Obtain expression for velocities of the two bodies after elastic collision in one dimension .

Explanation

Consider elastic collision in one dimension as shown in figure .
Suppose a ball of mass `m_(1)` moving with velocity `v_(1)` along X - axis undergoes an elastic collision with a ball of mass `m_(2)` moving with velocity `v_(2i) and v_(1i) gt v_(2i)`
Their final velocities are `v_(1f) and v_(2f)` respectively .
Momentum is conserved in collision .
`:. m_(1)v_(1i)+m_(2)v_(2i)=m_(1)v_(1f)+m_(2)v_(2f) " "...(1)`
` :. m_(1)(v_(1i)-v_(1f)) =m_(2)(v_(2f)-v_(2i))`
Collision is elastic collision so kinetic energy is also conserved .
`1/2 m_(1)v_(1i)^(2)+m_(2)v_(2i)^(2) =m_(1)v_(1f)^(2)+m_(2)v_(2f)^(2) ...." "(2)`
` :. m_(1)(v_(1i)-v_(1f))(v_(1i) +v_(1f))=m_(2)(v_(2f)-v_(2i)) (v_(2f)+v_(2i))`
From equation (2) ,
`v_(1i)+v_(1f)=v_(2f)+v_(2i)`
` :. v_(1f)=v_(2f)+v_(2i)-v_(1i) " "....(3)`
From equation (1) ,
`m_(1)v_(1i)+m_(2)v_(2i)=m_(1)(v_(2f)+v_(2i)-v_(1i))+m_(2)v_(2f)`
`:. m_(1)v_(1i)+m_(2)v_(2i)=m_(1)v_(2f)=m_(1)v_(2f)+m_(1)v_(2f)+m_(1)v_(2i)-m_(1)v_(1)+m_(2)v_(2f)`
`2m_(1)v_(1i)=m_(1)v_(2i)-m_(2)v_(2i)-m_(2)v_(2i)+m_(1)v_(2f)+m_(2)v_(2f)`
`2m_(1)v_(1i)=(m_(1)-m_(2))v_(2i)+(m_(1)+m_(2))v_(2f)`
If ball of mass `m_(2)` having intial velocity
`v_(2i)=0` then
`2m_(1)v_(1i) =(m_(1)+m_(2))v_(2f)`
` :. v_(2f) =(2m_(1)v_(1i))/(m_(1)+m_(2))`
`v_(1f)=v_(2f)+v_(2i)-v_(1i)`
` = (2m_(1)v_(1i))/(m_(1)+m_(2))-v_(1i) " " [ :. v_(2i)=0 " supposed"] `
`=(2m_(1)v_(1i)-m_(1)v_(1i)-m_(2)v_(1i))/(m_(1)+m_(2))`
` :. v_(1f) = (m_(1)v_(1i)-m_(2)v_(1i))/(m_(1)+m_(2))`
` :. v_(1f) = ((m_(1)-m_(2))v_(1i))/((m_(1)+m_(2)))`
In these equations if `m_(1),m_(2) and v_(1i) ` is given then the value of `v_(1f) and v_(2f)` can be calculated .