A body of mass `m` moving with velocity `v` collides head on with another body of mass `2 m` which is initially at rest. The ratio of K.E. of colliding body before and after collision will be

Options

A. `1:1`
B. `2:1`
C. `4:1`
D. `9:1`

Explanation

To solve the problem, we need to find the ratio of the kinetic energy of the colliding body (mass `m`) before and after the collision with another body of mass `2m`. We will assume the collision is elastic.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Mass of the first body (m1) = m, moving with velocity (u1) = v.
- Mass of the second body (m2) = 2m, initially at rest (u2 = 0).

2. **Calculate the initial kinetic energy (K.E.) of the colliding body (mass m):**
\[ \text{K.E.}_{\text{initial}} = \frac{1}{2} m u_1^2 = \frac{1}{2} m v^2 \]

3. **Apply the conservation of momentum:**
The total momentum before the collision equals the total momentum after the collision.
\[ m u_1 + 2m u_2 = m v_1 + 2m v_2 \]
Substituting the values:
\[ mv + 2m \cdot 0 = mv_1 + 2m v_2 \]
This simplifies to:
\[ mv = mv_1 + 2m v_2 \quad \text{(1)} \]

4. **Use the coefficient of restitution for elastic collision:**
The coefficient of restitution (e) is 1 for elastic collisions.
\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]
Substituting the values:
\[ 1 = \frac{v_2 - v_1}{v - 0} \]
This gives us:
\[ v_2 = v_1 + v \quad \text{(2)} \]

5. **Substitute equation (2) into equation (1):**
Replace \(v_2\) in equation (1):
\[ mv = mv_1 + 2m(v_1 + v) \]
Simplifying this:
\[ mv = mv_1 + 2mv_1 + 2mv \]
\[ mv = 3mv_1 + 2mv \]
Rearranging gives:
\[ mv - 2mv = 3mv_1 \]
\[ -mv = 3mv_1 \]
Therefore:
\[ v_1 = -\frac{v}{3} \]

6. **Calculate the final kinetic energy (K.E.) of the colliding body (mass m):**
\[ \text{K.E.}_{\text{final}} = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left(-\frac{v}{3}\right)^2 = \frac{1}{2} m \cdot \frac{v^2}{9} = \frac{mv^2}{18} \]

7. **Find the ratio of kinetic energy before and after the collision:**
\[ \text{Ratio} = \frac{\text{K.E.}_{\text{initial}}}{\text{K.E.}_{\text{final}}} = \frac{\frac{1}{2} mv^2}{\frac{mv^2}{18}} = \frac{\frac{1}{2}}{\frac{1}{18}} = \frac{18}{2} = 9 \]
Thus, the ratio of kinetic energy before and after the collision is:
\[ 9:1 \]

### Conclusion:
The ratio of the kinetic energy of the colliding body before and after the collision is **9:1**.