A uniform rod `AB` of mass `m` and length `l` is at rest on a smooth horizontal surface. An impulse `J` is applied to the end `B`, perpendicular to the rod in the horizontal direction. Speed of particlem `P` at a distance `(l)/(6)` from the centre towards `A` of the rod after time `t = (pi m l)/(12 J)` is.

Options

A. `2J/M`
B. `(J)/(sqrt2M)`
C. `J/M`
D. `sqrt2J/M`

Answer

To solve the problem, we need to analyze the motion of the uniform rod after an impulse is applied. The impulse causes both translational and rotational motion of the rod. Let's break down the steps to find the speed of the particle P after a given time.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a uniform rod `AB` of mass `m` and length `l` resting on a smooth horizontal surface.
- An impulse `J` is applied at the end `B` perpendicular to the rod.

2. **Finding the Velocity of the Center of Mass**:
- The impulse `J` causes a change in momentum. Since the rod is initially at rest, the change in momentum is equal to the impulse:
\[ J = m \cdot v_{cm} \]
- Therefore, the velocity of the center of mass (`v_{cm}`) of the rod after the impulse is:
\[ v_{cm} = \frac{J}{m} \]

3. **Calculating the Angular Velocity**:
- The impulse also causes the rod to rotate about its center of mass. The angular impulse is given by:
\[ \text{Angular Impulse} = J \cdot \frac{l}{2} \]
- The moment of inertia `I` of the rod about its center is:
\[ I = \frac{ml^2}{12} \]
- The change in angular momentum is equal to the angular impulse:
\[ J \cdot \frac{l}{2} = I \cdot \omega \]
- Substituting for `I`:
\[ J \cdot \frac{l}{2} = \frac{ml^2}{12} \cdot \omega \]
- Solving for angular velocity (`ω`):
\[ \omega = \frac{6J}{ml} \]

4. **Finding the Angle Rotated After Time `t`**:
- The time given is:
\[ t = \frac{\pi ml}{12J} \]
- The angle rotated (`θ`) in time `t` is:
\[ θ = \omega \cdot t = \left(\frac{6J}{ml}\right) \cdot \left(\frac{\pi ml}{12J}\right) = \frac{\pi}{2} \]
- This means the rod has rotated 90 degrees and is now horizontal.

5. **Finding the Velocity of Point P**:
- Point P is located at a distance of `l/6` from the center of the rod towards A.
- The linear velocity of point P due to the translational motion of the center of mass is:
\[ v_{P, cm} = v_{cm} = \frac{J}{m} \]
- The tangential velocity of point P due to rotation is:
\[ v_{P, rot} = \omega \cdot \left(\frac{l}{6}\right) = \left(\frac{6J}{ml}\right) \cdot \left(\frac{l}{6}\right) = \frac{J}{m} \]
- Since the rod is horizontal, the velocities are perpendicular. Therefore, the resultant velocity of point P is:
\[ v_P = \sqrt{(v_{P, cm})^2 + (v_{P, rot})^2} = \sqrt{\left(\frac{J}{m}\right)^2 + \left(\frac{J}{m}\right)^2} = \sqrt{2 \left(\frac{J}{m}\right)^2} = \frac{J}{m} \sqrt{2} \]

6. **Final Result**:
- The speed of particle P after time `t` is:
\[ v_P = \frac{J \sqrt{2}}{m} \]