Stationary particles of mass `m_(2)` is hit by another particles of mass `m_(1)`. The stationary particle deviates through `theta` and the other by `90^(circ)`. Find the value of `theta` if the collision is perfectly elastic.

Answer

`cos theta = sqrt((m_(1)+m_(2))/(2m_(2))`

Explanation

`{:(,,m_(1)u=m_(2)v_(2) cos theta,,),("From COM",,,,),(,,m_(1)v_(1)=m_(2)v_(2) sin theta,,):}}` ...(i)

From `COE`
`(1)/(2)m_(1)u^(2) = (1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2)`
`rArr m_(1)u^(2) = m_(1)v_(1)^(2)+m_(2)v_(2)^(2)` ...(ii)
From (i) `m_(2)^(2)v_(2)^(2) = m_(1)^(2)(u^(2)+v_(1)^(2))`
Putting the value of `m_(1)v_(1)^(2)` in (ii)
`m_(1)u^(2) = (m_(2)^(2)v_(2)^(2)-m_(1)^(2)u^(2))/(m_(1))+m_(2)v_(2)^(2)`
`rArr v_(1)^(2)u^(2) = m_(2)^(2)v_(2)^(2)-m_(1)^(2)u^(2)+m_(1)m_(2)v_(2)^(2)`
`rArr 2m_(1)^(2)u^(2) = m_(2)v_(2)^(2)(m_(1)+m_(2))`
`rArr v_(2)^(2) = (2m_(1)^(2))/(m_(2)(m_(1)+m_(2)))`
Putting in (i)
`m_(1)u = sqrt((2m_(1)^(2)m_(2))/((m_(1)+m_(2))))u cos theta rArr 1 = sqrt((2m_(2))/(m_(1)+m_(2))) cos theta`
`rArr cos theta = sqrt((m_(1)+m_(2))/(2m_(2)))`