Two particles A and B have the same mass m. A is moving along X-axis with a speed of `5 ms^(-1)` and B is at rest. After undergoing a perfectly elastic collision with B, particle A gets scat tered through an angle of `60^(@).` What is the direction of B, and the speeds of A and B, after the collision ?

Explanation

A & B have same mass and the collision is perfectly elastic
`therefore theta+phi=90^(@)`
`phi =90^(@)-60^(@)=30^(@)" "…(i)`
Using law of conservation of linear momentum
(a) For X-components
`u=5=v(1)cos60^(@)+v_(2)cos30^(@)`
`5=(v_(1))/(2)+v_(2)(sqrt3)/(2)`
`10=v_(1)+v_(2)sqrt3" "...(ii)`
(b) For Y-components
`0=v_(1)sin60^(@)-v_(2)sin30^(@)`
`v_(1)(sqrt3)/(2)=(v_(2))/(2)`
`v_(1)sqrt3=v_(2)" "...(iii)`
From equations (ii) & (iii)
`10=v_(1)+3v_(1)`
`v_(1)=2.5ms^(-1)`
`v_(2)=v_(1)sqrt3==2.5xx1:732`
`v_(2)=4.33ms^(-1)`