The Physics Question Bank

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Equation sheet: given in the exam

The equation

kinetic energy = 0.5 × mass × (speed)²    Eₖ = ½ m v²

Quantity	Symbol	Unit
kinetic energy	Eₖ	joules (J)
mass	m	kilograms (kg)
speed	v	metres per second (m/s)

Rearranged: m = 2Eₖ ÷ v²  |  v = √(2Eₖ ÷ m)  |  Square the speed before multiplying.

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.

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Questions

Q1 (F) A 60 kg runner jogs at 5 m/s. Calculate her kinetic energy.

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F: Eₖ = ½ m v²
I: Eₖ = 0.5 × 60 × 5²
F: Eₖ is already the subject
A: 0.5 × 60 × 25 = 750 J

Q2 (F) A netball thrown at 4 m/s has 12 J of kinetic energy. Calculate its mass.

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F: Eₖ = ½ m v²
I: 12 = 0.5 × m × 4², so 12 = 8 × m
F: m = 12 ÷ 8
A: 1.5 kg

Q3 (F) A 1000 kg car has 50 000 J of kinetic energy. Calculate its speed.

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F: Eₖ = ½ m v²
I: 50 000 = 0.5 × 1000 × v², so 50 000 = 500 × v²
F: v² = 100, so v = √100
A: 10 m/s

Q4 (F) A 20 g dart has 4.0 J of kinetic energy. Calculate its speed.

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Convert first: 20 g = 0.020 kg
F: Eₖ = ½ m v²
I: 4.0 = 0.5 × 0.020 × v², so 4.0 = 0.010 × v²
F: v² = 400, so v = √400
A: 20 m/s

Q5 (F) A 1500 kg car has 300 kJ of kinetic energy. Calculate its speed.

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Convert first: 300 kJ = 300 000 J
F: Eₖ = ½ m v²
I: 300 000 = 0.5 × 1500 × v², so 300 000 = 750 × v²
F: v² = 400, so v = √400
A: 20 m/s

Q6 (F/H) A 1200 kg car on a motorway has 375 kJ of kinetic energy. Calculate its speed in km/h.

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Convert first: 375 kJ = 375 000 J
F: Eₖ = ½ m v²
I: 375 000 = 0.5 × 1200 × v², so 375 000 = 600 × v²
F: v² = 625, so v = 25 m/s
A: 25 × 3600 ÷ 1000 = 90 km/h

Q7 (F/H) A 75 kg cyclist speeds up from 6 m/s to 10 m/s. Calculate the increase in her kinetic energy.

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F: Eₖ = ½ m v²
I: before = 0.5 × 75 × 6² = 1350 J; after = 0.5 × 75 × 10² = 3750 J
F: increase = 3750 − 1350
A: 2400 J
Watch out: you cannot use ½ × m × (change in speed)².

Q8 (F/H) A satellite with a mass of 1.2 × 10³ kg has 3.6 × 10¹⁰ J of kinetic energy. Calculate its speed, to 2 significant figures.

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F: Eₖ = ½ m v²
I: 3.6 × 10¹⁰ = 0.5 × 1.2 × 10³ × v²
F: v² = 3.6 × 10¹⁰ ÷ 600 = 6.0 × 10⁷, so v = √(6.0 × 10⁷) = 7746 m/s
A: 7.7 × 10³ m/s (2 s.f.)

Q9 (H) A 0.50 kg ball is dropped from a height of 5.0 m. Calculate its speed just before it hits the ground, to 2 significant figures. (g = 9.8 N/kg; ignore air resistance)

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Step 1 – GPE lost: Eₚ = m g h = 0.50 × 9.8 × 5.0 = 24.5 J = Eₖ gained
Step 2 – speed
F: Eₖ = ½ m v²
I: 24.5 = 0.5 × 0.50 × v²
F: v² = 24.5 ÷ 0.25 = 98, so v = √98 = 9.89… m/s
A: 9.9 m/s (2 s.f.)

Q10 (H) A 1100 kg car travelling at 25 m/s brakes. The brakes transfer 220 kJ of energy. Calculate the car’s new speed.

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Step 1 – starting Eₖ: 0.5 × 1100 × 25² = 343 750 J
Step 2 – remaining Eₖ: 343 750 − 220 000 = 123 750 J
F: Eₖ = ½ m v²
I: 123 750 = 0.5 × 1100 × v², so 123 750 = 550 × v²
F: v² = 225, so v = √225
A: 15 m/s