Explain the special caes of elastic collision in one dimension .

Explanation

Case 1 . If the two mases are equal `(m_(1)=m_(2)) v_(1f)=0, v_(2f)=v_(1i)`
The first mass comes to rest and pushes off the second mass with its initial speed on collision .
`:. ` Velocity of first mass after collision ,
`v_(1f)=((m_(1)-m_(2))/(m_(1)+m_(2)))v_(1i)`
`= (0/(2m_(1)))v_(1i)`
= 0 and velocity of second mass
`v_(2f) =((2m_(1)v_(1i))/(m_(1)+m_(2)))`
`=(2mv_(1i))/(m+m) " "[ :. m_(1)=m_(2)=m " supposed"] `
`v_(2f) =v_(1i)`
` :. ` The velocity of first ball after collision .
Case 2: If `m_(2) gt gt m_(1)` then second ball is more heavier then first one .
` :. ` The velocity of first after collision ,
`v_(1f) =((m_(1)-m_(2))/(m_(1)+m_(2)))v_(1i)`
Neglecting the `m_(1)` compared to `m_(2)`
`v_(1f) =((0-m_(2)))/((0+m_(2)))v_(1i)`.
` :. v_(1f) =-v_(1i)`
` :. ` First ball having same velocity after collision and
`v_(2f) =(2m_(1)v_(1i))/(m_(1)+m_(2)),` here `m_(1)=0`
` = (2xx0xxv_(1i))/(0+m_(2))=0`
Hence, there is no effect in velocity of heavier ball and so it remain static at its position .
Case 3 : If `m_(1) gt gt m_(2)`
The velocity after collision of first ball ,
`v_(if) = ((m_(1)-m_(2))/(m_(1)+m_(2)))v_(1i)` here `m_(2) = 0 `
`v_(1f) =((m_(1)-0)/(m_(1)+0))v_(1i) = v_(1i)`
velocity of second ball ,
`v_(2f) = (2m_(1)v_(1i))/(m_(1)+m_(2)) `,here `m_(2) = 0 `
`v_(2f) = 2v_(li)`