1. Initial data and setup

• Ring mass: $M=2\,\text{kg}$
• Ring radius: $R=0.5\,\text{m}$
• Initial speed of ring: $v=1\,\text{m s}^{-1}$
• Since it is rolling without slipping, $\omega = \frac{v}{R} = \frac{1}{0.5}=2\,\text{rad s}^{-1}$

Take rightward as positive $x$.

The ring moves to the right, so its angular velocity is clockwise. I will take counterclockwise as positive, hence initially $\omega_i=-2\,\text{rad s}^{-1}$

For a thin ring, $I_{\text{CM}}=MR^2=2(0.5)^2=0.5\,\text{kg m}^2$

The ball has:

• mass $m=0.1\,\text{kg}$
• initial velocity $-20\,\text{m s}^{-1}$ (to the left)
• final velocity vertically upward $=10\,\text{m s}^{-1}$, so final horizontal velocity is $0$.

The collision point is at height $0.75\,\text{m}$ above ground. Since the ring center is at height $R=0.5\,\text{m}$, the hit point is $y=0.75-0.5=0.25\,\text{m}$ above the center.

2. Horizontal impulse delivered by the ball to the ring

For the ball, horizontal momentum changes from $p_{x,i}=0.1(-20)=-2$ to $p_{x,f}=0$ So the change in horizontal momentum of the ball is $\Delta p_x=0-(-2)=+2\,\text{kg m s}^{-1}$

Thus the ring receives an equal and opposite horizontal impulse: $J=-2\,\text{N s}$ (i.e. toward the left).

This changes the translational momentum of the ring: $M V_f = M V_i + J$ $2V_f = 2(1) - 2 = 0$ Hence $V_f=0$

So immediately after collision, the center of mass of the ring is momentarily at rest.

Therefore, option B (“the ring comes to a complete stop”) is false, because it may still rotate.

3. Angular impulse about the center of the ring

The impulse acts horizontally leftward at height $y=0.25\,\text{m}$ above the center.

Torque impulse about CM: $\Delta L_{\text{CM}} = \vec r \times \vec J$ Here, $\vec r=(0,0.25), \quad \vec J=(-2,0)$ So, $\Delta L_z = xJ_y - yJ_x = 0 - (0.25)(-2)=+0.5\,\text{kg m}^2\text{/s}$ This is positive, i.e. counterclockwise.

Initial angular momentum of ring about CM: $L_i = I\omega_i = 0.5(-2)=-1\,\text{kg m}^2\text{/s}$

Thus final angular momentum: $L_f = L_i + \Delta L = -1 + 0.5 = -0.5\,\text{kg m}^2\text{/s}$ So $\omega_f = \frac{L_f}{I} = \frac{-0.5}{0.5}=-1\,\text{rad s}^{-1}$

Hence after collision the ring is still rotating clockwise with angular speed $1\,\text{rad s}^{-1}$, while its CM is stationary.

Therefore option A is true: the ring has pure rotation about its stationary CM.

4. Direction of friction just after collision

Immediately after collision:

• $V_{\text{CM}}=0$
• $\omega=-1\,\text{rad s}^{-1}$ (clockwise)

For pure rotation with clockwise angular velocity, the bottom point of the ring tends to move left relative to the center. Since the center is stationary, the point in contact tends to slip left relative to ground.

Friction opposes this tendency, so friction on the ring must act to the right, not to the left.

Hence option C is false.

Since there is a tendency to slip, friction is not zero. So option D is also false.

5. Final evaluation of options

• A: True
• B: False
• C: False
• D: False

So the correct answer is: $\boxed{A}$

6. Comparison with stored answer

Stored correct answer: A, C

My derived answer is A only.

Reason for disagreement: after collision the ring has $V_{\text{CM}}=0$ and clockwise rotation. The bottom point therefore tends to move left relative to ground, so friction on the ring must act to the right, not left. Hence option C should be false.