Conservation of Horizontal Momentum in Inelastic Collision

Steps:

• The cart moves with velocity $v$; rock dropped vertically has zero horizontal velocity, so initial horizontal momentum is $m v$ (cart mass $m$).
• Rock lands and stays in cart, forming inelastic collision; horizontal momentum conserved: final momentum $(m + M) v_f = m v$, so $v_f = \frac{m v}{m + M} < v$ (rock mass ).
• Final kinetic energy: $\frac{1}{2} (m + M) v_f^2 = \frac{1}{2} \frac{m^2 v^2}{m + M} < \frac{1}{2} m v^2$ (initial KE decreases due to inelastic nature).
• No external horizontal forces (negligible resistance), so momentum same, but KE not conserved.

Why B is correct:

• Momentum conserved by definition in isolated horizontal system; KE lost to deformation/heat in inelastic collision.

Why the others are wrong:

• A: Momentum same, not smaller (conservation law).
• C: KE decreases, not same (inelastic collision).
• D: KE smaller, not same (as above); momentum same, not smaller.

Final answer: B