Question 319: Work, Energy and Power & Circular Motion

Question

A thin ring of mass 2 kg and radius 0.5 m is rolling without on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision,

Accepted Answer

Let M be the mass of the ring and m that of the ball and 
 let V and $$\upsilon $$ be their velocity before collision. The initial momentum of the system (ring and ball) in the horizontal direction is
$${\overrightarrow p _i} = M\overrightarrow V  + m\overrightarrow \upsilon  $$
$$ = 2 	imes 1 + 0.1 	imes ( - 20)$$
$$ = 2 - 2 = 0$$
From conservation of momentum, the final momentum of the system $${\overrightarrow p _f} = 0$$ in the horizontal direction. Hence $${V_{cm}} = 0$$ for the ring, i.e. the ring has pure rotation about its centre of mass. So choice (a) is correct.
The total initial angular momentum of the system about the point of collision is
$${L_i} = m\upsilon r - I\omega $$
$$ = m\upsilon r - M{R^2}{V \over R}$$
$$ = m\upsilon r - MRV$$
$$ = 0.1 	imes 20 	imes 0.75 - 2 	imes 0.5 	imes 1$$
$$ = 1.5 - 1 = 0.5$$ kg m
2
s
$$-$$1
From the conservation of angular momentum, the final angular velocity must be anticlockwise. Hence the friction between the ring and the ground is to the left. So the correct choices are (a) and (c).

Answer

the ring has pure rotation about its stationary CM.

Answer

the ring comes to a complete stop.

Answer

friction between the ring and the ground is to the left.

Answer

there is no friction between the ring and the ground.

Work, Energy and Power & Circular Motion

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