A bar of mass $M = 1.00\,\text{kg}$ and length $L = 0.20\,\text{m}$ is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass $m = 0.10\,\text{kg}$ is moving on the same horizontal surface with $5.00\,\text{m/s}$ speed on a path perpendicular to the bar. It hits the bar at a distance $L/2$ from the pivoted end and returns back on the same path with speed $v$. After this elastic collision, the bar rotates with an angular velocity $\omega$. Which of the following statement is correct?

Step 1: Calculate the moment of inertia of the bar.

The bar is pivoted at one end $O$, so its moment of inertia $I$ about the pivot is given by the formula for a uniform rod pivoted at its end.

$$I = \frac{1}{3}ML^2$$

Substitute the given values $M = 1.00\text{ kg}$ and $L = 0.20\text{ m}$.

$$I = \frac{1}{3}(1.00)(0.20)^2$$ $$I \approx 1.33 \times 10^{-2}\text{ kg m}^2$$

Step 2: Apply conservation of angular momentum.

Since there are no external torques acting on the system about the vertical axis through pivot $O$, the angular momentum is conserved.

$$L_i = L_f$$ $$m u r = I\omega - m v r$$

Rearrange the equation to express $\omega$ in terms of $v$.

$$I\omega = m r (u + v)$$ $$\omega = \frac{m r}{I}(u + v)$$

Substitute the values $m = 0.10\text{ kg}$, $r = 0.10\text{ m}$, $u = 5.00\text{ m s}^{-1}$, and $I = 1.33 \times 10^{-2}\text{ kg m}^2$.

$$\omega = \frac{(0.10)(0.10)}{1.33 \times 10^{-2}}(5.00 + v)$$ $$\omega = 0.75(5.00 + v)$$ $$\omega = 3.75 + 0.75v$$

Step 3: Apply conservation of kinetic energy.

Since the collision is elastic, the total kinetic energy before the collision equals the total kinetic energy after the collision.

$$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Substitute the known values into the energy equation.

$$\frac{1}{2}(0.10)(5.00)^2 = \frac{1}{2}(0.10)v^2 + \frac{1}{2}(1.33 \times 10^{-2})\omega^2$$ $$1.25 = 0.05v^2 + 6.67 \times 10^{-3}\omega^2$$

Step 4: Solve for the final velocity $v$ and angular velocity $\omega$.

Substitute the expression for $\omega$ from Step 2 into the kinetic energy equation.

$$1.25 = 0.05v^2 + 6.67 \times 10^{-3}(3.75 + 0.75v)^2$$

Expand and simplify the terms to form a quadratic equation.

$$43v^2 + 30v - 925 = 0$$

Solve for $v$ using the quadratic formula.

$$v = \frac{-30 + \sqrt{30^2 - 4(43)(-925)}}{2(43)}$$ $$v = \frac{370}{86}$$ $$v \approx 4.30\text{ m s}^{-1}$$

Substitute $v$ back into the expression for $\omega$.

$$\omega = 3.75 + 0.75(4.30)$$ $$\omega \approx 6.98\text{ rad s}^{-1}$$

Answer: $\omega = 6.98\text{ rad s}^{-1}$ and $v = 4.30\text{ m s}^{-1}$