A vessel at rest explodes breaking it into three pieces. Two pieces having equal mass fly off perpendicular to one another with the same speed of `30 m//s`. The third piece has three times the mass of each of the other two pieces. What is the direction (w.r.t. the pieces having equal masses) and magnitude of its velocity immediately after the explosion?

Options

A. `10sqrt(2), 135^(@)`
B. `10sqrt(2), 90^(@)`
C. `10sqrt(2),60^(@)`
D. `10sqrt(2), 30^(@)`

Explanation

To solve the problem step by step, we will use the principle of conservation of momentum.

### Step 1: Understand the System
We have a vessel at rest that explodes into three pieces:
- Two pieces have equal mass \( m \) and move with a speed of \( 30 \, \text{m/s} \) perpendicular to each other.
- The third piece has a mass of \( 3m \).

### Step 2: Set Up the Coordinate System
Let's define our coordinate system:
- Let the first piece (mass \( m \)) move in the positive x-direction.
- Let the second piece (mass \( m \)) move in the positive y-direction.
- The third piece (mass \( 3m \)) will have an unknown velocity \( \mathbf{v} \) with components \( v_x \) and \( v_y \).

### Step 3: Apply Conservation of Momentum in the x-direction
The initial momentum of the system is zero since the vessel is at rest. Therefore, the total momentum after the explosion must also be zero.

In the x-direction:
\[ 0 = m(30) + m(0) - 3m(v_x) \]
This simplifies to:
\[ 0 = 30m - 3mv_x \]
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ 0 = 30 - 3v_x \implies v_x = 10 \, \text{m/s} \]

### Step 4: Apply Conservation of Momentum in the y-direction
Now, applying conservation of momentum in the y-direction:
\[ 0 = m(0) + m(30) - 3m(v_y) \]
This simplifies to:
\[ 0 = 30m - 3mv_y \]
Again, dividing through by \( m \):
\[ 0 = 30 - 3v_y \implies v_y = 10 \, \text{m/s} \]

### Step 5: Calculate the Magnitude of the Velocity of the Third Piece
Now we have the components of the velocity of the third piece:
- \( v_x = 10 \, \text{m/s} \)
- \( v_y = 10 \, \text{m/s} \)

The magnitude of the velocity \( v \) can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10)^2 + (10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \]

### Step 6: Calculate the Direction of the Velocity
To find the direction of the velocity, we can use the tangent function:
\[ \tan(\theta) = \frac{v_y}{v_x} = \frac{10}{10} = 1 \]
Thus, \( \theta = 45^\circ \).

Since the third piece moves in the negative x and negative y direction (to balance the momentum of the other two pieces), the angle with respect to the positive x-axis is:
\[ 180^\circ - 45^\circ = 135^\circ \]

### Final Answer
The magnitude of the velocity of the third piece is \( 10\sqrt{2} \, \text{m/s} \) and the direction is \( 135^\circ \) with respect to the positive x-axis.