A 0.50 kg ball moving at 6.0 m/s hits a wall and comes to rest in 0.020 s. What is the average force exerted by the wall?

Question

BlackTom AI · Accepted Answer

Consider the scenario: a 0.50 kg ball moves at 6.0 m/s and comes to rest after 0.020 s because of the wall.

First, determine the change in momentum (impulse): Δp = m(v_f − v_i) = 0.50 kg × (0 − 6.0 m/s) = −3.0 kg·m/s. The impulse is negative because the momentum direction reverses due to the wall.

Next, compute the average force using impulse–time relation: F_avg = Δp / Δt = (−3.0 kg·m/s) / (0.020 s) = −150 N. The negative sign indicates the force acts opposite to the initial motion; the question asks for average force exerted by the wall, so its magnitude is 150 N.

Now evaluate the answer choices:
- Option a: 150 N. This matches the magnitude calculated from impulse and time, so it is consistent with the physical result.
- Option b: 100 N. This would correspond to a Δp of 2.0 kg·m/s over 0.020 s, which is not the actual momentum change here.
- Option c: 125 N. This implies a Δp of 2.5 kg·m/s over 0.020 s, which is incorrect for the given initial velocity.
- Option d: 75 N. This would require a Δp of 1.5 kg·m/s in 0.020 s, far from the actual Δp calculated from the initial speed.

In summary, the computed average force magnitude is 150 N, arising from the momentum change of −3.0 kg·m/s over 0.020 s. The other options do not align with the momentum-change calculation for the given data.

A 0.50 kg ball moving at 6.0 m/s hits a wall and comes to rest in 0.020 s. What is the average force exerted by the wall?单项选择题

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