A sphere of mass m moves with a velocity 2v and collides inelastically with another identical sphere of mass m. After collision the first mass moves with velocity v in a direction perpendicular to the initial direction of motion . Find the speed of the second sphere after collision .

Options

A. v
B. `vsqrt(5)`
C. `2/sqrt3 v`
D. `v/sqrt3`

Answer

To solve the problem, we will use the principle of conservation of momentum. The steps are as follows:

### Step 1: Define the initial conditions
- Let sphere A have mass \( m \) and an initial velocity of \( 2v \) in the horizontal direction.
- Let sphere B also have mass \( m \) and be initially at rest, so its initial velocity is \( 0 \).

### Step 2: Set up the momentum conservation equations
Since the collision is inelastic, we can apply the conservation of momentum in both the x and y directions.

#### In the x-direction:
- Initial momentum of sphere A: \( m \cdot 2v \)
- Initial momentum of sphere B: \( 0 \)
- Final momentum of sphere A (after the collision): \( m \cdot 0 \) (since it moves vertically)
- Final momentum of sphere B: \( m \cdot v_B \cos \theta \)

Setting up the equation:
\[ m \cdot 2v + 0 = 0 + m \cdot v_B \cos \theta \]
This simplifies to:
\[ 2v = v_B \cos \theta \quad \text{(Equation 1)} \]

#### In the y-direction:
- Initial momentum of sphere A: \( 0 \)
- Initial momentum of sphere B: \( 0 \)
- Final momentum of sphere A: \( m \cdot v \) (moving vertically)
- Final momentum of sphere B: \( -m \cdot v_B \sin \theta \) (downward)

Setting up the equation:
\[ 0 + 0 = m \cdot v - m \cdot v_B \sin \theta \]
This simplifies to:
\[ 0 = m \cdot v - m \cdot v_B \sin \theta \]
Rearranging gives:
\[ v_B \sin \theta = v \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( v_B \cos \theta = 2v \)
2. \( v_B \sin \theta = v \)

We can square both equations and add them:
\[ (v_B \cos \theta)^2 + (v_B \sin \theta)^2 = (2v)^2 + v^2 \]
This gives:
\[ v_B^2 (\cos^2 \theta + \sin^2 \theta) = 4v^2 + v^2 \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ v_B^2 = 5v^2 \]
Taking the square root:
\[ v_B = \sqrt{5}v \]

### Conclusion
The speed of the second sphere after the collision is \( \sqrt{5}v \).

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