Question

A wooden block of mass $M$ rests on a horizontal surface. A bullet of mass $m$ moving in the horizontal direction strikes and gets embedded in it. The combined system covers a distance $x$ on the surface. If the coefficient of friction between wood and the surface is $\mu$, the speed of the bullet at the time of striking the block is (where $m$ is mass of the bullet)

• A. $\sqrt{\frac{2Mg}{\mu m}}$
• B. $\sqrt{\frac{2\mu mg}{Mx}}$
• C. $\sqrt{2\mu gx}\left(\frac{M+m}{m}\right)$
• D. $\sqrt{\frac{2\mu mx}{M+m}}$

Answer 1

The correct option is C $\sqrt{2\mu gx}\left(\frac{M+m}{m}\right)$

Conservation of momentum gives
$mu=(M+m)V\Rightarrow V=\frac{mu}{M+m}$

$K.E$ of the combined system is
$K.E=\frac{1}{2}(M+m)V^2=\frac{1}{2}(M+m)\frac{m^2{u}^2}{(M+m{)}^2}=\frac{m^2{u}^2}{2(M+m)}$

Once the system enters the rough surface, it loses all its kinetic energy to overcome the work done against friction. We have,

$K_i+U_i+W_{NC}=K_f+U_f$
$\Rightarrow \frac{m^2{u}^2}{2(M+m)}+0+f_{k}xcos{180}^{\circ }=0+0$
$\Rightarrow \frac{m^2{u}^2}{2(M+m)}=\mu (M+m)gx[\because f_k=\mu N]\Rightarrow u=\sqrt{2\mu gx}\left(\frac{M+m}{m}\right)$