Two identical balls each moving with speed v at right angle to each other collide perfectly inelastically. Their speed after collision is

Options

A. v
B. sqrt2v`
C. `(v)/(2)`
D. `(v)/(sqrt2)`

Answer

To solve the problem of two identical balls colliding perfectly inelastically, we will follow the steps below:

### Step 1: Understand the Initial Conditions
We have two identical balls, each with mass \( m \), moving with speed \( v \) at right angles to each other. Let's assume:
- Ball 1 is moving along the x-axis with velocity \( \vec{v_1} = v \hat{i} \)
- Ball 2 is moving along the y-axis with velocity \( \vec{v_2} = v \hat{j} \)

### Step 2: Apply Conservation of Momentum
Since the collision is perfectly inelastic, the two balls stick together after the collision. We will use the conservation of momentum in both the x and y directions.

#### In the x-direction:
Before the collision:
- Momentum of Ball 1: \( m \cdot v \)
- Momentum of Ball 2: \( 0 \)

Total initial momentum in the x-direction:
\[ p_{x, \text{initial}} = mv + 0 = mv \]

After the collision, let the common velocity of the combined mass (2m) be \( v_x \):
\[ p_{x, \text{final}} = 2m \cdot v_x \]

By conservation of momentum:
\[ mv = 2m \cdot v_x \]
\[ v_x = \frac{v}{2} \]

#### In the y-direction:
Before the collision:
- Momentum of Ball 1: \( 0 \)
- Momentum of Ball 2: \( m \cdot v \)

Total initial momentum in the y-direction:
\[ p_{y, \text{initial}} = 0 + mv = mv \]

After the collision, let the common velocity in the y-direction be \( v_y \):
\[ p_{y, \text{final}} = 2m \cdot v_y \]

By conservation of momentum:
\[ mv = 2m \cdot v_y \]
\[ v_y = \frac{v}{2} \]

### Step 3: Calculate the Resultant Velocity
Now we have the components of the velocity after the collision:
- \( v_x = \frac{v}{2} \)
- \( v_y = \frac{v}{2} \)

The magnitude of the resultant velocity \( v_f \) can be calculated using the Pythagorean theorem:
\[ v_f = \sqrt{v_x^2 + v_y^2} \]
\[ v_f = \sqrt{\left(\frac{v}{2}\right)^2 + \left(\frac{v}{2}\right)^2} \]
\[ v_f = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{2v^2}{4}} = \sqrt{\frac{v^2}{2}} = \frac{v}{\sqrt{2}} \]

### Conclusion
The speed of the two balls after the perfectly inelastic collision is:
\[ \frac{v}{\sqrt{2}} \]