Two equal spheres `B` and `C`, each of mass `m`, are in contact on a smooth horizontal table. A third sphere `A` of same size as that of `B` or `C` but mass `m//2` impinges symmetrically on them with a velocity `u` and is itself brought to rest.

The loss of `KE` during collision is

Options

A. `("mu"^(2))/8`
B. `("mu"^(2))/6`
C. `("mu"^(2))/3`
D. `("mu"^(2))/2`

Answer

B

Explanation

`u=`velocity of sphere `A` before impact. As the spheres are identical, the triangle `ABC` formed by joining their centres is equilateral. The spheres `B` and `C` will move in direction `AB` and `AC` after impact making can angle of `30^(@)` with the origina lines of motion of ball `A`.
Let `v` be the speed ofthe balls `B` and `C` after impact.
Momentum conservation gives
`(m/2)u=mvcos30^(@) +mvcos30^(@)`
`u=2sqrt(3)vimpliesv=u/(2sqrt(3))`.......i
From Newton's experimental law, for an oblique collision, we have take components along normal, i.e. along `AB` for balls `A` and `B`.
`v_(B)-v_(A)=-e(u_(B)-u_(A))`
`v-0=-e(0-ucos30^(@))`
`v=eucos30^(@)`.........ii
Combining eqn i and ii we get `e=1/3`
Loss in `KE =1/2 m/2 u^(2)-2(1/2mv^(2))`
`=1/4mu^(2)-m(u/(2sqrt(3)))^(2)=1/6mu^(2)`