Question

Two ball A and B are approaching each other with the velocities 4.5 m/s and 7.2 m/s respectively. The man of ball A is 3.2 kg

 While that of ball is 5.4 kg. Fine the velocity of the two bodies after the impact assuming that the collision is perfectly inelastic.

Answer 1

Let us choose the positive axis direction along the velocity of the ball B. Momentum before the impact is $m_B v_B - m_A v_A$, and after is $(m_A + m_B) v$, where $v$ is unknown. According to the conservation of momentum $m_B v_B - m_A v_A = (m_A + m_B) v$, from where $v = \frac{m_B v_B - m_A v_A}{m_A + m_B} \approx 2.84 \frac{m}{s}$