14. We choose our positive direction in the direction of the rebound (so the ball’s initial velocity is negative-

valued). We evaluate the integral J =

F dt by adding the appropriate areas (of a triangle, a rectangle,

and another triangle) shown in the graph (but with the t converted to seconds). With m = 0.058 kg and
v = 34 m/s, we apply the impulse-momentum theorem:



F

wall

dt

=

m

v

f

− m v

i



0.002

0

F dt +



0.004

0.002

F dt +



0.006

0.004

F dt

=

m(+v)

− m(−v)

1

2

F

max

(0.002 s) + F

max

(0.002 s) +

1

2

F

max

(0.002 s)

=

2mv

F

max

(0.004 s)

=

2 (0.05 8 kg) (34 m/s)

which yields F

max

= 9.9

× 10

2

N.