Two particles of masses `m_(1)` and `m_(2)` and velocities `u_(1)` and `alphau_(1)(alpha!=0)` make an elastic head on collision. If the initial kinetic energies of the two particles are equal and `m_(1)` comes to rest after collision, then

Options

A. `u_(1)/u_(2)=sqrt(2)+1`
B. `u_(1)/u_(2)=sqrt(2)-1`
C. `m_(2)/m_(1)=3+2sqrt(2)`
D. `m_2/m_1=3-2sqrt2`

Explanation

To solve the problem step by step, we will analyze the given information and apply the principles of conservation of momentum, conservation of energy, and the concept of elastic collisions.

### Step 1: Set up the equations for initial kinetic energies
Given that the initial kinetic energies of the two particles are equal, we can write:

\[ \frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_2 (\alpha u_1)^2 \]

### Step 2: Simplify the kinetic energy equation
Cancelling \(\frac{1}{2}\) from both sides gives:

\[ m_1 u_1^2 = m_2 \alpha^2 u_1^2 \]

Now, we can divide both sides by \(u_1^2\) (assuming \(u_1 \neq 0\)):

\[ m_1 = m_2 \alpha^2 \]

### Step 3: Express \(\alpha\) in terms of masses
From the equation \(m_1 = m_2 \alpha^2\), we can express \(\alpha\):

\[ \alpha = \sqrt{\frac{m_1}{m_2}} \]

### Step 4: Apply conservation of momentum
According to the conservation of momentum, we have:

\[ m_1 u_1 + m_2 (\alpha u_1) = m_2 v_2 + m_1 \cdot 0 \]

This simplifies to:

\[ m_1 u_1 + m_2 \alpha u_1 = m_2 v_2 \]

### Step 5: Substitute \(\alpha\) into the momentum equation
Substituting \(\alpha = \sqrt{\frac{m_1}{m_2}}\):

\[ m_1 u_1 + m_2 \left(\sqrt{\frac{m_1}{m_2}} u_1\right) = m_2 v_2 \]

This simplifies to:

\[ m_1 u_1 + \sqrt{m_1 m_2} u_1 = m_2 v_2 \]

### Step 6: Factor out \(u_1\)
Factoring out \(u_1\):

\[ u_1 (m_1 + \sqrt{m_1 m_2}) = m_2 v_2 \]

### Step 7: Solve for \(v_2\)
Rearranging gives:

\[ v_2 = \frac{u_1 (m_1 + \sqrt{m_1 m_2})}{m_2} \]

### Step 8: Apply the coefficient of restitution
The coefficient of restitution \(e\) for an elastic collision is given by:

\[ e = \frac{\text{velocity of separation}}{\text{velocity of approach}} \]

For our scenario, since \(m_1\) comes to rest after the collision, we have:

\[ 1 = \frac{v_2 - 0}{u_1 - \alpha u_1} \]

This simplifies to:

\[ v_2 = u_1(1 - \alpha) \]

### Step 9: Equate the two expressions for \(v_2\)
Now we have two expressions for \(v_2\):

1. \(v_2 = \frac{u_1 (m_1 + \sqrt{m_1 m_2})}{m_2}\)
2. \(v_2 = u_1(1 - \alpha)\)

Setting these equal gives:

\[ \frac{u_1 (m_1 + \sqrt{m_1 m_2})}{m_2} = u_1(1 - \alpha) \]

### Step 10: Cancel \(u_1\) and solve for \(\alpha\)
Assuming \(u_1 \neq 0\), we can cancel \(u_1\):

\[ \frac{m_1 + \sqrt{m_1 m_2}}{m_2} = 1 - \alpha \]

Substituting \(\alpha = \sqrt{\frac{m_1}{m_2}}\):

\[ \frac{m_1 + \sqrt{m_1 m_2}}{m_2} = 1 - \sqrt{\frac{m_1}{m_2}} \]

### Step 11: Solve for the ratio of masses
From this equation, we can derive the relationship between \(m_1\) and \(m_2\) to find the required ratios.