A thin rod of mass m and length l is free to rotate on a smooth horizontal plane about its one fixed end. When it is at rest, it receives a horizontal impulse J at its other end, at angle of `37^(@)` with the length. Choose incorrect option immediately after impact:

Options

A. Angular momentum of the rod about fixed axis of rotation is `0.6Jl`
B. Angular velocity of the rod fixed of rotation is `(1.8J)/(ml)`
C. Kinetic energy of the rod is `(0.2J^(2))/(m)`
D. Linear velocity of the centre of mass of the rod is `(9)/(10)(J)/(m)`

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Break down the impulse into components
The impulse \( J \) is applied at an angle of \( 37^\circ \) with respect to the length of the rod. We can resolve this impulse into two components:
- The horizontal component: \( J_x = J \cos(37^\circ) \)
- The vertical component: \( J_y = J \sin(37^\circ) \)

Using the trigonometric values:
- \( \cos(37^\circ) = \frac{4}{5} \)
- \( \sin(37^\circ) = \frac{3}{5} \)

Thus, we have:
- \( J_x = J \cdot \frac{4}{5} \)
- \( J_y = J \cdot \frac{3}{5} \)

### Step 2: Calculate the angular impulse
The angular impulse \( \Delta L \) about the fixed end of the rod can be calculated using the formula:
\[ \Delta L = r \cdot J_y \]
where \( r \) is the length of the rod \( l \) and \( J_y \) is the vertical component of the impulse.

Substituting the values:
\[ \Delta L = l \cdot J \cdot \frac{3}{5} \]

### Step 3: Calculate the moment of inertia
The moment of inertia \( I \) of a thin rod about one end is given by:
\[ I = \frac{1}{3} m l^2 \]

### Step 4: Relate angular impulse to angular momentum
Since the rod was initially at rest, the change in angular momentum is equal to the angular impulse:
\[ \Delta L = I \omega \]
where \( \omega \) is the angular velocity after the impulse.

Substituting the values:
\[ l \cdot J \cdot \frac{3}{5} = \frac{1}{3} m l^2 \cdot \omega \]

### Step 5: Solve for angular velocity \( \omega \)
Rearranging the equation to find \( \omega \):
\[ \omega = \frac{3J}{5m} \cdot \frac{1}{l} \]

### Step 6: Calculate the linear velocity of the center of mass
The linear velocity \( v_{cm} \) of the center of mass of the rod (which is at a distance \( \frac{l}{2} \) from the fixed end) can be calculated as:
\[ v_{cm} = \omega \cdot \frac{l}{2} \]
Substituting for \( \omega \):
\[ v_{cm} = \left(\frac{3J}{5m} \cdot \frac{1}{l}\right) \cdot \frac{l}{2} = \frac{3J}{10m} \]

### Step 7: Determine incorrect statement
Now we can analyze the statements provided in the question. The correct calculations lead us to the conclusion that the linear velocity of the center of mass is \( \frac{3J}{10m} \) and the angular velocity is \( \frac{3J}{5ml} \). Any statement that contradicts these results would be incorrect.