Download PWE 7-9: Follow

Example 7-9 Follow-through
Tennis players are taught to follow through as they serve the ball—that is, to continue to
swing the racket after first striking the ball. The rationale is that by increasing the time
that the ball and racket are in contact, the ball might have a larger speed as it leaves the
racket. Using her racket, a player can exert an average force of 560 N on a 57-g tennis ball during an overhand serve (Figure 7-18). At what speed does the ball leave her
racket if the contact time between them is 0.0050 s? What would the speed of the ball
be if she improved her follow-through and increased the contact time by a factor of 1.3
to 0.0065 s? During an overhand serve, the tennis ball is struck when it is essentially
motionless.
Figure 7-18 ​Tennis physics How does an improved follow-through affect the speed of a tennis ball?
Set Up
We are given the force that the racket exerts
on the ball and the contact time (the time
that the ball is touching the racket). We also
know that the ball starts at rest and so with
zero ­momentum. We can use Equation 7-24
to ­calculate the final momentum of the ball,
and then use Equation 7-5 to determine the
final speed.
Solve
The racket follows a curved path during the
serve. But since the collision between racket
and ball lasts such a short time, the ball is
in contact with the racket for just a short
segment of that path, which we can treat as a
straight line. We choose the positive x axis to
be in the direction of the motion of the ball.
Collision force, contact time, and
momentum change:
sf - p
si
Fscollision Dt = p
s = mv
s
p
contact time ∆t
(7-24)
Fcollision
m
(7-5)
pi = 0
The collision is one-dimensional, so we just
need the x component of Equation 7-24:
Fcollision, x Dt = pfx 2 pix
The ball is initially at rest, so vix = 0 and
pix = 0
m
+x
vix = 0
After the collision with the racket, the ball
has (unknown) velocity vfx, so
pfx = mvfx
Then the x component of Equation 7-24
becomes
Fcollision, x Dt = mvfx 2 0 = mvfx
Solve for the final velocity of the ball in each
case.
pf = mvf
Rewrite the x component of Equation 7-24 to solve for vfx:
vfx =
Fcollision, x Dt
m
In both cases, Fcollision, x = 560 N
m = 57 g a
1 kg
b = 0.057 kg
1000 g
vfx = ?
Calculate vfx if the contact time is 0.0050 s. Recall that
1 N = 1 kg # m>s 2:
vfx =
Fcollision, x Dt
m
=
1560 N2 10.0050 s2
0.057 kg
2
= 49
Ns 1 kgm>s
a
b = 49 m>s
kg
1N
With improved follow-through and a contact time of 0.0065 s,
vfx =
Reflect
Fcollision, x Dt
m
=
1560 N2 10.0065 s2
0.057 kg
= 64 m>s
For a given force, the change in momentum and therefore the change in velocity during a collision are directly proportional
to the contact time. Increasing the contact time Dt from 0.0050 s to (1.3)(0.0050 s) = 0.0065 s results in an increase in
the final speed of the tennis ball from 49 m>s to 11.32 149 m>s2 = 64 m>s.