IIT-JEE 2011 · Paper 2 · Physics Q32

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PhysicsMultiple correct+4Question 32 of 60

A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping 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

Ring on the ground moving left at 1 m/s; a ball moving right at 20 m/s strikes it at height 0.75 m and leaves vertically upward at 10 m/s.

the ring has pure rotation about its stationary CM.

the ring comes to a complete stop.

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

there is no friction between the ring and the ground.

Answer and worked solution

Official answer: A

Linear momentum (horizontal, collision impulse ≫ friction impulse). Taking right as positive: before, $2(-1)+0.1(20)=0$ . After, the ball has no horizontal velocity, so the ring's centre is at rest.

Angular momentum about the centre. The ball hits at height 0.75 m, 0.25 m above the centre, at horizontal offset $-\sqrt{0.5^2-0.25^2}=-0.433$ m. The ring receives impulse $(+2,-1)$ N s there, with angular impulse $(-0.433)(-1)-(0.25)(2)=-0.067$ kg m² s⁻¹. Initially $L=I\omega=2(0.5)^2(2)=1$ , so afterwards $L=0.93$ and the ring still spins (≈ 1.9 rad/s anticlockwise).

Result. Pure rotation about a stationary CM (A). The bottom point then slides to the right, so kinetic friction acts to the left (C). This is why the key accepts both A and AC.

(A) (key: A or AC)

Topics:Angular momentum and its conservation Conservation of linear momentum Rolling motion

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IIT-JEE 2011 · Paper 2, Physics question 32