In one dimeensional collision of two particles velocities are interchanged when
(a)Collision is elastic and masses are equal
(b)Collision is inelastic but masses are unequal

Options

A. Only 'a' is correct
B. Only 'b' is correct
C. a' and 'b' both are correct
D. None of these

Answer

To solve the question regarding one-dimensional collisions of two particles, we need to analyze both scenarios presented: (a) when the collision is elastic and the masses are equal, and (b) when the collision is inelastic and the masses are unequal.

### Step-by-Step Solution:

**Step 1: Analyze the first scenario (Elastic Collision with Equal Masses)**

1. **Define the initial conditions**:
- Let the mass of each particle be \( m \).
- Let the initial velocity of the first particle be \( u \) and the second particle be \( 0 \) (at rest).

2. **Apply the conservation of momentum**:
- The total initial momentum \( P_i \) is given by:
\[ P_i = mu + m \cdot 0 = mu \]

3. **Final velocities after collision**:
- Let the final velocity of the first particle be \( v_1 \) and the second particle be \( v_2 \).
- According to the conservation of momentum:
\[ mu = mv_1 + mv_2 \]
- Simplifying gives:
\[ u = v_1 + v_2 \quad \text{(1)} \]

4. **Apply the conservation of kinetic energy** (since the collision is elastic):
- The total initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2}mu^2 + \frac{1}{2}m \cdot 0^2 = \frac{1}{2}mu^2 \]
- The total final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 \]
- Setting \( KE_i = KE_f \):
\[ \frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 \]
- Simplifying gives:
\[ u^2 = v_1^2 + v_2^2 \quad \text{(2)} \]

5. **Solve equations (1) and (2)**:
- From equation (1), we can express \( v_2 \) as \( v_2 = u - v_1 \).
- Substitute \( v_2 \) into equation (2):
\[ u^2 = v_1^2 + (u - v_1)^2 \]
- Expanding gives:
\[ u^2 = v_1^2 + (u^2 - 2uv_1 + v_1^2) \]
- Combining terms:
\[ u^2 = 2v_1^2 - 2uv_1 + u^2 \]
- Cancelling \( u^2 \) from both sides:
\[ 0 = 2v_1^2 - 2uv_1 \]
- Factoring out \( 2v_1 \):
\[ 2v_1(v_1 - u) = 0 \]
- This gives \( v_1 = 0 \) or \( v_1 = u \). Thus, the velocities are interchanged:
\[ v_1 = 0, \quad v_2 = u \]

**Conclusion for (a)**: In an elastic collision with equal masses, the velocities are interchanged.

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**Step 2: Analyze the second scenario (Inelastic Collision with Unequal Masses)**

1. **Define the initial conditions**:
- Let the masses be \( m_1 \) and \( m_2 \) (where \( m_1 \neq m_2 \)).
- Let the initial velocities be \( u_1 \) and \( u_2 \).

2. **Apply the conservation of momentum**:
- The total initial momentum \( P_i \) is:
\[ P_i = m_1 u_1 + m_2 u_2 \]
- The final velocities after an inelastic collision (where the particles stick together) will be \( v \):
\[ P_f = (m_1 + m_2)v \]
- Setting initial momentum equal to final momentum:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \]

3. **Solve for final velocity \( v \)**:
- Rearranging gives:
\[ v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \]
- This shows that the final velocity is a weighted average of the initial velocities, and they do not simply interchange.

**Conclusion for (b)**: In an inelastic collision with unequal masses, the velocities are not interchanged.

### Final Answer:
(a) The velocities are interchanged when the collision is elastic and masses are equal.
(b) The velocities are not interchanged when the collision is inelastic and masses are unequal.

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