A body of mass M at rest explodes into three pieces, two of which of mass M//4 each are thrown off in perpendicular directions eith velocities of `3m//s` and `4m//s` respectively. The third piece will be thrown off with a velocity of

Options

A. `1.5 ms^(-1)`
B. `2.0 ms^(-1)`
C. `2.5 ms ^(-1)`
D. `3.0ms^(-1)`

Answer

To solve the problem, we need to apply the principle of conservation of momentum. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
Initially, the body of mass \( M \) is at rest. Therefore, the initial momentum of the system is zero.

### Step 2: Identify the Masses and Velocities
After the explosion, the body splits into three pieces:
- Two pieces each of mass \( \frac{M}{4} \).
- The first piece moves with a velocity of \( 3 \, \text{m/s} \) in the x-direction.
- The second piece moves with a velocity of \( 4 \, \text{m/s} \) in the y-direction.
- The third piece's mass will be \( M - \frac{M}{4} - \frac{M}{4} = \frac{M}{2} \).

### Step 3: Set Up the Momentum Conservation Equation
Since there are no external forces acting on the system, the total momentum before the explosion must equal the total momentum after the explosion.

The momentum of the first piece (moving in the x-direction):
\[ p_1 = \frac{M}{4} \cdot 3 \hat{i} = \frac{3M}{4} \hat{i} \]

The momentum of the second piece (moving in the y-direction):
\[ p_2 = \frac{M}{4} \cdot 4 \hat{j} = \frac{4M}{4} \hat{j} = M \hat{j} \]

Let the velocity of the third piece be \( \vec{v_0} \). Its momentum will be:
\[ p_3 = \frac{M}{2} \cdot \vec{v_0} \]

### Step 4: Write the Total Momentum Equation
The total momentum after the explosion must equal the initial momentum (which is zero):
\[ p_1 + p_2 + p_3 = 0 \]
Substituting the expressions for \( p_1 \), \( p_2 \), and \( p_3 \):
\[ \frac{3M}{4} \hat{i} + M \hat{j} + \frac{M}{2} \vec{v_0} = 0 \]

### Step 5: Solve for \( \vec{v_0} \)
Rearranging the equation gives:
\[ \frac{M}{2} \vec{v_0} = -\left(\frac{3M}{4} \hat{i} + M \hat{j}\right) \]
Dividing through by \( \frac{M}{2} \):
\[ \vec{v_0} = -\left(\frac{3}{2} \hat{i} + 2 \hat{j}\right) \]

### Step 6: Find the Magnitude of \( \vec{v_0} \)
The magnitude of \( \vec{v_0} \) is given by:
\[ |\vec{v_0}| = \sqrt{\left(-\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{9}{4} + \frac{16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} = 2.5 \, \text{m/s} \]

### Final Answer
The velocity of the third piece is \( 2.5 \, \text{m/s} \).

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