Conservation of momentum in inelastic collision

Steps:

• The collision is perfectly inelastic; the pellet embeds in the block.
• Initial momentum: $m v_0$ (block at rest).
• Final momentum immediately after impact: $(m + M) v$.
• Set equal: $m v_0 = (m + M) v$, so $v_0 = v \frac{m + M}{m}$.

Why C is correct:

• For equal masses ($M = m$, as implied by diagram), $v_0 = 2v$ from momentum conservation.

Why the others are wrong:

• A: $v$ ignores mass ratio; underestimates $v_0$.
• B: $v^2$ has incorrect units (not speed).
• D: $4v$ overestimates; applies to elastic collision with equal masses, not inelastic.

Final answer: C