A glass marble whose mass is $200 g$ falls from a height of $2.5 m$ and rebounds to a height of $1.6 m$ . Find the change in momentum due to its rebound.

4.9 N s

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8.8 N s

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1.57 N s

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2.54 N s

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Solution

The correct option is D $2.54 N s$
Let $h_{2}$ be 2.5 $m$ and $h_{1}$ be 1.6 $m$ ,
We know Impulse is change in momemtum and velocity of glass marble at the time of hitting the ground and rebound can be calculted by relation $v = \sqrt{2 g h}$
Hence,
$I = Δ p = m (\to v - \to u)$
$= m [\sqrt{(2 g h_{2}}) - \sqrt{(2 g h_{1}})]$
$= 0.2 [7 - (- 5.6)] = 2.54 N s$

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