mistake in GMP?

JEE Advanced 2015 Discussion Forum › Mechanics questions › mistake in GMP?

A circular ring of mass M and radius a is placed in a gravity free space. A small particle of mass m is placed on the axis of the ring at a distance âˆš3a from the centre of the ring is released from rest. the velocity of the particle when it crosses the centre of the ring?
the options are :

#Physics #Mechanics

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6 Answers

6

Aakash Sharawat ·2009-03-11 05:08:19

this is how i solved it and got the answer to be A but the answer is given as D

if you get D please work out the solution and also explain where i made the mistake!

cheers!

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39

Dr.House ·2009-03-11 05:08:20

wats gmp?

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6

Aakash Sharawat ·2009-03-11 05:20:18

its a grand master package for giving final touch to JEE preparations issued by FIITJEE!

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33

Abhishek Priyam ·2009-03-11 05:38:39

Ring is also free to move..

let velocity of ring be v₂ and particle v

then GMm/2a=mv²/2+Mv₁²/2 ...(i)

conservation of momentum mv=Mv₁
putting value of v₁ in (i)

we get v=\sqrt{\frac{GM}{a}(\frac{M}{M+m})}

hence (d)

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1

chemistry organic ·2009-03-11 05:56:20

one doubt have u taken p.e negative????

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Aakash Sharawat ·2009-03-11 20:26:14

thanks priyam!

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Your Answer

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