A glass marble whose mass is 20 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

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Solution

$I = Δ p = m (¯ v - ¯ u) = m [\sqrt{2 g h_{2}} - \sqrt{2 g h_{1}}] = 0.2 [7 - (- 5.6)] = 2.52 N - s$

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A glass marble whose mass is 20 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

I=Δp=m(¯v−¯u)=m[√2gh2−√2gh1]=0.2[7−(−5.6)]=2.52 N−s

$I = Δ p = m (¯ v - ¯ u) = m [\sqrt{2 g h_{2}} - \sqrt{2 g h_{1}}] = 0.2 [7 - (- 5.6)] = 2.52 N - s$