A particle of mass `4m` is projected from the ground at some angle with horizontal. Its horizontal range is `R`. At the highest point of its path it breaks into two pieces of masses `m` and `3m`, respectively, such that the smaller mass comes to rest. The larger mass finally falls at a distance `x` from the point of projection, where `x` is equal to

Options

A. `(2R)/3`
B. `(7R)/6`
C. `(5R)/4`
D. none of these

Explanation

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the Problem
A particle of mass `4m` is projected and reaches a horizontal range `R`. At the highest point of its trajectory, it breaks into two pieces of masses `m` and `3m`. The smaller mass `m` comes to rest, and we need to find the distance `x` where the larger mass `3m` lands.

### Step 2: Analyze the Situation at the Highest Point
At the highest point of the projectile's path, the horizontal distance traveled is `R/2`. The particle breaks into two masses: `m` (which comes to rest) and `3m` (which continues moving).

### Step 3: Use the Center of Mass Concept
The center of mass (CM) of the system will continue to move in the same horizontal line since there are no external horizontal forces acting on it. The position of the center of mass can be calculated using the formula:
\[ x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \]
where:
- \(m_1 = m\) (mass of the smaller piece),
- \(x_1 = \frac{R}{2}\) (position of the smaller piece, which comes to rest),
- \(m_2 = 3m\) (mass of the larger piece),
- \(x_2 = x\) (position where the larger mass lands).

### Step 4: Set Up the Equation for the Center of Mass
Since the center of mass must remain at the same horizontal position (which is `R`), we can set up the equation:
\[ R = \frac{m \cdot \frac{R}{2} + 3m \cdot x}{m + 3m} \]

### Step 5: Simplify the Equation
Substituting the values into the equation:
\[ R = \frac{m \cdot \frac{R}{2} + 3m \cdot x}{4m} \]
Cancelling out the mass `m` from the numerator and denominator gives:
\[ R = \frac{\frac{R}{2} + 3x}{4} \]

### Step 6: Multiply Both Sides by 4
To eliminate the fraction, multiply both sides by 4:
\[ 4R = \frac{R}{2} + 3x \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ 3x = 4R - \frac{R}{2} \]
Converting \(4R\) to a common denominator:
\[ 4R = \frac{8R}{2} \]
Thus,
\[ 3x = \frac{8R}{2} - \frac{R}{2} = \frac{7R}{2} \]

### Step 8: Solve for x
Now, divide both sides by 3:
\[ x = \frac{7R}{6} \]

### Final Answer
The distance `x` from the point of projection where the larger mass lands is:
\[ x = \frac{7R}{6} \]