Solution to Question 13
1. Work-Energy Principle
The work done by the rod on the particle is equal to the change in the kinetic energy of the particle (since gravity and normal force from the floor do no work).
$$ W = \Delta K = K_f – K_i $$
Initially, the particle is at rest, so $K_i = 0$. We need to find the final velocity of the particle in the ground frame.
2. Analyzing Velocity in the Rod’s Frame
Let the rod define a moving reference frame. The rod moves with velocity $\vec{V}_{rod} = v \hat{i}$.
- Initial State (Relative): The particle is stationary in the lab frame. In the rod’s frame, the particle approaches the rod with velocity $\vec{v}_{rel} = -v \hat{i}$.
- Interaction: The particle enters the curved section (quarter circle) tangentially at A (where the tangent is horizontal) and exits at B (where the tangent is vertical).
- Since the rod is frictionless, the speed of the particle in the rod’s frame remains constant. The direction changes by $90^\circ$.
- Final State (Relative): The particle exits the rod with velocity magnitude $v$, directed perpendicular to the initial relative velocity. Let’s say it exits in the $\hat{j}$ direction. So, $\vec{v}_{rel, final} = v \hat{j}$.
3. Determining Final Velocity in Lab Frame
We transform the final velocity back to the lab frame:
$$ \vec{v}_{final} = \vec{V}_{rod} + \vec{v}_{rel, final} $$
$$ \vec{v}_{final} = v \hat{i} + v \hat{j} $$
The magnitude of the final velocity is:
$$ |\vec{v}_{final}| = \sqrt{v^2 + v^2} = v\sqrt{2} $$
4. Calculating Work Done
$$ W = \Delta K = \frac{1}{2}m (v\sqrt{2})^2 – 0 $$
$$ W = \frac{1}{2}m (2v^2) = mv^2 $$
Correct Option: (b)