A bullet of mass m moving with velocity v str | vedclass

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(B) By the law of conservation of linear momentum during the collision:$mv + M(0) = (m + M)V$where $V$ is the common velocity of the bullet and the bl

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$\frac{(M + m)}{m}\sqrt{2gh}$

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(B) By the law of conservation of linear momentum during the collision:$mv + M(0) = (m + M)V$where $V$ is the common velocity of the bullet and the block immediately after the collision.$V = \frac{mv}{m + M}$Now,applying the law of conservation of mechanical energy for the block-bullet system as it rises to height $h$:$\frac{1}{2}(m + M)V^2 = (m + M)gh$$V^2 = 2gh$$V = \sqrt{2gh}$Equating the two expressions for $V$:$\frac{mv}{m + M} = \sqrt{2gh}$$v = \frac{(M + m)}{m}\sqrt{2gh}$

$A$ bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$,the initial velocity of the bullet will be

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