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. `2 (J)/(m)`
B. `(J)/(sqrt(2)m)`
C. `(J)/(m)`
D. `sqrt(2) (J)/(m)`

Explanation

To find the speed of the particle P at a distance \( \frac{l}{6} \) from the center towards A of the rod after time \( t = \frac{\pi m l}{12 J} \), we can follow these steps:

### Step 1: Determine the Linear Velocity of the Center of Mass (VCM)
The impulse \( J \) applied to the rod will impart a linear momentum to the center of mass. By the conservation of linear momentum, we have:

\[ J = m \cdot V_{CM} \]

From this, we can solve for \( V_{CM} \):

\[ V_{CM} = \frac{J}{m} \]

### Step 2: Determine the Angular Velocity (ω)
The impulse also causes the rod to rotate about its center of mass. The torque \( \tau \) due to the impulse about the center of mass is given by:

\[ \tau = J \cdot \frac{l}{2} \]

Using the relation for torque and angular momentum, we have:

\[ \tau = I \cdot \omega \]

where \( I \) is the moment of inertia of the rod about its center, given by:

\[ I = \frac{1}{12} m l^2 \]

Setting the two expressions for torque equal gives:

\[ J \cdot \frac{l}{2} = \frac{1}{12} m l^2 \cdot \omega \]

Solving for \( \omega \):

\[ \omega = \frac{6J}{ml} \]

### Step 3: Determine the Angle of Rotation (θ)
The angle of rotation \( \theta \) after time \( t \) can be found using the formula:

\[ \theta = \omega \cdot t \]

Substituting \( t = \frac{\pi m l}{12 J} \):

\[ \theta = \left(\frac{6J}{ml}\right) \cdot \left(\frac{\pi m l}{12 J}\right) = \frac{\pi}{2} \]

### Step 4: Determine the Velocity of Particle P
The particle P is located at a distance \( \frac{l}{6} \) from the center of mass. It has two components of velocity:

1. The translational velocity \( V_{CM} \) in the direction of motion.
2. The tangential velocity due to rotation, which is given by \( v_{tangential} = \omega \cdot r \), where \( r = \frac{l}{6} \).

Calculating the tangential velocity:

\[ v_{tangential} = \omega \cdot \frac{l}{6} = \left(\frac{6J}{ml}\right) \cdot \frac{l}{6} = \frac{J}{m} \]

### Step 5: Total Velocity of Particle P
The total velocity \( V_P \) of particle P is the vector sum of the translational and tangential velocities:

\[ V_P = \sqrt{V_{CM}^2 + v_{tangential}^2} \]

Substituting the values we found:

\[ V_P = \sqrt{\left(\frac{J}{m}\right)^2 + \left(\frac{J}{m}\right)^2} = \sqrt{2\left(\frac{J}{m}\right)^2} = \frac{J}{m} \sqrt{2} \]

### Final Answer
The speed of particle P at a distance \( \frac{l}{6} \) from the center towards A of the rod after time \( t = \frac{\pi m l}{12 J} \) is:

\[ V_P = \frac{J}{m} \sqrt{2} \]