With `u_(1)gtu_(2)`
Relative velocity of approach before collision `=u_(1)-u_(2)" "…(i)`
Let the veocity of A after collision be `v_(1)` and the velocity of B after collision be `v_(2).`
When `v_(2)gtv_(1),` teh bodies separate after collision.
Relative velocity of spearafion after collision `=v_(2)-v_(1)" "...(i)`
Linear momentum of the two balls before colision `=v_(2)-v_(1)" "...(ii)`
Linear momentum of the two balls before collision `=m_(1)r_(1)+m_(2)u_(2)`
Linear momentum of hte two balls after colision `=m_(1)v_(1)+m_(2)v_(2)`
`m_(1)v_(1)+m_(2)v_(2)=m_(1)u_(1)+m_(2)u_(2)" "...(iii)`
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 `=1/2m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)" "...(v)`
Total KE of the two balls after collision
`=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)" "...(vi)`
`1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)=1/2+1/2m_(2)u_(2)^(2)` According to the law of conservation of KE `1/2m_(2)(v_(2)^(2)-u_(2)^(2))=1/2(u_(1)^(2)-y_(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 (from equationa (i) & (ii))
From equation (viii) , `(v_(2)-v_(1))/(u_(1)-u_(2))=1`
By definition, `(v_(2)-v_(1))/(u_(1)-u_(2))=0=1`
Hence, coefficient fo restitution/resilience of a perfectly elastic collision in one dimension is unity From equation (viii),`" "...(ix)`
`v_(2)=u_(1)-u_(2)+v_(1)`
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_(2)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)` fro 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_(12))]+u_(2)[(2m_(2))/(m_(1)+m_(2))-1]`
`=u_(1)[(m_(1)+m_(2)+m_(1)-m_(2))/(m_(1)+m_(2))]+u_(2)[(2m_(2)-m_(1)-m_(2))/(m_(1)+m_(2))]`
`v_(2)=(2m_(1)u_(1))/(m_(1)+m_(2))+((m_(2)-m_(1))u_(2))/((m_(1)+m_(2)))" "...(x i)`
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).
