Conservation of momentum distinguishes collision types

Steps:

• For wooden block (inelastic collision): Ball embeds, so final velocity $V_w = \frac{m v}{m + M}$, where $m$ is ball mass, $M$ is block mass.
• For steel block: Ball rebounds at $v/2$, so $m(-v/2) + M V_s = m v$, solving gives $V_s = \frac{3 m v}{2 M}$.
• Compare: $V_s = \frac{3}{2} \frac{m}{M} v > \frac{m}{m + M} v = V_w$ since $\frac{3}{2} \frac{m}{M} > \frac{m}{m + M}$ for $m, M > 0$.
• Thus, steel block travels faster.

Why B is correct:

• Momentum conservation yields $V_s > V_w$ directly from collision outcomes, as rebound transfers more momentum to the steel block (Newton's third law in partial elastic interaction).

Why the others are wrong:

• A: Speeds differ due to inelastic vs. rebounding collision; $V_s > V_w$.
• C: Equal masses and given speeds suffice; no additional info needed.

Final answer: B