Answer:
[tex]v_o\approx0.7059\ m.s^{-1}[/tex]
Explanation:
Given:
mass of the boat, [tex]m=120\ kg[/tex]
uniform speed of the boat, [tex]v=1\ m.s^{-1}[/tex]
rate of accumulation of water mass in the boat, [tex]\dot m=100\ kg.hr^{-1}[/tex]
time of observation, [tex]t=0.5\ hr[/tex]
The mass of the boat after the observed time:
[tex]m_o=m+\dot m\times t[/tex]
[tex]m_o=120+100\times 0.5[/tex]
[tex]m_o=170\ kg[/tex]
Now using the conservation of momentum:
[tex]m.v=m_o.v_o[/tex]
[tex]120\times 1=170\times v_o[/tex]
[tex]v_o\approx0.7059\ m.s^{-1}[/tex]