A projectile of mass M is fired so that the horizontal range is 4 km. At the highest point the projectile explodes in two parts of masses M/4 and 3M/4 respectively and the heavier part starts falling down vertically with zero initial speed. The horizontal range (distance form point of fring) of the lighter part is :

Options

A. `16 km`
B. `1 km`
C. `10 km`
D. `2 km`

Answer

To solve the problem, we need to analyze the motion of the projectile and the effects of the explosion at its highest point. Here’s a step-by-step breakdown:

### Step 1: Understand the initial conditions
The projectile is fired with a horizontal range of 4 km. This means that if the projectile were to continue its motion without any explosions, it would land 4 km away from the firing point.

**Hint:** Remember that the horizontal range is the total distance traveled horizontally before landing.

### Step 2: Analyze the explosion
At the highest point of its trajectory, the projectile explodes into two parts: one with mass \( \frac{M}{4} \) (the lighter part) and the other with mass \( \frac{3M}{4} \) (the heavier part). The heavier part falls vertically with zero initial speed.

**Hint:** The explosion does not affect the horizontal motion of the center of mass of the system.

### Step 3: Apply the conservation of momentum
Since there are no external forces acting in the horizontal direction, the horizontal displacement of the center of mass remains unchanged. Thus, the horizontal range of the center of mass before and after the explosion is the same.

**Hint:** Use the principle of conservation of momentum to relate the masses and their respective displacements.

### Step 4: Set up the equation for horizontal displacement
Let \( x_1 \) be the horizontal range of the lighter part (\( \frac{M}{4} \)) after the explosion. The heavier part (\( \frac{3M}{4} \)) will not contribute to horizontal displacement since it falls straight down. The equation for the horizontal displacement of the center of mass can be written as:

\[ M \cdot R = \left(\frac{M}{4} \cdot x_1\right) + \left(\frac{3M}{4} \cdot 4\right) \]

Where \( R \) is the total horizontal range (4 km).

**Hint:** Factor out the mass \( M \) from the equation since it appears in all terms.

### Step 5: Simplify the equation
Cancelling \( M \) from both sides gives:

\[ 4 = \frac{1}{4} x_1 + 3 \]

### Step 6: Solve for \( x_1 \)
Now, rearranging the equation to solve for \( x_1 \):

\[ 4 - 3 = \frac{1}{4} x_1 \]

\[ 1 = \frac{1}{4} x_1 \]

Multiplying both sides by 4:

\[ x_1 = 4 \text{ km} \]

### Step 7: Conclusion
The horizontal range of the lighter part after the explosion is 4 km.

**Final Answer:** The horizontal range of the lighter part is 4 km.