The Physics Question Bank

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

The equation

elastic potential energy = 0.5 × spring constant × (extension)²    Eₑ = ½ k e²

Quantity	Symbol	Unit
elastic potential energy	Eₑ	joules (J)
spring constant	k	newtons per metre (N/m)
extension	e	metres (m)

Rearranged: k = 2Eₑ ÷ e²  |  e = √(2Eₑ ÷ k)  |  Square the extension 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 spring with a spring constant of 200 N/m is stretched by 0.10 m. Calculate the elastic potential energy stored.

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F: Eₑ = ½ k e²
I: Eₑ = 0.5 × 200 × 0.10²
F: Eₑ is already the subject
A: 0.5 × 200 × 0.01 = 1 J

Q2 (F) A catapult elastic stretched by 0.20 m stores 4.0 J. Calculate its spring constant.

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F: Eₑ = ½ k e²
I: 4.0 = 0.5 × k × 0.20², so 4.0 = 0.02 × k
F: k = 4.0 ÷ 0.02
A: 200 N/m

Q3 (F) A trampoline spring has a spring constant of 800 N/m and stores 16 J. Calculate its extension.

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F: Eₑ = ½ k e²
I: 16 = 0.5 × 800 × e², so 16 = 400 × e²
F: e² = 0.04, so e = √0.04
A: 0.2 m

Q4 (F) A toy launcher spring (k = 50 N/m) stores 0.09 J. How far is it compressed, in cm?

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F: Eₑ = ½ k e²
I: 0.09 = 0.5 × 50 × e², so 0.09 = 25 × e²
F: e² = 0.0036, so e = 0.06 m
A: 6 cm

Q5 (F) A car suspension spring (k = 40 000 N/m) stores 50 J when compressed. How far is it compressed, in cm?

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F: Eₑ = ½ k e²
I: 50 = 0.5 × 40 000 × e², so 50 = 20 000 × e²
F: e² = 0.0025, so e = 0.05 m
A: 5 cm

Q6 (F/H) An archer’s bow acts like a spring with k = 1500 N/m. When drawn back it stores 120 J. How far is the string pulled back?

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F: Eₑ = ½ k e²
I: 120 = 0.5 × 1500 × e², so 120 = 750 × e²
F: e² = 0.16, so e = √0.16
A: 0.40 m

Q7 (F/H) At the bottom of a jump, a bungee cord is stretched 20 m and stores 12 kJ. Calculate its spring constant.

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Convert first: 12 kJ = 12 000 J
F: Eₑ = ½ k e²
I: 12 000 = 0.5 × k × 20², so 12 000 = 200 × k
F: k = 12 000 ÷ 200
A: 60 N/m

Q8 (F/H) A train buffer spring has k = 2.0 × 10⁶ N/m and stores 1.0 × 10⁴ J. Calculate its compression.

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F: Eₑ = ½ k e²
I: 1.0 × 10⁴ = 0.5 × 2.0 × 10⁶ × e², so 1.0 × 10⁴ = 1.0 × 10⁶ × e²
F: e² = 0.01, so e = √0.01
A: 0.10 m (1.0 × 10⁻¹ m)

Q9 (H) A catapult (k = 400 N/m) is stretched 0.15 m and fires a 20 g stone. Assuming all the elastic energy becomes kinetic energy, calculate the stone’s speed to 2 significant figures.

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Step 1 – energy stored: Eₑ = ½ k e² = 0.5 × 400 × 0.15² = 4.5 J
Step 2 – speed
F: Eₖ = ½ m v²
I: 4.5 = 0.5 × 0.020 × v², so 4.5 = 0.010 × v²
F: v² = 450, so v = √450 = 21.2… m/s
A: 21 m/s (2 s.f.)

Q10 (H) A force of 30 N stretches a gym spring by 0.12 m. Calculate the elastic potential energy stored.

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Step 1 – spring constant
F: F = k e
I: 30 = k × 0.12
F: k = 30 ÷ 0.12 = 250 N/m
Step 2 – energy
F: Eₑ = ½ k e²
I: Eₑ = 0.5 × 250 × 0.12²
A: 1.8 J