A slender rod of mass `M` and length `L` rests on a horizontal frictionless surface. The rod is pivoted about one of ends. The impulse of the force exerted on the rod by the pivot when the rod is struck by a blow of impulse `J` perpendicular to the rod at other end is

Options

A. `J`
B. `J/2`
C. `J/3`
D. Information is insufficient

Explanation

To solve the problem, we will analyze the situation step by step, focusing on the impulse applied to the rod and the resulting motion.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a slender rod of mass \( M \) and length \( L \) resting on a frictionless horizontal surface.
- The rod is pivoted at one end, and an impulse \( J \) is applied perpendicular to the rod at the other end.

2. **Impulse and Momentum**:
- When the impulse \( J \) is applied, it will cause a change in momentum of the rod.
- The initial momentum of the rod is zero (since it is at rest). After the impulse is applied, the final momentum will be \( J \).

3. **Linear Momentum Equation**:
- The change in linear momentum can be expressed as:
\[ J = \Delta p = p_{\text{final}} - p_{\text{initial}} = mv_{\text{cm}} - 0 \]
- Therefore, the velocity of the center of mass \( v_{\text{cm}} \) can be expressed as:
\[ v_{\text{cm}} = \frac{J}{M} \]

4. **Angular Momentum Consideration**:
- The impulse \( J \) also creates an angular momentum about the pivot point.
- The torque \( \tau \) due to the impulse \( J \) is given by:
\[ \tau = J \cdot L \]
- The angular impulse is equal to the change in angular momentum:
\[ J \cdot L = I \cdot \omega \]
- The moment of inertia \( I \) of the rod about the pivot point is:
\[ I = \frac{1}{3}ML^2 \]

5. **Relating Angular Velocity and Linear Velocity**:
- The angular velocity \( \omega \) can be related to the linear velocity of the center of mass:
\[ v_{\text{cm}} = \omega \cdot \frac{L}{2} \]
- Substituting \( v_{\text{cm}} \) from step 3:
\[ \frac{J}{M} = \omega \cdot \frac{L}{2} \]
- Rearranging gives:
\[ \omega = \frac{2J}{ML} \]

6. **Using Angular Momentum Equation**:
- Substitute \( \omega \) back into the angular momentum equation:
\[ J \cdot L = \frac{1}{3}ML^2 \cdot \frac{2J}{ML} \]
- Simplifying this gives:
\[ J \cdot L = \frac{2}{3}JL \]
- This implies:
\[ J = \frac{2}{3}J \]

7. **Impulse by the Pivot**:
- The impulse \( J_0 \) exerted by the pivot can be found by considering the net impulse:
\[ J - J_0 = 0 \quad \Rightarrow \quad J_0 = \frac{1}{2}J \]
- Thus, the impulse of the force exerted on the rod by the pivot is:
\[ J_0 = -\frac{J}{2} \]
- The negative sign indicates that the direction of the impulse by the pivot is opposite to the direction of the applied impulse \( J \).

### Final Answer:
The impulse of the force exerted on the rod by the pivot when the rod is struck by a blow of impulse \( J \) is \( -\frac{J}{2} \).