Question

A bullet of mass m is fired at speed v0 into a wooden block of mass M. The bullet instantaneously comes to rest in the block. The block with the embedded bullet slides along a horizontal surface with a coefficient of kinetic friction m.

Answer 1

Let us determine the momentum of the bullet. It is $mv_0.$ Next, we should determine the velocity $v$ of the system "bullet+block". According to the momentum conservation law, we get $mv_0 = (M+m)v$, so $v = \dfrac{m}{m+M}v_0$ . Next, we should calculate the kinetic energy of the system "bullet+block" at the beginning of their simultaneous motion. The energy will be

$E_0 = \dfrac{(m+M)v^2}{2} = \dfrac{m^2v_0^2}{2(m+M)}.$

The system will stop when the modulus of the friction force work will be equal to $E_0,$ so $E_0=|A_{fr}|$ , where $|A_{fr}| = \mu NS = \mu (m+M)gS.$ Therefore, the distance, after that the system will stop, is $S = \dfrac{E_0}{\mu (m+M)g} = \dfrac{m^2v_0^2}{2\mu (m+M)^2g}$ .