A tennis ball has a mass of 0.057 kg. A professional tennis player hits the ball
ID: 1661140 • Letter: A
Question
A tennis ball has a mass of 0.057 kg. A professional tennis player hits the ball hard enough to give it a speed of 46 m/s (about 103 miles per hour.) The ball moves toward the left, hits a wal and bounces straight back to the right with almost the same speed (46 m/s). As indicated in the diagram below, high-speed photography shows that the ball is crushed about d 2.2 cm at the instant when its speed is momentarily zero, before rebounding Making the very rough approximation that the large force that the wall exerts on the ball is approximately constant during contact, determine the approximate magnitude of this force What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero? How much time elapses between first contact with the wal, and coming to a stop? what is the magnitude of the average force exerted by the wail on the ball during contact? In contrast, what is the magnitude of the gravitational force of the Earth on the ball? mvs seconds Additional Materials ul Section 22Explanation / Answer
consider the motion when the ball touch the wall to the instant when it momentarily stops
Vi = initial velocity = 46 m/s
Vf = final velocity = 0 m/s
a = acceleration
d = stopping distance = 2.2 cm = 0.022 m
using the equation
Vf2 = Vi2 + 2 a d
02 = 462 + 2 a (0.022)
a = - 4.8 x 104 m/s2
t = time taken
using the equation
Vf = Vi + a t
0 = 46 + (- 4.8 x 104) t
t = 9.6 x 10-4 sec
average speed is given as
Vavg = d/t = 0.022/(9.6 x 10-4) = 22.92 m/s
t = time between first touching to ball coming to stop = 9.6 x 10-4 sec
consider the collision
Vi = initial velocity = - 46 m/s
Vf = final velocity = 46 m/s
m = mass = 0.057 kg
F = force
t = time = 9.6 x 10-4 sec
using impulse -change in momentum equation
F t = m (Vf - Vi)
F (9.6 x 10-4) = (0.057) (46 - (-46))
F = 5462.5 N
gravitational force is given as
F = mg = (0.057) (9.8) = 0.56 N
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