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## Answers

Let us consider two bodies of masses m1 and m2 moving in straight line in the same direction with initial velocities u1 and u2. They collide for a short time ∆t. After collision, they move with velocities v1 and v2.From 2nd law of motion,Force applied by A on B = Rate of change of momentum of B FAB = (m2v2-m2u2)/∆tSimilarly,Force applied by B on A = Rate of change of momentum of AFBA = (m1v1-m1u1)/∆tFrom Newton’s 3rd law of motion,FAB = -FBAOr, (m2v2-m2u2)/∆t = -(m1v1-m1u1)/∆tOr, m2v2-m2u2 = -m1v1+m1u1Or, m1u1 + m2u2 = m1v1 + m2v2This means the total momentum before collision is equal to total momentum after collision. This proves the principle of co conservation of linear momentum.