An object of mass 5 kg is projecte with a velocity of `20ms^(-1)` at an angle of `60^(@)` to the horizontal. At the highest point of its path , the projectile explodes and breaks up into two fragments of masses 1kg and 4 kg. The fragments separate horizontally after the explosion, which releases internal energy such that `K.E.` of the system at the highest point is doubled. Calculate the separation betweent the two fragments when they reach the ground.

Options

A. 11 m
B. 22 m
C. 44 m
D. 66 m

Explanation

To solve the problem step by step, we will follow the outlined process:

### Step 1: Determine the horizontal and vertical components of the initial velocity.

1. **Given Data:**
- Mass of the object, \( m = 5 \, \text{kg} \)
- Initial velocity, \( v = 20 \, \text{m/s} \)
- Angle of projection, \( \theta = 60^\circ \)

2. **Horizontal Component of Velocity (\( V_x \)):**
\[ V_x = V \cos \theta = 20 \cos 60^\circ = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

3. **Vertical Component of Velocity (\( V_y \)):**
\[ V_y = V \sin \theta = 20 \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 2: Calculate the time taken to reach the highest point.

1. **Time to reach the highest point (\( t \)):**
\[ t = \frac{V_y}{g} = \frac{10\sqrt{3}}{9.8} \approx 1.77 \, \text{s} \]

### Step 3: Analyze the explosion at the highest point.

1. **At the highest point, the object splits into two fragments:**
- Mass of fragment 1, \( m_1 = 1 \, \text{kg} \)
- Mass of fragment 2, \( m_2 = 4 \, \text{kg} \)

2. **Conservation of linear momentum:**
- Before explosion: Total horizontal momentum = \( MV_x = 5 \times 10 = 50 \, \text{kg m/s} \)
- After explosion:
\[ m_1 v_1 + m_2 v_2 = 50 \]
\[ 1 v_1 + 4 v_2 = 50 \quad \text{(Equation 1)} \]

### Step 4: Calculate the initial kinetic energy and the final kinetic energy after the explosion.

1. **Initial Kinetic Energy (\( KE_i \)):**
\[ KE_i = \frac{1}{2} m V^2 = \frac{1}{2} \times 5 \times (10)^2 = 250 \, \text{J} \]

2. **Final Kinetic Energy (\( KE_f \)):**
\[ KE_f = 2 \times KE_i = 2 \times 250 = 500 \, \text{J} \]

3. **Final Kinetic Energy equation:**
\[ KE_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
\[ \frac{1}{2} \times 1 v_1^2 + \frac{1}{2} \times 4 v_2^2 = 500 \]
\[ v_1^2 + 4 v_2^2 = 1000 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations to find \( v_1 \) and \( v_2 \).

1. **From Equation 1:**
\[ v_1 = 50 - 4v_2 \]

2. **Substituting into Equation 2:**
\[ (50 - 4v_2)^2 + 4v_2^2 = 1000 \]
Expanding and simplifying gives:
\[ 2500 - 400v_2 + 16v_2^2 + 4v_2^2 = 1000 \]
\[ 20v_2^2 - 400v_2 + 1500 = 0 \]
Dividing through by 20:
\[ v_2^2 - 20v_2 + 75 = 0 \]

3. **Using the quadratic formula:**
\[ v_2 = \frac{20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot 75}}{2 \cdot 1} = \frac{20 \pm \sqrt{400 - 300}}{2} = \frac{20 \pm \sqrt{100}}{2} \]
\[ v_2 = \frac{20 \pm 10}{2} \]
This gives \( v_2 = 15 \, \text{m/s} \) or \( v_2 = 5 \, \text{m/s} \).

4. **Finding \( v_1 \):**
- If \( v_2 = 5 \, \text{m/s} \):
\[ v_1 = 50 - 4 \times 5 = 30 \, \text{m/s} \]

### Step 6: Calculate the separation between the two fragments when they reach the ground.

1. **Horizontal separation (\( S \)):**
\[ S = (v_1 - v_2) \times t = (30 - 5) \times 1.77 = 25 \times 1.77 = 44.25 \, \text{m} \]

### Final Answer:
The separation between the two fragments when they reach the ground is approximately **44.25 meters**.

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