A bomb is rest at the summit of a cliff breaks into two equal fragments.One of the fragments attains a horizontal velocity of `20sqrt3 ms^(-1)`.The horizontal distance between the two fragments, when their displacement vectors is inclined at `60^(@)` relative to each other is `(g=10ms^(-2))`

Options

A. `40sqrt3`
B. `80sqrt3`
C. `120sqrt3`
D. `480sqrt3`

Explanation

To solve the problem step by step, we will analyze the motion of the two fragments after the bomb explodes.

### Step 1: Understand the situation
A bomb at rest on a cliff breaks into two equal fragments. One fragment moves horizontally with a velocity of \(20\sqrt{3} \, \text{m/s}\). The other fragment will have a velocity that we need to determine based on the conservation of momentum.

### Step 2: Apply conservation of momentum
Since the bomb is at rest before the explosion, the total momentum before the explosion is zero. After the explosion, if one fragment has a momentum of \(m \cdot 20\sqrt{3}\) (where \(m\) is the mass of one fragment), the other fragment must have momentum of \(-m \cdot 20\sqrt{3}\) to conserve momentum.

### Step 3: Determine the velocities of the fragments
Let the velocity of the second fragment be \(-20\sqrt{3} \, \text{m/s}\) (in the opposite direction). Thus, we have:
- Fragment 1: \( \vec{v_1} = 20\sqrt{3} \, \hat{i} \)
- Fragment 2: \( \vec{v_2} = -20\sqrt{3} \, \hat{i} \)

### Step 4: Analyze the motion of the fragments
Both fragments will undergo projectile motion after the explosion. The vertical motion will be influenced by gravity, while the horizontal motion remains constant.

### Step 5: Write the displacement equations
For the first fragment:
- Horizontal displacement: \( x_1 = 20\sqrt{3} \cdot t \)
- Vertical displacement: \( y_1 = -\frac{1}{2} g t^2 = -5t^2 \)

For the second fragment:
- Horizontal displacement: \( x_2 = -20\sqrt{3} \cdot t \)
- Vertical displacement: \( y_2 = -\frac{1}{2} g t^2 = -5t^2 \)

### Step 6: Find the angle between the displacement vectors
The angle between the displacement vectors is given as \(60^\circ\). We can use the dot product to find the relationship:
\[ \cos(60^\circ) = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} \]
Where:
- \(\vec{d_1} = (20\sqrt{3} t, -5t^2)\)
- \(\vec{d_2} = (-20\sqrt{3} t, -5t^2)\)

### Step 7: Calculate the dot product
\[ \vec{d_1} \cdot \vec{d_2} = (20\sqrt{3} t)(-20\sqrt{3} t) + (-5t^2)(-5t^2) = -1200t^2 + 25t^4 \]

### Step 8: Calculate the magnitudes
\[ |\vec{d_1}| = \sqrt{(20\sqrt{3} t)^2 + (-5t^2)^2} = \sqrt{1200t^2 + 25t^4} \]
\[ |\vec{d_2}| = \sqrt{(-20\sqrt{3} t)^2 + (-5t^2)^2} = \sqrt{1200t^2 + 25t^4} \]

### Step 9: Set up the equation using the cosine
Using \(\cos(60^\circ) = \frac{1}{2}\):
\[ \frac{-1200t^2 + 25t^4}{(1200t^2 + 25t^4)} = \frac{1}{2} \]

### Step 10: Solve for \(t\)
Cross-multiplying and simplifying leads to:
\[ -2400t^2 + 50t^4 = 1200t^2 + 25t^4 \]
This simplifies to:
\[ 25t^4 - 3600t^2 = 0 \]
Factoring out \(t^2\):
\[ t^2(25t^2 - 3600) = 0 \]
Thus, \(t^2 = 144\) which gives \(t = 12 \, \text{s}\).

### Step 11: Calculate the horizontal distance
The total horizontal distance between the two fragments is:
\[ \text{Total horizontal distance} = x_1 + |x_2| = 20\sqrt{3} \cdot 12 + 20\sqrt{3} \cdot 12 = 40\sqrt{3} \cdot 12 = 480\sqrt{3} \, \text{m} \]

### Final Answer
The horizontal distance between the two fragments when their displacement vectors are inclined at \(60^\circ\) relative to each other is \(480\sqrt{3} \, \text{m}\).

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