A particle of mass m collides with a stationary particle and continues to move at an angle of `45^(@)` with respect to the original direction. The second particle also recoils at an angle of `45^(@)` to this direction. The mass of the second particle is (collision is elastic)

Options

A. m
B. `sqrt(2)m`
C. `(m)/(sqrt(2))`
D. `(m)/(2)`

Answer

To solve the problem, we will use the principles of conservation of momentum and conservation of kinetic energy, as the collision is elastic.

### Step 1: Define the initial conditions
- Let the mass of the first particle be \( m \) and its initial velocity be \( u \).
- The second particle is stationary, so its initial velocity is \( 0 \).
- After the collision, the first particle moves at an angle of \( 45^\circ \) with respect to the original direction, and let its velocity be \( u_1 \).
- The second particle recoils at an angle of \( 45^\circ \) to this direction with a velocity \( u_2 \).

### Step 2: Apply conservation of momentum in the x-direction
The total momentum before the collision in the x-direction is:
\[ p_{initial,x} = m \cdot u + 0 = mu \]
After the collision, the momentum in the x-direction is:
\[ p_{final,x} = m \cdot u_1 \cos(45^\circ) + m_2 \cdot u_2 \cos(45^\circ) \]
Setting these equal gives:
\[ mu = m \cdot u_1 \cdot \frac{1}{\sqrt{2}} + m_2 \cdot u_2 \cdot \frac{1}{\sqrt{2}} \]
Multiplying through by \( \sqrt{2} \):
\[ \sqrt{2}mu = mu_1 + m_2u_2 \]

### Step 3: Apply conservation of momentum in the y-direction
The total momentum before the collision in the y-direction is:
\[ p_{initial,y} = 0 \]
After the collision, the momentum in the y-direction is:
\[ p_{final,y} = m \cdot u_1 \sin(45^\circ) - m_2 \cdot u_2 \sin(45^\circ) \]
Setting these equal gives:
\[ 0 = m \cdot u_1 \cdot \frac{1}{\sqrt{2}} - m_2 \cdot u_2 \cdot \frac{1}{\sqrt{2}} \]
Multiplying through by \( \sqrt{2} \):
\[ m u_1 = m_2 u_2 \]

### Step 4: Relate the velocities using kinetic energy conservation
Since the collision is elastic, the initial kinetic energy is equal to the final kinetic energy:
\[ \frac{1}{2} mu^2 = \frac{1}{2} mu_1^2 + \frac{1}{2} m_2 u_2^2 \]
Cancelling \( \frac{1}{2} \) from both sides:
\[ mu^2 = mu_1^2 + m_2 u_2^2 \]

### Step 5: Substitute \( m_2 \) from momentum equations
From the y-direction momentum equation, we can express \( m_2 \) in terms of \( m \), \( u_1 \), and \( u_2 \):
\[ m_2 = \frac{m u_1}{u_2} \]
Substituting this into the kinetic energy equation:
\[ mu^2 = mu_1^2 + \left(\frac{m u_1}{u_2}\right) u_2^2 \]
This simplifies to:
\[ mu^2 = mu_1^2 + mu_1 u_2 \]
Dividing through by \( m \):
\[ u^2 = u_1^2 + u_1 u_2 \]

### Step 6: Solve for \( m_2 \)
From the momentum equations, we have:
1. \( \sqrt{2}mu = mu_1 + m_2u_2 \)
2. \( mu_1 = m_2u_2 \)

Substituting \( u_2 = \frac{mu_1}{m_2} \) into the first equation and solving gives:
\[ m_2 = m \]

### Final Answer
The mass of the second particle \( m_2 \) is equal to the mass of the first particle \( m \).