Q. 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)
Explanation
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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
$\frac{1}{2}\frac{m^{2}v^{2}}{(m + M)} = \mu(m + M)g \times x \Rightarrow v^{2} = 2\mu gx\left( \frac{m + M}{m} \right)^{2}$
$\therefore v = \sqrt{2\mu gx}\left( \frac{m + m}{m} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.