Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Equation sheet: given in the exam
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/s²) |
Rearranged: a = F ÷ m | m = F ÷ a | Resultant force = forward force − backward force
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.
Foundation: aim for Q1–8. Higher: aim for Q4–10. Most questions need rearranging.
Questions
Q1 (F) A cyclist and bike have a total mass of 60 kg and accelerate at 1.5 m/s². Calculate the resultant force.
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F: F = m a
I: F = 60 × 1.5
F: F is already the subject – no rearranging needed
A: 90 N
Q2 (F) A 1500 kg car has a resultant force of 4500 N. Calculate its acceleration.
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F: F = m a
I: 4500 = 1500 × a
F: a = 4500 ÷ 1500
A: 3 m/s²
Q3 (F) A resultant force of 30 N makes a loaded shopping trolley accelerate at 0.6 m/s². Calculate the mass of the trolley.
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F: F = m a
I: 30 = m × 0.6
F: m = 30 ÷ 0.6
A: 50 kg
Q4 (F) A 1200 kg car has a resultant force of 3.6 kN. Calculate its acceleration.
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Convert first: 3.6 kN = 3600 N
F: F = m a
I: 3600 = 1200 × a
F: a = 3600 ÷ 1200
A: 3 m/s²
Q5 (F) A tennis racket exerts a force of 29 N on a 58 g ball. Calculate the ball’s acceleration.
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Convert first: 58 g = 0.058 kg
F: F = m a
I: 29 = 0.058 × a
F: a = 29 ÷ 0.058
A: 500 m/s²
Q6 (F/H) A 1100 kg car has a driving force of 3000 N and air resistance and friction of 800 N. Calculate its acceleration.
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Resultant force: 3000 − 800 = 2200 N
F: F = m a
I: 2200 = 1100 × a
F: a = 2200 ÷ 1100
A: 2 m/s²
Q7 (F/H) A 900 kg car accelerates at 2.0 m/s² against resistive forces of 600 N. Calculate the driving force from the engine.
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F: F = m a
I: F = 900 × 2.0 = 1800 N (resultant)
F: driving force − 600 = 1800, so driving force = 1800 + 600
A: 2400 N
Q8 (F/H) A resultant force of 4.0 × 106 N gives a cargo ship an acceleration of 0.020 m/s². Calculate the mass of the ship in standard form.
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F: F = m a
I: 4.0 × 106 = m × 0.020
F: m = 4.0 × 106 ÷ 0.020
A: 2.0 × 108 kg
Q9 (H) A braking force of 6000 N stops a car travelling at 20 m/s in 4.0 s. Calculate the mass of the car.
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Step 1 – deceleration: a = Δv / t = 20 ÷ 4.0 = 5 m/s²
Step 2 – mass
F: F = m a
I: 6000 = m × 5
F: m = 6000 ÷ 5
A: 1200 kg
Q10 (H) A lift of mass 850 kg accelerates upwards at 1.2 m/s². Calculate the tension in the lift cable, to 2 significant figures. (g = 9.8 N/kg)
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Weight: W = m g = 850 × 9.8 = 8330 N
F: F = m a
I: tension − 8330 = 850 × 1.2 = 1020
F: tension = 1020 + 8330 = 9350 N
A: 9400 N (2 s.f.)
