A rod of mass m and length l in placed on a smooth table. An another particle of same mass m strikes the rod with velocity `v_(0)` in a direction perpendicular to the rod at distance `x( lt l//2)` from its centre . Particle sticks to the end. Let `omega` be the angular speed of system after collision , then
Find the maximum possible value of impulse (by varying x) that can be imparted to the particle during collision. Particle still sticks to the rod.

Options

A. `(mv_(0))/(2)`
B. `(2mv_(0))/(3)`
C. `(3mv_(0))/(4)`
D. `(4mv_(0))/(5)`

Answer

To solve the problem, we need to analyze the situation step by step. We have a rod of mass \( m \) and length \( l \) on a smooth table, and a particle of the same mass \( m \) strikes the rod perpendicularly at a distance \( x \) from the center of the rod. The particle sticks to the end of the rod after the collision. We need to find the maximum possible impulse imparted to the particle during the collision by varying \( x \).

### Step 1: Understand the System
We have a rod and a particle, both of mass \( m \). The particle strikes the rod with an initial velocity \( v_0 \) at a distance \( x \) from the center of the rod. After the collision, the particle sticks to the rod, and we need to find the angular speed \( \omega \) of the system.

### Step 2: Conservation of Momentum
Before the collision, the momentum of the system is entirely due to the particle:
\[ p_{\text{initial}} = mv_0 \]
After the collision, the total momentum of the system can be expressed in terms of the linear velocity \( v' \) of the center of mass and the angular velocity \( \omega \):
\[ p_{\text{final}} = mv' + m(v' + \omega l/2) \]
However, since the particle sticks to the end of the rod, we can simplify the analysis by focusing on the angular momentum about the center of the rod.

### Step 3: Angular Momentum Conservation
The angular momentum before the collision about the center of the rod is:
\[ L_{\text{initial}} = mv_0 \cdot x \]
After the collision, the angular momentum is:
\[ L_{\text{final}} = I \omega \]
where \( I \) is the moment of inertia of the system. The moment of inertia of the rod about its center is \( \frac{1}{12}ml^2 \), and the moment of inertia of the particle at a distance \( \frac{l}{2} \) from the center is \( m \left(\frac{l}{2}\right)^2 = \frac{1}{4}ml^2 \). Therefore, the total moment of inertia \( I \) is:
\[ I = \frac{1}{12}ml^2 + \frac{1}{4}ml^2 = \frac{1}{3}ml^2 \]

### Step 4: Equate Angular Momenta
Setting the initial and final angular momentum equal gives:
\[ mv_0 \cdot x = \frac{1}{3}ml^2 \omega \]
From this, we can solve for \( \omega \):
\[ \omega = \frac{3v_0 x}{l^2} \]

### Step 5: Impulse Calculation
The impulse \( J \) imparted to the particle during the collision can be calculated as the change in momentum of the particle:
\[ J = mv' - mv_0 \]
The final velocity \( v' \) of the particle after the collision can be expressed in terms of \( \omega \) and the distance from the center:
\[ v' = v_0 - \omega x \]
Substituting \( \omega \) from Step 4:
\[ v' = v_0 - \frac{3v_0 x^2}{l^2} \]
Thus, the impulse becomes:
\[ J = m\left(v_0 - \frac{3v_0 x^2}{l^2}\right) - mv_0 = -m\frac{3v_0 x^2}{l^2} \]

### Step 6: Maximizing the Impulse
To find the maximum impulse, we need to maximize \( -\frac{3v_0 x^2}{l^2} \). Since \( x < \frac{l}{2} \), the maximum value occurs at the maximum value of \( x \), which is \( x = \frac{l}{2} \):
\[ J_{\text{max}} = -m\frac{3v_0 \left(\frac{l}{2}\right)^2}{l^2} = -m\frac{3v_0 \frac{l^2}{4}}{l^2} = -\frac{3mv_0}{4} \]

### Final Answer
The maximum possible value of the impulse imparted to the particle during the collision is:
\[ J_{\text{max}} = \frac{3mv_0}{4} \]