A neutron travelling with a velocity `v` and kinetic energy `E` collides perfectly elastically head on with the nucleus of an atom of mass number `A` at rest. The fraction of the total kinetic energy retained by the neutron is

Options

A. `((A-1)/(A+1))^(2)`
B. `((A+1)/(A-1))^(2)`
C. `((A-1)/(A))^(2)`
D. `((A+1)/(A))^(2)`

Explanation

To solve the problem of the fraction of total kinetic energy retained by the neutron after a perfectly elastic collision with a nucleus at rest, we can follow these steps:

### Step 1: Understand the Initial Conditions
- A neutron of mass \( m \) is traveling with a velocity \( v \) and has a kinetic energy \( E \).
- The nucleus of an atom with mass number \( A \) is at rest.

### Step 2: Write the Expression for Initial Kinetic Energy
The initial kinetic energy \( E \) of the neutron can be expressed as:
\[ E = \frac{1}{2} m v^2 \]

### Step 3: Apply Conservation of Momentum
In a perfectly elastic collision, momentum is conserved. The initial momentum of the system is:
\[ p_{\text{initial}} = mv + 0 = mv \]
After the collision, let the neutron have a velocity \( v_1 \) and the nucleus have a velocity \( v_2 \). The final momentum is:
\[ p_{\text{final}} = mv_1 + (Am)v_2 \]
Setting the initial momentum equal to the final momentum gives:
\[ mv = mv_1 + Am v_2 \quad \text{(Equation 1)} \]

### Step 4: Apply Conservation of Kinetic Energy
In a perfectly elastic collision, the total kinetic energy before and after the collision is also conserved. Thus:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mv_1^2 + \frac{1}{2} (Am)v_2^2 \]
This simplifies to:
\[ mv^2 = mv_1^2 + Amv_2^2 \quad \text{(Equation 2)} \]

### Step 5: Solve the Equations
From Equation 1, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = v - Av_2 \quad \text{(from rearranging Equation 1)} \]

Substituting \( v_1 \) into Equation 2:
\[ mv^2 = m(v - Av_2)^2 + Amv_2^2 \]
Expanding and simplifying:
\[ mv^2 = m(v^2 - 2Av v_2 + A^2 v_2^2) + Amv_2^2 \]
\[ mv^2 = mv^2 - 2Amv v_2 + (mA + Am)v_2^2 \]
Rearranging gives:
\[ 2Amv v_2 = (mA + Am)v_2^2 \]
This can be factored to find \( v_2 \):
\[ v_2(2Av - (A + 1)v_2) = 0 \]
Thus, either \( v_2 = 0 \) (not possible since it’s a collision) or:
\[ v_2 = \frac{2v}{A + 1} \]

### Step 6: Find \( v_1 \)
Substituting \( v_2 \) back into the equation for \( v_1 \):
\[ v_1 = v - A\left(\frac{2v}{A + 1}\right) = v\left(1 - \frac{2A}{A + 1}\right) = v\left(\frac{1 - A}{A + 1}\right) \]

### Step 7: Calculate the Kinetic Energy Retained by the Neutron
The kinetic energy of the neutron after the collision is:
\[ K = \frac{1}{2} mv_1^2 = \frac{1}{2} m\left(v\left(\frac{1 - A}{A + 1}\right)\right)^2 \]
Substituting \( E = \frac{1}{2} mv^2 \):
\[ K = E \left(\frac{(1 - A)^2}{(A + 1)^2}\right) \]

### Step 8: Calculate the Fraction of Kinetic Energy Retained
The fraction of the total kinetic energy retained by the neutron is:
\[ \frac{K}{E} = \frac{(1 - A)^2}{(A + 1)^2} \]

### Final Answer
The fraction of the total kinetic energy retained by the neutron is:
\[ \frac{(1 - A)^2}{(A + 1)^2} \]