A glass marble whose mass is $200 g m$ , 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 fall.

5.65 N s

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25.4 N s

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56.5 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 $u$ be the velocity with which it hits the ground and $v$ be the velocity with which it rebounds.
Then, $u = - \sqrt{2 g h_{1}} = - \sqrt{50} m / s$ and
$v = \sqrt{2 g h_{2}} = \sqrt{32} m / s$
(taking upward direction as +ve)

Then change in momentum
$Δ p = m (v - u) = m (\sqrt{32} - (- \sqrt{50}))$
$= 0.2 (5.65 - (- 7.07)) = 2.54 N s$

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A glass marble whose mass is 200 gm, 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 fall.

A glass marble whose mass is $200 g m$ , 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 fall.