A ball of mass 0.2 kg is dropped from a height of 5 m. It hits the ground and rebounds to the same height of 5 m. Calculate the magnitude of the impulse exerted by the ground on the ball in Ns. (Assum

Question

A ball of mass 0.2 kg is dropped from a height of 5 m. It hits the ground and rebounds to the same height of 5 m. Calculate the magnitude of the impulse exerted by the ground on the ball in Ns. (Assume g = 10 m / s 2 ).

TestHub · Accepted Answer

The velocity of the ball just before hitting the ground can be found using $v^{2} = u^{2} + 2 g h$ . Since it's dropped, $u = 0$ . So, $v_{d o w n} = 2 g h = 2 \times 10 \times 5 = 100 = 10 m / s$ (downwards).

Since it rebounds to the same height, its velocity just after hitting the ground is $v_{u p} = 2 g h = 10 m / s$ (upwards). Impulse $J = Δ p = m (v_{f ina l} - v_{ini t ia l})$ . Taking upwards as positive, $v_{ini t ia l} = - 10 m / s$ and $v_{f ina l} = + 10 m / s$ . $J = 0.2 \times (10 - (- 10)) = 0.2 \times 20 = 4 N s$

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