A body oa mass M at rest explodes into three pieces, two of which of mass (M/4) each are thrown- off in perpendicular directions with velocities of ` 6 ms^(-1)` and `8 ms^(-1)` respectively. The third piece will be thrown-off with a velocity of `

Options

A. `1.5 ms^(-1)`
B. `2 ms^(-1)`
C. `5 ms^(-1)`
D. `3 ms^(-1)`

Explanation

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

### Step 1: Understand the Problem
We have a body of mass \( M \) that is initially at rest. Upon explosion, it breaks into three pieces:
- Two pieces of mass \( \frac{M}{4} \) each.
- One piece of mass \( \frac{M}{2} \) (since \( M - 2 \times \frac{M}{4} = \frac{M}{2} \)).

The two pieces are thrown off in perpendicular directions with velocities \( 6 \, \text{m/s} \) and \( 8 \, \text{m/s} \).

### Step 2: Set Up the Conservation of Momentum Equation
Since the initial momentum of the system is zero (the body is at rest), the final momentum must also equal zero:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
\[ 0 = M_1 \vec{V_1} + M_2 \vec{V_2} + M_3 \vec{V_3} \]

Where:
- \( M_1 = \frac{M}{4} \), \( \vec{V_1} = 6 \, \text{m/s} \, \hat{i} \)
- \( M_2 = \frac{M}{4} \), \( \vec{V_2} = 8 \, \text{m/s} \, \hat{j} \)
- \( M_3 = \frac{M}{2} \), \( \vec{V_3} \) is unknown.

### Step 3: Write the Momentum Vectors
Substituting the values into the momentum equation:
\[ 0 = \left(\frac{M}{4} \cdot 6 \hat{i}\right) + \left(\frac{M}{4} \cdot 8 \hat{j}\right) + \left(\frac{M}{2} \vec{V_3}\right) \]
This simplifies to:
\[ 0 = \frac{6M}{4} \hat{i} + \frac{8M}{4} \hat{j} + \frac{M}{2} \vec{V_3} \]

### Step 4: Simplify the Equation
Now, we can simplify the equation:
\[ 0 = \frac{3M}{2} \hat{i} + 2M \hat{j} + \frac{M}{2} \vec{V_3} \]

### Step 5: Isolate \( \vec{V_3} \)
Rearranging gives:
\[ \frac{M}{2} \vec{V_3} = -\frac{3M}{2} \hat{i} - 2M \hat{j} \]
Dividing through by \( \frac{M}{2} \) (assuming \( M \neq 0 \)):
\[ \vec{V_3} = -3 \hat{i} - 4 \hat{j} \]

### Step 6: Calculate the Magnitude of \( \vec{V_3} \)
To find the magnitude of \( \vec{V_3} \):
\[ |\vec{V_3}| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s} \]

### Conclusion
The third piece will be thrown off with a velocity of \( 5 \, \text{m/s} \).

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