Cambridge IGCSE0972

Motion

Physics 0972 Chapter Notes

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Motion
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In this chapter

  1. 1Speed, Velocity, and Distance
  2. 2Understanding Acceleration
  3. 3Analysing Distance-Time Graphs
  4. 4Mastering Speed-Time Graphs
  5. 5Acceleration Due to Gravity
  6. 6Momentum and Impulse
  7. 7Conservation of Momentum

1. Speed, Velocity, and Distance

In physics, we describe how things move. 'Distance' is the total length of the path travelled, for example, 200 metres. 'Speed' is how fast you cover that distance. If you run 200 metres in 25 seconds, your speed is 200/25 = 8 m/s. These are 'scalar' quantities, as they only have a size (magnitude). However, sometimes direction matters. 'Displacement' is the distance in a straight line from the start point to the end point, including the direction. 'Velocity' is the speed in a specific direction. So, 10 m/s is a speed, but 10 m/s North is a velocity. These are 'vector' quantities.

speed = distance / time

average velocity = displacement / time

Key term

Vector vs Scalar: A scalar quantity has only magnitude (e.g., speed, distance, mass), while a vector quantity has both magnitude and direction (e.g., velocity, displacement, force).

Examiner insight

Examiners often test the distinction between speed and velocity. Be prepared to explain the difference and use the correct terms in your answers.

Common pitfall

Confusing distance with displacement. For a journey that ends at the starting point (like a lap of a track), the distance is the length of the lap, but the final displacement is zero.

Worked example 14 marks

A runner jogs 400 m North, then turns and jogs 300 m East. The entire run takes 100 seconds. Calculate:(a) the total distance travelled,(b) the average speed,(c) the magnitude of the final displacement, and(d) the magnitude of the average velocity.

  1. 1

    a) Total distance is the sum of the individual paths: Distance = 400 m + 300 m = 700 m.

  2. 2

    b) Average speed = total distance / total time. Speed = 700 m / 100 s = 7.0 m/s.

  3. 3

    c) The displacement is the straight line from start to finish. The paths North and East form a right-angled triangle. Using Pythagoras' theorem: Displacement² = 400² + 300² = 160000 + 90000 = 250000. Displacement = √250000 = 500 m.

  4. 4

    d) Average velocity = displacement / time. Velocity = 500 m / 100 s = 5.0 m/s.

Recap

  • Distance is how far an object has moved, a scalar quantity.
  • Speed is the rate of change of distance, also a scalar.
  • Displacement is the object's change in position, a vector quantity.
  • Velocity is the rate of change of displacement, also a vector.
  • The formula linking speed, distance, and time is fundamental.

Quick check

  1. A car travels once around a 2 km circular track. What is its final displacement?1 mark
  2. Is temperature a scalar or a vector quantity?1 mark

2. Understanding Acceleration

Acceleration is the measure of how quickly an object's velocity changes. It's not just about speeding up! Since velocity includes direction, you are accelerating if you speed up, slow down, or change direction. Speeding up is positive acceleration. Slowing down is called deceleration, or negative acceleration. Acceleration is measured in metres per second squared (m/s²). An acceleration of 2 m/s² means that for every second that passes, the object's velocity increases by 2 m/s.

acceleration (a) = (final velocity (v) - initial velocity (u)) / time taken (t)

a = (v - u) / t

Key term

Acceleration: The rate of change of velocity, measured in metres per second squared (m/s²).

Examiner insight

Always show your formula and substitution. Marks are often given for the correct formula and correct substitution, even if the final calculation is wrong. Don't forget the units (m/s²).

Common pitfall

Forgetting that 'at rest' or 'starts from stationary' means the initial velocity (u) is 0 m/s, and 'comes to a stop' means the final velocity (v) is 0 m/s.

Worked example 12 marks

A sprinter starts from rest and reaches a top speed of 10 m/s in 2.5 seconds. What is her average acceleration?

  1. 1

    Identify the given values: Initial velocity(u) = 0 m/s (from rest), Final velocity(v) = 10 m/s, Time(t) = 2.5 s.

  2. 2

    Use the formula for acceleration: a = (v -u) / t.

  3. 3

    Substitute the values: a = (10 - 0) / 2.5.

  4. 4

    Calculate the result: a = 4 m/s².

Worked example 22 marks

A car travelling at 20 m/s applies its brakes and comes to a stop in 4 seconds. Calculate its deceleration.

  1. 1

    Identify the given values: Initial velocity(u) = 20 m/s, Final velocity(v) = 0 m/s (comes to a stop), Time(t) = 4 s.

  2. 2

    Use the formula for acceleration: a = (v -u) / t.

  3. 3

    Substitute the values: a = (0 - 20) / 4.

  4. 4

    Calculate the result: a = -5 m/s². The negative sign indicates deceleration. The deceleration is 5 m/s².

Recap

  • Acceleration is the rate of change of velocity.
  • The unit for acceleration is m/s².
  • Deceleration is negative acceleration, meaning the object is slowing down.
  • An object moving at a constant velocity has zero acceleration.
  • A change in direction at constant speed is also an acceleration.

Quick check

  1. An object's velocity changes from 15 m/s to 5 m/s in 2 seconds. What is its acceleration?1 mark

3. Analysing Distance-Time Graphs

A distance-time graph is a simple way to visualise an object's journey. Time is always plotted on the horizontal(x) axis, and the distance from the starting point is on the vertical(y) axis. The shape of the line tells you about the object's speed. A horizontal line means the distance isn't changing, so the object is stationary. A straight, sloped line shows the object is moving at a constant speed. A steeper line means a higher speed. A curved line indicates that the speed is changing, which means the object is accelerating or decelerating.

Speed = Gradient of the distance-time graph

Gradient = Change in vertical axis / Change in horizontal axis = Δd / Δt

Key term

Gradient: On a distance-time graph, the gradient (steepness) of the line represents the speed of the object.

Common pitfall

Confusing a distance-time graph with a speed-time graph. A horizontal line on a distance-time graph means zero speed, whereas on a speed-time graph it means constant speed.

Worked example 13 marks

The graph shows the journey of a cyclist.(a) What is the total distance travelled?(b) Describe the motion between 20 s and 40 s.(c) Calculate the speed during the first 20 seconds.

  1. 1

    a) The highest point on the distance axis is 300 m. The cyclist travels 300 m away and then 300 m back. Total distance = 300 m + 300 m = 600 m. (Note: The question is often interpreted as distance from start, which would be 300m. Be careful with wording. Let's assume it means the furthest distance from start). The furthest distance from the start is 300 m.

  2. 2

    b) Between 20 s and 40 s, the distance from the start is constant at 300 m. The line is horizontal, so the cyclist is stationary.

  3. 3

    c) Speed is the gradient of the first section. Gradient = rise / run = (300 m - 0m) / (20 s - 0 s). Speed = 300 / 20 = 15 m/s.

Recap

  • On a distance-time graph, the y-axis shows distance from the start.
  • The x-axis shows the time elapsed.
  • A horizontal line means the object is stationary (speed = 0).
  • A straight, sloped line means constant speed.
  • The gradient of the line is equal to the speed.
  • A curve shows acceleration or deceleration.

Quick check

  1. On a distance-time graph, what does a line sloping downwards represent?1 mark

4. Mastering Speed-Time Graphs

Speed-time graphs (or velocity-time graphs) are powerful tools. They show how an object's speed changes over time. Time is on the x-axis and speed is on the y-axis. A horizontal line means constant speed (zero acceleration). A straight, sloped line shows constant acceleration. A line sloping upwards means positive acceleration, while a line sloping downwards means negative acceleration (deceleration). The gradient of the line tells you the acceleration, and the area under the entire graph tells you the total distance travelled.

Acceleration = Gradient of the speed-time graph = Δv / Δt

Distance travelled = Area under the speed-time graph

Key term

Area under the graph: On a speed-time graph, the area under the line represents the total distance travelled by the object.

Examiner insight

Examiners frequently ask for the total distance from a speed-time graph. Show your working clearly by splitting the area into shapes (triangles, rectangles) and calculating each one separately before adding them up.

Fun fact

The shape of the area under a speed-time graph is called a trapezoid or trapezium. You can calculate the total distance in the worked example in one step using the formula for the area of a trapezium: Area = 1/2 × (a+b) × h = 1/2 × (20+50) × 20 = 700 m. (Here 'a' and 'b' are the parallel sides).

Worked example 14 marks

A car's journey is shown on the speed-time graph. Calculate:(a) the acceleration during the first 10 seconds, and(b) the total distance travelled in 50 seconds.

  1. 1

    a) Acceleration is the gradient of the first section. a = rise / run = (20 m/s - 0 m/s) / (10 s - 0 s). a = 20 / 10 = 2 m/s².

  2. 2

    b) Total distance is the area under the graph. We can split this into a triangle (0-10s), a rectangle (10-30s), and another triangle (30-50s).

  3. 3

    Area of first triangle = 1/2 × base × height = 1/2 × 10 × 20 = 100 m.

  4. 4

    Area of rectangle = base × height = (30 - 10) × 20 = 20 × 20 = 400 m.

  5. 5

    Area of second triangle = 1/2 × base × height = 1/2 × (50 - 30) × 20 = 1/2 × 20 × 20 = 200 m.

  6. 6

    Total distance = 100 m + 400 m + 200 m = 700 m.

Recap

  • On a speed-time graph, the gradient is the acceleration.
  • The area under the line is the distance travelled.
  • A horizontal line means constant speed (zero acceleration).
  • A straight, sloped line means constant acceleration.
  • An upward slope is acceleration, a downward slope is deceleration.

Quick check

  1. On a speed-time graph, what does a horizontal line at 10 m/s mean?1 mark
  2. How would you find the deceleration from a speed-time graph?1 mark

5. Acceleration Due to Gravity

When you drop an object, it accelerates downwards because of the force of gravity. Near the Earth's surface, all objects fall with the same acceleration if we ignore air resistance. This is called the 'acceleration of free fall', and its symbol is 'g'. For IGCSE, you can usually approximate its value as g = 10 m/s². This means for every second an object falls, its downward velocity increases by 10 m/s. In reality, air resistance opposes the motion. As an object falls faster, air resistance increases until it balances the force of gravity. At this point, the net force is zero, the object stops accelerating and falls at a constant 'terminal velocity'.

Weight (W) = mass (m) × gravitational field strength (g)

For free fall (no air resistance): a = g ≈ 10 m/s²

Key term

Terminal Velocity: The constant speed that a freely falling object eventually reaches when the force of air resistance equals the force of gravity.

Fun fact

On the Moon, where there is virtually no atmosphere and thus no air resistance, astronaut David Scott famously dropped a hammer and a feather in 1971. They both hit the ground at the exact same time, proving Galileo's theory from centuries earlier.

Worked example 14 marks

A stone is dropped from a high cliff. Ignoring air resistance, calculate:(a) its speed after 3 seconds, and(b) the distance it has fallen in that time. (Use g = 10 m/s²).

  1. 1

    a) We can use the equation v = u + at. Here, u = 0 (dropped), a = g = 10 m/s², and t = 3 s. So, v = 0 + (10 × 3) = 30 m/s.

  2. 2

    b) We can find the distance from the area under a speed-time graph. The graph would be a straight line from (0,0) to (3,30). The area is a triangle. Distance = Area = 1/2 × base × height = 1/2 × 3 s × 30 m/s = 45 m.

Recap

  • The acceleration of free fall (g) is approximately 10 m/s² near the Earth.
  • In the absence of air resistance, all objects fall with the same acceleration.
  • Air resistance is a frictional force that opposes motion through the air.
  • Terminal velocity is reached when the force of gravity equals the force of air resistance.
  • A skydiver has a high terminal velocity, but opening a parachute increases air resistance and reduces it to a safe level.

Quick check

  1. What is the approximate value of 'g' on Earth?1 mark
  2. What is the acceleration of an object falling at its terminal velocity?1 mark

6. Momentum and Impulse

Momentum is a measure of an object's 'quantity of motion'. A fast-moving car has more momentum than a slow-moving one, and a heavy truck has more momentum than a light car moving at the same speed. Momentum is a vector quantity, so its direction is important. It is calculated by multiplying an object's mass by its velocity. A change in momentum is called an 'impulse'. According to Newton's second law, the force acting on an object is equal to the rate of change of its momentum. This means a large force can cause a rapid change in momentum. This is why car airbags work: they increase the time taken for your head's momentum to become zero, which reduces the force on your head.

momentum (p) = mass (m) × velocity (v)

Force (F) = change in momentum (Δp) / time (t)

Key term

Momentum: A vector quantity that is the product of an object's mass and velocity (p = mv).

Examiner insight

When calculating a change in momentum, be careful with directions. If an object bounces, its final velocity is in the opposite direction to its initial velocity, so you must use a negative sign for one of them.

Worked example 14 marks

A 1200 kg car is travelling at 20 m/s.(a) Calculate its momentum.(b) The driver applies the brakes, and the car stops in 5 seconds. Calculate the average braking force.

  1. 1

    a) Use the formula p = m × v. p = 1200 kg × 20 m/s = 24000 kg m/s.

  2. 2

    b) First, find the change in momentum (Δp). Initial momentum = 24000 kg m/s. Final momentum = 1200 kg × 0 m/s = 0 kg m/s. Δp = final p - initial p = 0 - 24000 = -24000 kg m/s.

  3. 3

    Now use the force formula: F = Δp / t. F = -24000 kg m/s / 5 s = -4800 N.

  4. 4

    The negative sign indicates the force is in the opposite direction to the initial motion. The average braking force is 4800 N.

Recap

  • Momentum is a product of mass and velocity.
  • The unit of momentum is the kilogram-metre per second (kg m/s).
  • Momentum is a vector; direction is crucial.
  • Force is the rate of change of momentum.
  • To reduce the force of an impact, you can increase the time over which the momentum changes.

Quick check

  1. What is the momentum of a 5 kg object that is stationary?1 mark
  2. A 2 kg ball moving at 3 m/s has its direction reversed to 3 m/s. What is the change in momentum?2 marks

7. Conservation of Momentum

The principle of conservation of momentum is a very important rule in physics. It states that in any collision or explosion, as long as no external forces are acting on the system, the total momentum before the event is equal to the total momentum after the event. Momentum isn't lost; it's just transferred between objects. This principle applies to snooker balls colliding, a rocket launching, or two ice skaters pushing off from each other. When solving problems, you calculate the total momentum of all objects before the collision and set it equal to the total momentum of all objects after.

Total momentum before = Total momentum after

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Key term

Conservation of Momentum: For a system of interacting objects in the absence of external forces, the total momentum remains constant.

Common pitfall

Forgetting to assign a negative sign to a velocity that is in the opposite direction. If you define 'right' as positive, any velocity to the 'left' must be negative.

Worked example 13 marks

A 2 kg trolley moving at 3 m/s to the right collides with a stationary 1 kg trolley. After the collision, they stick together. What is their velocity after the collision?

  1. 1

    State the principle: Total momentum before = Total momentum after.

  2. 2

    Calculate total momentum before: p_before = (m₁u₁) + (m₂u₂) = (2 kg × 3 m/s) + (1 kg × 0 m/s) = 6 kg m/s.

  3. 3

    Define momentum after: The trolleys stick together, so they have a combined mass (m₁ + m₂) and a common velocity (v). p_after = (m₁ + m₂) × v = (2 kg + 1 kg) × v = 3v.

  4. 4

    Equate before and after: 6 kg m/s = 3v.

  5. 5

    Solve for v: v = 6 / 3 = 2 m/s. Their velocity is 2 m/s to the right.

Worked example 23 marks

A 50 kg girl and a 70 kg boy are on ice skates, initially at rest. The girl pushes the boy, who moves away at 1.5 m/s. What is the girl's recoil velocity?

  1. 1

    State the principle: Total momentum before = Total momentum after.

  2. 2

    Calculate total momentum before: Since they are at rest, p_before = 0.

  3. 3

    Define momentum after: Let the boy's direction be positive. p_after = (m_boy × v_boy) + (m_girl × v_girl) = (70 kg × 1.5 m/s) + (50 kg × v_girl).

  4. 4

    Equate before and after: 0 = (70 × 1.5) + (50 × v_girl).

  5. 5

    Solve for v_girl: 0 = 105 + 50v_girl. -105 = 50v_girl. v_girl = -105 / 50 = -2.1 m/s.

  6. 6

    The girl's velocity is 2.1 m/s in the opposite direction to the boy.

Recap

  • In a closed system, total momentum is always conserved.
  • Total momentum before an event equals total momentum after the event.
  • This principle applies to both collisions and explosions.
  • Remember to assign positive and negative signs to velocities to indicate direction.
  • If objects stick together after a collision, their masses add up.

Quick check

  1. Two objects of equal mass collide head-on with equal and opposite velocities. If they stick together, what is their final velocity?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A cheetah can run at 32 m/s. If it maintains this speed, how far will it travel in 15 seconds?2 marks
  2. A train accelerates uniformly from 4 m/s to 20 m/s in 80 seconds. Calculate its acceleration.2 marks
  3. Describe the motion of an object represented by a straight horizontal line on (a) a distance-time graph, and (b) a speed-time graph.2 marks
  4. A cyclist accelerates from rest to 8 m/s in 4 s, travels at a constant speed for 10 s, and then brakes to a stop in 2 s. Draw a speed-time graph for the journey and use it to find the total distance travelled.5 marks
  5. A 2 kg ball is dropped from a tall building. Ignoring air resistance, what is its velocity after 2.5 s? (Use g = 10 m/s²).2 marks
  6. A 0.05 kg tennis ball hits a wall at 30 m/s and bounces back at 20 m/s. Calculate the change in momentum (impulse) of the ball.3 marks
  7. Explain, in terms of momentum, how a safety helmet protects a cyclist's head in a crash.3 marks
  8. A 4 kg cannonball is fired from a 500 kg cannon. The cannonball moves forward with a velocity of 80 m/s. Calculate the recoil velocity of the cannon.3 marks
  9. A speed-time graph shows a straight line passing through the points (t=2s, v=5m/s) and (t=12s, v=15m/s). Calculate (a) the acceleration, and (b) the distance travelled between t=2s and t=12s.4 marks
  10. A 3 kg trolley moving at 4 m/s collides with a 1 kg trolley moving at 2 m/s in the same direction. They stick together. Calculate their common velocity after the collision.3 marks

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