Specification: AQA A-level Physics (7408) 3.4.1.8 | 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
Kinetic energy: the energy an object has because of its motion, equal to the work that would be done to accelerate it from rest to its speed: Eₖ = ½mv².
Gravitational potential energy: the energy an object has because of its position in a gravitational field. Near the Earth’s surface, where g is uniform, the change is ΔEₚ = mgΔh.
Principle of conservation of energy: energy cannot be created or destroyed, only transferred from one form to another. The total energy of a closed system stays constant; work done against resistive forces is transferred to internal (thermal) energy.
The equation
kinetic energy = ½ × mass × (speed)² | change in gravitational potential energy = mass × gravitational field strength × change in height Eₖ = ½mv² ΔEₚ = mgΔh With resistive forces: ΔEₚ lost = Eₖ gained + work done against resistance (F × d)
| Quantity | Symbol | Unit |
|---|---|---|
| kinetic energy | Eₖ | joules (J) |
| change in gravitational potential energy | ΔEₚ | joules (J) |
| mass | m | kilograms (kg) |
| speed | v | metres per second (m s⁻¹) |
| gravitational field strength | g | newtons per kilogram (N kg⁻¹), 9.81 N kg⁻¹ |
| change in height (vertical) | Δh | metres (m) |
| resistive force | F | newtons (N) |
| distance moved against the resistive force | d | metres (m) |
Rearranged: v = √(2Eₖ ÷ m) | m = 2Eₖ ÷ v² | Δh = ΔEₚ ÷ mg | with no resistance, ½mv² = mgΔh so v = √(2gΔh) and the mass cancels | Δh is always the vertical height change, never the length of a slope.
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 0.057 kg tennis ball is served at 50.0 m s⁻¹. Calculate its kinetic energy, to 3 s.f.
Show solution
F: Eₖ = ½mv²
I: Eₖ = 0.5 × 0.057 × 50.0²
F: Eₖ is already the subject
A: 0.5 × 0.057 × 2500 = 71.25 = 71.3 J (3 s.f.)
Q2 (Core) A 65.0 kg climber climbs a vertical height of 450 m. Calculate the increase in her gravitational potential energy, to 3 s.f. (g = 9.81 N kg⁻¹)
Show solution
F: ΔEₚ = mgΔh
I: ΔEₚ = 65.0 × 9.81 × 450
F: ΔEₚ is already the subject
A: ΔEₚ = 286 942.5 J = 2.87 × 105 J (3 s.f.)
Q3 (Core) A 57 g tennis ball has 76 J of kinetic energy. Calculate its speed, to 3 s.f.
Show solution
Convert first: 57 g = 0.057 kg
F: Eₖ = ½mv²
I: 76 = 0.5 × 0.057 × v²
F: v² = 2 × 76 ÷ 0.057 = 2666.6…, so v = √2666.6…
A: v = 51.63… = 51.6 m s⁻¹ (3 s.f.)
Watch out: leaving the mass in grams gives an answer √1000 ≈ 32 times too small.
Q4 (Core) A car travelling at 90 km h⁻¹ has 375 kJ of kinetic energy. Calculate the mass of the car.
Show solution
Convert first: 90 km h⁻¹ = 90 000 m ÷ 3600 s = 25 m s⁻¹; 375 kJ = 375 000 J
F: Eₖ = ½mv²
I: 375 000 = 0.5 × m × 25², so 375 000 = 312.5m
F: m = 375 000 ÷ 312.5
A: 1200 kg
Q5 (Exam) A pendulum bob is pulled aside so that it is raised 12.0 cm above its lowest point, then released from rest. Calculate its speed at the lowest point, to 3 s.f. (g = 9.81 N kg⁻¹; ignore air resistance)
Show solution
Convert first: 12.0 cm = 0.120 m
F: Eₖ gained = ΔEₚ lost, so ½mv² = mgΔh, giving v = √(2gΔh)
I: v = √(2 × 9.81 × 0.120) = √2.3544
F: v is already the subject
A: v = 1.534… = 1.53 m s⁻¹ (3 s.f.)
Watch out: you do not need the mass – it cancels. Δh is the vertical rise of the bob, not the length of the string.
Q6 (Exam) A roller-coaster car passes over the top of a hill at 3.0 m s⁻¹, then drops through a vertical height of 37.0 m. Assuming no energy is lost to friction or air resistance, calculate its speed at the bottom, to 3 s.f. (g = 9.81 N kg⁻¹)
Show solution
F: ½mv² = ½mu² + mgΔh, so v² = u² + 2gΔh (the mass cancels)
I: v² = 3.0² + 2 × 9.81 × 37.0 = 9.0 + 725.94
F: v² = 734.94, so v = √734.94
A: v = 27.10… = 27.1 m s⁻¹ (3 s.f.)
Watch out: the car’s starting kinetic energy must be included. Using v = √(2gΔh) ignores the 3.0 m s⁻¹ at the top.
Q7 (Exam) A 0.145 kg baseball is thrown vertically upwards at 25.0 m s⁻¹ and reaches a maximum height of 28.0 m. Calculate (a) the energy transferred to the surroundings by air resistance, and (b) the average air resistance force on the ball. Give both answers to 3 s.f. (g = 9.81 N kg⁻¹)
Show solution
Step 1 – energies
F: Eₖ = ½mv² and ΔEₚ = mgΔh
I: Eₖ = 0.5 × 0.145 × 25.0² = 45.3125 J; ΔEₚ = 0.145 × 9.81 × 28.0 = 39.8286 J
F: energy lost = Eₖ − ΔEₚ = 45.3125 − 39.8286
A: (a) energy lost = 5.4839 = 5.48 J (3 s.f.)
Step 2 – average force
F: work done = F × d, so F = W ÷ d
I: F = 5.4839 ÷ 28.0
A: (b) F = 0.1958… = 0.196 N (3 s.f.)
Q8 (Exam) A 70.0 kg skier starts from rest and skis down a slope 120 m long, descending a vertical height of 25.0 m. Her speed at the bottom is 18.0 m s⁻¹. Calculate (a) the energy transferred by resistive forces, to 3 s.f., and (b) the average resistive force acting along the slope, to 3 s.f. (g = 9.81 N kg⁻¹)
Show solution
Step 1 – energy changes
I: ΔEₚ lost = mgΔh = 70.0 × 9.81 × 25.0 = 17 167.5 J; Eₖ gained = ½mv² = 0.5 × 70.0 × 18.0² = 11 340 J
F: energy lost = ΔEₚ − Eₖ = 17 167.5 − 11 340
A: (a) 5827.5 J = 5.83 × 103 J (3 s.f.)
Step 2 – average force
F: work done = F × d, so F = 5827.5 ÷ 120
A: (b) F = 48.56… = 48.6 N (3 s.f.)
Watch out: use the vertical drop (25.0 m) for ΔEₚ but the distance along the slope (120 m) for the work done against friction.
Q9 (Stretch) An 800 kg roller-coaster car starts from rest at the top of a drop of vertical height 45.0 m. It travels 150 m along the track to the bottom against an average resistive force of 1200 N. Calculate, to 3 s.f., (a) its speed at the bottom and (b) the speed it would have with no resistive forces. (c) The track then rises over a hill 20.0 m high, 60.0 m further along the track, with the same resistive force. Show whether the car reaches the top of this hill. (g = 9.81 N kg⁻¹)
Show solution
Step 1 – energy at the bottom
I: ΔEₚ lost = 800 × 9.81 × 45.0 = 353 160 J; work against resistance = 1200 × 150 = 180 000 J
F: Eₖ = 353 160 − 180 000 = 173 160 J, and v = √(2Eₖ ÷ m)
I: v = √(2 × 173 160 ÷ 800) = √432.9
A: (a) v = 20.80… = 20.8 m s⁻¹ (3 s.f.)
Step 2 – no resistance: v = √(2gΔh) = √(2 × 9.81 × 45.0) = √882.9
A: (b) v = 29.71… = 29.7 m s⁻¹ (3 s.f.)
Step 3 – the hill: energy needed = mgΔh + F × d = 800 × 9.81 × 20.0 + 1200 × 60.0 = 156 960 + 72 000 = 228 960 J
A: (c) 228 960 J > 173 160 J available, so the car does not reach the top of the hill
Watch out: in (c) both the gain in Eₚ and the work against resistance on the climb must come from the car’s kinetic energy.
Q10 (Stretch) A 5.00 kg box is released from rest at the top of a ramp 3.00 m long, inclined at 25.0° to the horizontal. A constant friction force of 8.00 N acts on it along the ramp. (a) Calculate the speed of the box at the bottom of the ramp, to 3 s.f. (b) The box then slides onto a rough horizontal floor where the friction force is 15.0 N. Calculate how far it slides before stopping, to 3 s.f. (g = 9.81 N kg⁻¹)
Show solution
Step 1 – vertical drop: Δh = 3.00 × sin 25.0° = 1.267… m
Step 2 – energy
I: ΔEₚ lost = 5.00 × 9.81 × 1.267… = 62.18… J; work against friction = 8.00 × 3.00 = 24.0 J
F: Eₖ = 62.18… − 24.0 = 38.18… J, and v = √(2Eₖ ÷ m)
I: v = √(2 × 38.18… ÷ 5.00)
A: (a) v = 3.908… = 3.91 m s⁻¹ (3 s.f.)
Step 3 – sliding on the floor: all the kinetic energy is done as work against friction, so Eₖ = F × d
I: 38.18… = 15.0 × d
A: (b) d = 2.545… = 2.55 m (3 s.f.)
Watch out: Δh = L sin θ, not L cos θ. Draw the right-angled triangle: the height is opposite the angle of the slope.
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