Explain how momentum gets conserved in collision of two bodies.
Consider two bodies i.e., balls A and B, the mass and initial velocities are mAuA and mBuB respectively before collision. The two bodies collide and force is exerted by each body. There is change in their velocities due to collision.
The momentum of ball A before collision is mAuA and final momentum is mAuA.
The rate of change of momentum, during the collision for 'A' will be \(m_A\frac{(v_A - u_A)}{t}\).
Initial momentum of B is mBuB and final momentum is mBuB.
Rate of change of momentum of B will be \(m_B\frac{(v_B - u_B)}{t}\)
According to the third law of motion, the force \(F_{AB}\) exerted by ball A on ball B and the force \(F_{BA}\) exerted by the ball B on ball A must be equal and opposite to each other. Therefore
\(F_{AB}\) - \(F_{BA}\)
or \(m_A\frac{(v_A - u_A)}{t}\) = \(m_B\frac{(v_B - u_B)}{t}\)
mAuA + mBuB = mAvA + mBvB
∴ (mAuA + mBuB) is the total momentum of the two balls A and B before collision and (mAvA + mBvB) is their total momentum after collision.
∴ The total momentum of the two balls remains unchanged or conserved provided no other external force acts.