Explanation:
Let [tex]u_{1}[/tex] be the velocity of the bowling ball before collision.
Let [tex]u_{2}[/tex] be the velocity of the pin before collision.
Let [tex]v_{1}[/tex] be the velocity of the bowling ball after collision.
Let [tex]v_{2}[/tex] be the velocity of the pin after collision.
Let [tex]m_{1}[/tex] be the mass of the bowling ball.
Let [tex]m_{2}[/tex] be the mass of the pin.
According to the conservation of momentum,
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
Given,
[tex]m_{1}=7Kg\\m_{2}=2Kg\\u_{1}=9ms^{-1}\\u_{2}=0ms^{-1}\\v_{2}=14ms^{-1}[/tex]
[tex]7(9)+2(0)=7(v_{1})+2(14)[/tex]
[tex]7v_{1}=35\\v_{1}=5ms^{-1}[/tex]
New velocity of bowling ball is [tex]5ms^{-1}[/tex]
