A ball of mass m falls vertically to the ground from a height $h_{1}$ and rebound to a height $h_{2}$ . The change in momentum of the ball on striking the ground is

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Solution

The correct option is B

When ball falls vertically downward from height $h_{1}$ its velocity ${\to v}_{1} = - \sqrt{2 g h_{1}}$ and its velocity after collision ${\to v}_{2} = \sqrt{2 g h_{2}}$

Change in momentum $Δ \to P = m ({\to v}_{2} - {\to v}_{1}) = m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$

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A ball of mass m falls vertically to the ground from a height h1 and rebound to a height h2. The change in momentum of the ball on striking the ground is

The correct option is B When ball falls vertically downward from height h1 its velocity →v1=−√2 gh1 and its velocity after collision →v2=√2 gh2 Change in momentum Δ→P=m(→v2−→v1)=m(√2gh1+√2gh2)

A ball of mass m falls vertically to the ground from a height $h_{1}$ and rebound to a height $h_{2}$ . The change in momentum of the ball on striking the ground is

The correct option is B

When ball falls vertically downward from height $h_{1}$ its velocity ${\to v}_{1} = - \sqrt{2 g h_{1}}$ and its velocity after collision ${\to v}_{2} = \sqrt{2 g h_{2}}$

Change in momentum $Δ \to P = m ({\to v}_{2} - {\to v}_{1}) = m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$