Question

A smooth narrow tube in the form of an arc AB of a circle of centre O and radius r is fixed so that A is vertically above O and OB is horizontal. Particles P of mass m and Q of mass 2 m with a light inextensible string of length (πr/2) connecting them are placed inside the tube with P at A and Q at B and released from rest. Assuming the string remains taut during motion, find the speed of particles when P reaches B.



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Answer 1

Answer:

The speed of particles when P reaches B is v = √ 2/3 (1 + π)gr

Explanation:

Mechanical energy of the system will be conserved, as all the surfaces of the system are smooth.

Decrease in potential energy of the blocks = Increase in kinetic energy of both the Blocks

When P reaches B it has lost height = r

and B will loose height = πr/2  (length of string)

mgr + 2mg (πr/2) = 12 (m + 2m)v^2

v = √ 2/3 (1 + π)gr

Hence the speed of particles when P reaches B is v = √ 2/3 (1 + π)gr