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$$^{-1}$$ 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?

The rod is hinged at one end, therefore the pivot supplies no external impulse about itself during the very short collision. Hence both angular momentum and kinetic energy of the two-body system are conserved.

Data: $$M = 1.00\;\text{kg},\; L = 0.20\;\text{m},\; m = 0.10\;\text{kg},\; u = 5.00\;\text{m s}^{-1},\; r = \frac{L}{2}=0.10\;\text{m}$$

Moment of inertia of a uniform rod about one end $$I = \frac{1}{3}ML^{2} = \frac{1}{3}\times 1.00\times (0.20)^{2} = 0.01333\;\text{kg m}^{2}$$ $$-(1)$$

Conservation of angular momentum about the pivot:
Initial angular momentum $$L_i = m u r$$ (mass moves perpendicular to the rod).
After collision the mass rebounds, so its angular momentum is $$-m v r$$ and the rod carries $$I\omega$$. $$m u r = I\omega - m v r$$ $$-(2)$$

Conservation of kinetic energy (elastic impact):
$$\frac{1}{2}m u^{2} = \frac{1}{2}I\omega^{2} + \frac{1}{2}m v^{2}$$ Multiplying by 2, $$m u^{2} = I\omega^{2} + m v^{2}$$ $$-(3)$$

Substituting known numbers in $$-(2)$$:
$$0.10 \times 5.00 \times 0.10 = 0.01333\,\omega - 0.10\,v \times 0.10$$
$$0.05 = 0.01333\,\omega - 0.01\,v$$
$$\omega = 75\,(0.05 + 0.01\,v) = 3.75 + 0.75\,v$$ $$-(4)$$

Put $$-(4)$$ in the energy equation $$-(3)$$:
$$0.10\,(5.00)^{2} = 0.01333\,(3.75 + 0.75\,v)^{2} + 0.10\,v^{2}$$
$$2.50 = \frac{1}{75}(3.75 + 0.75\,v)^{2} + 0.10\,v^{2}$$

Multiplying by 75 to clear the denominator:
$$187.5 = (3.75 + 0.75\,v)^{2} + 7.5\,v^{2}$$

Expanding and simplifying gives the quadratic
$$8.0625\,v^{2} + 5.625\,v - 173.4375 = 0$$

Discriminant $$D = 5.625^{2} + 4\times 8.0625 \times 173.4375 = 75^{2}$$ $$\sqrt{D} = 75$$

Therefore $$v = \frac{-5.625 + 75}{2\times 8.0625} = 4.30\;\text{m s}^{-1}$$ (negative root would give a negative speed and is rejected).

From $$-(4)$$, $$\omega = 3.75 + 0.75 \times 4.30 = 6.98\;\text{rad s}^{-1}$$

Thus the bar acquires an angular speed of $$6.98\;\text{rad s}^{-1}$$ and the small mass rebounds with $$4.30\;\text{m s}^{-1}$$, matching

Option A which is: $$\omega = 6.98\;\text{rad s}^{-1}$$ and $$v = 4.30\;\text{m s}^{-1}$$