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. `1.5m//s`
B. `2.0m//s`
C. `2.5m//s`
D. `3.0m//s`

Explanation

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

### Step 1: Understand the problem
A body of mass \( M \) explodes into three pieces. Two pieces have masses \( \frac{M}{4} \) each and are thrown off in perpendicular directions with velocities \( 3 \, \text{m/s} \) and \( 4 \, \text{m/s} \) respectively. We need to find the velocity of the third piece.

### Step 2: Set up the momentum conservation equation
Since the body is initially at rest, the total initial momentum is zero. Therefore, the total final momentum must also equal zero.

Let:
- \( P_1 \) be the momentum of the first piece,
- \( P_2 \) be the momentum of the second piece,
- \( P_3 \) be the momentum of the third piece.

The momentum of each piece can be calculated as:
- \( P_1 = \frac{M}{4} \times 3 \) (in the x-direction)
- \( P_2 = \frac{M}{4} \times 4 \) (in the y-direction)

### Step 3: Calculate the momenta of the first two pieces
Calculating \( P_1 \) and \( P_2 \):
- \( P_1 = \frac{M}{4} \times 3 = \frac{3M}{4} \) (in the x-direction)
- \( P_2 = \frac{M}{4} \times 4 = \frac{4M}{4} = M \) (in the y-direction)

### Step 4: Find the resultant momentum of the first two pieces
Since \( P_1 \) and \( P_2 \) are perpendicular, we can find the resultant momentum \( P_R \) using the Pythagorean theorem:
\[ P_R = \sqrt{P_1^2 + P_2^2} = \sqrt{\left(\frac{3M}{4}\right)^2 + (M)^2} \]
Calculating this gives:
\[ P_R = \sqrt{\frac{9M^2}{16} + M^2} = \sqrt{\frac{9M^2}{16} + \frac{16M^2}{16}} = \sqrt{\frac{25M^2}{16}} = \frac{5M}{4} \]

### Step 5: Set up the equation for the third piece
Since the total momentum must equal zero, the momentum of the third piece \( P_3 \) must equal the negative of the resultant momentum of the first two pieces:
\[ P_3 = -P_R = -\frac{5M}{4} \]
The mass of the third piece is:
\[ M_3 = M - \frac{M}{4} - \frac{M}{4} = \frac{M}{2} \]
Thus, the momentum of the third piece can be expressed as:
\[ P_3 = \frac{M}{2} \times v_3 \]
Setting the two expressions for \( P_3 \) equal gives:
\[ \frac{M}{2} \times v_3 = -\frac{5M}{4} \]

### Step 6: Solve for the velocity of the third piece
Dividing both sides by \( \frac{M}{2} \):
\[ v_3 = -\frac{5M}{4} \times \frac{2}{M} = -\frac{5}{2} = -2.5 \, \text{m/s} \]
The negative sign indicates the direction of the velocity is opposite to the resultant momentum of the first two pieces.

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

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