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1B

2The elastic energy is four times as large and so is the kinetic energy of the ball. Hence the speed is twice as large.

3It is the same since the vertical height is the same.

4The boy raises his centre of mass by a total of 1.3 m and so the work done by the muscles is mgh = 60 × 10 × 1.3 = 780 J.

After losing contact with the ground he moves up a distance of 0.80 m so his launch speed is v=2 g h=4.0 m s−1.

His acceleration when in contact with the ground is then found from 4.02 = 0 + 2a × 0.5 ⇒ a = 16 m s−2

so he was in contact with the ground for 4.0 = 0 + 16 × t ⇒ t = 0.25 s.

Hence the average power developed is 780 0.25−3120≈3 kW.

Equivalently, with an acceleration of a = 16 m s−2 the force from the ground F is found from F − mg = ma and so F = 60 × 10 + 60 × 16 = 1560 N.

The average power is then P¯=F u+v 2=1560×0+4.0 2=3120 W.

5D (kinetic energy is constant)

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