1.3 Kinetic and gravitational potential energy
9 exam questions worth 23 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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Calculate the kinetic energy of a 1200 kg car travelling at 10 m/s.
Mark scheme
Ek = ½ × 1200 × 102 (1); 60 000 J (1)
Model answer
Ek = ½mv² = 0.5 × 1200 × 10² = 60 000 J.
A child of mass 30 kg climbs to the top of a slide 2.5 m high. Calculate the gravitational potential energy gained. Gravitational field strength = 9.8 N/kg.
Mark scheme
Ep = 30 × 9.8 × 2.5 (1); 735 J (1)
Model answer
Ep = mgh = 30 × 9.8 × 2.5 = 735 J.
The speed of a moving object is doubled. What happens to its kinetic energy?
Choose one answer.
Correct Not quite: the answer is It is 4 times bigger
Mark scheme
C: 4 times bigger
Model answer
It becomes 4 times bigger (C), because kinetic energy depends on the speed squared.
A ball has 8.0 J of kinetic energy and a mass of 0.25 kg. Calculate its speed.
Mark scheme
v2 = 2 × 8.0 / 0.25 (1); = 64 (1); v = 8 m/s (1)
Model answer
v² = 2Ek / m = 16 / 0.25 = 64, so v = 8 m/s.
A stone of mass 0.40 kg is dropped from a height of 5.0 m. Ignore air resistance. Calculate the speed of the stone just before it hits the ground. Gravitational field strength = 9.8 N/kg.
Mark scheme
Ep = 0.40 × 9.8 × 5.0 = 19.6 J (1); Ek = Ep / gain in Ek equals loss in Ep (1); v2 = 2 × 19.6 / 0.40 = 98 (1); v = 9.9 m/s (1)
Model answer
Ep lost = mgh = 0.40 × 9.8 × 5.0 = 19.6 J. This all becomes kinetic energy, so ½ × 0.40 × v² = 19.6, v² = 98 and v = 9.9 m/s.
Describe the energy changes of a pendulum as it swings from its highest point to its lowest point.
Mark scheme
gravitational potential energy decreases (1); kinetic energy increases (1); (energy is transferred from the gravitational potential store to the kinetic store)
Model answer
Gravitational potential energy decreases and kinetic energy increases, so the pendulum moves fastest at the lowest point.
A roller coaster car of mass 500 kg starts from rest at the top of a hill 30 m above the lowest point of the track. Ignore friction. Calculate the speed of the car at the lowest point. (g = 9.8 N/kg)
Mark scheme
E_p lost = m g h = 500 × 9.8 × 30 = 147 000 J (1); E_k = E_p (1); v² = 2 × 147 000 ÷ 500 = 588 (1); v = 24 m/s (1)
Model answer
E_p = 500 × 9.8 × 30 = 147 000 J. This becomes kinetic energy, so ½ × 500 × v² = 147 000, giving v² = 588 and v = 24 m/s.
A ball is thrown vertically upwards at 12 m/s. Ignore air resistance. Calculate the maximum height it reaches. (g = 9.8 N/kg)
Mark scheme
kinetic energy at the start is transferred to gravitational potential energy: ½ m v² = m g h (1); h = v² ÷ 2g = 144 ÷ 19.6 (1); 7.3 m (1)
Model answer
½ m v² = m g h, so h = v² / 2g = 144 / 19.6 = 7.3 m.
A car crashes at 30 m/s. Another identical car crashes at 15 m/s. Explain why the damage is likely to be much greater in the first crash.
Mark scheme
kinetic energy depends on the speed squared, so doubling the speed gives four times the kinetic energy (1); so there is four times as much energy to be transferred in the crash (1)
Model answer
Doubling the speed makes the kinetic energy four times bigger, so there is much more energy transferred to deform the car.
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