A mass m moves with a velocity v and collides inelastically with another identical mass. After collision the 1st mass moves with velocity `v/(sqrt3)` in a direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision

Options

A. v
B. `sqrt3v`
C. `2/(sqrt3)`v
D. `v/(sqrt3)`

Answer

To solve the problem, we will use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision since the collision is inelastic.

### Step-by-Step Solution:

1. **Identify the initial momentum**:
The initial momentum of the system can be calculated using the mass and velocity of the first mass (m) before the collision. The second mass is initially at rest.
\[ \text{Initial momentum} = m \cdot v + m \cdot 0 = mv \]

2. **Determine the final velocities**:
After the collision, the first mass moves with a velocity of \( \frac{v}{\sqrt{3}} \) in a direction perpendicular to its initial direction. Let’s denote the velocity of the second mass after the collision as \( \mathbf{u} \).

3. **Set up the momentum conservation equation**:
Since the first mass is moving in the y-direction after the collision, we can express the final momentum in vector form:
\[ \text{Final momentum} = m \left( \frac{v}{\sqrt{3}} \mathbf{j} + \mathbf{u} \right) \]
Here, \( \mathbf{u} \) can be expressed in terms of its components \( u_x \) and \( u_y \).

4. **Apply conservation of momentum in the x and y directions**:
- In the x-direction:
\[ mv = mu_x \quad \Rightarrow \quad u_x = v \]
- In the y-direction:
\[ 0 = m \frac{v}{\sqrt{3}} + mu_y \quad \Rightarrow \quad u_y = -\frac{v}{\sqrt{3}} \]

5. **Calculate the magnitude of the velocity of the second mass**:
The magnitude of the velocity \( \mathbf{u} \) can be calculated using the Pythagorean theorem:
\[ |\mathbf{u}| = \sqrt{u_x^2 + u_y^2} = \sqrt{v^2 + \left(-\frac{v}{\sqrt{3}}\right)^2} \]
\[ |\mathbf{u}| = \sqrt{v^2 + \frac{v^2}{3}} = \sqrt{\frac{3v^2}{3} + \frac{v^2}{3}} = \sqrt{\frac{4v^2}{3}} = \frac{2v}{\sqrt{3}} \]

6. **Final result**:
The speed of the second mass after the collision is:
\[ |\mathbf{u}| = \frac{2v}{\sqrt{3}} \]

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