A bag of sand of mass M is suspended by a str | vedclass

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Question

(d) Initial kinetic energy of bullet = $\frac{1}{2}m{v^2}$

After inelastic collision system moves with velocity $V$

By the conservation of momen

Vedclass · Accepted Answer

$\frac{1}{2}m{v^2}\left( {\frac{M}{{M + m}}} \right)$

Vedclass · Answer

(d) Initial kinetic energy of bullet = $\frac{1}{2}m{v^2}$

After inelastic collision system moves with velocity $V$

By the conservation of momentum

$mv + 0 = (m + M)\,V$ ==> $V = \frac{{mv}}{{m + M}}$

Kinetic energy of system = $\frac{1}{2}(m + M)\;{V^2}$

= $\frac{1}{2}(m + M)\;{\left( {\frac{{mv}}{{m + M}}} \right)^2}$

Loss of kinetic energy = $\frac{1}{2}m{v^2} - \frac{1}{2}(m + M)\;{\left( {\frac{{mv}}{{m + M}}} \right)^2}$

=$\frac{1}{2}m{v^2}\left( {\frac{M}{{m + M}}} \right)$

A bag of sand of mass $M$ is suspended by a string. A bullet of mass $m$ is fired at it with velocity $v$ and gets embedded into it. The loss of kinetic energy in this process is

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