- Home
- Standard 11
- Physics
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
$\frac{1}{2}m{v^2}$
$\frac{1}{2}m{v^2} \times \frac{1}{{M + m}}$
$\frac{1}{2}m{v^2} \times \frac{M}{m}$
$\frac{1}{2}m{v^2}\left( {\frac{M}{{M + m}}} \right)$
Solution
(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 = $\frac{1}{2}m{v^2} – \frac{1}{2}(m + M)\;{\left( {\frac{{mv}}{{m + M}}} \right)^2}$
=$\frac{1}{2}m{v^2}\left( {\frac{M}{{m + M}}} \right)$
