Forces Topic Test – Higher

✏️ Paper first! Work out every question on paper before you tap Show solution. Write down every step – the equation, the numbers with units, the rearranging and the answer with its unit. In the exam, if your final answer is wrong you can still get marks for correct working, but only if the examiner can see it.

AQA-style topic test · Forces · Higher tier

Time allowed: 50 minutes
Total: Separate Physics 42 marks · Combined Science 30 marks (skip Questions 06 and 07)
You need: a calculator, a ruler and the equation sheet.

Instructions: Answer all questions on paper. Show your working in calculations. Give answers to an appropriate number of significant figures. Gravitational field strength = 9.8 N/kg. Open the mark schemes only when you have finished the whole test.

Question 01

01.1 Two forces act on a boat. A force of 30 N acts east and a force of 40 N acts north.
Determine the size of the resultant force. You may use a scale drawing.    [2 marks]

Mark scheme

correct scale drawing (forces tip-to-tail at right angles) or √(30² + 40²) (1)
50 N (1) – allow 49–51 N from a scale drawing

01.2 A worker pushes a box with a force of 250 N for 4.0 m across a floor.
Calculate the work done.    [2 marks]

Mark scheme

W = 250 × 4.0 (1)
W = 1000 J (1)

01.3 What is meant by 1 joule of work done?    [1 mark]

Mark scheme

a force of 1 N moves an object through 1 m (in the direction of the force) (1)

Question 01 total: 5 marks

Question 02

A car starts from rest and accelerates steadily to 20 m/s in 8.0 s. It then travels at a constant 20 m/s for 12 s.

02.1 Calculate the acceleration of the car in the first 8.0 s.    [2 marks]

Mark scheme

a = 20 ÷ 8.0 (1)
a = 2.5 m/s² (1)

02.2 Use the area under a velocity–time graph to calculate the total distance travelled in the 20 s.    [3 marks]

Mark scheme

triangle: ½ × 8.0 × 20 = 80 m (1)
rectangle: 12 × 20 = 240 m (1)
total = 320 m (1)

02.3 A train travelling at 30 m/s brakes steadily and stops in a distance of 450 m.
Calculate the deceleration of the train.
Use the equation: v² − u² = 2 a s    [3 marks]

Mark scheme

0² − 30² = 2 × a × 450 (1)
a = −900 ÷ 900 (1)
deceleration = 1.0 m/s² (1) – allow a = −1.0 m/s²

Question 02 total: 8 marks

Question 03

A student investigates how the force on a trolley affects its acceleration. A string attached to the trolley runs over a pulley to a hanging mass holder. Light gates measure the acceleration.

03.1 Describe how the student can change the force while keeping the total mass constant.    [2 marks]

Mark scheme

start with masses stacked on the trolley (1)
move them one at a time from the trolley to the hanger (1)

03.2 Describe the graph of acceleration against force that the student should expect.    [1 mark]

Mark scheme

a straight line through the origin / acceleration directly proportional to force (1)

03.3 The measured accelerations are all slightly lower than expected. Suggest a reason and how to reduce the effect.    [2 marks]

Mark scheme

friction acts on the trolley / at the pulley, so the resultant force is smaller (1)
tilt the runway slightly so the trolley just keeps moving at a steady speed (1)

03.4 Define inertial mass.    [1 mark]

Mark scheme

a measure of how difficult it is to change an object’s velocity / force ÷ acceleration (1)

Question 03 total: 6 marks

Question 04

04.1 A car of mass 1200 kg travels at 15 m/s. Calculate its momentum.    [2 marks]

Mark scheme

p = 1200 × 15 (1)
p = 18 000 kg m/s (1)

04.2 A 2.0 kg trolley moving at 3.0 m/s hits a stationary 1.0 kg trolley. They stick together.
Calculate their velocity after the collision.    [3 marks]

Mark scheme

momentum before = 2.0 × 3.0 = 6.0 kg m/s (1)
6.0 = 3.0 × v (1)
v = 2.0 m/s (1)

Question 04 total: 5 marks

Question 05

05.1 Explain the factors that affect the stopping distance of a car. Include why a very large deceleration can be dangerous.    [6 marks]

Mark scheme

Level 3 (5–6 marks): factors are clearly linked to thinking distance or braking distance, with explanations, and the dangers of large decelerations are explained.
Level 2 (3–4 marks): factors for both thinking and braking distance are given, with some explanation.
Level 1 (1–2 marks): some factors are listed with little or no explanation.
0 marks: no relevant content.

Indicative content:
• stopping distance = thinking distance + braking distance
• tiredness, alcohol, drugs and distractions increase reaction time, so thinking distance increases
• higher speed increases both thinking and braking distance
• wet or icy roads and worn tyres reduce friction, so braking distance increases
• worn brakes reduce the braking force
• braking: friction does work, transferring kinetic energy to the thermal store of the brakes
• large decelerations need large braking forces, which can make the brakes overheat and the driver lose control / skid

Question 05 total: 6 marks

🔷 SEPARATE PHYSICS ONLY – Combined Science students skip Questions 06 and 07
(Moments, pressure in fluids, and force as rate of change of momentum are only in Separate Physics.)

Question 06

06.1 A child of weight 300 N sits 2.0 m from the pivot of a seesaw. A second child of weight 400 N sits on the other side.
Calculate how far from the pivot the second child must sit for the seesaw to balance.    [3 marks]

Mark scheme

clockwise moment = anticlockwise moment (1)
300 × 2.0 = 400 × d (1)
d = 1.5 m (1)

06.2 Calculate the pressure due to the water at a depth of 10 m in a lake.
Density of water = 1000 kg/m³
Use the equation: p = h ρ g    [2 marks]

Mark scheme

p = 10 × 1000 × 9.8 (1)
p = 98 000 Pa (1)

06.3 Explain why an object submerged in water experiences an upthrust.    [2 marks]

Mark scheme

pressure increases with depth, so the pressure on the bottom surface is greater than on the top surface (1)
so there is a resultant upward force (1)

Question 06 total: 7 marks

Question 07

07.1 In a crash test, a car of mass 1000 kg travelling at 15 m/s is brought to rest in 0.10 s.
Calculate the average force on the car.
Use the equation: F = m Δv ÷ Δt    [2 marks]

Mark scheme

F = 1000 × 15 ÷ 0.10 (1)
F = 150 000 N (1) – allow 1.5 × 105 N

07.2 Explain how an air bag reduces the force on a driver in a crash.    [3 marks]

Mark scheme

the air bag increases the time taken for the driver to stop (1)
so the rate of change of momentum is smaller (1)
force = rate of change of momentum, so the force on the driver is smaller (1)

Question 07 total: 5 marks


END OF TEST. Total: Separate Physics 42 marks · Combined Science 30 marks.

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