1.

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. Find the velocity acquired by each of the spheres B and C after collision.

Answer»

`u/sqrt(3)`
`(2u)/sqrt(3)`
`u/(2sqrt(3))`
`u/2`

Solution :`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 conservationgives
`(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)`


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