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Question: Answered & Verified by Expert
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

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Options:
  • A 10 m s-1
  • B 4 m s-1
  • C 5 m s-1
  • D 10 m s-1
Solution:
2552 Upvotes Verified Answer
The correct answer is: 10 m s-1
Let the speed of the block be v. Then from conservation of momentum, the velocity of the cylinder will be 2v in the opposite direction m=M2

Now from conservation of energy

mgh=12Mv2+12m2v2, h=R-r=1 m

so v=10 m s-1

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