Question

A particle of mass $2m$ is connected by an in extensible string of length $1.2m$ to a ring of mass $m$ which is free to slide on a horizontal smooth rod. Initially the ring and the particle are at the same level with the string taut. Both are then released simultaneously. The distance in meters moved by the ring when the string becomes vertical is:

• A. $0$
• B. $0.4$
• C. $0.8$
• D. $1.2$

Answer 1

The correct option is C $0.8$

Center of mass of $x$-coordinate does not change.

Let it and O margin ring move $x$ then,

$0=mx-(1.2-x)2m$

$3mx=2\times 1.2$m

$x=0.8$