Specification: AQA A-level Physics (7408) 3.4.1.6 | Paper: 1 (and 2 as assumed knowledge) | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Impulse: the product of a force and the time for which it acts, FΔt. Impulse is equal to the change in momentum of the object and is a vector in the direction of the force.
Area under a force–time graph: the area under a force–time graph is equal to the impulse, and so to the change in momentum, even when the force is not constant.
Safety features: crumple zones, airbags and seat belts increase the time over which a passenger’s momentum changes. For the same change in momentum this reduces the force on the passenger.
The equation
impulse = force × time = change in momentum FΔt = Δ(mv)
| Quantity | Symbol | Unit |
|---|---|---|
| impulse (= change in momentum) | FΔt or Δ(mv) | newton seconds (N s), equivalent to kg m s⁻¹ |
| force | F | newtons (N) |
| time for which the force acts | Δt | seconds (s) |
| mass | m | kilograms (kg) |
| velocity | v | metres per second (m s⁻¹) |
Rearranged: F = Δ(mv) ÷ Δt | Δt = Δ(mv) ÷ F | Δv = FΔt ÷ m | Δ(mv) = mv − mu: if the object changes direction, one velocity is negative, so the change is bigger than either momentum.
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.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A constant force of 50 N acts on a box for 0.20 s. Calculate the impulse on the box.
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F: impulse = FΔt
I: impulse = 50 × 0.20
F: impulse is already the subject
A: 10 N s
Q2 (Core) A 0.50 kg ball at rest is given an impulse of 6.0 N s. Calculate the speed of the ball afterwards.
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F: FΔt = Δ(mv) = mv − mu
I: 6.0 = 0.50 × v − 0
F: v = 6.0 ÷ 0.50
A: 12 m s⁻¹
Q3 (Core) A 430 g football at rest is kicked with an average force of 1.2 kN. The foot is in contact with the ball for 9.0 ms. Calculate the speed of the ball as it leaves the foot.
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Convert first: 430 g = 0.430 kg; 1.2 kN = 1200 N; 9.0 ms = 9.0 × 10−3 s
F: FΔt = mv − mu
I: 1200 × 9.0 × 10−3 = 0.430 × v − 0, so 10.8 = 0.430v
F: v = 10.8 ÷ 0.430 = 25.1… m s⁻¹
A: 25 m s⁻¹ (2 s.f.)
Q4 (Core) A 58 g tennis ball is served from rest and leaves the racket at 50 m s⁻¹. The ball is in contact with the racket for 4.0 ms. Calculate the average force on the ball.
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Convert first: 58 g = 0.058 kg; 4.0 ms = 4.0 × 10−3 s
F: FΔt = Δ(mv)
I: F × 4.0 × 10−3 = 0.058 × 50 − 0 = 2.9 N s
F: F = 2.9 ÷ 4.0 × 10−3 = 725 N
A: 730 N (2 s.f.)
Q5 (Exam) A force–time graph for a hockey stick striking a stationary 0.16 kg ball is a triangle. The force rises steadily from 0 to a peak of 2.4 kN at 2.0 ms, then falls steadily back to 0 at 5.0 ms. Calculate the speed of the ball as it leaves the stick.
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Step 1 – impulse = area under the graph
area of triangle = ½ × base × height = ½ × 5.0 × 10−3 × 2400 = 6.0 N s
Step 2 – speed
F: FΔt = mv − mu
I: 6.0 = 0.16 × v − 0
F: v = 6.0 ÷ 0.16 = 37.5 m s⁻¹
A: 38 m s⁻¹ (2 s.f.)
Watch out: the base of the triangle is the whole contact time (5.0 ms), not the time to the peak.
Q6 (Exam) A 0.20 kg ball hits a wall at 8.0 m s⁻¹ and rebounds along the same line at 6.0 m s⁻¹. It is in contact with the wall for 0.050 s. Calculate the change in momentum of the ball and the average force the wall exerts on it.
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Take the direction towards the wall as positive, so u = +8.0 m s⁻¹ and v = −6.0 m s⁻¹
F: Δ(mv) = mv − mu; F = Δ(mv) ÷ Δt
I: Δ(mv) = 0.20 × (−6.0) − 0.20 × 8.0 = −1.2 − 1.6
F: Δ(mv) = −2.8 kg m s⁻¹, so F = −2.8 ÷ 0.050
A: change in momentum = 2.8 kg m s⁻¹ away from the wall; average force = 56 N away from the wall
Watch out: subtracting the speeds (8.0 − 6.0) gives 0.40 kg m s⁻¹ and 8.0 N. When the ball changes direction the momenta add – this is one of the most common errors in examiners’ reports.
Q7 (Exam) In a crash a 70 kg driver travelling at 20 m s⁻¹ is brought to rest. With an airbag the driver stops in 0.12 s; without one the driver hits the steering wheel and stops in 0.015 s. Calculate the average force on the driver in each case and explain the difference.
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F: F = Δ(mv) ÷ Δt
I: Δ(mv) = 70 × 20 − 0 = 1400 kg m s⁻¹
F: with airbag: F = 1400 ÷ 0.12 = 11 666.7 N; without: F = 1400 ÷ 0.015 = 93 333.3 N
A: with airbag 1.2 × 104 N; without 9.3 × 104 N (2 s.f.)
Explain: the change in momentum is the same in both cases, but the airbag increases the stopping time by a factor of 8, so the force is 8 times smaller
Q8 (Exam) A student pushes a 40 kg trolley that is already moving at 1.5 m s⁻¹. The force–time graph is a trapezium: the force rises steadily from 0 to 400 N in 0.10 s, stays at 400 N for 0.30 s, then falls steadily to 0 in 0.10 s. The force is in the direction of motion. Calculate the final speed of the trolley (ignore friction) and show that 1 N s = 1 kg m s⁻¹.
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Step 1 – impulse = area under the graph
area of trapezium = ½ × (sum of parallel sides) × height = ½ × (0.50 + 0.30) × 400 = 160 N s
Step 2 – final speed
F: FΔt = mv − mu
I: 160 = 40 × v − 40 × 1.5
F: v = (160 + 60) ÷ 40
A: 5.5 m s⁻¹
Units: 1 N = 1 kg m s⁻², so 1 N s = 1 kg m s⁻² × s = 1 kg m s⁻¹
Q9 (Stretch) A 0.50 kg ball is dropped from a height of 1.80 m onto a hard floor and rebounds to a height of 1.25 m. It is in contact with the floor for 25 ms. Calculate the average force the floor exerts on the ball, to 2 significant figures. (g = 9.81 N kg⁻¹; ignore air resistance)
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Step 1 – speeds using energy (or v² = u² + 2as)
impact speed = √(2gh) = √(2 × 9.81 × 1.80) = 5.94… m s⁻¹ downwards; rebound speed = √(2 × 9.81 × 1.25) = 4.95… m s⁻¹ upwards
Step 2 – change in momentum (up positive)
Δ(mv) = 0.50 × 4.95… − 0.50 × (−5.94…) = 5.447… kg m s⁻¹ upwards
Step 3 – resultant force
F = Δ(mv) ÷ Δt = 5.447… ÷ 0.025 = 217.9… N upwards
Step 4 – force from the floor
the resultant force = floor force − weight, so floor force = 217.9… + 0.50 × 9.81 = 222.8… N
A: 220 N upwards (2 s.f.)
Watch out: FΔt = Δ(mv) gives the resultant force. Forgetting to add the weight, or using a negative rebound velocity in the wrong place, are both common.
Q10 (Stretch) A car crashes into a wall at 15 m s⁻¹. Its crumple zone shortens by 0.60 m as the car stops, with uniform deceleration. Calculate the average force on a 65 kg passenger held by a seat belt, to 2 significant figures, and state the effect on this force of a crumple zone that shortens by 1.20 m instead.
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Step 1 – stopping time (suvat)
s = ½(u + v)t, so 0.60 = ½ × (15 + 0) × t, giving t = 1.20 ÷ 15 = 0.080 s
Step 2 – force
F: F = Δ(mv) ÷ Δt
I: F = (65 × 15 − 0) ÷ 0.080 = 975 ÷ 0.080
F: F = 12 187.5 N
A: 1.2 × 104 N (2 s.f.)
Longer crumple zone: doubling the distance doubles the stopping time (0.16 s) for the same change in momentum, so the force halves to about 6.1 × 10³ N
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