A wooden block of mass M is suspended b | vedclass

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Question

$mv + M 	imes 0 = m\frac{v}{2} + M 	imes V$

==> $V = \frac{m}{{2M}}v$

$h = \frac{{{V^2}}}{{2g}} = \frac{{{{(mv/2M)}^2}}}{{2g}} = \frac{{{m^

Vedclass · Accepted Answer

${m^2}{v^2}/8{M^2}g$

Vedclass · Answer

$mv + M 	imes 0 = m\frac{v}{2} + M 	imes V$

==> $V = \frac{m}{{2M}}v$

$h = \frac{{{V^2}}}{{2g}} = \frac{{{{(mv/2M)}^2}}}{{2g}} = \frac{{{m^2}{v^2}}}{{8{M^2}g}}$.

A wooden block of mass $M$ is suspended by a cord and is at rest. A bullet of mass $m,$ moving with a velocity $v$ passes through the block and comes out with a velocity $v/2$ in the same direction. If there is no loss in kinetic energy, then upto what height the block will rise

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