Respuesta :
Given data:
* The initial velocity of the man is u_1 = 0 m/s.
* The initial velocity of the ball is u_2 = 0 m/s.
* The final velocity of the ball is v_2 = 9 m/s.
* The mass of the man is m_1 = 65 kg.
* The mass of the ball is m_2 = 3 kg.
Solution:
The initial momentum of the system is,
[tex]p_1=m_1u_1+m_2u_2[/tex]Substituting the known values,
[tex]\begin{gathered} p_1=0+0 \\ p_1=0\text{ kgm/s} \end{gathered}[/tex]The final momentum of the system is,
[tex]p_2=m_1v_1+m_2v_2[/tex]where v_1 is the final velocity of the man,
Substituting the known values,
[tex]\begin{gathered} p_2=65v_1+3\times9 \\ p_2=65v_1+27 \end{gathered}[/tex]According to the law of conservation of momentum,
[tex]\begin{gathered} p_1=p_2 \\ 0=65v_1+27 \\ 65v_1=-27 \\ v_1=\frac{-27}{65} \end{gathered}[/tex]By simplifying,
[tex]v_1=-0.42\text{ m/s}[/tex]Here, the negative sign indicates the direction of motion of man is opposite to the direction of motion of ball.
Thus, the speed of the man is 0.42 meter
