A mortar fires a shell of mass M which explodes into two pieces of mass `M/5` and `(4M)/(5)` at the top of the trajectory. The smaller mass falls very close to the mortar. In the same time bigger piece lands a distance D from the mortar. The shell would have fallen at a distance R from the mortar if there was no explosion. The value of D is (neglect air resistance)

Options

A. (a) `(3R)/(2)`
B. (b) `(4R)/(3)`
C. (c) `(5R)/(4)`
D. (d) None of these

Explanation

To solve the problem step by step, we will analyze the situation involving the mortar, the shell, and the resulting pieces after the explosion.

### Step 1: Understand the Problem
A mortar fires a shell of mass \( M \) that explodes into two pieces at the top of its trajectory. The masses of the pieces are \( \frac{M}{5} \) (smaller mass) and \( \frac{4M}{5} \) (larger mass). The smaller mass falls very close to the mortar, while the larger mass lands a distance \( D \) from the mortar. We need to find the value of \( D \) in terms of \( R \), the distance the shell would have fallen if there was no explosion.

### Step 2: Analyze the Center of Mass
Since there is no external horizontal force acting on the system, the center of mass of the shell before and after the explosion will remain in the same horizontal position. Thus, the range covered by the center of mass before the explosion will equal the range covered by the center of mass after the explosion.

### Step 3: Define the Center of Mass
The center of mass \( x_{CM} \) for the two pieces after the explosion can be calculated using the formula:
\[ x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
where:
- \( m_1 = \frac{M}{5} \) (smaller mass)
- \( x_1 = 0 \) (the smaller mass falls very close to the mortar)
- \( m_2 = \frac{4M}{5} \) (larger mass)
- \( x_2 = D \) (the distance where the larger mass lands)

### Step 4: Set Up the Equation
Substituting the values into the center of mass equation:
\[ x_{CM} = \frac{\left(\frac{M}{5} \cdot 0\right) + \left(\frac{4M}{5} \cdot D\right)}{\frac{M}{5} + \frac{4M}{5}} = \frac{\frac{4M}{5} D}{M} = \frac{4D}{5} \]

### Step 5: Relate to the Original Range \( R \)
Before the explosion, the range \( R \) is simply the distance that the shell would have traveled. Since the center of mass does not change its horizontal position, we have:
\[ R = x_{CM} = \frac{4D}{5} \]

### Step 6: Solve for \( D \)
Now, we can rearrange the equation to solve for \( D \):
\[ D = \frac{5R}{4} \]

### Conclusion
Thus, the value of \( D \) in terms of \( R \) is:
\[ D = \frac{5R}{4} \]