Answer
(a) The magnitude of the impulse is equal to the magnitude of the change of momentum, which is 14.5 kg m/s.
(b) The magnitude of the average force applied by the bat is 7250 N.
Work Step by Step
(a) $J = \Delta p = m \Delta v$
$J = (0.145~kg)[55.0~m/s-(-45.0~m/s)]$
$J = (0.145~kg)(100.0~m/s)$
$J = 14.5~kg~m/s$
The magnitude of the impulse is equal to the magnitude of the change of momentum, which is 14.5 kg m/s
(b) $F~t = J$
$F = \frac{J}{t} = \frac{14.5~kg~m/s}{0.00200~s}$
$F = 7250~N$
The magnitude of the average force applied by the bat is 7250 N.
