Answer to Question #180335 in Physics for Jaylee

Question #180335

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.


1
Expert's answer
2021-04-13T06:30:24-0400

Let us choose the positive axis direction along the velocity of the ball B. Momentum before the impact is mBvBmAvAm_B v_B - m_A v_A, and after is (mA+mB)v(m_A + m_B) v, where vv is unknown. According to the conservation of momentum mBvBmAvA=(mA+mB)vm_B v_B - m_A v_A = (m_A + m_B) v, from where v=mBvBmAvAmA+mB2.84msv = \frac{m_B v_B - m_A v_A}{m_A + m_B} \approx 2.84 \frac{m}{s}


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