A shell is fired from a cannon with a velocity V at an angle `theta` with the horizontal direction. A the highest point i its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is

Options

A. `3Vcostheta`
B. `2Vcostheta`
C. `3/2Vcostheta`
D. `Vcostheta`

Answer

To solve the problem step by step, we need to analyze the situation involving the shell fired from the cannon, its trajectory, and the explosion at the highest point.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- A shell is fired with an initial velocity \( V \) at an angle \( \theta \) with the horizontal.
- At the highest point of its trajectory, the vertical component of the velocity is zero, and only the horizontal component remains, which is \( V_x = V \cos \theta \).

2. **Explosion at the Highest Point**:
- At the highest point, the shell explodes into two pieces of equal mass. Let's denote the mass of the shell as \( m \), so each piece has a mass of \( \frac{m}{2} \).

3. **Movement of the First Piece**:
- One of the pieces retraces its path back to the cannon. This means it must have the same horizontal velocity \( V \cos \theta \) but in the opposite direction (to the left if we consider right as positive).
- Therefore, the velocity of the first piece is \( -V \cos \theta \).

4. **Conservation of Momentum**:
- According to the law of conservation of momentum, the total momentum before the explosion must equal the total momentum after the explosion.
- Before the explosion, the total momentum \( p_{initial} \) is:
\[ p_{initial} = m \cdot V \cos \theta \]
- After the explosion, the momentum \( p_{final} \) can be expressed as:
\[ p_{final} = \left(\frac{m}{2}\right)(-V \cos \theta) + \left(\frac{m}{2}\right)V' \]
where \( V' \) is the velocity of the second piece.

5. **Setting Up the Equation**:
- Setting the initial momentum equal to the final momentum:
\[ m V \cos \theta = \left(\frac{m}{2}\right)(-V \cos \theta) + \left(\frac{m}{2}\right)V' \]

6. **Simplifying the Equation**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ V \cos \theta = \left(-\frac{1}{2} V \cos \theta\right) + \frac{1}{2} V' \]
- Rearranging gives:
\[ V \cos \theta + \frac{1}{2} V \cos \theta = \frac{1}{2} V' \]
- This simplifies to:
\[ \frac{3}{2} V \cos \theta = \frac{1}{2} V' \]

7. **Finding the Velocity of the Second Piece**:
- Multiply both sides by 2 to solve for \( V' \):
\[ V' = 3 V \cos \theta \]

### Final Answer:
The speed of the other piece immediately after the explosion is \( V' = 3 V \cos \theta \).

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