Step 1: Analyzing the Angular Momentum

Initially, the sphere rotates about the $x$-axis. The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.

$$ \vec{L}_i = I \omega_0 \hat{i} $$

After the interaction, the sphere rotates about the $y$-axis with angular velocity $\omega_0$. To determine the direction (positive or negative $y$), we observe point P. Point P is on the negative $x$-axis (at $-R \hat{i}$). It is observed to move in the positive $z$-direction.

The velocity of P is given by $\vec{v}_P = \vec{\omega}_f \times \vec{r}_P$. If $\vec{\omega}_f = \omega_0 \hat{j}$:

$$ \vec{v}_P = (\omega_0 \hat{j}) \times (-R \hat{i}) = -\omega_0 R (\hat{j} \times \hat{i}) = -\omega_0 R (-\hat{k}) = \omega_0 R \hat{k} $$

This corresponds to the positive $z$-direction, matching the observation. Thus:

$$ \vec{L}_f = I \omega_0 \hat{j} $$

Step 2: The Impulse-Momentum Relation

The change in angular momentum is caused by the torque of the impulsive force $\vec{J}$. The bullet moves in the $+z$ direction and stops, exerting an impulse $\vec{J} = J \hat{k}$ on the sphere.

$$ \Delta \vec{L} = \vec{L}_f – \vec{L}_i = I\omega_0 (\hat{j} – \hat{i}) $$

Also, $\Delta \vec{L} = \vec{r}_{imp} \times \vec{J}$, where $\vec{r}_{imp} = x\hat{i} + y\hat{j}$ is the point of impact on the sphere.

$$ \vec{r}_{imp} \times (J \hat{k}) = (x\hat{i} + y\hat{j}) \times J\hat{k} = xJ(-\hat{j}) + yJ(\hat{i}) $$

Comparing the components:

• $\hat{i}$ component: $-I\omega_0 = yJ \implies y = -\frac{I\omega_0}{J}$
• $\hat{j}$ component: $I\omega_0 = -xJ \implies x = -\frac{I\omega_0}{J}$

Thus, $|x| = |y|$.

Step 3: Calculating Velocity of Center of Mass

The problem states the impact is at a distance $d = 0.2R$ from the $z$-axis. Since the impact is on the surface, the distance from the $z$-axis in the $xy$-plane is $\sqrt{x^2 + y^2}$.

$$ \sqrt{x^2 + y^2} = 0.2R $$

Substituting $|x| = |y|$:

$$ \sqrt{2x^2} = 0.2R \implies |x|\sqrt{2} = \frac{R}{5} \implies |x| = \frac{R}{5\sqrt{2}} $$

From the impulse relation $J = \frac{I\omega_0}{|x|}$:

$$ J = \frac{\frac{2}{5}MR^2 \omega_0}{\frac{R}{5\sqrt{2}}} = \frac{2}{5}MR^2 \omega_0 \cdot \frac{5\sqrt{2}}{R} = 2\sqrt{2} MR \omega_0 $$

The velocity of the center of mass $v_{cm}$ is given by the linear impulse equation $J = M v_{cm}$:

$$ v_{cm} = \frac{J}{M} = \frac{2\sqrt{2} MR \omega_0}{M} = 2\sqrt{2} R \omega_0 $$