A and B are two particles having the same mass m. A is moving along X-axis with a speed of `10ms^(-1)` and B is at rest. After undergoing a perfectly elastic collision with B, particle A gets scattered through an angle of `30^(@)`. What is th edirection of motion of B, and the speeds of A and B, after the collision?

Explanation

Here, `u=10m//s, theta=30^(@), upsilon_(1)=?upsilon_(2)=?`
As `theta+phi=90^(@), phi=90^(@)-theta=90^(@)-30^(@)=60^(@)`
From `u=upsilon_(1)cos theta+upsilon_(2)cos phi`
`10=upsilon_(1)cos 30^(@)+upsilon_(2)cos60^(@)=(upsilon_(1)sqrt(3))/(2)+(upsilon_(2))/(2) or upsilon_(1)sqrt(3)+upsilon_(2)=20` ...(i)
Again , from `0=upsilonsintheta-upsilon_(2)sin phi`
`upsilon_(2)sinphi=upsilon_(1)sin theta`
`upsilon_(2)sin 60^(@)=upsilon_(1)sin 30^(@) or upsilon_(2)(sqrt(3))/(2)=(upsilon_(1))/(2) or upsilon_(1)=sqrt(3)upsilon_(2)` ...(ii)
From (i), `sqrt(3)upsilon_(2)sqrt(3)+upsilon_(2)=20, upsilon_(2)=(20)/(4)=5ms^(-1)`
From (ii) `upsilon_(1)=sqrt(3)upsilon_(2)=sqrt(3)xx5=1.732xx5=8.66m//s`