Question

In the figure shown, initially the spring is in relaxed position. A bullet of mass $m$ hits the block $A$ with horizontal speed $v_0$ and gets embedded into it. Find the maximum compression in the spring. Stiffness of the spring is $K$. Neglect any friction and deformation in the spring during collision.

Answer 1

Since no external force associated with the system $(M+M+m)$ conservation of horizontal momentum of the system yields
$m{v}_0=(M+m)v_1=(2M+m)v_2$
Where $v_1=$ velocity of $(M+m)$ just after the impact and $v_2=$ velocity of $(M+M+m)$ just of the maximum compression of the spring
$\Rightarrow v_1=\frac{m{v}_0}{M+m};v_2=\frac{m{v}_0}{2M+m}$
The loss of KE between the two portion (a) and (b) is given as
$\Delta KE=\frac{1}{2}(M+m)v_1^2-\frac{1}{2}(2M+m)v_2^2$
$\Rightarrow \Delta KE=\frac{m^2{v}_0^2}{2}[\frac{1}{(M+m)}-\frac{1}{(2M+m)}]=\frac{m^2{Mv}_0^2}{2(M+m)(2M+m)}$
From conservation of mechanical energy
$\Rightarrow \frac{1}{2}k{x}^2=\frac{m^2{Mv}_0^2}{2(M+m)(2M+m)}\Rightarrow x=\sqrt{\frac{m^2{Mv}_0^2}{2(M+m)(2M+m)}}=m{v}_0\sqrt{\frac{M}{2(M+m)(2M+m)}}$