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 √3a 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 :

6 Answers

6
Aakash Sharawat ·


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!

39
Dr.House ·

wats gmp?

6
Aakash Sharawat ·

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

33
Abhishek Priyam ·

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)

1
chemistry organic ·

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

6
Aakash Sharawat ·

thanks priyam!

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