A 4.0 kg mess kit sliding on a frictionless surface explodes into two 2.0 kg parts: 3.0 m/s, due north, and 6.0 m/s, `30^@` north of east. What is the original speed of the mess kit ?

Answer

To solve the problem of finding the original speed of the mess kit after it explodes into two parts, we will use the principle of conservation of momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
The mess kit has a mass of 4.0 kg and is moving on a frictionless surface. It explodes into two parts, each with a mass of 2.0 kg. One part moves at 3.0 m/s due north, and the other moves at 6.0 m/s at an angle of 30° north of east. We need to find the original speed of the mess kit before the explosion.

### Step 2: Set Up the Momentum Conservation Equation
According to the conservation of momentum, the total momentum before the explosion (initial momentum) must equal the total momentum after the explosion (final momentum).

Let \( \vec{P}_{initial} \) be the initial momentum of the mess kit and \( \vec{P}_{final} \) be the total final momentum of the two parts.

\[ \vec{P}_{initial} = \vec{P}_{final} \]

### Step 3: Calculate the Final Momentum of Each Part
1. **Momentum of the first part (2.0 kg, 3.0 m/s due north)**:
- Mass \( m_1 = 2.0 \, \text{kg} \)
- Velocity \( \vec{v}_1 = 3.0 \, \text{m/s} \, \hat{j} \)
- Momentum \( \vec{P}_1 = m_1 \vec{v}_1 = 2.0 \times 3.0 \, \hat{j} = 6.0 \, \hat{j} \, \text{kg m/s} \)

2. **Momentum of the second part (2.0 kg, 6.0 m/s at 30° north of east)**:
- Mass \( m_2 = 2.0 \, \text{kg} \)
- Velocity components:
- \( v_{2x} = 6.0 \cos(30°) = 6.0 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \, \text{m/s} \)
- \( v_{2y} = 6.0 \sin(30°) = 6.0 \times \frac{1}{2} = 3.0 \, \text{m/s} \)
- Momentum \( \vec{P}_2 = m_2 \vec{v}_2 = 2.0 \times (3\sqrt{3} \, \hat{i} + 3.0 \, \hat{j}) = 6\sqrt{3} \, \hat{i} + 6.0 \, \hat{j} \, \text{kg m/s} \)

### Step 4: Calculate Total Final Momentum
Now, we can find the total final momentum \( \vec{P}_{final} \) by adding \( \vec{P}_1 \) and \( \vec{P}_2 \):

\[ \vec{P}_{final} = \vec{P}_1 + \vec{P}_2 = (6\sqrt{3} \, \hat{i} + 6.0 \, \hat{j}) + (0 \, \hat{i} + 6.0 \, \hat{j}) = 6\sqrt{3} \, \hat{i} + 12.0 \, \hat{j} \, \text{kg m/s} \]

### Step 5: Calculate the Initial Momentum
The initial momentum of the mess kit can be expressed as:

\[ \vec{P}_{initial} = m_{total} \cdot \vec{v}_{initial} = 4.0 \, \text{kg} \cdot \vec{v}_{initial} \]

### Step 6: Set Initial Momentum Equal to Final Momentum
Since momentum is conserved:

\[ 4.0 \cdot \vec{v}_{initial} = 6\sqrt{3} \, \hat{i} + 12.0 \, \hat{j} \]

### Step 7: Solve for \( \vec{v}_{initial} \)
To find \( \vec{v}_{initial} \):

\[ \vec{v}_{initial} = \frac{6\sqrt{3}}{4.0} \, \hat{i} + \frac{12.0}{4.0} \, \hat{j} = \frac{3\sqrt{3}}{2} \, \hat{i} + 3.0 \, \hat{j} \]

### Step 8: Calculate the Magnitude of \( \vec{v}_{initial} \)
The magnitude of the initial velocity is:

\[ |\vec{v}_{initial}| = \sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2 + (3.0)^2} \]

Calculating:

\[ |\vec{v}_{initial}| = \sqrt{\frac{27}{4} + 9} = \sqrt{\frac{27}{4} + \frac{36}{4}} = \sqrt{\frac{63}{4}} = \frac{\sqrt{63}}{2} \approx 3.96 \, \text{m/s} \]

### Conclusion
The original speed of the mess kit before the explosion is approximately **3.96 m/s**.

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