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 :
[tex]mu+m'u'=(m+m')V\\\\V=\dfrac{mu+m'u'}{(m+m')}\\\\V=\dfrac{0.425\times 12+0}{(0.425+68)}\\\\V=0.0745\ m/s[/tex]
(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,
[tex]mu+m'u'=mv+m'v'[/tex]
v = -9 m/s
[tex]mu=mv+m'v'\\\\v'=\dfrac{m(u-v)}{m'}\\\\v'=\dfrac{0.427\times (12-(-9))}{68}\\\\v'=0.131\ m/s[/tex]
Hence, this is the required solution.