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Two equal spheres of mass m are in conta...

Two equal spheres of mass `m` are in contact on a smooth horizontal table. A third identical sphere impinges symmetrically on them and reduces to rest. Then:

A

Coefficient of restitution is `e = (2)/(3)`

B

Loss of kinetic energy `(1)/(6) m u^(2)` where `u` is velocity before impact

C

After the collision, velocity of equal mass sphere is `(u)/(sqrt(3))`

D

Loss of kinetic energy `(1)/(3) m u^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
From momentum conservation
`m u = m v cos 30 + m v cos 30 , v = (u)/(sqrt(3))`
So, choice (c ) is correct.
For an oblique collision, we have to take components along normal i.e., along `AB` for sphere `A` and `B`.
`v_(B) = v_(A) = e (u_(A)-u_(B))` , `v-0 = e[u cos 30-0]`
`v = e u xx (sqrt(3))/(2)` , `v = e.v sqrt(3).(sqrt(3))/2`, `e = (2)/(3)`
So, choice (a) is correct. Also, loss of kinetic energy
`Delta K = (1)/(2)m u^(2) - 2((1)/(2) m v^(2))`
`= (1)/(2)m u^(2) - 2(1)/(2)m[((u)/(sqrt(3)))^(2)] = (1)/(6)m u^(2)`
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