Alex throws a 0.250-kg rubber ball down onto the floor.The ball's speed just before impact is 5.30 m/s,and just after is 3.71 m/s.If the ball is in contact with the floor for 0.028 s,what is the magnitude of the average force applied by the floor on the ball?
A)80.4 N
B)14.2 N
C)47.3 N
D)33.1 N
E)2.25 N

Question

Quizplus · Accepted Answer

The magnitude of the average force applied by the floor on the ball can be calculated using the change in momentum over time, which is given by the formula $ F_{avg} = \frac{\Delta p}{\Delta t} $, where $ \Delta p $ is the change in momentum and $ \Delta t $ is the time interval. The change in momentum $ \Delta p = m(v_f - v_i) $, where $ m $ is the mass of the ball, $ v_f $ is the final velocity (upwards, so we take it as positive 3.71 m/s), and $ v_i $ is the initial velocity (downwards, so we take it as negative -5.30 m/s). Plugging in the values, $ \Delta p = 0.250 \times (3.71 - (-5.30)) = 0.250 \times 9.01 = 2.2525 $ kg·m/s. The time interval $ \Delta t = 0.028 $ s. Therefore, $ F_{avg} = \frac{2.2525}{0.028} \approx 80.4 $ N.

Alex Throws a 0