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Edexcel International GCSE Physics · Topic Test
Forces and Motion
FREE Online Physics GCSE Course35minutes
40marks
7questions
How it works
- Answer each question. Type into the boxes or pick an option, and use paper for working and drawings.
- Press Check my answers at the end of the question. Tick-box and number answers are marked for you.
- For written answers, compare yours with the mark scheme and tick each point you made. We highlight the key words we spot, but you decide, just like an examiner.
You will need: paper, a pen, a pencil, a ruler, a protractor and a calculator. Use g = 10 N/kg. Your answers save automatically in this browser.
1
Question 1
6 marksFigure 1 shows a skier using a drag lift. The drag lift pulls the skier from the bottom to the top of a ski slope.
The arrows, A, B, C and D, represent the forces acting on the skier and her skis.
Figure 1
(a)
Which arrow represents the force pulling the skier up the slope?
(1 mark)
Mark scheme
Answer: D
(b)
Which arrow represents the normal contact force?
(1 mark)
Mark scheme
Answer: C
(c)
What is the unit for momentum?
(1 mark)
Mark scheme
Answer: B kg m/s
(d)
In the formula F = m a, what is the best description of F?
(1 mark)
Mark scheme
Answer: B The resultant force acting on an object
(e)
Near the surface of the Earth, any object falling freely under gravity (when the only force acting is the object’s weight) has an acceleration of:
(1 mark)
Mark scheme
Answer: A 10 m/s²
(f)
A child drops a small ball which hits the ground at 10 m/s. The air resistance is negligible. What height was the ball dropped from, to 1 significant figure? (u = 0 m/s, v = 10 m/s, g = 10 m/s²)
(1 mark)
Mark scheme
v² = u² + 2 g h: 100 = 0 + 2 × 10 × h, so h = 5 m
Answer: A Roughly 5 m
2
Question 2
5 marksThe figure below shows a child who is balancing on a pogo stick. The child and pogo stick are not moving. The child weighs 350 N.
Figure 2 shows how the compression of the spring changes with the force on it.
The pogo stick
Figure 2 · the pogo stick’s spring
(a)
State Hooke’s law.
(1 mark)
Mark scheme
Tick each marking point your answer makes:
⚠️ “The extension increases as the force increases” isn’t enough on its own. Say they are directly proportional.
(b)
Explain how Figure 2 shows that the spring obeys Hooke’s law.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
(c)
Use Figure 2 to find the compression of the spring when the child stands on the pogo stick.
cm
(1 mark)
Mark scheme
Compression: 7.0 cm (accept 6.8 – 7.2 cm)
💡 Right value, but the graph is in cm: give it in cm.
💡 Read across from 350 N on the force axis, up to the line, then across to the compression.
(d)
When no force acts on it, the spring is 30 cm long. Calculate its length when the child stands on the pogo stick.
cm
(1 mark)
Mark scheme
Length: 30 − 7.0 = 23 cm (error carried forward from (c))
💡 The spring is compressed, so it gets shorter: 30 − 7.
3
Question 3
8 marksA train travels from town A to town B. Figure 1 shows the route taken by the train. Figure 1 has been drawn to scale.
Figure 1 · drawn to scale
(a)
The distance the train travels between A and B is not the same as the displacement of the train. What is the difference between distance and displacement?
(1 mark)
Mark scheme
Tick each marking point your answer makes:
(b)
Use Figure 1 to determine the displacement of the train in travelling from A to B. Show how you obtain your answer.
On a screen, Figure 1 may not be shown at its true size. Measure the 1 cm scale bar with your ruler first, then scale up your measurement from A to B.
km
(2 marks)
Mark scheme
Displacement: 37.5 km (accept 36.0 – 38.0 km)
Direction: 062° or N62°E (accept 60° – 64° clockwise from north)
(c)
There are places on the journey where the train accelerates without changing speed. Explain how this can happen.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
(d)
Figure 2 shows how the velocity of the train changes with time as the train travels along a straight section of the journey.
Figure 2
Estimate the distance travelled by the train along the section of the journey shown in Figure 2. Show how you worked out your answer.
m
(3 marks)
Mark scheme
Distance: about 8500 m (17 squares × 500 m; accept 16 – 18 squares)
💡 That looks like a number of squares. What distance does one square represent? (100 s × 5 m/s)
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
4
Question 4
Physics only4 marksTwo children, A and B, sit on a seesaw. Child A weighs 300 N and sits 2.0 m from the pivot.
The seesaw (not to scale)
(a)
State the condition for there to be no turning effect on the seesaw, so that it balances.
(1 mark)
Mark scheme
Tick each marking point your answer makes:
⚠️ It’s the moments that must balance, not the forces: a lighter child further out can balance a heavier one.
(b)
Child B weighs 400 N. Calculate how far from the pivot child B must sit for the seesaw to balance.
m
(3 marks)
Mark scheme
Distance: 1.5 m
💡 Upside down. 400 × d = 300 × 2.0, so d = 600 ÷ 400.
💡 600 N m is the moment of child A. Now divide by B’s weight: 600 ÷ 400.
💡 B is heavier, so B must sit closer to the pivot than A.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
5
Question 5
5 marksThe graph shows changes in the velocity of a racing car.
Velocity–time graph
(a)
Describe the motion of the racing car during the period labelled W.
(1 mark)
Mark scheme
Tick each marking point your answer makes:
⚠️ “Acceleration increases” does not score: W is a straight line, so the acceleration is constant. The car is speeding up.
(b)
Calculate the acceleration of the racing car during the period labelled X. Show clearly how you work out your answer and give the unit.
(4 marks)
Mark scheme
Acceleration: 10
💡 Use only period X: from 2 s (10 m/s) to 6 s (50 m/s).
💡 The change in velocity is 50 − 10 = 40 m/s, not 50 m/s.
💡 X lasts from 2 s to 6 s, which is 4 s.
Unit: m/s² or m s⁻² (reject m s²)
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
6
Question 6
6 marks(a)
State one factor, apart from speed, which would increase a car’s braking distance.
(1 mark)
Mark scheme
Any one of these scores (max 1):
⚠️ Tiredness, alcohol, drugs and distractions affect thinking distance, not braking distance, so they don’t score here.
(b)
The diagram shows a 1000 kg car before and during a crash test. The car hits the wall at 14 metres per second (m/s) and takes 0.25 seconds (s) to stop.
Crash test
In an accident the crumple zone at the front of a car collapses progressively. This increases the time it takes the car to stop. In a front-end collision the injury to the car passengers should be reduced. Explain why.
By increasing the time it takes for the car to stop, the …
(2 marks)
Mark scheme
Tick each marking point your answer makes:
(c)
Calculate the impact force on the car.
N
(3 marks)
Mark scheme
Impact force: 56 000 N
💡 14 000 kg m/s is the change in momentum. Now divide by the time: F = change in momentum ÷ t.
💡 You multiplied by 0.25 s. Divide by the time instead.
💡 56 m/s² is the deceleration. Multiply by the mass: F = m a.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
7
Question 7
6 marksThe diagram shows the forces acting on a parachutist in free fall.
The graph shows how the vertical velocity of a parachutist changes from the moment the parachutist jumps from the aircraft until landing on the ground.
Free fall
Velocity–time graph
In this question you will be assessed on using good English, organising information clearly and using specialist terms where appropriate.
Using the idea of forces, explain why the parachutist reaches a terminal velocity and why opening the parachute reduces the terminal velocity.
(6 marks)
Mark scheme
Why they reach terminal velocity: tick each point you made
Why opening the parachute lowers it: tick each point you made
| 0 | No relevant content. | 0 |
|---|---|---|
| Level 1 | A brief attempt to explain why the velocity / speed of the parachutist changes, or the effect of opening the parachute on velocity / speed is given. | 1–2 |
| Level 2 | The change in velocity / speed is clearly explained in terms of force(s), or there is a reasoned argument for the open parachute producing a lower speed. | 3–4 |
| Level 3 | A clear and detailed explanation of why the parachutist reaches terminal velocity and a reasoned argument for the open parachute producing a lower speed. | 5–6 |
Your mark:
Show a model answer
When the parachutist leaves the plane, the only force acting is their weight, so the resultant force is downwards and they accelerate. As their speed increases, the air resistance increases, so the resultant force gets smaller and the acceleration decreases. When the air resistance equals the weight, the forces are balanced, the resultant force is zero and they fall at a constant speed: the terminal velocity (60 m/s).
Opening the parachute increases the surface area, so the air resistance increases and becomes greater than the weight. The resultant force now acts upwards, so the parachutist decelerates. As they slow down the air resistance decreases, until it equals the weight again. The forces are balanced at a new, lower terminal velocity (about 8 m/s).
Your results
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| Question | Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Total |
|---|---|---|---|---|---|---|---|---|
| Out of | 6 | 5 | 8 | 4 | 5 | 6 | 6 | 40 |
| Your mark | – | – | – | – | – | – | – | 0 |
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