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

Step 1: Identify the initial parameters of the ring and the ball.

The ring rolls without slipping, so its initial angular velocity is related to its linear velocity.

$$\omega_i = \frac{v}{R} = \frac{1}{0.5} = 2\,\text{rad/s}$$

The moment of inertia of the ring about its center of mass is:

$$I_{\text{cm}} = MR^2 = 2 \times (0.5)^2 = 0.5\,\text{kg}\cdot\text{m}^2$$

Step 2: Analyze the geometry of the impact.

The ball hits the ring at a height $h = 0.75\,\text{m}$. Since the center of the ring is at height $R = 0.5\,\text{m}$, the impact point is $0.25\,\text{m}$ above the center.

The horizontal distance $x$ of the impact point from the vertical centerline of the ring is:

$$x = \sqrt{R^2 - 0.25^2} = \sqrt{0.5^2 - 0.25^2} \approx 0.433\,\text{m}$$

Step 3: Determine the velocity of the ring's center of mass immediately after the collision.

Using conservation of linear momentum in the horizontal direction during the collision:

$$P_i = -M v + m u$$ $$P_i = -(2)(1) + (0.1)(20) = 0$$

Since the ball moves vertically upward after the collision, its final horizontal velocity is zero. Thus, the final velocity of the ring's center of mass $V_f$ must also be zero:

$$M V_f = 0$$ $$V_f = 0$$

Step 4: Determine the angular velocity of the ring immediately after the collision.

The initial angular momentum of the ring about the contact point $O$ on the ground is:

$$L_i = 2MRv = 2(2)(0.5)(1) = 2\,\text{kg}\cdot\text{m}^2/\text{s}$$

The angular impulse $\tau_O$ exerted by the ball on the ring about $O$ is:

$$\tau_O = m v_y' x - m u h$$ $$\tau_O = (0.1)(10)(0.433) - (0.1)(20)(0.75)$$ $$\tau_O = 0.433 - 1.50 = -1.07\,\text{N}\cdot\text{s}\cdot\text{m}$$

The final angular momentum about $O$ is:

$$L_f = L_i + \tau_O$$ $$L_f = 2 - 1.07 = 0.93\,\text{kg}\cdot\text{m}^2/\text{s}$$

Since the center of mass is stationary ($V_f = 0$), the final angular momentum is purely due to rotation about the center of mass:

$$I_{\text{cm}}\omega_f = 0.93$$ $$0.5\,\omega_f = 0.93$$ $$\omega_f = 1.87\,\text{rad/s}$$

Step 5: Determine the direction of the friction force.

The velocity of the contact point of the ring with the ground immediately after the collision is:

$$v_O = V_f + \omega_f R$$ $$v_O = 0 + (1.87)(0.5) = 0.935\,\text{m/s}$$

Since the contact point of the ring slips to the right, the kinetic friction force on the ring acts to the left.

Answer: Immediately after the collision, the ring is in pure rotation about its stationary center of mass and the friction force acts to the left.