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.

类似问题

A 4.6 kg rock is dropped from a clifftop and falls 7.1 m to the ground. How much momentum did it have just before it hit the ground? Give your answer to the nearest whole number.

How is change of momentum calculated?

What is the momentum of a rhinoceros of mass 1147 kg charging at 14 km/h? Give your answer to the nearest whole number.

What is the speed (in km/h) of a table tennis ball of mass 2.7 g if it has momentum of 0.06 kg ms-1? Give your answer to the nearest whole number.

Which of the following represents the correct formula for momentum?

A 75-g ball strikes a wall horizontally at 12 m/s and rebounds in the opposite direction at 10 m/s. The ball’s change of momentum has magnitude

What is the speed (in km/h) of a table tennis ball of mass 2.9 g if it has momentum of 0.08 kg ms-1? Give your answer to the nearest whole number.

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单