A block of mass M = 2 kg with a semicircular track of radius R = 1.1 m rests on a horizontal frictionless surface. A uniform cylinder of radius r = 10 cm and mass m = 1.0 kg is released from rest from the top point A. The cylinder slips on the semicircular frictionless track. The speed of the block when the cylinder reaches the bottom of the track at B is (g = 10 m/s 2 )
m/s
m/s
m/s
m/s
Text Solution
Verified by ExpertsA
Let the speed of block is v. Then from conservation of
momentum, velocity of cylinder will be 2v in opposite
direction 
Now from conservation of energy
mgh = 
Mv 2 + 
m(2v) 2 , h = R – r = 1 m
so, v = 
m/sec
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