A particle of mass `m` travelling with velocity `v` and kinetic energy `E` collides elastically to another particle of mass `nm`, at rest. What is the fraction of total energy retained by the particle of mass `m`?

Options

A. `((n+1)/n)^(2)`
B. `((n+1)/(n-1)^(2))`
C. `((n-1)/(n+1))^(2)`
D. none of these

Explanation

To solve the problem, we need to analyze the elastic collision between two particles: one of mass \( m \) moving with velocity \( v \) and kinetic energy \( E \), and another of mass \( nm \) at rest. We will find the fraction of total energy retained by the particle of mass \( m \) after the collision.

### Step-by-Step Solution:

1. **Initial Kinetic Energy**:
The initial kinetic energy \( E \) of the particle of mass \( m \) is given by:
\[ E = \frac{1}{2} mv^2 \]

2. **Conservation of Momentum**:
In an elastic collision, momentum is conserved. The initial momentum of the system is:
\[ p_{\text{initial}} = mv + 0 = mv \]
After the collision, let the velocities of the two particles be \( v_1 \) (for mass \( m \)) and \( v_2 \) (for mass \( nm \)). The final momentum is:
\[ p_{\text{final}} = mv_1 + nm v_2 \]
Setting the initial and final momentum equal gives:
\[ mv = mv_1 + nm v_2 \quad \text{(1)} \]

3. **Relative Velocity in Elastic Collision**:
For elastic collisions, the relative velocity of separation is equal to the relative velocity of approach:
\[ v_2 - v_1 = -v \quad \text{(2)} \]

4. **Solving the Equations**:
From equation (2), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_1 - v \]
Substituting this into equation (1):
\[ mv = mv_1 + nm(v_1 - v) \]
Expanding and simplifying:
\[ mv = mv_1 + nm v_1 - nm v \]
\[ mv = (m + nm)v_1 - nm v \]
Rearranging gives:
\[ mv + nm v = (m + nm)v_1 \]
\[ v_1 = \frac{(m + nm)v}{m + nm} = \frac{v}{1 + n} \quad \text{(3)} \]

5. **Finding \( v_2 \)**:
Now substituting \( v_1 \) back into equation (2):
\[ v_2 = v_1 - v = \frac{v}{1 + n} - v = \frac{v - v(1 + n)}{1 + n} = \frac{-nv}{1 + n} \quad \text{(4)} \]

6. **Calculating Kinetic Energies After Collision**:
The kinetic energy of mass \( m \) after the collision is:
\[ KE_m = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left(\frac{v}{1 + n}\right)^2 = \frac{mv^2}{2(1 + n)^2} \]
The total initial kinetic energy is \( E = \frac{1}{2} mv^2 \).

7. **Fraction of Total Energy Retained**:
The fraction of total energy retained by the particle of mass \( m \) is:
\[ \text{Fraction} = \frac{KE_m}{E} = \frac{\frac{mv^2}{2(1 + n)^2}}{\frac{1}{2} mv^2} = \frac{1}{(1 + n)^2} \]

### Final Answer:
The fraction of total energy retained by the particle of mass \( m \) after the collision is:
\[ \frac{1}{(1 + n)^2} \]