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

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Question

(A) Let $V$ be the velocity of the block (with the bullet embedded in it) immediately after the collision.By the law of conservation of energy,the kin

Vedclass · Accepted Answer

$\sqrt{2gh}$

Vedclass · Answer

(A) Let $V$ be the velocity of the block (with the bullet embedded in it) immediately after the collision.By the law of conservation of energy,the kinetic energy of the system $(m+M)$ just after the collision is converted into potential energy at the maximum height $h$.$\frac{1}{2}(m+M)V^2 = (m+M)gh$Canceling $(m+M)$ from both sides:$\frac{1}{2}V^2 = gh$$V^2 = 2gh$$V = \sqrt{2gh}$Thus,the initial velocity of the block immediately after the collision is $\sqrt{2gh}$.

$A$ 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 block immediately after the collision will be

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