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Three particles each of mass m can slide...

Three particles each of mass m can slide on fixed frictionless horizontal circular tracks in the same horizontal plane as shown in the figure. The coefficient of restitution being `e = 0.5`. Assuming that `m_(2)` and `m_(3)` are at rest initially and lie along a radial line before impact and the string is initially unstretched, then maximum extension in spring in subsequent motion

A

`(3)/(4)V_(0) sqrt((m)/(k))`

B

`(3)/(4)V_(0) sqrt((m)/(5k))`

C

`(3)/(4) V_(0)sqrt((2m)/(5k))`

D

`(3)/(5)V_(0)sqrt((m)/(k))`

Text Solution

Verified by Experts

The correct Answer is:
B
Velocity just after collision
`V_(0)=V_(1)+V_(2)` ….(A)
`(1)/(2)=(-V_(1)+V_(2))/(V_(2))-V_(1)+V_(2)=(V_(0))/(2)`....(B)
from (A) & (B)
`rArr V_(1)=(V_(0))/(4)& V_(2)=(3V_(0))/(4)`
When maximum extension occur's then angular speed with rest to centre for mass `m_(2)` & `m_(3)` are same Using angular momentum conservation about centre of circle.

`m=(3)/(4)V_(0)(2R)+0=mR^(2)omega+m(2R)^(2)omega`
`omega = (3)/(10)(V_(0))/(R)`
So velocity of `m_(2) = (3)/(5) V_(0)`
& velocity of `m_(3)=(3)/(10)V_(0)`
When extension is maximum using energy conservation
`(1)/(2)m((3)/(4)V_(0))^(2)`
`=(1)/(2)m((3)/(5)V_(0))^(2)+(1)/(2)m((3)/(10)V_(0))^(2)+(1)/(2)kx_(max)^(2)`
`rArr X_(max)=(3)/(4)V_(0)sqrt((m)/(5k))`.
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