Obtain an expression for final velocities of two colliding bodes initially in motion.

Explanation

From the law of conservation of linear momentum,
`m_(1)v_(1i)+m_(2)u_(2i)=m_(1)v_(1f)+m_(2)v_(2f)`
From the law of conservation of total kinetic energy,
`(1)/(2)m_(1)v_(1i)+(1)/(2)m_(2)v_(2i)=(1)/(2)m_(1)v_(1f)^(2)+(1)/(2)m_(2)v_(2f)^(2)`
i.e., `m_(1)v_(1i)^(2)+m_(2)v_(2i)^(2)=m_(1)v_(1f)^(2)+m_(2)v_(2f)^(2)`
`m_(2)(v_(2i)+v_(2f))(v_(2i)-v_(2f))=m_(1)(v_(1f)-v_(1i))(v_(1f)+v_(1i))`
`m_(2)(v_(2i)+v_(2f))(v_(2i)-v_(2f))=m_(1)(v_(1f)-v_(1i))(v_(1f)+v_(1i))`
From (1) `m_(1)(v_(1f)-v_(1i))=m_(2)(v_(2i)-v_(2f))`,
using (5) in (1) we get,
`m_(1)v_(1i)+m_(2)v_(2i)=m_(1)v_(1f)+m_(2)(v_(1f)+v_(1i)-v_(2i))`
`m_(1)v_(1i)+m_(2)v_(2i)=m_(1)v_(1f)+m_(2)v_(1f)+m_(2)v_(1i)-m_(2)v_(2i)`
i.e., `v_(1i)(m_(1)-m_(2))+2v_(2i)(m_(2))=v_(i f)(m_(1)+m_(2))`
`therefore v_(1f)=((m_(1)-m_(2))/(m_(1)+m_(2)))v_(1i)+((2m_(2)v_(2i))/(m_(1)+m_(2)))`
Similarly `v_(2f)=((m_(2)-m_(1))/(m_(1)+m_(2)))v_(2i)+((2m_(1)v_(1i))/(m_(1)+m_(2)))`