A cannon ball is fired with a velocity 200m / sec at an angle of 60° with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity 100 m / sec , the second one falling vertically downwards with a velocity 100 m / sec . The third fragment will be moving with a velocity

Options

A. `100 ms^(-1)` in the horizontal direction
B. `300 ms^(-1)` in the horizontal direction
C. `300 ms^(-1)` in a direction making an angle of `60^(@)` with the horizontal
D. `200 ms^(-1)` in a direction making an angle of `60^(@0` with the horizontal

Explanation

To solve the problem step by step, we will analyze the motion of the cannonball and apply the principle of conservation of momentum.

### Step 1: Determine the initial velocity components of the cannonball.

The cannonball is fired with a velocity of \(200 \, \text{m/s}\) at an angle of \(60^\circ\) with the horizontal. We can break this velocity into its horizontal and vertical components.

- **Horizontal component (\(u_x\))**:
\[ u_x = 200 \cos(60^\circ) = 200 \times \frac{1}{2} = 100 \, \text{m/s} \]

- **Vertical component (\(u_y\))**:
\[ u_y = 200 \sin(60^\circ) = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \, \text{m/s} \]

### Step 2: Analyze the motion at the highest point.

At the highest point of its flight, the vertical component of the velocity becomes \(0\) (as it momentarily stops rising before falling). Therefore, the velocity of the cannonball at the highest point is purely horizontal:

\[ \vec{V} = 100 \, \hat{i} \, \text{m/s} \]

### Step 3: Apply conservation of momentum during the explosion.

When the cannonball explodes into three equal fragments, we need to apply the conservation of momentum. The initial momentum of the cannonball before the explosion is:

\[ \vec{P}_{initial} = m \cdot \vec{V} = m \cdot (100 \hat{i}) \, \text{kg m/s} \]

After the explosion, the momentum of the three fragments can be expressed as follows:

1. The first fragment goes vertically upwards with a velocity of \(100 \, \text{m/s}\):
\[ \vec{P}_1 = \frac{m}{3} \cdot (100 \hat{j}) \, \text{kg m/s} \]

2. The second fragment falls vertically downwards with a velocity of \(100 \, \text{m/s}\):
\[ \vec{P}_2 = \frac{m}{3} \cdot (-100 \hat{j}) \, \text{kg m/s} \]

3. The third fragment has an unknown velocity \(\vec{V}_3\):
\[ \vec{P}_3 = \frac{m}{3} \cdot \vec{V}_3 \, \text{kg m/s} \]

### Step 4: Set up the momentum conservation equation.

According to the conservation of momentum, the total initial momentum must equal the total final momentum:

\[ m \cdot (100 \hat{i}) = \frac{m}{3} \cdot (100 \hat{j}) + \frac{m}{3} \cdot (-100 \hat{j}) + \frac{m}{3} \cdot \vec{V}_3 \]

### Step 5: Simplify the equation.

The vertical components from the first two fragments cancel each other out:

\[ m \cdot (100 \hat{i}) = \frac{m}{3} \cdot \vec{V}_3 \]

### Step 6: Solve for \(\vec{V}_3\).

We can cancel \(m\) from both sides (assuming \(m \neq 0\)):

\[ 100 \hat{i} = \frac{1}{3} \vec{V}_3 \]

Multiplying both sides by \(3\):

\[ \vec{V}_3 = 300 \hat{i} \, \text{m/s} \]

### Conclusion

The third fragment moves with a velocity of \(300 \, \text{m/s}\) in the horizontal direction.