A ball of mass 3 kg, is dropped onto the ground. The velocity of the ball just before hitting the ground is 2.2 ms-1 and it bounces back up with 60 % of the magnitude of the downward velocity.Taking downward direction as positive, what is the change in momentum (kg m/s) of the ball immediately after bouncing back up again?

Question

BlackTom AI · Accepted Answer

We start by noting the given data and the chosen sign convention: downward is positive, so the velocity just before impact is +2.2 m/s.
Option A: -13.20 kg m/s. To get such a value, one would have to assume a much larger change in velocity or a different final velocity. If we incorrectly took the rebound velocity as +1.32 m/s (same direction), the calculation m(v_final − v_initial) would not match this negative value. This option does not align with the given rebound speed being 60% of the downward speed in the opposite direction.
Option B: -10.56 kg m/s. The rebound velocity magnitude is 60% of the downward speed, so v_final = −(0.60 × 2.2) = −1.32 m/s. Change in momentum Δp = m(v_final − v_initial) = 3(−1.32 − 2.2) = 3(−3.52) = −10.56 kg m/s. This matches the sign convention (negative, since the momentum flips direction and reduces in the downward-positive frame) and the magnitudes involved.
Option C: 2.64 kg m/s. This would correspond to a small positive change in momentum, which would imply either a final velocity only slightly different from the initial one or a miscalculation (e.g., using +1.32 m/s for the final velocity while keeping +2.2 m/s for the initial). It ignores the correct direction reversal and the appropriate subtraction in Δp.
Option D: 13.20 kg m/s. Achieving a magnitude of +13.20 would require a much larger net change in velocity (or a different sign convention) than provided. Specifically, to reach 13.20 with m = 3, the velocity change would need to be 4.4 m/s, which does not correspond to the 2.2 m/s incoming and 1.32 m/s outgoing speeds with the given directions.

Putting it together, the physically consistent calculation uses the specified rebound speed (60% of 2.2 m/s) in the opposite direction and the chosen downward-positive convention, yielding Δp = −10.56 kg m/s. The other options arise from incorrect assumptions about rebound velocity, direction, or arithmetic errors in computing Δp.

MUF0121 Physics Unit 1 - Semester 2, 2025

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