Course: Separate Physics only | Tier: Higher only | Equation sheet: given in the exam
The equation
force = change in momentum ÷ time taken F = m Δv / Δt
| Quantity | Symbol | Unit |
|---|---|---|
| force | F | newtons (N) |
| mass | m | kilograms (kg) |
| change in velocity | Δv | metres per second (m/s) |
| time taken | Δt | seconds (s) |
Rearranged: Δt = m Δv ÷ F | Δv = F Δt ÷ m | m = F Δt ÷ Δv
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Higher tier only: work through all 10. Most questions need rearranging.
Questions
Q1 A 0.45 kg football is kicked from rest to 20 m/s. The boot is in contact for 0.05 s. Calculate the average force on the ball.
Show solution
F: F = m Δv / Δt
I: F = 0.45 × 20 ÷ 0.05
F: F is already the subject
A: 180 N
Q2 A braking force of 9000 N stops a 1200 kg car travelling at 15 m/s. How long does it take to stop?
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F: F = m Δv / Δt
I: 9000 = 1200 × 15 ÷ Δt
F: Δt = 18 000 ÷ 9000
A: 2 s
Q3 A 70 kg sprinter pushes off the starting blocks with an average force of 350 N for 0.4 s. Calculate her change in velocity.
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F: F = m Δv / Δt
I: 350 = 70 × Δv ÷ 0.4
F: Δv = 350 × 0.4 ÷ 70
A: 2 m/s
Q4 A 58 g tennis ball is served from rest to 50 m/s. The racket is in contact for 5 ms. Calculate the average force.
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Convert first: 58 g = 0.058 kg; 5 ms = 0.005 s
F: F = m Δv / Δt
I: F = 0.058 × 50 ÷ 0.005
F: F is already the subject
A: 580 N
Q5 An average force of 2.4 kN speeds up a motorbike and rider by 12 m/s in 1.5 s. Calculate their total mass.
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Convert first: 2.4 kN = 2400 N
F: F = m Δv / Δt
I: 2400 = m × 12 ÷ 1.5
F: m = 2400 × 1.5 ÷ 12
A: 300 kg
Q6 A seatbelt can safely exert a maximum force of 7000 N. A 70 kg driver is travelling at 20 m/s. What is the minimum time the seatbelt must take to stop the driver?
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F: F = m Δv / Δt
I: 7000 = 70 × 20 ÷ Δt
F: Δt = 1400 ÷ 7000
A: 0.2 s
Q7 A 0.60 kg basketball hits the floor at 8 m/s and bounces back up at 6 m/s. It is in contact with the floor for 0.035 s. Calculate the average force from the floor.
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Change in velocity: it reverses direction, so Δv = 8 + 6 = 14 m/s
F: F = m Δv / Δt
I: F = 0.60 × 14 ÷ 0.035
F: F is already the subject
A: 240 N
Q8 A tugboat exerts 2.5 × 103 N to slow a 2.0 × 105 kg ship by 0.50 m/s. How long does this take? Give your answer in standard form.
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F: F = m Δv / Δt
I: 2.5 × 103 = 2.0 × 105 × 0.50 ÷ Δt
F: Δt = 1.0 × 105 ÷ 2.5 × 103
A: 4.0 × 101 s (40 s)
Q9 A 1000 kg car hits a wall at 15 m/s. Without a crumple zone it stops in 0.10 s. With a crumple zone it stops in 0.25 s. By how much does the crumple zone reduce the average force?
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F: F = m Δv / Δt
I: without: F = 1000 × 15 ÷ 0.10 = 150 000 N
I: with: F = 1000 × 15 ÷ 0.25 = 60 000 N
A: reduced by 150 000 − 60 000 = 90 000 N
Longer stopping time → smaller rate of change of momentum → smaller force.
Q10 A fielder catches a 0.16 kg cricket ball travelling at 32 m/s. By pulling her hands back, she keeps the force on her hands to 85 N. Calculate the minimum time the catch takes, to 2 significant figures.
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F: F = m Δv / Δt
I: 85 = 0.16 × 32 ÷ Δt
F: Δt = 5.12 ÷ 85 = 0.0602… s
A: 0.060 s (2 s.f.)
