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. `(2v)/(sqrt3)`
B. `v/(sqrt3)`
C. `v`
D. `sqrt(3)v`

Explanation

To solve the problem, we need to analyze the momentum before and after the collision. Here’s a step-by-step solution:

### Step 1: Define the initial momentum
The first mass \( m \) is moving with a velocity \( v \) along the x-axis. Therefore, the initial momentum \( \vec{p}_{\text{initial}} \) of the system can be expressed as:
\[ \vec{p}_{\text{initial}} = m \vec{v} = m v \hat{i} \]
where \( \hat{i} \) is the unit vector in the x-direction.

### Step 2: Define the final momentum of the first mass
After the collision, the first mass moves with a velocity of \( \frac{v}{\sqrt{3}} \) in the y-direction. Thus, the final momentum \( \vec{p}_{1} \) of the first mass is:
\[ \vec{p}_{1} = m \left(\frac{v}{\sqrt{3}} \hat{j}\right) \]
where \( \hat{j} \) is the unit vector in the y-direction.

### Step 3: Define the final momentum of the second mass
Let the velocity of the second mass after the collision be \( \vec{v}_{2} \). The momentum of the second mass \( \vec{p}_{2} \) can be expressed as:
\[ \vec{p}_{2} = m \vec{v}_{2} \]

### Step 4: Apply conservation of momentum
Since the collision is inelastic and there are no external forces acting on the system, the total initial momentum must equal the total final momentum:
\[ \vec{p}_{\text{initial}} = \vec{p}_{1} + \vec{p}_{2} \]
Substituting the expressions we have:
\[ m v \hat{i} = m \left(\frac{v}{\sqrt{3}} \hat{j}\right) + m \vec{v}_{2} \]

### Step 5: Cancel the mass \( m \)
We can divide the entire equation by \( m \) (assuming \( m \neq 0 \)):
\[ v \hat{i} = \frac{v}{\sqrt{3}} \hat{j} + \vec{v}_{2} \]

### Step 6: Rearranging to find \( \vec{v}_{2} \)
Rearranging the equation gives us:
\[ \vec{v}_{2} = v \hat{i} - \frac{v}{\sqrt{3}} \hat{j} \]

### Step 7: Calculate the magnitude of \( \vec{v}_{2} \)
To find the speed of the second mass, we need to calculate the magnitude of \( \vec{v}_{2} \):
\[ |\vec{v}_{2}| = \sqrt{(v)^2 + \left(-\frac{v}{\sqrt{3}}\right)^2} \]
Calculating the second term:
\[ \left(-\frac{v}{\sqrt{3}}\right)^2 = \frac{v^2}{3} \]
Thus,
\[ |\vec{v}_{2}| = \sqrt{v^2 + \frac{v^2}{3}} = \sqrt{\frac{3v^2}{3} + \frac{v^2}{3}} = \sqrt{\frac{4v^2}{3}} = \frac{2v}{\sqrt{3}} \]

### Final Answer
The speed of the second mass after the collision is:
\[ \boxed{\frac{2v}{\sqrt{3}}} \]