Two particles of equal mass m at A and B are connected by a rigid light rod AB lying on a smooth horizontal table. An impulse J is applied at A in the plane of the table and perpendicular at AB. Then the velocity of particle at A is :

Options

A. `(J)/(2m)`
B. `(J)/(m)`
C. `(2J)/(m)`
D. zero

Answer

To solve the problem, we need to analyze the situation involving two particles connected by a rigid rod and the effect of the applied impulse. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two particles of equal mass \( m \) located at points \( A \) and \( B \).
- They are connected by a rigid light rod \( AB \) on a smooth horizontal table.
- An impulse \( J \) is applied at point \( A \) perpendicular to the rod \( AB \).

2. **Defining the Impulse**:
- The impulse \( J \) causes a change in momentum for the particle at \( A \).
- Since the impulse is applied perpendicular to the rod, it will also create an angular motion about the center of mass of the system.

3. **Calculating Angular Impulse**:
- The angular impulse \( \tau \) about the center of mass \( C \) can be calculated as:
\[ \tau = J \cdot d \]
where \( d \) is the distance from the center of mass to point \( A \). Since the masses are equal and the rod is uniform, the center of mass is at the midpoint, so \( d = \frac{l}{2} \).

4. **Relating Angular Impulse to Angular Momentum**:
- The change in angular momentum \( \Delta L \) is equal to the angular impulse:
\[ \Delta L = J \cdot \frac{l}{2} \]

5. **Moment of Inertia Calculation**:
- The moment of inertia \( I \) of the system about the center of mass is given by:
\[ I = m \left(\frac{l}{2}\right)^2 + m \left(\frac{l}{2}\right)^2 = 2m \left(\frac{l}{2}\right)^2 = \frac{ml^2}{2} \]

6. **Using the Relation of Angular Momentum**:
- The angular momentum \( L \) can also be expressed as:
\[ L = I \omega \]
- Setting the angular impulse equal to the change in angular momentum gives:
\[ J \cdot \frac{l}{2} = \frac{ml^2}{2} \omega \]

7. **Solving for Angular Velocity \( \omega \)**:
- Rearranging the equation to solve for \( \omega \):
\[ \omega = \frac{J}{ml} \]

8. **Finding the Velocity of Particle at A**:
- The linear velocity \( v_A \) of particle \( A \) can be related to the angular velocity \( \omega \) by:
\[ v_A = \frac{l}{2} \omega \]
- Substituting for \( \omega \):
\[ v_A = \frac{l}{2} \cdot \frac{J}{ml} = \frac{J}{2m} \]

9. **Final Result**:
- Therefore, the velocity of particle at \( A \) is:
\[ v_A = \frac{J}{2m} \]