A 0.065-kg tennis ball moving to the right with a speed of 15 m/s is struck by a tennis racket, causing it to move to the left with a speed of 15 m/s. If the ball remains in contact with the racquet for 0.020 s, what is the magnitude of the average force exerted on the ball?
A)zero newtons
B)98 N
C)160 N
D)240 N
E)320 N

Question

Quizplus · Accepted Answer

The change in momentum of the ball is:$$\Delta p = m(v_f - v_i) = (0.065 	ext{ kg})(-15 	ext{ m/s} - 15 	ext{ m/s}) = -1.95 	ext{ kg m/s}$$The average force exerted on the ball is:$$F = \frac{\Delta p}{\Delta t} = \frac{-1.95 	ext{ kg m/s}}{0.020 	ext{ s}} = -97.5 	ext{ N}$$The magnitude of the average force is:$$|F| = |-97.5 	ext{ N}| = 97.5 	ext{ N}$$Therefore, the magnitude of the average force exerted on the ball is 98 N.

A 0065-Kg Tennis Ball Moving to the Right with a Speed