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

$\sqrt{2\mu gx}\left(\frac{M+m}{m}\right)$

Let speed of the bullet = v

Speed of the system after the collision = V

By conservation of momentum mv = (m + M)V

$\Rightarrow V=\frac{mv}{M+m}$

So the initial K.E. acquired by the system

$=\frac{1}{2}(M+m)V^2=\frac{1}{2}(m+M){\left(\frac{mv}{M+m}\right)}^2=\frac{1}{2}\frac{m^2{v}^2}{(m+M)}$

This kinetic energy goes against friction work done by friction $=\mu R\times x=\mu (m+M)g\times x$

By the law of conservation of energy

$\frac{1}{2}\frac{m^2{v}^2}{(m+M)}=\mu (m+M)g\times x\Rightarrow v^2=2\mu gx{\left(\frac{m+M}{m}\right)}^2\therefore v=\sqrt{2\mu gx}\left(\frac{M+m}{m}\right)$