Question
A 80 kg man skating with a speed of 10 m/s on a smooth horizontal surface collides with a 20 kg skater at rest, as a result the two cling to each other. What percentage of initial kinetic energy of skaters is lost during the collision? (Ignore any friction)
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Solution
Sol. Key concept: Momentum conservation in a perfectly inelastic collision.
There is no external horizontal force, so horizontal momentum is conserved. The two skaters move together after collision.
$$v = \frac{80(10) + 20(0)}{80 + 20} = 8\ m\ s^{- 1}$$
$$K_{i} = \frac{1}{2}(80)(10)^{2} = 4000\ J$$
$$K_{f} = \frac{1}{2}(100)(8)^{2} = 3200\ J$$
$$\%\ loss = \frac{4000 - 3200}{4000} \times 100 = 20\%$$
Momentum is conserved, but kinetic energy is not conserved when the skaters cling together. Hence (c).
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