Two balls of equal masses moving with equal speed in mutually perpendicular directions, stick together after collision. If the balls were initially moving with a speed of `30/√2` ms^(-1) each, the speed of their combined mass after collision is

Options

A. 30 `ms^(-1)`
B. `15 ms^(-1)`
C. `30/√2 ms^(-1)`
D. `30sqrt2 ms^(-1)`

Answer

To solve the problem step by step, we will apply the principles of conservation of momentum. Here’s how to approach the problem:

### Step 1: Understand the scenario
We have two balls of equal mass (let's denote the mass of each ball as \( m \)) moving with equal speed in mutually perpendicular directions. The speed of each ball is given as \( \frac{30}{\sqrt{2}} \, \text{ms}^{-1} \).

### Step 2: Set up the momentum before the collision
Since the balls are moving in perpendicular directions, we can analyze their momentum components separately.

- For Ball 1 moving in the x-direction:
\[ p_{1x} = m \cdot v_1 = m \cdot \frac{30}{\sqrt{2}} \]

- For Ball 2 moving in the y-direction:
\[ p_{2y} = m \cdot v_2 = m \cdot \frac{30}{\sqrt{2}} \]

### Step 3: Calculate the total momentum before the collision
The total momentum of the system before the collision can be represented as a vector sum. Since the balls are moving at right angles, we can use the Pythagorean theorem to find the magnitude of the total momentum.

The total momentum in the x-direction is:
\[ P_x = m \cdot \frac{30}{\sqrt{2}} \]

The total momentum in the y-direction is:
\[ P_y = m \cdot \frac{30}{\sqrt{2}} \]

The magnitude of the total momentum \( P \) before the collision is:
\[ P = \sqrt{P_x^2 + P_y^2} = \sqrt{\left(m \cdot \frac{30}{\sqrt{2}}\right)^2 + \left(m \cdot \frac{30}{\sqrt{2}}\right)^2} \]
\[ = \sqrt{2 \left(m \cdot \frac{30}{\sqrt{2}}\right)^2} = m \cdot \frac{30}{\sqrt{2}} \cdot \sqrt{2} = m \cdot 30 \]

### Step 4: Apply conservation of momentum
After the collision, the two balls stick together, forming a combined mass of \( 2m \). Let \( v' \) be the speed of the combined mass after the collision.

Using conservation of momentum:
\[ \text{Total momentum before} = \text{Total momentum after} \]
\[ m \cdot 30 = 2m \cdot v' \]

### Step 5: Solve for \( v' \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 30 = 2v' \]
\[ v' = \frac{30}{2} = 15 \, \text{ms}^{-1} \]

### Conclusion
The speed of the combined mass after the collision is \( 15 \, \text{ms}^{-1} \).

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