lawn bowling momentum

Question

You throw a 2.6 kg lawn bowling ball forward (that is in the positive direction) at 14.7 m/s and it hits a 1.1 kg "pin" that is at rest. If the pin flies forward at 29.6 m/s, what is the ball's new velocity?

Answer 1

Assuming no forces external to the two objects, the momentum is conserved in the collision:

m_Bv_BBef = m_Pv_Paff + m_Bv_Baft where B = Ball, P=Pin and the velocities are before and after the collision.

You know everything but v_Baft. Solve.

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