Given:
The mass of the bowling ball is m1 = 1.6 kg
The mass of the pin is m2 = 0.68 kg
The initial velocity of the ball is
[tex]v_i=\text{ 6 m/s}[/tex]
Required: Velocity of the bowling pin after the collision.
Explanation:
According to the conservation of momentum, the velocity after the collision will be
[tex]\begin{gathered} m1v_i+m2\times0\text{ =\lparen m1+m2\rparen v}_f \\ v_f=\frac{m1v_i}{m1+m2} \\ =\frac{1.6\times6}{1.6+0.68} \\ =4.21\text{ m/s} \end{gathered}[/tex]
Final Answer: The velocity of the bowling pin after the collision is 4.21 m/s
