AQA-style topic test · Forces · Foundation tier
Time allowed: 40 minutes
Total: Separate Physics 33 marks · Combined Science 26 marks (skip Question 06)
You need: a calculator, a ruler and the equation sheet.
Instructions: Answer all questions on paper. Show your working in calculations – you can get marks for working even if your final answer is wrong. Open the mark schemes only when you have finished the whole test.
Question 01
01.1 Which two quantities are vectors?
Tick (✓) two boxes.
☐ distance ☐ force ☐ mass ☐ speed ☐ velocity [2 marks]
Mark scheme
force (1)
velocity (1)
more than two boxes ticked: deduct 1 mark for each extra tick
01.2 Name one non-contact force. [1 mark]
Mark scheme
gravitational / weight / magnetic / electrostatic (1)
01.3 A student has a mass of 60 kg.
Gravitational field strength = 9.8 N/kg
Calculate the student’s weight.
Use the equation: W = m g [2 marks]
Mark scheme
W = 60 × 9.8 (1)
W = 588 N (1)
Question 01 total: 5 marks
Question 02
A student hangs masses on a spring. The spring is 5.0 cm long with no mass on it. With a 6.0 N weight on it, the spring is 8.0 cm long.
02.1 Calculate the extension of the spring. [1 mark]
Mark scheme
8.0 − 5.0 = 3.0 cm (1)
02.2 Calculate the spring constant in N/m.
Use the equation: F = k e [3 marks]
Mark scheme
3.0 cm = 0.030 m (1)
6.0 = k × 0.030 (1)
k = 200 N/m (1)
02.3 What is meant by the limit of proportionality? [1 mark]
Mark scheme
the point beyond which the extension is no longer directly proportional to the force (1)
Question 02 total: 5 marks
Question 03
03.1 A person walks 600 m in 400 s. Calculate their average speed.
Use the equation: distance = speed × time [2 marks]
Mark scheme
600 = v × 400 (1)
v = 1.5 m/s (1)
03.2 What does a horizontal line on a distance–time graph show? [1 mark]
Mark scheme
the object is stationary / not moving (1)
03.3 A car accelerates from rest to 12 m/s in 4.0 s. Calculate its acceleration.
Use the equation: a = Δv ÷ t [2 marks]
Mark scheme
a = 12 ÷ 4.0 (1)
a = 3.0 m/s² (1)
Question 03 total: 5 marks
Question 04
A car has a mass of 1000 kg. The driving force is 2000 N. Air resistance and friction together are 500 N.
04.1 Calculate the resultant force on the car. [1 mark]
Mark scheme
2000 − 500 = 1500 N (forwards) (1)
04.2 Calculate the acceleration of the car.
Use the equation: F = m a [2 marks]
Mark scheme
1500 = 1000 × a (1)
a = 1.5 m/s² (1) – allow error carried forward from 04.1
04.3 Later, the driving force is equal to the air resistance and friction. Describe the motion of the car. [1 mark]
Mark scheme
it moves at a constant speed (in a straight line) (1)
do not accept: it stops / stays still
04.4 The car’s tyres push backwards on the road. According to Newton’s third law, what force does the road exert on the tyres? [1 mark]
Mark scheme
an equal force forwards (in the opposite direction) (1)
Question 04 total: 5 marks
Question 05
This question is about the stopping distance of a car.
05.1 Complete the sentence: stopping distance = thinking distance + ___________ [1 mark]
Mark scheme
braking distance (1)
05.2 Give two factors that increase a driver’s thinking distance. [2 marks]
Mark scheme
any two from (1 each): tiredness · alcohol · drugs · distractions (e.g. using a phone) · higher speed
05.3 Give one factor that increases braking distance. [1 mark]
Mark scheme
any one from (1): wet or icy roads · worn tyres · worn brakes · higher speed
do not accept: “condition of the tyres” without saying worn / poor
05.4 A car travels at 20 m/s. The driver’s reaction time is 0.70 s.
Calculate the thinking distance. [2 marks]
Mark scheme
distance = 20 × 0.70 (1)
distance = 14 m (1)
Question 05 total: 6 marks
🔷 SEPARATE PHYSICS ONLY – Combined Science students skip Question 06
(Moments and pressure are only in Separate Physics.)
Question 06
06.1 A mechanic pushes on a spanner with a force of 20 N. Her hand is 0.25 m from the nut.
Calculate the moment of the force.
Use the equation: M = F d [2 marks]
Mark scheme
M = 20 × 0.25 (1)
M = 5.0 N m (1)
06.2 Explain why a longer spanner makes it easier to undo a tight nut. [1 mark]
Mark scheme
a larger distance from the pivot gives a larger moment for the same force (1)
06.3 A person with a weight of 600 N stands on snow. The area of their boots in contact with the snow is 0.030 m².
Calculate the pressure on the snow.
Use the equation: p = F ÷ A [2 marks]
Mark scheme
p = 600 ÷ 0.030 (1)
p = 20 000 Pa (1)
06.4 The wall of a dam is much thicker at the bottom than at the top. Explain why. [2 marks]
Mark scheme
the pressure in a liquid increases with depth (1)
so there is a larger force on the bottom of the wall (1)
Question 06 total: 7 marks
END OF TEST. Total: Separate Physics 33 marks · Combined Science 26 marks.
Next: Forces – Higher test · Forces topic page · All topic tests
