A body of mass 2 kg is moving with velocity 10 m / s towards east. Another body of same mass and same velocity moving towards north collides with former and coalsces and moves towards northeast. Its velocity is

Options

A. 10 m/s
B. 5 m/s
C. 2.5 m/s
D. `5sqrt(2)m//s`

Answer

To solve the problem, we will use the principle of conservation of momentum. Here's a step-by-step breakdown of the solution:

### Step 1: Identify the initial momentum of both bodies
- The first body (mass \( m_1 = 2 \, \text{kg} \)) is moving east with a velocity \( v_1 = 10 \, \text{m/s} \).
- The second body (mass \( m_2 = 2 \, \text{kg} \)) is moving north with a velocity \( v_2 = 10 \, \text{m/s} \).

The momentum of the first body (east direction):
\[ \text{Momentum}_1 = m_1 \cdot v_1 = 2 \, \text{kg} \cdot 10 \, \text{m/s} = 20 \, \text{kg m/s} \, \text{(east)} \]

The momentum of the second body (north direction):
\[ \text{Momentum}_2 = m_2 \cdot v_2 = 2 \, \text{kg} \cdot 10 \, \text{m/s} = 20 \, \text{kg m/s} \, \text{(north)} \]

### Step 2: Calculate the total initial momentum
The total initial momentum is the vector sum of both momenta:
\[ \text{Total Initial Momentum} = \text{Momentum}_1 + \text{Momentum}_2 = 20 \, \text{kg m/s} \, \hat{i} + 20 \, \text{kg m/s} \, \hat{j} \]
Where \( \hat{i} \) represents the east direction and \( \hat{j} \) represents the north direction.

### Step 3: Determine the final mass after collision
After the collision, the two bodies coalesce, so the total mass \( m_f \) becomes:
\[ m_f = m_1 + m_2 = 2 \, \text{kg} + 2 \, \text{kg} = 4 \, \text{kg} \]

### Step 4: Apply conservation of momentum
According to the conservation of momentum:
\[ \text{Total Initial Momentum} = \text{Total Final Momentum} \]
Let \( \vec{V} \) be the final velocity of the combined mass moving towards the northeast. The final momentum can be expressed as:
\[ \text{Total Final Momentum} = m_f \cdot \vec{V} \]

### Step 5: Set up the equation
Using the components of the momentum:
\[ 20 \, \hat{i} + 20 \, \hat{j} = 4 \, \text{kg} \cdot \vec{V} \]
We can express \( \vec{V} \) in terms of its components:
\[ \vec{V} = V_x \hat{i} + V_y \hat{j} \]
Since the final velocity is towards the northeast, \( V_x = V_y \).

### Step 6: Solve for the final velocity
Substituting \( V_x = V_y = V \):
\[ 20 \, \hat{i} + 20 \, \hat{j} = 4 \, \text{kg} \cdot (V \hat{i} + V \hat{j}) \]
This gives us two equations:
1. \( 20 = 4V \) (for the east component)
2. \( 20 = 4V \) (for the north component)

From either equation, we can solve for \( V \):
\[ V = \frac{20}{4} = 5 \, \text{m/s} \]

### Step 7: Find the magnitude of the final velocity
Since the final velocity is directed towards the northeast, the magnitude of the velocity can be calculated using Pythagoras' theorem:
\[ V_{final} = \sqrt{V_x^2 + V_y^2} = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \, \text{m/s} \]

### Final Answer
The final velocity of the coalesced body moving towards the northeast is:
\[ V_{final} = 5\sqrt{2} \, \text{m/s} \approx 7.07 \, \text{m/s} \]

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