Consider objects of masses `m_(1)and m_(2)` moving initially along the same straight line with velocities `u_(1)and u_(2)` respectively. Considering a perfectly elastic collision (with `u_(1)gt u_(2))` derive expressions for their velocities after collision.

Explanation

With `u_(1) gtu_(2)`
Relative velocity of approach before collision `=u_(1)-u_(2)" …(i)"`
Let the velocity of A after collision by `v_(1)` and the velocity of B after collision be `v_(2)`
When `v_(2) gt v_(1)`, the bodies separate after collision
Relative velocity of separation after collision `=v_(2)-v_(1)" ....(ii)"`
Linear momentum of the two balls before collision `=m_(1)u_(1)+m_(2)u_(2)`
Linear momentum of the two balls after collision `=m_(1)v_(1)+m_(2)v_(2)`
`m_(1)v_(1)+m_(2)v_(2)=m_(1)u_(1)+m_(2)u_(2)`
According to the law of conservation of linear momentum
`"or "m_(2)(v_(2)-u_(2))=m_(1)(u_(1)-v_(1))" .....(iv)"`
`"Total KE of the two balls before collision"=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)" ...(v)"`
`"Total KE of the two balls after collision"=(1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2)" ...(vi)"`
`(1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2)=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)" According to the law of conservation of KE"`
`(1)/(2)m_(2)(v_(2)^(2)-u_(2)^(2))=(1)/(2)m_(1)(u_(1)^(2)-v_(1)^(2))`
`"or "m_(2)(v_(2)^(2)-u_(2)^(2))=m_(1)(u_(1)^(2)-v_(1)^(2))" ....(vii)"`
Dividing equation (vii) by equation (iv)
`(m_(2)(v_(2)^(2)-u_(2)^(2)))/(m_(2)(v_(2)-u_(2)))=(m_(1)(u_(1)^(2)-v_(1)^(2)))/(m_(1)(u_(1)-v_(1)))`
`((v_(2)+u_(2))(v_(2)-u_(2)))/((v_(2)-u_(2)))=((u_(1)+v_(1))(u_(1)-v_(1)))/((u_(1)-v_(1)))`
`"or "v_(2)+u_(2)=u_(1)+v_(1)`
`"or "v_(2)-v_(1)=u_(1)-u_(2)" ...(viii)"`
Hence, in one - dimensional elastic collision, relative velocity to separation after collision is equal to relative velocity of approach before collision (from equations (i) & (ii))
From equation (viii), `(v_(2)-v_(1))/(u_(1)(-u_(2))=1`
By definition,`" "(v_(2)-v_(1))/(u_(1)-u_(2))=e=1`
Hence, coefficient of restitution/resilience of a perfectly elastic collision in one dimension is unity.
From equation (viii),
`v_(2)=u_(1)-u_(2)+v_(1)" ....(ix)"`
Put this value in equation (iii)
`m_(1)v_(1)+m_(2)(u_(1)-u_(2)+v_(1))=m_(1)u_(1)+m_(2)u_(2)`
`m_(1)v_(1)+m_(2)u_(1)-m_(2)u_(2)+m_(2)v_(1)=m_(1)u_(1)+m_(2)u_(2)`
`v_(1)(m_(1)+m_(2))=(m_(1)-m_(2))u_(1)+2m_(1)u_(2)`
`v_(1)=((m_(1)-m_(2))u_(1))/((m_(1)+m_(2)))+(2m_(2)u_(2))/(m_(1)+m_(2))" ...(x)"`
Put this value of `v_(1)` from equation (x) in equation (ix)
`v_(2)=u_(1)-u_(2)+((m_(1)-m_(2))u_(1))/(m_(1)+m_(2))+(2m_(2)u_(2))/(m_(1)+m_(2))`
`=u_(1)[1+(m_(1)-m_(2))/(m_(1)+m_(2))]+u^(2)[(2m_(2))/(m_(1)+m_(2))-1]`
`v_(2)=(2m_(1)u_(2))/(m_(1)+m_(2))+((m_(2)-m_(1))u_(2))/((m_(1)+m_(2)))" ...(xi)"`
Note : You can get exp. for `v_(2)` from equation (x) by interchanging `m_(1) and m_(2) and u_(1) and u_(2)`. You can also get equation (x) from equation (xi).