A uniform rod of length 8 a and mass 6 m lies on a smooth horizontal surface. Two point masses m and 2 m moving in the same plane with speed 2 v and v respectively strike the rod perpendicular at distances a and 2a from the mid point of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is

Options

A. `(6v)/(32a)`
B. `(6v)/(33a)`
C. `(6v)/(40a)`
D. `(6v)/(41a)`

Answer

To solve the problem step by step, we will use the principles of conservation of angular momentum and the calculation of moment of inertia.

### Step 1: Understand the System
We have a uniform rod of length \(8a\) and mass \(6m\) lying on a smooth horizontal surface. Two point masses \(m\) and \(2m\) are moving towards the rod with speeds \(2v\) and \(v\) respectively. The mass \(m\) strikes the rod at a distance \(a\) from the midpoint, and the mass \(2m\) strikes at a distance \(2a\) from the midpoint in the opposite direction.

### Step 2: Calculate Initial Angular Momentum
The angular momentum \(L\) of a point mass about a point is given by the product of its linear momentum and the perpendicular distance from the line of action of the force to the point about which we are calculating the angular momentum.

1. For mass \(m\) moving with speed \(2v\) at a distance \(a\):
\[ L_1 = m \cdot 2v \cdot a = 2mva \]

2. For mass \(2m\) moving with speed \(v\) at a distance \(2a\):
\[ L_2 = 2m \cdot v \cdot 2a = 4mva \]

Since both angular momenta are in the same direction (counterclockwise), we can add them:
\[ L_{\text{initial}} = L_1 + L_2 = 2mva + 4mva = 6mva \]

### Step 3: Calculate Moment of Inertia of the System
The moment of inertia \(I\) of the rod about its center is given by:
\[ I_{\text{rod}} = \frac{1}{12} m L^2 = \frac{1}{12} \cdot 6m \cdot (8a)^2 = \frac{1}{12} \cdot 6m \cdot 64a^2 = 32ma^2 \]

Next, we need to add the contributions from the two masses that stick to the rod.

1. For mass \(m\) at a distance \(a\):
\[ I_1 = m \cdot a^2 = m \cdot a^2 \]

2. For mass \(2m\) at a distance \(2a\):
\[ I_2 = 2m \cdot (2a)^2 = 2m \cdot 4a^2 = 8ma^2 \]

Now, the total moment of inertia of the system after the collision is:
\[ I_{\text{final}} = I_{\text{rod}} + I_1 + I_2 = 32ma^2 + ma^2 + 8ma^2 = 41ma^2 \]

### Step 4: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_{\text{initial}} = L_{\text{final}} \]
The final angular momentum can be expressed as:
\[ L_{\text{final}} = I_{\text{final}} \cdot \omega \]
Setting the two expressions for angular momentum equal gives:
\[ 6mva = 41ma^2 \cdot \omega \]

### Step 5: Solve for Angular Velocity \(\omega\)
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ 6va = 41a^2 \cdot \omega \]
Dividing both sides by \(41a^2\):
\[ \omega = \frac{6v}{41a} \]

### Final Answer
The angular velocity of the system immediately after the collision is:
\[ \omega = \frac{6v}{41a} \]