# A bullet hits a solid target lying on a frictionless surface and gets embeded in it.what is conserved(kinetic energy, momentum,both or none)?

Aug 9, 2018

Let us consider the followings

$\text{mass of the bullet} = m$

$\text{initial velocity of the bullet} = v$

$\text{mass of the target} = M$

$\text{initial velocity of solid target} = 0$

$\text{final velocity of the combination} = V$

Here the collision is inelastic one as the bullet gets embedded in the target. Conservation of momentum occurs in all types of collision.

So by law conservation of momentum we get

$\left(M + m\right) V = m v$

$\implies V = \frac{m v}{M + m}$

Change kinetic energy due to collision

$\text{final KE" -"initial KE}$

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

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

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

$= - \frac{1}{2} m {v}^{2} \cdot \frac{M}{M + m} < 0$

So there is net loss in KE. It is not conserved here.