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Two masses 'm' and '2m' are placed in fi...

Two masses `'m'` and `'2m'` are placed in fixed horizontal circular smooth hollow tube as shown. The mass `'m'` is moving with speed `'u'` and the amss `'2m'` is stationary . After their first collision, the time elapsed for next collision. `(` coefficient of restitution `e=1//2)`

A

`(2pir)/(u)`

B

`(4pir)/(u)`

C

`(3pir)/(u)`

D

`(12pir)/(u)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(B)` Let the speeds of balls of mass `m` and `2m` after collision be `v_(1)` and `v_(2)` as shown in figure.
Applying conservation of momentum
`mv_(1)+2mv_(2)=m u ` and `-v_(1)+v_(2)=(u)/(2)`
solving we get `v_(1)=0` and `v_(2)=(u)/(2)`
Hence the ball of mass `m` comes to rest and ball of mass `2m` moves with speed `(u)/(2)`.
`t=(2pir)/(u//2)=(4pir)/(u)`
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