A shell is fired from a gun and it explodes in two pieces at the highest point of trajectory . One piece retraces its apth to gun . What will be the velocity of other part just after explosion if shell is fired with a speed u at an angle `theta ` with the horizontal ?

Options

A. `u cos theta `
B. `2 u cos theta `
C. `3 u cos theta `
D. `4 u cos theta `

Explanation

To solve the problem step by step, we will use the principle of conservation of momentum. Here’s how we can approach it:

### Step 1: Understand the scenario
The shell is fired with an initial speed \( u \) at an angle \( \theta \). At the highest point of its trajectory, it explodes into two pieces. One piece retraces its path back to the gun, while we need to find the velocity of the other piece just after the explosion.

### Step 2: Determine the initial momentum
At the highest point of the trajectory, the vertical component of the velocity is zero. The horizontal component of the velocity is given by:
\[ u_x = u \cos \theta \]
Since the mass of the shell is \( 2m \) (it splits into two pieces of mass \( m \) each), the initial momentum \( p_{initial} \) of the shell can be calculated as:
\[ p_{initial} = \text{mass} \times \text{velocity} = 2m \cdot u \cos \theta \]

### Step 3: Analyze the explosion
After the explosion, one piece retraces its path back to the gun with a velocity of \( -u \cos \theta \) (the negative sign indicates that it is moving in the opposite direction). The momentum of this piece is:
\[ p_{1} = m \cdot (-u \cos \theta) = -m u \cos \theta \]

### Step 4: Apply conservation of momentum
According to the conservation of momentum, the total momentum before the explosion must equal the total momentum after the explosion. Therefore, we have:
\[ p_{initial} = p_{1} + p_{2} \]
Where \( p_{2} \) is the momentum of the second piece. Substituting the values we have:
\[ 2m u \cos \theta = -m u \cos \theta + m v \]
Where \( v \) is the velocity of the second piece after the explosion.

### Step 5: Solve for the velocity of the second piece
Rearranging the equation gives:
\[ 2m u \cos \theta + m u \cos \theta = m v \]
\[ 3m u \cos \theta = m v \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ v = 3u \cos \theta \]

### Conclusion
The velocity of the other part just after the explosion is:
\[ v = 3u \cos \theta \]