Answer :
Answer:
The final velocity of the thrower is [tex]\bf{3.88~m/s}[/tex] and the final velocity of the catcher is [tex]\bf{0.029~m/s}[/tex].
Explanation:
Given:
The mass of the thrower, [tex]m_{t} = 70~Kg[/tex].
The mass of the catcher, [tex]m_{c} = 55~Kg[/tex].
The mass of the ball, [tex]m_{b} = 0.0480~Kg[/tex].
Initial velocity of the thrower, [tex]v_{it} = 3.90~m/s[/tex]
Final velocity of the ball, [tex]v_{fb} = 33.5~m/s[/tex]
Initial velocity of the catcher, [tex]v_{ic} = 0~m/s[/tex]
Consider that the final velocity of the thrower is [tex]v_{ft}[/tex]. From the conservation of momentum,
[tex]&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s[/tex]
Consider that the final velocity of the catcher is [tex]v_{fc}[/tex]. From the conservation of momentum,
[tex]&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s[/tex]
Thus, the final velocity of thrower is [tex]3.88~m/s[/tex] and that for the catcher is [tex]0.029~m/s[/tex].