system, the total energy remains constant.
Closed system: matter cannot be exchanged with the surroundings, but energy can be.
Open system: energy and matter can be exchanged with the surroundings.

Wasted energy is simply defined as energy that is not useful. Wasted forms of energy typically include heat, light and sound.

A diagram illustrating the flow of energy. It shows Electric energy leading to Useful kinetic energy, with arrows pointing to Wasted heat energy and Wasted sound energy.

For instance, in any real mechanical system, useful energy output is always less than the total energy input because friction between a machine’s moving parts generates heat and sound.

Question walkthrough

Finding Work Done Against Friction

Uses the work-energy principle to find the energy lost to friction when a force accelerates an object over a known distance, then finds the corresponding average frictional force.

The exchange between kinetic energy and gravitational potential energy occurs for an object moving in a gravitational field. Examples include:

An object that falls from the roof of a building loses gravitational potential energy and gains kinetic energy.
A swing at its highest position, moving towards its lowest position, loses gravitational potential energy while gaining kinetic energy.
When a ball is thrown up with an initial velocity, it loses kinetic energy as it ascends and gains gravitational potential energy.

For an object moving in a gravitational field, in the absence of frictional forces, the gain in one form of energy is equal to the loss in the other:

A tall building with multiple windows is depicted. There are two red points labeled A and B, with point A located at the top of the building and point B at the bottom. A dashed line connects point A to the top of the building.

If a ball at A possesses of gravitational potential energy and falls, it loses of by the time it reaches B. This loss in is converted to kinetic energy. Therefore, the ball possesses of when it reaches the ground.

Although systems in a gravitational field may look different, the way that their energy changes between kinetic and gravitational in the absence of frictional forces is the same.

General statements can be made about objects in a gravitational field released from a raised level or projected upwards from a lower level:

At maximum height, an object’s kinetic energy is minimal, and its gravitational potential energy is at its maximum.
At ground level, an object’s gravitational potential energy is minimal, and its kinetic energy is at its maximum.
At intermediate heights, kinetic energy and gravitational potential energy will vary, but their sum remains constant, which can be written as:

This is a statement of the law of energy conservation.

Question walkthrough

Energy Conservation on a Frictionless Slope

Uses GPE = mgh and KE = ½mv² to track energy conversion for a frictionless skier's descent, then finds the resulting speed at the bottom of the slope.

Near the Earth’s surface, gravitational potential energy can be approximated using:

Where:

is the mass of the object ()
is the gravitational field strength
is the height above a reference level often taken as the ground ().

The approximation is valid because Earth’s gravitational field is nearly uniform near its surface. This means remains approximately constant.

The gravitational potential energy is defined as zero at Earth’s surface. As an object is lifted, work is done against gravity, and its gravitational potential energy increases.

It is useful to note that the choice of zero at the surface is a convention that simplifies calculations when dealing with heights relative to the ground.

The gravitational field strength can be approximated as constant () near the Earth’s surface. The gravitational field strength on the surface of other celestial bodies can generally be considered constant, too:

However, the simplified equation above only applies only near the Earth’s surface because at greater distances, the field is no longer uniform but radial, and varies with distance.

Left side: Uniform gravitational field with arrows pointing downwards, labeled 'Surface of body' and equations 'ΔE = mgΔh' and 'ΔE = GMm(1/r1 - 1/r2)' with a red cross over the second equation. Right side: Radial gravitational field with arrows pointing outwards from a globe, labeled 'Far from body' and equations 'ΔE = ngΔh' and 'ΔE = GMm(1/r1 - 1/r2)' with a red cross over the first equation.

Use the following formula for cases where the object is far from the massive body and the gravitational field is radial:

On this page

Scalars and vectors (3.4.1.1)

Question walkthrough

Triangle of Forces for a Box in Equilibrium

Moments (3.4.1.2)

Question walkthrough

Moment of an Angled Force on a Rod

Question walkthrough

Identifying a Couple from Force Pairs

Question walkthrough

Balancing Moments on a Pivoted Beam

Question walkthrough

Proving a Beam is in Equilibrium

Motion along a straight line: Kinematics (3.4.1.3)

Question walkthrough

Distance vs Displacement on a Track

Question walkthrough

Finding Displacement from a Velocity-Time Graph

Projectile motion: Horizontal and vertical components (3.4.1.4)

Question walkthrough

Resolving Velocity into Horizontal and Vertical Components

Question walkthrough

Time of Flight for Same-Height Projectile Launch

Question walkthrough

Maximum Height of a Projectile at an Angle

Question walkthrough

Time of Flight from Projectile Range

Question walkthrough

Time, Height and Range for an Elevated Launch

Motion along a straight line: Equations of motion (3.4.1.3)

Question walkthrough

Maximum Height of a Ball Thrown Upwards

Question walkthrough

Time of Flight for a Ball Thrown Upwards

Question walkthrough

Finding Acceleration Graphically from s and t²

Question walkthrough

Finding g Graphically from a Pendulum

Projectile motion: Friction, lift and drag forces (3.4.1.4)

Question walkthrough

Finding a Mouse's Terminal Velocity

Newton’s laws of motion (3.4.1.5)

Question walkthrough

Finding Net Force and Acceleration

Question walkthrough

Finding g from an Elevator's Scale

Question walkthrough

Finding Normal Force on an Incline

Momentum: Linear momentum and impulse (3.4.1.6)

Question walkthrough

Finding Mass Using Force and Momentum Change

Question walkthrough

Impulse Delivered to a Returning Tennis Ball

Question walkthrough

Finding impulse via integration

Momentum: Collisions (3.4.1.6)

Question walkthrough

Proving a Collision is Elastic

Question walkthrough

Classifying a Collision as Elastic or Inelastic

Work, energy and power (3.4.1.7)

Question walkthrough

Calculating Work Done Lifting an Object

Question walkthrough

Verifying Work Done via Two Methods

Question walkthrough

Calculating Work Done at an Angle

Question walkthrough

Finding Height Gained from Energy Transfer

Question walkthrough

Calculating a Lift's Work and Power

Question walkthrough

Finding Engine Power from Resistive Forces

Question walkthrough

Calculating Elevator Motor Power and Efficiency

Conservation of energy (3.4.1.8)

Question walkthrough

Finding Work Done Against Friction

Question walkthrough

Energy Conservation on a Frictionless Slope