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

Options

A. `1.5 ms^(-1)`
B. `2 ms^(-1)`
C. `2.5 ms^(-1)`
D. `3 ms^(-1)`

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.

### Step-by-Step Solution:

1. **Identify the Masses and Velocities:**
- The total mass of the body is \( M \).
- Two pieces have a mass of \( \frac{M}{4} \) each.
- The third piece will have a mass of \( M - \frac{M}{4} - \frac{M}{4} = \frac{M}{2} \).
- The velocities of the two pieces are \( 3 \, \text{m/s} \) and \( 4 \, \text{m/s} \), thrown off in perpendicular directions.

2. **Calculate the Momentum of Each Piece:**
- For the first piece (mass \( \frac{M}{4} \) moving at \( 3 \, \text{m/s} \)):
\[ p_1 = \frac{M}{4} \times 3 = \frac{3M}{4} \]
- For the second piece (mass \( \frac{M}{4} \) moving at \( 4 \, \text{m/s} \)):
\[ p_2 = \frac{M}{4} \times 4 = M \]

3. **Determine the Resultant Momentum:**
- Since the two pieces are moving in perpendicular directions, we can find the resultant momentum using the Pythagorean theorem:
\[ p_{total} = \sqrt{p_1^2 + p_2^2} = \sqrt{\left(\frac{3M}{4}\right)^2 + (M)^2} \]
- Calculating:
\[ p_{total} = \sqrt{\left(\frac{9M^2}{16}\right) + (M^2)} = \sqrt{\left(\frac{9M^2}{16} + \frac{16M^2}{16}\right)} = \sqrt{\frac{25M^2}{16}} = \frac{5M}{4} \]

4. **Apply Conservation of Momentum:**
- The total momentum before the explosion is zero (since the body was at rest). Therefore, the total momentum after the explosion must also be zero:
\[ p_1 + p_2 + p_3 = 0 \]
- Rearranging gives:
\[ p_3 = - (p_1 + p_2) = -\frac{5M}{4} \]
- Since the third piece has a mass of \( \frac{M}{2} \), we can express its momentum as:
\[ p_3 = \frac{M}{2} \cdot v_3 \]
- Setting the two expressions for \( p_3 \) equal:
\[ \frac{M}{2} \cdot v_3 = -\frac{5M}{4} \]

5. **Solve for the Velocity of the Third Piece:**
- Cancel \( M \) from both sides (assuming \( M \neq 0 \)):
\[ \frac{1}{2} v_3 = -\frac{5}{4} \]
- Multiply both sides by 2:
\[ v_3 = -\frac{5}{2} = -2.5 \, \text{m/s} \]

### Final Answer:
The velocity of the third piece is \( -2.5 \, \text{m/s} \) (indicating it moves in the opposite direction to the resultant momentum of the first two pieces).

---