Two particles connected by a rigid light rod AB, lying on a smooth horizontal table. An impulse J is applied at A in the plane of thetable 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 motion of the two particles connected by a rigid light rod when an impulse \( J \) is applied at point \( A \). The steps to find the velocity of particle \( A \) are as follows:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two particles connected by a rigid light rod \( AB \) on a smooth horizontal table.
- An impulse \( J \) is applied at point \( A \) perpendicular to the rod.

2. **Impulse and Change in Momentum**:
- The impulse \( J \) causes a change in momentum. Since the initial momentum is zero, the change in momentum is equal to the impulse:
\[ \Delta p = J \]

3. **Velocity of the Center of Mass (CM)**:
- The total mass of the system is \( 2m \) (assuming each particle has mass \( m \)).
- The velocity of the center of mass \( V_{CM} \) can be calculated using the formula:
\[ V_{CM} = \frac{\Delta p}{\text{Total mass}} = \frac{J}{2m} \]

4. **Angular Motion**:
- The impulse also causes the rod to rotate about the center of mass. We need to find the angular velocity \( \omega \) of the rod.
- The angular impulse is given by:
\[ \text{Angular Impulse} = J \cdot \frac{L}{2} \]
where \( L \) is the length of the rod.

5. **Moment of Inertia**:
- The moment of inertia \( I \) of the system about the center of mass is:
\[ I = m \left(\frac{L}{2}\right)^2 + m \left(\frac{L}{2}\right)^2 = \frac{mL^2}{4} + \frac{mL^2}{4} = \frac{mL^2}{2} \]

6. **Relating Angular Impulse to Angular Momentum**:
- The angular impulse is equal to the change in angular momentum:
\[ J \cdot \frac{L}{2} = I \cdot \omega \]
- Substituting for \( I \):
\[ J \cdot \frac{L}{2} = \frac{mL^2}{2} \cdot \omega \]
- Solving for \( \omega \):
\[ \omega = \frac{J}{mL} \]

7. **Velocity of Point A**:
- The total velocity of point \( A \) is the sum of the translational velocity of the center of mass and the rotational velocity due to angular motion:
\[ V_A = V_{CM} + \omega \cdot \frac{L}{2} \]
- Substituting the values:
\[ V_A = \frac{J}{2m} + \left(\frac{J}{mL}\right) \cdot \frac{L}{2} \]
- Simplifying:
\[ V_A = \frac{J}{2m} + \frac{J}{2m} = \frac{J}{m} \]

### Final Answer:
The velocity of particle \( A \) is:
\[ V_A = \frac{J}{m} \]