Difficulty: Easy
Correct Answer: Change in momentum suffered by the lead ball is less than that of the rubber ball
Explanation:
Introduction / Context:Momentum change in collisions depends on whether the object rebounds or stops. The rebound reverses the normal component of velocity, increasing the total change in momentum compared to a case where the body merely comes to rest against the surface.
Given Data / Assumptions:
Concept / Approach:Change in momentum Δp along the line of impact equals m * (v_after − v_before). If the body stops, v_after ≈ 0 → Δp ≈ −m * v_before (magnitude m * v). If it rebounds with nearly equal speed in the opposite direction, v_after ≈ −v_before → Δp ≈ −2 m * v_before (magnitude 2 m * v), which is larger.
Step-by-Step Solution:
Lead ball: v_after ≈ 0 → |Δp_lead| = m * v.Rubber ball: v_after ≈ −k v, with 0 < k ≤ 1 → |Δp_rubber| = m * v * (1 + k) ≥ m * v.For nearly elastic rebound (k close to 1), |Δp_rubber| ≈ 2 m * v > |Δp_lead|.Verification / Alternative check:Impulse J equals change in momentum. A larger rebound speed implies a larger impulse from the wall, consistent with higher Δp for the rubber ball.
Why Other Options Are Wrong:
Common Pitfalls:Comparing momenta instead of changes in momentum; ignoring direction when evaluating Δp vectors.
Final Answer:Change in momentum suffered by the lead ball is less than that of the rubber ball
Discussion & Comments