Question

A 7.4 kg bowling ball traveling at 12.91 m/s strikes a 2.9 kg bowling pin. The impact slows the bowling ball to 9.67 m/s.



Determine the impulse given to the pin.

Answer 1

Let $v_1 = 12.91\frac{m}{s}, v_2 = 0 \frac{m}{s}$ be the initial speeds of the bowling ball and bowling pin respectively, and $v_1' = 9.67 \frac{m}{s}$ be the speed of the ball after the impact.

Using law of conservation of momentum: $m_1 v_1 + 0 = m_1 v_1' + m_2 v_2'$. The impulse given to the pin is $m_2 v_2' = m_1 v_1 - m_1 v_1' = m_1(v_1 - v_1') \approx 23.98 kg \cdot \frac{m}{s}$.