A bar of mass M = 1.00 kg and length L = 0.20 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 kg is moving on the same horizontal surface with 5.00 m/s speed on a path perpendicular to the bar. It hits the bar at a distance $\frac{L}{2}$ 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 ω. Which of the following statement(s) is correct?

The correct option is ω = 6.98 rad/s and v = 4.30 m/s.

1. Given Parameters:
Mass of the bar, M = 1.00 kg
Length of the bar, L = 0.20 m
Mass of the small particle, m = 0.10 kg
Initial speed of the particle, u = 5.00 m/s
Distance of impact from the pivot, $r=\frac{L}{2}=\frac{0.20}{2}=0.10\text{m}$
Moment of inertia of the bar about its pivoted end:

$I=\frac{1}{3}M{L}^2=\frac{1}{3}\times 1.00\times {\left(0.20\right)}^2=\frac{0.04}{3}\text{kg·m}2$

2. Conservation of Angular Momentum:
Since no net external torque acts on the system about the pivot point, angular momentum about the pivot is conserved.

Initial angular momentum about the pivot, $L_i=mur$
Final angular momentum about the pivot, $L_f=-mvr+I\omega$ (taking initial direction as positive)

$mur=-mvr+I\omega$

$I\omega =mr(u+v)$    —   (Equation 1)

3. Conservation of Kinetic Energy:
Since the collision is perfectly elastic, kinetic energy is conserved.

$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$m(u^2-v^2)=I{\omega }^2$

$m(u-v)(u+v)=I{\omega }^2$    —   (Equation 2)

4. Solving for Final Velocity v:
Substituting $\omega =\frac{mr(u+v)}{I}$ from Equation 1 into Equation 2:

$m(u-v)(u+v)=I{\left[\frac{mr(u+v)}{I}\right]}^2=\frac{m^2{r}^2{(u+v)}^2}{I}$

Dividing both sides by $m(u+v)$:

$u-v=\frac{m{r}^2}{I}(u+v)$

Substitute the numerical values:

$\frac{m{r}^2}{I}=\frac{0.10\times {\left(0.10\right)}^2}{0.04/3}=\frac{0.001}{0.013333}=0.075$

$5.00-v=0.075(5.00+v)$

$5.00-v=0.375+0.075v$

$1.075v=4.625$

$v=\frac{4.625}{1.075}\approx 4.30\text{m/s}$

5. Solving for Angular Velocity ω:
Using Equation 1:

$\omega =\frac{mr(u+v)}{I}=\frac{0.10\times 0.10\times (5.00+4.3023)}{0.04/3}=\frac{0.01\times 9.3023}{0.013333}\approx 6.98\text{rad/s}$

Thus, the correct values are ω = 6.98 rad/s and v = 4.30 m/s.