Question

Two identical blocks A and B of mass M each are kept on each other on a smooth horizontal plane. There exists friction between A and B. If a bullet of mass m hits the lower block with a horizontal velocity v and gets embedded into it, the work done by friction between A and B is

• A. - $\frac{M{m}^2{v}^2}{2(m+2M)(m+M)}$
• B. - $\frac{M{m}^2{v}^2}{(m+2M)(m+2M)}$
• C. - $\frac{M{m}^2{v}^2}{2(m+2M)(m+M)}$
• D. - $\frac{M{m}^2{v}^2}{2(m+2M)(2m+M)}$

Answer 1

The correct option is C

- $\frac{M{m}^2{v}^2}{2(m+2M)(m+M)}$

Let the velocity of lower block just after collision be $V_1$

$\therefore$ Applying conservation of liner momentum between bullet and lower block as impluse due to friction is negligible.

$\therefore$ mv = (M + m )$V_1$ $\Rightarrow$ $V_1$ = ( $\frac{m}{M+m}$) V

After collision frictional force acts on both and will act till velocities of both blocks become equal.

$\therefore$ (M + m )$V_1$ = (2M + m )$V_2$ $\Rightarrow$ $V_2$ ( $\frac{m}{2M+m}$) V

Applying work energy theorem ,$W_{Friction}$ = $K_f$ - $K_i$ = $\frac{1}{2}$ [2 M + m] ${V_2}^2$

$\Rightarrow$ $K_{friction}$ = $\frac{-M{m}^2{v}^2}{2(m+2M)(m+M)}$