Two equal masses `m_1` and `m_2` moving along the same straight line with velocites `+3 m//s` and `- 5 m//s` respectively collide elastically. Their velocities after the collision will be respectively.

Options

A. `+4m//s` for both
B. `-3m//s" and "+5m//s`
C. `-4m//s "and" +4m//s`
D. `-5m//s" and "+3m//s`

Answer

To solve the problem of two equal masses \( m_1 \) and \( m_2 \) colliding elastically, we can follow these steps:

### Step 1: Understand the Initial Conditions
- We have two equal masses \( m_1 \) and \( m_2 \).
- The initial velocity of mass \( m_1 \) is \( v_{1i} = +3 \, \text{m/s} \).
- The initial velocity of mass \( m_2 \) is \( v_{2i} = -5 \, \text{m/s} \).

### Step 2: Use the Conservation of Momentum
For elastic collisions, both momentum and kinetic energy are conserved. However, since the masses are equal, we can use a simplified approach. The conservation of momentum states:
\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]
Since \( m_1 = m_2 = m \), we can simplify this to:
\[ m v_{1i} + m v_{2i} = m v_{1f} + m v_{2f} \]
Dividing through by \( m \):
\[ v_{1i} + v_{2i} = v_{1f} + v_{2f} \tag{1} \]

### Step 3: Use the Conservation of Kinetic Energy
For elastic collisions, the kinetic energy before and after the collision is also conserved:
\[ \frac{1}{2} m v_{1i}^2 + \frac{1}{2} m v_{2i}^2 = \frac{1}{2} m v_{1f}^2 + \frac{1}{2} m v_{2f}^2 \]
Again, dividing through by \( \frac{1}{2} m \):
\[ v_{1i}^2 + v_{2i}^2 = v_{1f}^2 + v_{2f}^2 \tag{2} \]

### Step 4: Substitute Initial Velocities
Substituting the initial velocities into equations (1) and (2):
1. From equation (1):
\[ 3 + (-5) = v_{1f} + v_{2f} \implies -2 = v_{1f} + v_{2f} \tag{3} \]

2. From equation (2):
\[ 3^2 + (-5)^2 = v_{1f}^2 + v_{2f}^2 \implies 9 + 25 = v_{1f}^2 + v_{2f}^2 \implies 34 = v_{1f}^2 + v_{2f}^2 \tag{4} \]

### Step 5: Solve the System of Equations
Now we have two equations (3) and (4):
- From (3): \( v_{1f} = -2 - v_{2f} \)
- Substitute \( v_{1f} \) into (4):
\[ (-2 - v_{2f})^2 + v_{2f}^2 = 34 \]
Expanding this:
\[ (4 + 4v_{2f} + v_{2f}^2) + v_{2f}^2 = 34 \]
\[ 2v_{2f}^2 + 4v_{2f} + 4 - 34 = 0 \]
\[ 2v_{2f}^2 + 4v_{2f} - 30 = 0 \implies v_{2f}^2 + 2v_{2f} - 15 = 0 \]
Factoring:
\[ (v_{2f} + 5)(v_{2f} - 3) = 0 \]
Thus, \( v_{2f} = -5 \) or \( v_{2f} = 3 \).

### Step 6: Find Final Velocities
- If \( v_{2f} = 3 \), then from (3):
\[ v_{1f} = -2 - 3 = -5 \]
- If \( v_{2f} = -5 \), then from (3):
\[ v_{1f} = -2 - (-5) = 3 \]

### Final Answer
Thus, the final velocities after the collision are:
- \( v_{1f} = -5 \, \text{m/s} \)
- \( v_{2f} = 3 \, \text{m/s} \)

### Summary
The final velocities of the two masses after the elastic collision are:
- Mass \( m_1 \) has a velocity of \( -5 \, \text{m/s} \)
- Mass \( m_2 \) has a velocity of \( 3 \, \text{m/s} \)