Answer:
0.734 m/s
Explanation:
Since there's no external force outside the system, by the law of momentum conservation, momentum before the throw (which is 0) must be the same as after the throw:
[tex]0 = m_pv_p + (m_c + m_b)v_b[/tex]
where m_p = 5.8 kg is the mass of the package
v_p = 10m/s is the velocity of the package
m_c = 29 kg is the mass of the child
m_b = 50 kg is the mass of the boat
As the package and the boat has opposite directions of motion, they should have opposite velocity:
[tex]5.8*10 + (29 + 50)v_b = 0[/tex]
[tex]b_b = \frac{-58}{79} = -0.734 m/s[/tex]
So the magnitude of the boat velocity is 0.734 m/s
