Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Equation sheet: given in the exam
Equations used: Eₚ = mgh | Eₖ = ½mv² | Eₑ = ½ke² | W = Fs | P = E/t | efficiency = useful output ÷ total input
Key idea: energy transferred out of one store goes into another. For example, GPE lost = KE gained (if no energy is wasted).
Watch the units in Q6–10: convert to metres (cm ÷ 100, mm ÷ 1000, km × 1000), seconds (minutes × 60, hours × 3600) and m/s (km/h ÷ 3.6) before you use an equation. Some answers need standard form.
Foundation: aim for Q1–8. Higher: aim for Q4–10. Use FIFA for every step and don’t round until the end. g = 9.8 N/kg.
Questions
Q1 (F) A 2.0 kg rock falls 5.0 m from a cliff. Calculate the GPE it loses, then its speed just before it hits the ground. (Ignore air resistance.)
Show solution
Step 1 – GPE lost
F: Eₚ = m g h
I: Eₚ = 2.0 × 9.8 × 5.0
A: Eₚ = 98 J
Step 2 – speed (GPE lost = KE gained = 98 J)
F: Eₖ = ½ m v²
I: 98 = 0.5 × 2.0 × v², so 98 = 1.0 × v²
F: v = √98
A: 9.9 m/s (2 s.f.)
Q2 (F) A 0.50 kg ball is thrown straight up at 14 m/s. Calculate its kinetic energy, then the maximum height it reaches.
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Step 1 – kinetic energy
F: Eₖ = ½ m v²
I: Eₖ = 0.5 × 0.50 × 14²
A: Eₖ = 49 J
Step 2 – height (KE lost = GPE gained = 49 J)
F: Eₚ = m g h
I: 49 = 0.50 × 9.8 × h, so 49 = 4.9 × h
F: h = 49 ÷ 4.9
A: 10 m
Q3 (F) A 500 kg rollercoaster car starts from rest at the top of a 20 m drop. Calculate its speed at the bottom. (Ignore friction.)
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Step 1 – GPE lost
F: Eₚ = m g h
I: Eₚ = 500 × 9.8 × 20
A: Eₚ = 98 000 J
Step 2 – speed
F: Eₖ = ½ m v²
I: 98 000 = 0.5 × 500 × v², so 98 000 = 250 × v²
F: v² = 392, so v = √392
A: 19.8 m/s (3 s.f.)
Q4 (F) A catapult elastic has a spring constant of 800 N/m. It is stretched 0.10 m and fires a 20 g stone. Calculate the energy stored, then the launch speed. (Assume all the energy goes to the stone.)
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Step 1 – energy stored
F: Eₑ = ½ k e²
I: Eₑ = 0.5 × 800 × 0.10²
A: Eₑ = 4.0 J
Step 2 – speed (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 60 kg skier moving at 12 m/s coasts up a slope until she stops. Calculate how much height she gains. (Ignore friction.)
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Step 1 – kinetic energy
F: Eₖ = ½ m v²
I: Eₖ = 0.5 × 60 × 12²
A: Eₖ = 4320 J
Step 2 – height
F: Eₚ = m g h
I: 4320 = 60 × 9.8 × h, so 4320 = 588 × h
F: h = 4320 ÷ 588
A: 7.3 m (2 s.f.)
Q6 (F/H) A 1.5 kg hammer head moving at 8.0 m/s drives a nail 12 mm into a piece of wood. Calculate the average force on the nail.
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Convert first: 12 mm = 12 ÷ 1000 = 0.012 m
Step 1 – kinetic energy of the hammer
F: Eₖ = ½ m v²
I: Eₖ = 0.5 × 1.5 × 8.0²
A: Eₖ = 48 J
Step 2 – force (work done on nail = KE lost)
F: W = F s
I: 48 = F × 0.012
F: F = 48 ÷ 0.012
A: 4000 N
Q7 (F/H) A crane lifts a 400 kg load 15 m in 1.5 minutes. Calculate the useful power output of the crane, to 3 significant figures.
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Convert first: 1.5 minutes = 1.5 × 60 = 90 s
Step 1 – GPE gained
F: Eₚ = m g h
I: Eₚ = 400 × 9.8 × 15
A: Eₚ = 58 800 J
Step 2 – power
F: P = E / t
I: P = 58 800 ÷ 90 = 653.3… W
A: 653 W (3 s.f.)
Q8 (F/H) A rocket accelerates a 2.0 × 103 kg satellite from rest to 7.5 km/s in 8.0 minutes. Calculate the kinetic energy gained and the average useful power. Give both answers in standard form to 2 significant figures.
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Convert first: 7.5 km/s = 7.5 × 1000 = 7500 m/s; 8.0 minutes = 480 s
Step 1 – kinetic energy
F: Eₖ = ½ m v²
I: Eₖ = 0.5 × 2.0 × 103 × 7500²
A: Eₖ = 5.625 × 1010 J = 5.6 × 1010 J (2 s.f.)
Step 2 – average power
F: P = E / t
I: P = 5.625 × 1010 ÷ 480 = 1.17… × 108 W
A: 1.2 × 108 W (2 s.f.)
Use the unrounded KE in step 2.
Q9 (H) A 50 kg child starts from rest at the top of a water slide 800 cm high and reaches the bottom at 36 km/h. Calculate the energy wasted and the efficiency of the slide as a percentage, to 2 significant figures.
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Convert first: 800 cm = 800 ÷ 100 = 8.0 m; 36 km/h = 36 000 m ÷ 3600 s = 10 m/s
Step 1 – GPE lost (total input)
F: Eₚ = m g h
I: Eₚ = 50 × 9.8 × 8.0
A: Eₚ = 3920 J
Step 2 – KE at bottom (useful output)
F: Eₖ = ½ m v²
I: Eₖ = 0.5 × 50 × 10²
A: Eₖ = 2500 J
Step 3 – wasted energy and efficiency
Wasted = 3920 − 2500 = 1420 J
F: efficiency = useful output ÷ total input
I: efficiency = 2500 ÷ 3920 = 0.637…
A: 64% (2 s.f.)
Q10 (H) At a hydroelectric power station, 3.0 × 104 kg of water falls 0.12 km through the turbines every minute. The turbines are 85% efficient. Calculate the useful power output, in standard form to 2 significant figures.
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Convert first: 0.12 km = 0.12 × 1000 = 120 m; 1 minute = 60 s
Step 1 – GPE lost each minute
F: Eₚ = m g h
I: Eₚ = 3.0 × 104 × 9.8 × 120
A: Eₚ = 3.528 × 107 J
Step 2 – input power
F: P = E / t
I: P = 3.528 × 107 ÷ 60
A: P = 5.88 × 105 W
Step 3 – useful power
F: efficiency = useful output ÷ total input
I: 0.85 = useful ÷ 5.88 × 105
F: useful = 0.85 × 5.88 × 105 = 4.998 × 105 W
A: 5.0 × 105 W (2 s.f.)
