An object of mass 5 kg is projected with a velocity of 20 m/s at angle of 60° to the horizontal. At the highest point of its path, the projectile explodes and breaks up into two fragments of masses 1 kg and 4 kg. The fragments separate horizontally after the explosion. The explosion releases internal energy such that the kinetic energy of the system at the highest point is doubled. If the separation between the two fragments when they reach the ground is x, then find the value of `x/(5sqrt(3))`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial conditions at the highest point
The object of mass 5 kg is projected with a velocity of 20 m/s at an angle of 60° to the horizontal. At the highest point, the vertical component of the velocity is 0, and the horizontal component can be calculated as follows:

\[ V_{x} = V \cdot \cos(\theta) = 20 \cdot \cos(60°) = 20 \cdot \frac{1}{2} = 10 \text{ m/s} \]

### Step 2: Calculate the initial momentum
The total initial momentum of the system at the highest point is:

\[ P_{initial} = m \cdot V_{x} = 5 \cdot 10 = 50 \text{ kg m/s} \]

### Step 3: Apply conservation of momentum after the explosion
After the explosion, the object splits into two fragments of masses 1 kg and 4 kg. Let \( V_1 \) be the velocity of the 4 kg fragment and \( V_2 \) be the velocity of the 1 kg fragment. Since they separate horizontally, we can write:

\[ P_{initial} = P_{final} \]
\[ 50 = 4V_1 + 1(-V_2) \]
\[ 4V_1 - V_2 = 50 \quad \text{(Equation 1)} \]

### Step 4: Calculate the kinetic energy before and after the explosion
The initial kinetic energy (KE) at the highest point is:

\[ KE_{initial} = \frac{1}{2} m V^2 = \frac{1}{2} \cdot 5 \cdot (10)^2 = 250 \text{ J} \]

The problem states that the kinetic energy is doubled after the explosion:

\[ KE_{final} = 2 \cdot KE_{initial} = 500 \text{ J} \]

The kinetic energy after the explosion can be expressed as:

\[ KE_{final} = \frac{1}{2} \cdot 4V_1^2 + \frac{1}{2} \cdot 1V_2^2 \]
\[ 4V_1^2 + \frac{1}{2}V_2^2 = 1000 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations simultaneously
From Equation 1, we can express \( V_2 \) in terms of \( V_1 \):

\[ V_2 = 4V_1 - 50 \]

Substituting \( V_2 \) in Equation 2:

\[ 4V_1^2 + \frac{1}{2}(4V_1 - 50)^2 = 1000 \]

Expanding and simplifying:

\[ 4V_1^2 + \frac{1}{2}(16V_1^2 - 400V_1 + 2500) = 1000 \]
\[ 4V_1^2 + 8V_1^2 - 200V_1 + 1250 = 1000 \]
\[ 12V_1^2 - 200V_1 + 250 = 0 \]

### Step 6: Solve the quadratic equation
Dividing the entire equation by 2:

\[ 6V_1^2 - 100V_1 + 125 = 0 \]

Using the quadratic formula \( V_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ V_1 = \frac{100 \pm \sqrt{(-100)^2 - 4 \cdot 6 \cdot 125}}{2 \cdot 6} \]
\[ V_1 = \frac{100 \pm \sqrt{10000 - 3000}}{12} \]
\[ V_1 = \frac{100 \pm \sqrt{7000}}{12} \]

Calculating the roots gives us two possible values for \( V_1 \).

### Step 7: Calculate the time of flight
The maximum height \( H_{max} \) can be calculated using:

\[ H_{max} = \frac{V^2 \sin^2(\theta)}{2g} = \frac{(20)^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2 \cdot 10} = \frac{400 \cdot \frac{3}{4}}{20} = 15 \text{ m} \]

The time to reach the ground from the maximum height is given by:

\[ t = \sqrt{\frac{2H_{max}}{g}} = \sqrt{\frac{2 \cdot 15}{10}} = \sqrt{3} \text{ s} \]

### Step 8: Calculate the separation distance
Using the relative velocity, we can find the separation distance \( x \):

\[ x = (V_1 + V_2) \cdot t \]

Substituting the values of \( V_1 \) and \( V_2 \) calculated earlier, we can find \( x \).

### Step 9: Find \( \frac{x}{5\sqrt{3}} \)
Finally, we compute \( \frac{x}{5\sqrt{3}} \).