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)

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- 2) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-a48zgkqnnlrh.png" /> (A) √(10/3 ) m s- 1 (B) √(4/3)ms- 1 (C) √(5/2)ms- 1 (D) √10ms- 1

Tardigrade Admin · Accepted Answer

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 = (M/2)) Now from conservation of energy mgh=(1/2)Mv2+(1/2)m(2 v)2, h=R-r=1 m so v=√(10/3 ) m s- 1

$\frac{10}{3} m s^{- 1}$

$\frac{4}{3} m s^{- 1}$

$\frac{5}{2} m s^{- 1}$

$10 m s^{- 1}$

Let the speed of the block be $v$ . Then from conservation of momentum, the velocity of the cylinder will be $2 v$ in the opposite direction $(m = \frac{M}{2})$
Now from conservation of energy
$m g h = \frac{1}{2} M v^{2} + \frac{1}{2} m (2 v)^{2}, h = R - r = 1 m$
so $v = \frac{10}{3} m s^{- 1}$

310​​ms−1

34​​ms−1

25​​ms−1

10​ms−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=2M​) Now from conservation of energy mgh=21​Mv2+21​m(2v)2,h=R−r=1m so v=310​​ms−1

$\frac{10}{3} m s^{- 1}$

$\frac{4}{3} m s^{- 1}$

$\frac{5}{2} m s^{- 1}$

$10 m s^{- 1}$

Let the speed of the block be $v$ . Then from conservation of momentum, the velocity of the cylinder will be $2 v$ in the opposite direction $(m = \frac{M}{2})$
Now from conservation of energy
$m g h = \frac{1}{2} M v^{2} + \frac{1}{2} m (2 v)^{2}, h = R - r = 1 m$
so $v = \frac{10}{3} m s^{- 1}$