Question

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)​

Answer 1

Answer:

The answer will be 1.82 m/s

Explanation:

According to the problem the mass of the block M and radius of the semicircular track, R is given and it is also said the the radius of the cylinder , r = 10 cm and the mass of it . m = 1.0 kg

Let the V1 be the velocity of block and V2 be the velocity of the cylinder.

As the cylinder is rotating when it reaches at point B

The momentum of the conserved, mV1= mV2

Now applying the energy conservation law,

1/2MV1^2 +1/2mV2^2 = mg(R-r)

MV1^2 +m(M/m)^2V1^2 = 2mg(R-r) [ as V2= MV1/m]

Therefore,

V1 = m√2g(R-r)/M(m+M) = 1.0√2 x 10 (1.1-0.1)/2(1.0+2) = 1.82 m/s