A shell is fired from a cannon with a velocity `v (m//sec.)` at an angle `theta` with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (in `m//sec.`) of the other piece immediately after the explosion is

Options

A. `3v cos theta`
B. `2v cos theta`
C. `3/2v cos theta`
D. `sqrt(3/2)v cos theta`

Explanation

To solve the problem, we need to analyze the situation using the principle of conservation of momentum. Here are the steps to find the speed of the piece of the shell that does not retrace its path:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The shell is fired with an initial velocity \( v \) at an angle \( \theta \).
- At the highest point of its trajectory, the vertical component of the velocity is zero, and the horizontal component is \( v \cos \theta \).

2. **Define Masses**:
- Let the total mass of the shell be \( 2m \). After the explosion, it splits into two equal masses, each of mass \( m \).

3. **Momentum Conservation**:
- According to the conservation of momentum, the total momentum before the explosion must equal the total momentum after the explosion.
- Before the explosion, the momentum in the horizontal direction is:
\[ P_{\text{initial}} = (2m)(v \cos \theta) \]
- After the explosion, one piece retraces its path back to the cannon with a velocity of \( -v \cos \theta \) (negative because it is in the opposite direction), and the other piece has an unknown velocity \( v' \).

4. **Set Up the Momentum Equation**:
- The momentum after the explosion can be expressed as:
\[ P_{\text{final}} = m(-v \cos \theta) + m(v') \]
- Setting the initial momentum equal to the final momentum gives:
\[ (2m)(v \cos \theta) = m(-v \cos \theta) + m(v') \]

5. **Simplify the Equation**:
- Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ 2(v \cos \theta) = -v \cos \theta + v' \]
- Rearranging the equation:
\[ v' = 2(v \cos \theta) + v \cos \theta \]
- Therefore:
\[ v' = 3(v \cos \theta) \]

6. **Final Result**:
- The speed of the other piece immediately after the explosion is:
\[ v' = 3v \cos \theta \]