A bomb at rest explodes into three fragments of equal massses Two fragments fly off at right angles to each other with velocities of `9m//s` and `12//s` Calculate the speed of the third fragment .

Explanation

To solve the problem of finding the speed of the third fragment after the bomb explodes, we can follow these steps:

### Step 1: Understand the Conservation of Momentum
Since the bomb is initially at rest, the total momentum before the explosion is zero. According to the law of conservation of momentum, the total momentum after the explosion must also equal zero.

### Step 2: Define the Fragments and Their Velocities
Let the mass of each fragment be \( m \). After the explosion, we have:
- Fragment 1: mass \( m/3 \) moving with velocity \( 9 \, \text{m/s} \) (let's assume it moves in the positive x-direction).
- Fragment 2: mass \( m/3 \) moving with velocity \( 12 \, \text{m/s} \) (let's assume it moves in the positive y-direction).
- Fragment 3: mass \( m/3 \) moving with an unknown velocity \( V \) in a direction that will balance the momentum to zero.

### Step 3: Calculate the Momentum of the First Two Fragments
The momentum of Fragment 1 in the x-direction:
\[ p_1 = \frac{m}{3} \cdot 9 = 3m \]
The momentum of Fragment 2 in the y-direction:
\[ p_2 = \frac{m}{3} \cdot 12 = 4m \]

### Step 4: Determine the Resultant Momentum
The two fragments are moving at right angles to each other. To find the resultant momentum, we can use the Pythagorean theorem:
\[ P_{resultant} = \sqrt{(p_1)^2 + (p_2)^2} = \sqrt{(3m)^2 + (4m)^2} = \sqrt{9m^2 + 16m^2} = \sqrt{25m^2} = 5m \]

### Step 5: Set Up the Momentum Equation for the Third Fragment
Since the total momentum must be zero, the momentum of the third fragment must equal the negative of the resultant momentum of the first two fragments:
\[ p_3 = -5m \]
The momentum of the third fragment is given by:
\[ p_3 = \frac{m}{3} \cdot V \]
Setting these equal gives:
\[ \frac{m}{3} \cdot V = -5m \]

### Step 6: Solve for the Speed of the Third Fragment
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{3} V = -5 \]
Multiplying both sides by 3:
\[ V = -15 \, \text{m/s} \]
The negative sign indicates that the third fragment moves in the opposite direction to the resultant momentum of the first two fragments.

### Final Answer
The speed of the third fragment is \( 15 \, \text{m/s} \).

---