A uniform rod of length `l` and mass 2 m rests on a smooth horizontal table. A point mass `m` moving horizontally at right angles to the rod with velocity `v` collides with one end of the rod and sticks it. Then

Options

A. `1/6mv^(2)`
B. `1/3mv^(2)`
C. `mv^(2)`
D. `(mv^(2))/5`

Explanation

To solve the problem step by step, we will analyze the scenario of a point mass colliding with a uniform rod and calculate the loss in kinetic energy.

### Step 1: Understand the System
We have a uniform rod of length \( l \) and mass \( 2m \) resting on a smooth horizontal table. A point mass \( m \) is moving horizontally with velocity \( v \) and collides with one end of the rod.

### Step 2: Apply Conservation of Linear Momentum
Before the collision, the linear momentum of the system is only due to the point mass \( m \):
\[ p_{\text{initial}} = mv \]
After the collision, the point mass sticks to the rod, and the system (rod + point mass) will move with some angular velocity \( \omega \). The total mass after the collision is \( 2m + m = 3m \).

Using conservation of linear momentum:
\[ mv = (3m)v_{\text{final}} \]
This gives us:
\[ v_{\text{final}} = \frac{v}{3} \]

### Step 3: Calculate the Moment of Inertia
The moment of inertia \( I \) of the rod about its center is:
\[ I_{\text{rod}} = \frac{1}{3}(2m)l^2 = \frac{2ml^2}{3} \]
The moment of inertia of the point mass \( m \) at a distance \( l \) from the center of the rod is:
\[ I_{\text{point mass}} = m \cdot l^2 \]
Thus, the total moment of inertia \( I_{\text{total}} \) of the system after the collision is:
\[ I_{\text{total}} = \frac{2ml^2}{3} + ml^2 = \frac{5ml^2}{3} \]

### Step 4: Relate Linear and Angular Velocity
The linear velocity \( v_{\text{final}} \) is related to the angular velocity \( \omega \) by:
\[ v_{\text{final}} = \frac{l}{2} \omega \]
Substituting \( v_{\text{final}} = \frac{v}{3} \):
\[ \frac{v}{3} = \frac{l}{2} \omega \implies \omega = \frac{2v}{3l} \]

### Step 5: Calculate the Kinetic Energy After Collision
The total kinetic energy \( KE \) of the system after the collision is the sum of translational and rotational kinetic energy:
\[ KE = \frac{1}{2}(3m)v_{\text{final}}^2 + \frac{1}{2}I_{\text{total}}\omega^2 \]
Substituting \( v_{\text{final}} = \frac{v}{3} \) and \( I_{\text{total}} = \frac{5ml^2}{3} \):
\[ KE = \frac{1}{2}(3m)\left(\frac{v}{3}\right)^2 + \frac{1}{2}\left(\frac{5ml^2}{3}\right)\left(\frac{2v}{3l}\right)^2 \]
Calculating each term:
1. Translational kinetic energy:
\[ KE_{\text{trans}} = \frac{1}{2}(3m)\left(\frac{v^2}{9}\right) = \frac{mv^2}{6} \]
2. Rotational kinetic energy:
\[ KE_{\text{rot}} = \frac{1}{2}\left(\frac{5ml^2}{3}\right)\left(\frac{4v^2}{9l^2}\right) = \frac{10mv^2}{54} = \frac{5mv^2}{27} \]
Thus, the total kinetic energy after the collision is:
\[ KE = \frac{mv^2}{6} + \frac{5mv^2}{27} \]

### Step 6: Calculate the Initial Kinetic Energy
The initial kinetic energy of the point mass \( m \) before the collision is:
\[ KE_{\text{initial}} = \frac{1}{2}mv^2 \]

### Step 7: Calculate the Loss in Kinetic Energy
The loss in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2}mv^2 - \left(\frac{mv^2}{6} + \frac{5mv^2}{27}\right) \]
Calculating \( KE_{\text{final}} \):
\[ KE_{\text{final}} = \frac{mv^2}{6} + \frac{5mv^2}{27} = \frac{9mv^2}{54} + \frac{10mv^2}{54} = \frac{19mv^2}{54} \]
Thus, the loss in kinetic energy becomes:
\[ \Delta KE = \frac{27mv^2}{54} - \frac{19mv^2}{54} = \frac{8mv^2}{54} = \frac{4mv^2}{27} \]

### Final Result
The loss in kinetic energy due to the impact is:
\[ \Delta KE = \frac{4mv^2}{27} \]