On a friction surface a block a mass `M` moving at speed `v` collides elastic with another block of same mass `M` which is initially at rest . After collision the first block moves at an angle `theta` to its initial direction and has a speed `(v)/(3)`. The second block's speed after the collision is

Options

A. `(2sqrt2)/(3)v`
B. `(3)/(4)v`
C. `(3)/(sqrt2)v`
D. `(sqrt3)/(2)v`

Explanation

To solve the problem of two blocks colliding elastically, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have two blocks of mass \( M \). Block A is moving with an initial speed \( v \) while Block B is at rest.

### Step 2: Write the Initial Kinetic Energy
The initial kinetic energy \( KE_{\text{initial}} \) of the system can be calculated as:
\[ KE_{\text{initial}} = \frac{1}{2} M v^2 + 0 = \frac{1}{2} M v^2 \]

### Step 3: Write the Final Kinetic Energy
After the collision, Block A moves at an angle \( \theta \) with a speed of \( \frac{v}{3} \), and Block B moves with an unknown speed \( v_B \). The final kinetic energy \( KE_{\text{final}} \) can be expressed as:
\[ KE_{\text{final}} = \frac{1}{2} M \left(\frac{v}{3}\right)^2 + \frac{1}{2} M v_B^2 \]
\[ KE_{\text{final}} = \frac{1}{2} M \frac{v^2}{9} + \frac{1}{2} M v_B^2 \]

### Step 4: Set Initial Kinetic Energy Equal to Final Kinetic Energy
Since the collision is elastic, the initial kinetic energy equals the final kinetic energy:
\[ \frac{1}{2} M v^2 = \frac{1}{2} M \frac{v^2}{9} + \frac{1}{2} M v_B^2 \]

### Step 5: Cancel Out Common Terms
We can cancel \( \frac{1}{2} M \) from both sides:
\[ v^2 = \frac{v^2}{9} + v_B^2 \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ v_B^2 = v^2 - \frac{v^2}{9} \]
\[ v_B^2 = \frac{9v^2}{9} - \frac{v^2}{9} = \frac{8v^2}{9} \]

### Step 7: Solve for \( v_B \)
Taking the square root of both sides gives:
\[ v_B = \sqrt{\frac{8v^2}{9}} = \frac{v \sqrt{8}}{3} = \frac{2\sqrt{2}v}{3} \]

### Final Answer
The speed of Block B after the collision is:
\[ v_B = \frac{2\sqrt{2}v}{3} \]

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