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# A smooth narrow tube in the form of an arc AB of a circle of center O and radius r is fixed so that A is vertically above O and OB is horizontal. Particles P and Q of mass m and 2m respectively. with a light in extensible string connecting them (length $\large\frac{\pi\;r}{2}$) are placed inside the tube with P at A and Q at B. and released from rest. Assuming that the string remains during motion, find speed of particles when P reaches B

$(a)\;\sqrt {(1+\pi) gr} \quad (b)\; \sqrt {\frac{2}{3}(1+\pi) gr} \quad (c)\;\sqrt {2 \pi gr} \quad (d)\; \sqrt {\frac{1}{3} \pi g r}$

All surfaces are smooth so, mechanical energy of the system will be conserved.
Decrease in P.E of both the block= Increase in KE of both the Blocks
When P reaches B it has lost height=r
and B will loose height $=\large\frac{\pi r}{2}$, length of string
$\therefore mgr+2mg \bigg(\large\frac{\pi r}{2}\bigg)=\frac{1}{2}$$(m+2m)v^2$
$v= \sqrt {\large\frac{2}{3} \normalsize (1+\pi) gr}$
Hence b is the correct answer
edited Jun 13, 2014
Can u explain to me why B will lose height pie*r/2..i cant understand,plz help!

+1 vote