## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 7 - Problems - Page 268: 84

#### Answer

The magnitude of the average force exerted on the ball by the floor is $33.8~N$

#### Work Step by Step

Any change in the horizontal component of momentum is negligible compared to the change in the vertical component of momentum so we can ignore any horizontal change in momentum. We can find the change in the vertical component velocity: $\Delta v = v_f-v_0 = (53~m/s)~sin~18^{\circ}-(-54~m/s)~sin~22^{\circ} = 36.6~m/s$ We can use the magnitude of the change in velocity to find the magnitude of the change in the ball's momentum: $\Delta p = m~\Delta v$ $\Delta p = (0.060~kg)(36.6~m/s)$ $\Delta p = 2.196~kg~m/s$ The impulse exerted on the ball by the floor is equal to the ball's change in momentum. We can use the impulse to find the average force exerted on the ball by the floor: $F~t = J$ $F = \frac{J}{t}$ $F = \frac{2.196~kg~m/s}{0.065~s}$ $F = 33.8~N$ The magnitude of the average force exerted on the ball by the floor is $33.8~N$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.