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5.Work, Energy, Power and Collision
medium
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
A
${m^2}{v^2}/2{M^2}g$
B
${m^2}{v^2}/8{M^2}g$
C
${m^2}{v^2}/4Mg$
D
${m^2}{v^2}/2Mg$
Solution
$mv + M \times 0 = m\frac{v}{2} + M \times V$
==> $V = \frac{m}{{2M}}v$
$h = \frac{{{V^2}}}{{2g}} = \frac{{{{(mv/2M)}^2}}}{{2g}} = \frac{{{m^2}{v^2}}}{{8{M^2}g}}$.
Standard 11
Physics
