Specification: AQA A-level Physics (7408) 3.4.1.5 | 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
Newton’s second law: the resultant force on an object is equal to the rate of change of its momentum. For an object of constant mass this becomes F = m a, and the acceleration is in the same direction as the resultant force.
Resultant force: the single force that has the same effect as all the forces acting on an object combined. It is the vector sum of all the forces.
Terminal velocity: the maximum velocity reached when the resistive forces (drag) on a falling or driven object become equal and opposite to the driving force, so the resultant force and the acceleration are zero.
The equation
resultant force = mass × acceleration F = m a
| Quantity | Symbol | Unit |
|---|---|---|
| resultant force | F | newtons (N) |
| mass | m | kilograms (kg) |
| acceleration | a | metres per second squared (m s⁻²) |
Rearranged: a = F ÷ m | m = F ÷ a | F must be the resultant force: add forces in the direction of motion and subtract those against it before you use the equation. The data booklet gives the full form F = Δ(mv) ÷ Δt, which is covered on the momentum page; F = m a is the special case for constant mass.
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 car of mass 1200 kg accelerates at 2.5 m s⁻². Calculate the resultant force on the car.
Show solution
F: F = m a
I: F = 1200 × 2.5
F: F is already the subject
A: 3000 N (2 s.f.)
Q2 (Core) A resultant force of 6.0 N acts on a 0.40 kg trolley. Calculate its acceleration.
Show solution
F: F = m a
I: 6.0 = 0.40 × a
F: a = 6.0 ÷ 0.40
A: 15 m s⁻² (2 s.f.)
Q3 (Core) A resultant force of 4.5 kN gives a car an acceleration of 3.0 m s⁻². Calculate the mass of the car.
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Convert first: 4.5 kN = 4500 N
F: F = m a
I: 4500 = m × 3.0
F: m = 4500 ÷ 3.0
A: 1500 kg (2 s.f.)
Q4 (Core) A tennis ball of mass 58 g is hit by a racket with an average resultant force of 87 N. Calculate the average acceleration of the ball.
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Convert first: 58 g = 0.058 kg
F: F = m a
I: 87 = 0.058 × a
F: a = 87 ÷ 0.058
A: 1500 m s⁻² (2 s.f.)
Watch out: forgetting to convert grams to kilograms gives an answer 1000 times too small.
Q5 (Exam) A car of mass 1100 kg has a driving force of 3.2 kN. The total resistive force on it is 900 N. Calculate the acceleration of the car, to 2 s.f.
Show solution
Convert first: 3.2 kN = 3200 N
Step 1 – resultant force: F = 3200 − 900 = 2300 N forwards
Step 2 – acceleration
F: F = m a
I: 2300 = 1100 × a
F: a = 2300 ÷ 1100 = 2.0909… m s⁻²
A: 2.1 m s⁻² (2 s.f.)
Watch out: using the driving force instead of the resultant force is one of the most common errors in examiners’ reports.
Q6 (Exam) A car of mass 1400 kg is travelling at 25 m s⁻¹. It brakes with constant deceleration and comes to rest in 5.0 s. Calculate the resultant force on the car while it is braking.
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Step 1 – acceleration (suvat): v = u + a t, so 0 = 25 + a × 5.0, giving a = −25 ÷ 5.0 = −5.0 m s⁻²
Step 2 – force
F: F = m a
I: F = 1400 × (−5.0)
F: F is already the subject
A: F = −7000 N, so the resultant force is 7000 N (2 s.f.) opposite to the direction of motion
Watch out: the minus sign tells you the direction. Say “opposite to the motion” rather than leaving a bare negative.
Q7 (Exam) A person of mass 65 kg stands on bathroom scales in a lift. Calculate the reading on the scales (the normal contact force) when the lift (a) accelerates upwards at 1.2 m s⁻² and (b) accelerates downwards at 1.2 m s⁻². Give both to 3 s.f. (g = 9.81 N kg⁻¹)
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Step 1 – weight: W = m g = 65 × 9.81 = 637.65 N
Step 2 – resultant force (take upwards as positive)
F: R − W = m a
I: (a) R − 637.65 = 65 × 1.2 = 78; (b) R − 637.65 = 65 × (−1.2) = −78
F: (a) R = 637.65 + 78 = 715.65 N; (b) R = 637.65 − 78 = 559.65 N
A: (a) 716 N (b) 560 N (3 s.f.)
Watch out: the scales read the normal contact force (the apparent weight), not m g. The person’s true weight is the same in both cases.
Q8 (Exam) A box of mass 5.0 kg slides down a slope inclined at 30° to the horizontal. A constant frictional force of 12 N acts up the slope. Calculate the acceleration of the box down the slope, to 2 s.f. (g = 9.81 N kg⁻¹)
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Step 1 – component of weight down the slope: m g sin θ = 5.0 × 9.81 × sin 30° = 24.525 N
Step 2 – resultant force down the slope: F = 24.525 − 12 = 12.525 N
F: F = m a
I: 12.525 = 5.0 × a
F: a = 12.525 ÷ 5.0 = 2.505 m s⁻²
A: 2.5 m s⁻² (2 s.f.)
Watch out: the component along the slope is m g sin θ, not m g cos θ. The m g cos θ component is balanced by the normal contact force.
Q9 (Stretch) A car of mass 1200 kg tows a trailer of mass 400 kg along a straight, level road using a light, rigid tow bar. The driving force on the car is 2400 N. The resistive forces are 300 N on the car and 100 N on the trailer. Calculate (a) the acceleration, (b) the tension in the tow bar and (c) the time taken to reach 15 m s⁻¹ from rest.
Show solution
Step 1 – treat car and trailer as one system: resultant force = 2400 − 300 − 100 = 2000 N; total mass = 1200 + 400 = 1600 kg
F: F = m a
I: 2000 = 1600 × a
F: a = 2000 ÷ 1600 = 1.25 m s⁻²
Step 2 – trailer on its own: the only forward force on it is the tension T, so T − 100 = 400 × 1.25 = 500, giving T = 600 N
Step 3 – suvat: v = u + a t, so 15 = 0 + 1.25 × t, giving t = 15 ÷ 1.25 = 12 s
A: (a) 1.25 m s⁻² (3 s.f., exact) (b) 600 N (2 s.f.) (c) 12 s (2 s.f.)
Watch out: when finding the tension, use only the trailer’s mass and only the forces acting on the trailer.
Q10 (Stretch) A skydiver of total mass 80 kg falls vertically. The air resistance on her is given by D = k v², where k = 0.25 kg m⁻¹. Calculate (a) her terminal velocity and (b) her acceleration when her speed is 40 m s⁻¹. Give both to 2 s.f. (g = 9.81 N kg⁻¹)
Show solution
Step 1 – weight: W = m g = 80 × 9.81 = 784.8 N
Step 2 – terminal velocity: the resultant force is zero, so D = W: 0.25 × v² = 784.8, giving v² = 3139.2 and v = √3139.2 = 56.028… m s⁻¹
Step 3 – at 40 m s⁻¹: D = 0.25 × 40² = 400 N, so the resultant force = 784.8 − 400 = 384.8 N downwards
F: F = m a
I: 384.8 = 80 × a
F: a = 384.8 ÷ 80 = 4.81 m s⁻²
A: (a) 56 m s⁻¹ (b) 4.8 m s⁻² downwards (2 s.f.)
Watch out: at terminal velocity the acceleration is zero, not g. As speed rises, drag rises, so the acceleration falls gradually from 9.81 m s⁻² to zero.
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