Question

A bullet of mass $m=50gm$ strikes a sand bag of mass $M=5kg$ hanging from a fixed point with a horizontal velocity ${\overrightarrow{v}}_p$. If bullet sticks to the sand bag, what fraction of K.E is lost during impact?

Answer 1

Since the bullet strikes the sand bag with a horizontal velocity of magnitude $v_p$. Therefore, its momentum just before the impact $=p_i=m{v}_p$
Let the speed of the combination (sand bad+bullet) just after the impact be $v^{\prime }$. Therefore the momentum of the system (sand bag+bullet) just after the impact $=p_f=(M+m)v^{\prime }$. Since no net horizontal external force acts on the system $(M+m)$ during the collision, its net momentum remains conserved during the impact
$\Rightarrow p_i=p_f\Rightarrow mv=(M+m)v^{\prime }\Rightarrow v^{\prime }=\left(\frac{m}{M+m}\right)v$
Since $m<<M,v^{\prime }=\frac{mv}{M}....\left(1\right)$
Fraction of KE lost be the bullet during impact
$\eta ={\left(\frac{v^{\prime }}{v}\right)}^2.....\left(2\right)$
Using (1) and (2) we obtain
$\eta ={\left(\frac{m}{M}\right)}^2={\left(\frac{1/20}{5}\right)}^2={10}^{-4}$