Question

A bullet of mass $m$ moving with a velocity $u$ gets embedded in a block of mass $M$ initially at rest. If $\mu$ is the coefficient of friction between the block and the ground, then the displacement of the block due to the impact is:

• A. $\dfrac{(Mu)^2}{2\mu g}$
• B. $\dfrac{(mu)^2}{2\mu g(M+m)^2}$
• C. $\dfrac{(mu)^2}{2\mu Mg}$
• D. $\dfrac{(mu)^2}{2\mu (M+m)g}$

Hint:

In inelastic collisions, momentum is conserved but kinetic energy is partly lost as heat or deformation. Always apply momentum conservation first, then energy considerations with friction. Use work–energy theorem to find stopping distance for sliding motion. Units of displacement remain consistent if all quantities are in SI (m, kg, s).

Answer 1

The Correct Option is B

• The bullet gets embedded — this is a case of a perfectly inelastic collision. Momentum is conserved, but kinetic energy is not.
• By conservation of momentum: \[ mu = (M+m)V \Rightarrow V = \frac{mu}{M+m} \] • Immediately after the impact, the block–bullet system slides against friction until it stops. The work done by friction equals the kinetic energy lost: \[ \frac{1}{2}(M+m)V^2 = \mu (M+m) g S \] • Solving for displacement $S$: \[ S = \frac{V^2}{2\mu g} = \frac{(mu)^2}{2\mu g(M+m)^2} \] • Hence, the displacement of the block due to the impact is $\dfrac{(mu)^2{2\mu g(M+m)^2}$}.