Question

A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and the ball continues forward at 4.0 m/s. What was the original velocity of the ball?

Answer 1

The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

$P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}$

Where:

• mb is the mass of the bowling ball
• mp the mass of the pin
• $v_{0b}\quad and\quad v_{0p}$ the initial velocities of the bowling ball and the pin.
• $v_{1b}\quad and\quad v_{1p}$ the final velocities of the bowling ball and the pin.

Solving for v0b:

$v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}$

R/ The original velocity of the ball was 5.71 m/s.