Step-by-step explanation:
Given that,
Mass of ball, m = 0.425 kg
Initial speed of the ball, u = 12 m/s
Initial speed of a person, u' = 0
Mass of a person, m' = 68 kg
(a) Let V is the combined speed of the person and the ball. Using conservation of momentum as :
![mu+m'u'=(m+m')V\\\\V=(mu+m'u')/((m+m'))\\\\V=(0.425* 12+0)/((0.425+68))\\\\V=0.0745\ m/s](https://img.qammunity.org/2021/formulas/physics/high-school/g5f7ackfwtjukyuhzyqv2pr5dhoqxjspiq.png)
(b) If the ball hits the person and bounces off his chest, so afterwards it is moving horizontally at 9.00 m/s in the opposite direction,. Let v' is the speed of the person after the collision. So,
![mu+m'u'=mv+m'v'](https://img.qammunity.org/2021/formulas/physics/high-school/kf14pent92q6m6hsw0vish7knt9mew4ka7.png)
v = -9 m/s
![mu=mv+m'v'\\\\v'=(m(u-v))/(m')\\\\v'=(0.427* (12-(-9)))/(68)\\\\v'=0.131\ m/s](https://img.qammunity.org/2021/formulas/physics/high-school/gc3vwufc6fzcva1aax5zbck3fjqaqzkzfl.png)
Hence, this is the required solution.