Forces Quiz

✏️ 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.

Course: Combined Science + Separate Physics  |  20 questions  |  g = 9.8 N/kg  |  ← All quizzes

Write A, B, C or D for each question, then tap Show answer to mark it.

Q1 (F, recall) Which of these is a vector quantity?
A. Speed
B. Mass
C. Velocity
D. Distance

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✅ C. Velocity
A vector has magnitude and direction. Velocity is speed in a given direction.

Q2 (F, calculation) What is the weight of a 50 kg person on Earth?
A. 5.1 N
B. 490 N
C. 50 N
D. 59.8 N

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✅ B. 490 N
F: W = m g
I: W = 50 × 9.8
A: 490 N
Examiner tip: mass is in kg, weight is a force in N – they are not the same thing.

Q3 (F, calculation) A cyclist travels at a constant 3.0 m/s for 2.0 minutes. How far does she travel?
A. 6.0 m
B. 40 m
C. 0.025 m
D. 360 m

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✅ D. 360 m
Convert: 2.0 minutes = 120 s
F: s = v t
I: s = 3.0 × 120
A: 360 m
Examiner tip: 6.0 m comes from leaving the time in minutes.

Q4 (F, calculation) A car of mass 1500 kg accelerates at 2.0 m/s². What resultant force acts on it?
A. 3000 N
B. 750 N
C. 1502 N
D. 0.0013 N

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✅ A. 3000 N
F: F = m a
I: F = 1500 × 2.0
A: 3000 N

Q5 (F, application) A car has a driving force of 800 N forwards and a resistive force of 200 N backwards. What is the resultant force?
A. 1000 N forwards
B. 200 N backwards
C. 600 N forwards
D. 0 N

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✅ C. 600 N forwards
Forces in opposite directions subtract: 800 − 200 = 600 N, in the direction of the larger force.
Examiner tip: always give the direction of a resultant force.

Q6 (F, recall) Which statement is Newton’s third law?
A. Force equals mass times acceleration
B. When two objects interact, they exert equal and opposite forces on each other
C. An object stays still unless a resultant force acts on it
D. Every action is followed by a bigger reaction

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✅ B
The two forces are the same type and act on different objects. A is the second law; C is part of the first law.

Q7 (F/H, definition) What is terminal velocity?
A. The speed at which an object hits the ground
B. The top speed of a car
C. The speed an object has when it starts falling
D. The constant maximum velocity reached when the resistive force equals the weight

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✅ D
The resultant force is zero, so the object stops accelerating.
Examiner tip: air resistance increases with speed; gravity does not get stronger as the object falls.

Q8 (F/H, graph) A velocity–time graph is a straight line from 0 m/s at 0 s to 12 m/s at 4.0 s. What is the acceleration?
A. 3.0 m/s²
B. 48 m/s²
C. 0.33 m/s²
D. 12 m/s²

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✅ A. 3.0 m/s²
Acceleration = gradient of a velocity–time graph
= change in velocity ÷ time = 12 ÷ 4.0 = 3.0 m/s²

Q9 (F/H, graph) For the same graph as Q8, how far does the object travel in the 4.0 s?
A. 48 m
B. 3.0 m
C. 24 m
D. 16 m

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✅ C. 24 m
Distance = area under a velocity–time graph
Area of triangle = ½ × base × height = ½ × 4.0 × 12 = 24 m
Examiner tip: 48 m means you forgot the ½.

Q10 (F/H, calculation) A car speeds up from 5.0 m/s to 25 m/s in 8.0 s. What is its acceleration?
A. 3.1 m/s²
B. 2.5 m/s²
C. 160 m/s²
D. 0.40 m/s²

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✅ B. 2.5 m/s²
F: a = Δv ÷ t
I: a = (25 − 5.0) ÷ 8.0
F: a = 20 ÷ 8.0
A: 2.5 m/s²
Examiner tip: 3.1 m/s² comes from using the final velocity instead of the change in velocity.

Q11 (F/H, application) A driver is very tired. Which part of the stopping distance increases?
A. The braking distance
B. Neither
C. Both, equally
D. The thinking distance

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✅ D. The thinking distance
Tiredness increases the driver’s reaction time, so the car travels further before the brakes are applied.
Examiner tip: braking distance is affected by the road, tyres, brakes and speed.

Q12 (F/H, calculation) A 40 N force pushes a box 3.0 m across the floor. How much work is done?
A. 120 J
B. 13 J
C. 43 J
D. 0.075 J

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✅ A. 120 J
F: W = F s
I: W = 40 × 3.0
A: 120 J

Q13 (F/H, calculation) A spring has a spring constant of 25 N/m. What force stretches it by 0.20 m?
A. 125 N
B. 500 N
C. 5.0 N
D. 0.0080 N

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✅ C. 5.0 N
F: F = k e
I: F = 25 × 0.20
A: 5.0 N

Q14 (F/H, practical) In the force and extension practical, a spring is 12.0 cm long with no load and 15.5 cm long with a weight attached. What is the extension?
A. 15.5 cm
B. 3.5 cm
C. 27.5 cm
D. 12.0 cm

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✅ B. 3.5 cm
Extension = new length − original length = 15.5 − 12.0 = 3.5 cm
Examiner tip: plotting the total length instead of the extension is a very common error. Read the ruler at eye level to avoid parallax.

Q15 (F/H, calculation) A trolley accelerates uniformly from rest at 2.0 m/s² over a distance of 25 m. What is its final velocity?
A. 100 m/s
B. 50 m/s
C. 7.1 m/s
D. 10 m/s

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✅ D. 10 m/s
F: v² − u² = 2 a s
I: v² − 0² = 2 × 2.0 × 25
F: v² = 100, so v = √100
A: 10 m/s
Examiner tip: 100 m/s means you forgot to square-root. 7.1 m/s means you missed the 2.

Q16 (H, calculation) A 0.15 kg ball moves at 20 m/s. What is its momentum?
A. 3.0 kg m/s
B. 133 kg m/s
C. 0.0075 kg m/s
D. 30 kg m/s

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✅ A. 3.0 kg m/s
F: p = m v
I: p = 0.15 × 20
A: 3.0 kg m/s

Q17 (H, calculation) A 2.0 kg trolley moving at 4.0 m/s collides with a stationary 2.0 kg trolley. They stick together. What is their velocity after the collision?
A. 4.0 m/s
B. 8.0 m/s
C. 2.0 m/s
D. 1.0 m/s

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✅ C. 2.0 m/s
Momentum before = 2.0 × 4.0 + 2.0 × 0 = 8.0 kg m/s
Momentum after = (2.0 + 2.0) × v
8.0 = 4.0 × v, so v = 2.0 m/s
Examiner tip: in a closed system, total momentum before = total momentum after.

Q18 (S only, calculation) A force of 30 N is applied 0.40 m from a pivot. What is the moment?
A. 75 Nm
B. 12 Nm
C. 0.013 Nm
D. 30.4 Nm

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✅ B. 12 Nm
F: M = F d
I: M = 30 × 0.40
A: 12 Nm
Examiner tip: d is the perpendicular distance from the pivot to the line of action of the force.

Q19 (S only, H, calculation) What is the pressure due to the water at a depth of 3.0 m? (Density of water = 1000 kg/m³)
A. 3000 Pa
B. 2940 Pa
C. 29.4 Pa
D. 29 400 Pa

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✅ D. 29 400 Pa
F: p = h ρ g
I: p = 3.0 × 1000 × 9.8
A: 29 400 Pa

Q20 (S only, H, calculation) A 0.060 kg tennis ball is hit from rest to 40 m/s. The racket is in contact with the ball for 0.012 s. What is the average force on the ball?
A. 200 N
B. 2.4 N
C. 0.018 N
D. 2000 N

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✅ A. 200 N
F: F = m Δv ÷ Δt
I: F = 0.060 × 40 ÷ 0.012
F: F = 2.4 ÷ 0.012
A: 200 N
Examiner tip: 2.4 N is the change in momentum (in kg m/s), not the force. Divide by the time.


Your score

Add up your marks out of 20 and multiply by 5 to get a percentage. Rough guide (not an official grade): 18–20 excellent, grade 8–9 standard  |  14–17 grade 6–7  |  10–13 grade 4–5  |  under 10 revise the notes and try again.

Answer key (for teachers and printing)

1 C   2 B   3 D   4 A   5 C   6 B   7 D   8 A   9 C   10 B   11 D   12 A   13 C   14 B   15 D   16 A   17 C   18 B   19 D   20 A

Revise: Forces topic page  |  Fill the gaps: Forces  |  Which equation? Forces