A mass ‘m’ moves with a velocity ‘v’ and collides inelastically with another identical mass at rest. After collision the 1st mass moves with velocity `(v)/(sqrt(3))` in a direction perpendicular to the initial direction of motion. Find the speed of the `2^(nd)` mass after collision :

Options

A. v
B. `sqrt(3)v`
C. `(2)/(sqrt(3)` v
D. `(v)/(sqrt(3))`

Answer

To solve the problem, we will use the principles of conservation of momentum.

### Step-by-step Solution:

1. **Identify the initial conditions:**
- Mass \( m_1 = m \) is moving with velocity \( \vec{v_1} = v \) in the positive x-direction.
- Mass \( m_2 = m \) is at rest, so \( \vec{v_2} = 0 \).

2. **Write the initial momentum:**
- The total initial momentum \( \vec{P_{initial}} \) is given by:
\[ \vec{P_{initial}} = m \vec{v_1} + m \vec{v_2} = m v \hat{i} + 0 = m v \hat{i} \]

3. **Post-collision conditions:**
- After the collision, mass \( m_1 \) moves with velocity \( \vec{v_1'} = \frac{v}{\sqrt{3}} \) in a direction perpendicular to its initial motion (which we can assume to be in the positive y-direction).
- Therefore, \( \vec{v_1'} = \frac{v}{\sqrt{3}} \hat{j} \).

4. **Let the velocity of mass \( m_2 \) after collision be \( \vec{v_2'} \).**
- We can express \( \vec{v_2'} \) in terms of its components:
\[ \vec{v_2'} = v_{2x} \hat{i} + v_{2y} \hat{j} \]

5. **Apply conservation of momentum in the x-direction:**
- The total momentum in the x-direction before the collision must equal the total momentum in the x-direction after the collision:
\[ m v = m v_{2x} + 0 \]
- This simplifies to:
\[ v_{2x} = v \]

6. **Apply conservation of momentum in the y-direction:**
- The total momentum in the y-direction before the collision is zero, and after the collision, it must also equal zero:
\[ 0 = m \left(\frac{v}{\sqrt{3}}\right) + m v_{2y} \]
- This simplifies to:
\[ v_{2y} = -\frac{v}{\sqrt{3}} \]

7. **Find the magnitude of the velocity \( \vec{v_2'} \):**
- Now we can find the magnitude of \( \vec{v_2'} \):
\[ |\vec{v_2'}| = \sqrt{(v_{2x})^2 + (v_{2y})^2} = \sqrt{v^2 + \left(-\frac{v}{\sqrt{3}}\right)^2} \]
- This simplifies to:
\[ |\vec{v_2'}| = \sqrt{v^2 + \frac{v^2}{3}} = \sqrt{\frac{3v^2}{3} + \frac{v^2}{3}} = \sqrt{\frac{4v^2}{3}} = \frac{2v}{\sqrt{3}} \]

### Final Answer:
The speed of the second mass after the collision is \( \frac{2v}{\sqrt{3}} \).