Two ice skaters A and B approach each other at right angles. Skater A has a mass 30 kg and velocity 1 m/s and skater B has a mass 20 kg and velocity 2 m/s. They meet and cling together. The final velocity of the couple is

Options

A. 2 m/s
B. 1.5 m/s
C. 1 m/s
D. 2.5 m/s

Explanation

To solve the problem of the two ice skaters A and B colliding and sticking together, we will use the principle of conservation of momentum. Here are the steps to find the final velocity of the couple:

### Step 1: Identify the masses and velocities of the skaters
- Mass of skater A, \( m_A = 30 \, \text{kg} \)
- Velocity of skater A, \( v_A = 1 \, \text{m/s} \)
- Mass of skater B, \( m_B = 20 \, \text{kg} \)
- Velocity of skater B, \( v_B = 2 \, \text{m/s} \)

### Step 2: Calculate the momentum of each skater
Momentum is given by the formula:
\[ p = m \cdot v \]
- Momentum of skater A, \( p_A = m_A \cdot v_A = 30 \, \text{kg} \cdot 1 \, \text{m/s} = 30 \, \text{kg m/s} \)
- Momentum of skater B, \( p_B = m_B \cdot v_B = 20 \, \text{kg} \cdot 2 \, \text{m/s} = 40 \, \text{kg m/s} \)

### Step 3: Determine the direction of the momentum vectors
Since the skaters approach each other at right angles, we can treat their momenta as vectors:
- \( p_A \) is in one direction (let's say along the x-axis).
- \( p_B \) is in a perpendicular direction (along the y-axis).

### Step 4: Calculate the net momentum using the Pythagorean theorem
The net momentum \( P_{net} \) can be calculated as:
\[ P_{net} = \sqrt{p_A^2 + p_B^2} \]
Substituting the values:
\[ P_{net} = \sqrt{(30 \, \text{kg m/s})^2 + (40 \, \text{kg m/s})^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{kg m/s} \]

### Step 5: Calculate the total mass after the collision
When skaters A and B cling together, their total mass \( m_{total} \) is:
\[ m_{total} = m_A + m_B = 30 \, \text{kg} + 20 \, \text{kg} = 50 \, \text{kg} \]

### Step 6: Use the conservation of momentum to find the final velocity
According to the conservation of momentum:
\[ P_{net} = m_{total} \cdot v_{net} \]
Substituting the known values:
\[ 50 \, \text{kg m/s} = 50 \, \text{kg} \cdot v_{net} \]
Solving for \( v_{net} \):
\[ v_{net} = \frac{50 \, \text{kg m/s}}{50 \, \text{kg}} = 1 \, \text{m/s} \]

### Final Answer
The final velocity of the couple after they cling together is \( 1 \, \text{m/s} \).

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