A shell at rest at the origin explodes into three fragments of masses 1kg, 2kg and m kg. The 1kg and 2kg pieces fly off with speeds of `5ms^(-1)` along x-axis and `6 ms^(-1)` along y-axis respectively . If the m kg piece flies off with a speed of `6.5ms^(-1)`, the total mass of the shell must be .

Options

A. 4 kg
B. 5 kg
C. 3.5 kg
D. 4.5 kg

Explanation

To solve the problem, we will use the principle of conservation of momentum. The total momentum before the explosion must equal the total momentum after the explosion. Here are the steps to find the total mass of the shell:

### Step 1: Understand the initial conditions
The shell is initially at rest at the origin. Therefore, the initial momentum of the shell is zero.

### Step 2: Define the masses and velocities of the fragments
- Mass of the first fragment, \( m_1 = 1 \, \text{kg} \) with velocity \( v_1 = 5 \, \text{m/s} \) along the x-axis.
- Mass of the second fragment, \( m_2 = 2 \, \text{kg} \) with velocity \( v_2 = 6 \, \text{m/s} \) along the y-axis.
- Mass of the third fragment, \( m_3 = m \, \text{kg} \) with velocity \( v_3 = 6.5 \, \text{m/s} \) (direction unknown).

### Step 3: Write down the momentum conservation equations
Since the initial momentum is zero, the total momentum after the explosion must also be zero:

\[ \text{Total Momentum} = m_1 v_1 + m_2 v_2 + m_3 v_3 = 0 \]

### Step 4: Break down the momentum into components
1. For the x-direction:
\[ m_1 v_1 + m_3 v_{3x} = 0 \]
where \( v_{3x} \) is the x-component of the velocity of the third fragment.

2. For the y-direction:
\[ m_2 v_2 + m_3 v_{3y} = 0 \]
where \( v_{3y} \) is the y-component of the velocity of the third fragment.

### Step 5: Calculate the components of the third fragment's velocity
Assuming the third fragment moves in a direction that balances the momentum:
- The x-component of the velocity of the third fragment can be found from the x-momentum equation:
\[ 1 \cdot 5 + m \cdot v_{3x} = 0 \implies v_{3x} = -\frac{5}{m} \]

- The y-component of the velocity of the third fragment can be found from the y-momentum equation:
\[ 2 \cdot 6 + m \cdot v_{3y} = 0 \implies v_{3y} = -\frac{12}{m} \]

### Step 6: Use the speed of the third fragment to find its mass
The speed of the third fragment is given as \( 6.5 \, \text{m/s} \):
\[ \sqrt{v_{3x}^2 + v_{3y}^2} = 6.5 \]

Substituting the expressions for \( v_{3x} \) and \( v_{3y} \):
\[ \sqrt{\left(-\frac{5}{m}\right)^2 + \left(-\frac{12}{m}\right)^2} = 6.5 \]

### Step 7: Solve for \( m \)
Squaring both sides:
\[ \left(-\frac{5}{m}\right)^2 + \left(-\frac{12}{m}\right)^2 = (6.5)^2 \]
\[ \frac{25}{m^2} + \frac{144}{m^2} = 42.25 \]
\[ \frac{169}{m^2} = 42.25 \]
\[ m^2 = \frac{169}{42.25} \]
\[ m^2 = 4 \implies m = 2 \, \text{kg} \]

### Step 8: Calculate the total mass of the shell
The total mass of the shell is:
\[ \text{Total mass} = m_1 + m_2 + m_3 = 1 + 2 + 2 = 5 \, \text{kg} \]

### Final Answer
The total mass of the shell must be \( 5 \, \text{kg} \).

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