A particle of mass `m_0`, travelling at speed `v_0`. Strikes a stationary particle of mass `2m_0`. As a result of the particle of mass `m_0` is deflected through `45^@` and has a final speed of `v_0/sqrt2`. Then the speed of the particle of mass `2m_0` after this collision is

Options

A. (a) `v_0/2`
B. (b) `(v_0)/(2sqrt2)`
C. (c) `sqrt2v_0`
D. (d) `(v_0)/(sqrt2)`

Explanation

To solve the problem, we will use the principle of conservation of linear momentum. Let's break it down step by step.

### Step 1: Identify the initial momentum
Before the collision, we have:
- The momentum of the particle of mass \( m_0 \) moving with speed \( v_0 \):
\[ \text{Initial momentum of } m_0 = m_0 v_0 \]
- The stationary particle of mass \( 2m_0 \) has no momentum:
\[ \text{Initial momentum of } 2m_0 = 0 \]

Thus, the total initial momentum \( P_{\text{initial}} \) is:
\[ P_{\text{initial}} = m_0 v_0 + 0 = m_0 v_0 \]

### Step 2: Identify the final momentum
After the collision:
- The particle of mass \( m_0 \) is deflected at an angle of \( 45^\circ \) with a final speed of \( \frac{v_0}{\sqrt{2}} \). We can resolve this velocity into its components:
- \( v_{x1} = \frac{v_0}{\sqrt{2}} \cos(45^\circ) = \frac{v_0}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{v_0}{2} \)
- \( v_{y1} = \frac{v_0}{\sqrt{2}} \sin(45^\circ) = \frac{v_0}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{v_0}{2} \)

Thus, the final momentum of the particle of mass \( m_0 \) is:
\[ \text{Final momentum of } m_0 = m_0 \left( \frac{v_0}{2} \hat{i} + \frac{v_0}{2} \hat{j} \right) \]

Let the velocity of the particle of mass \( 2m_0 \) after the collision be \( \vec{V} = V_x \hat{i} + V_y \hat{j} \). The final momentum of the particle of mass \( 2m_0 \) is:
\[ \text{Final momentum of } 2m_0 = 2m_0 \vec{V} \]

### Step 3: Apply conservation of momentum
According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]
This gives us:
\[ m_0 v_0 = m_0 \left( \frac{v_0}{2} \hat{i} + \frac{v_0}{2} \hat{j} \right) + 2m_0 \vec{V} \]

### Step 4: Separate into components
Separating into \( \hat{i} \) and \( \hat{j} \) components, we have:
1. For the \( \hat{i} \) component:
\[ m_0 v_0 = m_0 \frac{v_0}{2} + 2m_0 V_x \]
Simplifying gives:
\[ v_0 = \frac{v_0}{2} + 2V_x \implies 2V_x = \frac{v_0}{2} \implies V_x = \frac{v_0}{4} \]

2. For the \( \hat{j} \) component:
\[ 0 = m_0 \frac{v_0}{2} + 2m_0 V_y \]
Simplifying gives:
\[ 0 = \frac{v_0}{2} + 2V_y \implies 2V_y = -\frac{v_0}{2} \implies V_y = -\frac{v_0}{4} \]

### Step 5: Calculate the magnitude of velocity \( \vec{V} \)
Now we have:
\[ \vec{V} = \frac{v_0}{4} \hat{i} - \frac{v_0}{4} \hat{j} \]

The magnitude of \( \vec{V} \) is:
\[ |\vec{V}| = \sqrt{ \left( \frac{v_0}{4} \right)^2 + \left( -\frac{v_0}{4} \right)^2 } = \sqrt{ \frac{v_0^2}{16} + \frac{v_0^2}{16} } = \sqrt{ \frac{2v_0^2}{16} } = \sqrt{ \frac{v_0^2}{8} } = \frac{v_0}{2\sqrt{2}} \]

### Final Answer
Thus, the speed of the particle of mass \( 2m_0 \) after the collision is:
\[ \frac{v_0}{2\sqrt{2}} \]

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