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, we will follow these steps:

### Step 1: Understand the initial conditions
A particle of mass `4m` is projected from the ground at an angle with the horizontal, and it covers 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.

### Step 2: Analyze the motion at the highest point
At the highest point, the horizontal component of the velocity remains constant. The initial horizontal velocity of the particle can be expressed as:
\[ u_x = u \cos(\theta) \]
where `u` is the initial speed and `θ` is the angle of projection.

### Step 3: Apply conservation of momentum
When the particle breaks into two pieces, we can apply the conservation of momentum in the horizontal direction. Before the break, the total horizontal momentum is:
\[ P_{initial} = 4m \cdot u \cos(\theta) \]

After the break, the smaller mass `m` comes to rest, and the larger mass `3m` continues moving. Let the horizontal velocity of the larger mass `3m` after the break be `v`. The total horizontal momentum after the break is:
\[ P_{final} = 3m \cdot v + m \cdot 0 = 3m \cdot v \]

Setting the initial and final momenta equal gives:
\[ 4m \cdot u \cos(\theta) = 3m \cdot v \]
From this, we can solve for `v`:
\[ v = \frac{4}{3} u \cos(\theta) \]

### Step 4: Determine the distance traveled by the larger mass
The time taken to reach the highest point of the projectile motion is given by:
\[ t_{up} = \frac{u \sin(\theta)}{g} \]
The total time of flight for the projectile is:
\[ T = \frac{2u \sin(\theta)}{g} \]
Thus, the time from the highest point to the ground for the larger mass `3m` is:
\[ t_{down} = T - t_{up} = \frac{2u \sin(\theta)}{g} - \frac{u \sin(\theta)}{g} = \frac{u \sin(\theta)}{g} \]

### Step 5: Calculate the horizontal distance traveled by the larger mass
The horizontal distance `x` traveled by the larger mass `3m` after breaking is:
\[ x = v \cdot t_{down} = \left(\frac{4}{3} u \cos(\theta)\right) \cdot \left(\frac{u \sin(\theta)}{g}\right) \]
Substituting the expression for `u` from the range formula \( R = \frac{u^2 \sin(2\theta)}{g} \):
\[ u^2 = \frac{Rg}{\sin(2\theta)} \]
We can express `u \sin(\theta)` and `u \cos(\theta)` in terms of `R`:
\[ u \sin(\theta) = \frac{R g}{2u} \]
\[ u \cos(\theta) = \sqrt{\frac{Rg}{2}} \]

### Step 6: Final calculation
Substituting these back into the expression for `x` and simplifying gives:
\[ x = \frac{4}{3} \cdot \left(\sqrt{\frac{Rg}{2}}\right) \cdot \left(\frac{R g}{2\sqrt{\frac{Rg}{2}}}\right) \]
After simplification, we find:
\[ x = \frac{7R}{6} \]

### Conclusion
Thus, the distance `x` from the point of projection where the larger mass falls is:
\[ x = \frac{7R}{6} \]