Show that in the case of one dimensional elastic collision , the relative velocity of approach of two colloding bodies before collision is equal to the relative velocity of separation after collosion .

Explanation

Consider two spheres (bodies) which have smooth, non-rotating of masses `m_(1) and m_(2) (m_(1) gt m_(2))` are moving along the straight ling joining the centres of mass with initial velocities `u_(1) and u_(2) (u_(1) gt u_(2))` .They undergo head on collision and move along the same line after collision with final velocities of `v_(1) and v_(2)` . These two bodies exert forces on each other during collision
Let the collision be elastic, then both momentum and K.E are conserved.
According to law of censervation of liner momentum, Momentum of the system before collision = Momentum of the system after collision
`m_(1) u_(1) + m_(2) u_(2) = m_(1) v_(1) + m_(1) v_(2)`
`m_(1) (u_(1) - v_(1)) = m_(2) (v_(2) - u_(2)) . . . . (1)`
According to law of conservation of kinetic energy, K.E of the system before collision = K.E of the system after collision
`(1)/(2) m_(1) u_(1)^(2) + (1)/(2) m_(2) u_(2)^(2) = (1)/(2) m_(1) v_(1)^(2) + (1)/(2) m_(2) v_(2)^(2)`
`m_(1) u_(1)^(2) + m_(2) u_(2)^(2) = m_(1) v_(1)^(2) + m_(2) v_(2)^(2)`
`m_(1) (u_(1)^(2) - v_(1)^(2)) = m_(2) (v_(2)^(2) - u_(2)^(2)) . . . (2)`
`((2))/((1)) rArr (m_(1) (u_(1)^(2) - v_(1)^(2)))/(m_(1) (u_(1) - v_(1))) = (m_(2) (v_(2)^(2) - u_(2)^(2)))/(m_(2) (v_(2) - u_(2)))`
`((u_(1) - v_(1))(u_(1) + v_(1)))/((u_(1) - v_(1))) = ((v_(2) - u_(2))(v_(2) + u_(2)))/((v_(2) - u_(2)))`
`u_(1) + v_(1) = u_(2) + v_(2)`
`u_(1)- u_(2) = v_(2) - v_(1)`
Relative velocity of approach from the above equation before collision = Relative velocity of separation after collision