A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron in a head-on collision?

Explanation

To solve the problem of finding the fraction of kinetic energy lost by the neutron in a head-on elastic collision with a stationary deuteron, we will follow these steps:

### Step 1: Define the masses and initial velocities
Let:
- Mass of the neutron, \( m_1 = m \)
- Mass of the deuteron, \( m_2 = 2m \) (since a deuteron consists of one proton and one neutron)
- Initial velocity of the neutron, \( u_1 = u \)
- Initial velocity of the deuteron, \( u_2 = 0 \) (since it is stationary)

### Step 2: Apply conservation of momentum
According to the conservation of momentum:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the values:
\[ m u + 2m \cdot 0 = m v_1 + 2m v_2 \]
This simplifies to:
\[ u = v_1 + 2v_2 \quad \text{(1)} \]

### Step 3: Apply conservation of kinetic energy
According to the conservation of kinetic energy:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the values:
\[ \frac{1}{2} m u^2 + 0 = \frac{1}{2} m v_1^2 + \frac{1}{2} (2m) v_2^2 \]
This simplifies to:
\[ u^2 = v_1^2 + 2v_2^2 \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (1):
\[ v_1 = u - 2v_2 \]
Substituting \( v_1 \) in equation (2):
\[ u^2 = (u - 2v_2)^2 + 2v_2^2 \]
Expanding the square:
\[ u^2 = u^2 - 4uv_2 + 4v_2^2 + 2v_2^2 \]
This simplifies to:
\[ 0 = -4uv_2 + 6v_2^2 \]
Factoring out \( v_2 \):
\[ v_2(6v_2 - 4u) = 0 \]
Since \( v_2 \neq 0 \):
\[ 6v_2 = 4u \implies v_2 = \frac{2u}{3} \]

### Step 5: Find \( v_1 \)
Substituting \( v_2 \) back into equation (1):
\[ v_1 = u - 2\left(\frac{2u}{3}\right) = u - \frac{4u}{3} = -\frac{u}{3} \]

### Step 6: Calculate the kinetic energy loss
The initial kinetic energy of the neutron:
\[ KE_{initial} = \frac{1}{2} m u^2 \]
The final kinetic energy of the neutron:
\[ KE_{final} = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left(-\frac{u}{3}\right)^2 = \frac{1}{2} m \frac{u^2}{9} = \frac{m u^2}{18} \]
The kinetic energy lost by the neutron:
\[ \Delta KE = KE_{initial} - KE_{final} = \frac{1}{2} m u^2 - \frac{m u^2}{18} \]
Finding a common denominator (which is 18):
\[ \Delta KE = \frac{9}{18} m u^2 - \frac{1}{18} m u^2 = \frac{8}{18} m u^2 = \frac{4}{9} m u^2 \]

### Step 7: Calculate the fraction of kinetic energy lost
The fraction of kinetic energy lost by the neutron is:
\[ \text{Fraction} = \frac{\Delta KE}{KE_{initial}} = \frac{\frac{4}{9} m u^2}{\frac{1}{2} m u^2} = \frac{\frac{4}{9}}{\frac{1}{2}} = \frac{4}{9} \cdot \frac{2}{1} = \frac{8}{9} \]

### Final Answer
The fraction of the kinetic energy lost by the neutron in the collision is \( \frac{8}{9} \).

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