Answer
(a) The impulse delivered to the ball during impact is 0.474 kg m/s.
(b) The average force on the ball during impact is 237 N.
Work Step by Step
(a) We can find the speed $v_1$ after falling 2.00 meters.
$v_1^2 = v_0^2+2ay = 0 + 2ay$
$v_1 = \sqrt{2ay} = \sqrt{(2)(9.80~m/s^2)(2.00~m)}$
$v_1 = 6.26~m/s$
We can find the speed $v_2$ after the rebound.
$v_2^2 = v^2-2ay = 0 - 2ay$
$v_2 = \sqrt{-2ay} = \sqrt{-(2)(-9.80~m/s^2)(1.60~m)}$
$v_2 = 5.60~m/s$
$J = \Delta p = m \Delta v$
$J = (0.0400~kg)(11.86~m/s)$
$J = 0.474~kg~m/s$
The impulse delivered to the ball during impact is 0.474 kg m/s.
(b) $F~t = J$
$F = \frac{J}{t} = \frac{0.474~kg~m/s}{0.00200~s}$
$F = 237~N$
The average force on the ball during impact is 237 N.
