A particle of mass `m`, kinetic energy `K` and momentum `p` collides head on elastically with another particle of mass `2 m` at rest. After collision, :
`{:(,"Column I",,"Column II",),((A),"Momentum of first particle",(p),3//4 p,),((B),"Momentum of second particle",(q),-K//9,),((C ),"Kinetic energy of first particle",(r ),-p//3,),((D),"Kinetic energy of second particle",(s),(8K)/(9),),(,,(t),"None",):}`

Explanation

To solve the problem, we will analyze the elastic collision between two particles: one with mass `m`, kinetic energy `K`, and momentum `p`, and another with mass `2m` at rest. We will use the principles of conservation of momentum and conservation of kinetic energy.

### Step 1: Conservation of Momentum
The initial momentum of the system is given by the momentum of the first particle since the second particle is at rest:
\[ p_{\text{initial}} = mv + 0 = mv \]
After the collision, let the velocities of the first and second particles be \(v_1\) and \(v_2\) respectively. The final momentum is:
\[ p_{\text{final}} = mv_1 + 2mv_2 \]
By conservation of momentum:
\[ mv = mv_1 + 2mv_2 \]
Dividing through by \(m\):
\[ v = v_1 + 2v_2 \quad \text{(1)} \]

### Step 2: Conservation of Kinetic Energy
For an elastic collision, the total kinetic energy before and after the collision is conserved. The initial kinetic energy is:
\[ K_{\text{initial}} = \frac{1}{2}mv^2 \]
The final kinetic energy is:
\[ K_{\text{final}} = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2 = \frac{1}{2}mv_1^2 + mv_2^2 \]
By conservation of kinetic energy:
\[ \frac{1}{2}mv^2 = \frac{1}{2}mv_1^2 + mv_2^2 \]
Dividing through by \(m\):
\[ \frac{1}{2}v^2 = \frac{1}{2}v_1^2 + v_2^2 \quad \text{(2)} \]

### Step 3: Solve the Equations
From equation (1), we can express \(v_1\) in terms of \(v_2\):
\[ v_1 = v - 2v_2 \]
Substituting this into equation (2):
\[ \frac{1}{2}v^2 = \frac{1}{2}(v - 2v_2)^2 + v_2^2 \]
Expanding the square:
\[ \frac{1}{2}v^2 = \frac{1}{2}(v^2 - 4vv_2 + 4v_2^2) + v_2^2 \]
Combining terms:
\[ \frac{1}{2}v^2 = \frac{1}{2}v^2 - 2vv_2 + 2v_2^2 \]
Cancelling \(\frac{1}{2}v^2\) from both sides:
\[ 0 = -2vv_2 + 2v_2^2 \]
Factoring out \(2v_2\):
\[ 0 = 2v_2(v_2 - v) \]
This gives us two solutions:
1. \(v_2 = 0\) (not possible since the second particle is at rest)
2. \(v_2 = v\)

### Step 4: Find Velocities
Substituting \(v_2 = v\) back into equation (1):
\[ v = v_1 + 2v \implies v_1 = v - 2v = -v \]

### Step 5: Calculate Momentum and Kinetic Energy
Now we can find the momentum and kinetic energy for both particles after the collision:
- **Momentum of first particle**:
\[ p_1 = mv_1 = m(-v) = -mv = -p \]
- **Momentum of second particle**:
\[ p_2 = 2mv_2 = 2m(v) = 2mv = 2p \]

- **Kinetic energy of first particle**:
\[ K_1 = \frac{1}{2}m(v_1^2) = \frac{1}{2}m(v^2) = \frac{1}{2}mv^2 = K \]
- **Kinetic energy of second particle**:
\[ K_2 = \frac{1}{2}(2m)(v_2^2) = mv^2 = 2K \]

### Final Results
- Momentum of first particle: \( -p \)
- Momentum of second particle: \( 2p \)
- Kinetic energy of first particle: \( K \)
- Kinetic energy of second particle: \( 2K \)

### Matching with Options
- **(A)** Momentum of first particle: \( -p \) (not in options)
- **(B)** Momentum of second particle: \( 2p \) (not in options)
- **(C)** Kinetic energy of first particle: \( K \) (not in options)
- **(D)** Kinetic energy of second particle: \( 2K \) (not in options)