A particle is projected such that its horizontal range would be R. At the highest point, the particle breaks into two identical parts P and Q. If P comes to rest, the horizontal distance of point from the point of projection (i.e. origin) where the particle Q lands on the ground is

Options

A. 1.5 R
B. R
C. 2.5 R
D. 2 R

Answer

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

### Step 1: Understand the Initial Conditions
A particle is projected with a horizontal range \( R \). At the highest point of its trajectory, the particle breaks into two identical parts, \( P \) and \( Q \). The part \( P \) comes to rest, while part \( Q \) continues moving.

### Step 2: Determine the Velocity at the Highest Point
At the highest point, the vertical component of the velocity is zero, and the horizontal component remains \( v \cos \theta \), where \( v \) is the initial velocity and \( \theta \) is the angle of projection.

### Step 3: Apply Conservation of Momentum
Before the explosion, the total momentum in the horizontal direction is:
\[ m \cdot v \cos \theta \]
After the explosion, part \( P \) has a mass of \( \frac{m}{2} \) and comes to rest, while part \( Q \) has a mass of \( \frac{m}{2} \) and moves with a velocity \( v_1 \). By conservation of momentum:
\[ m \cdot v \cos \theta = 0 + \frac{m}{2} \cdot v_1 \]
This simplifies to:
\[ v \cos \theta = \frac{1}{2} v_1 \]
Thus,
\[ v_1 = 2 v \cos \theta \]

### Step 4: Calculate the Time of Flight for Part \( Q \)
The maximum height \( h_{\text{max}} \) reached by the particle can be expressed as:
\[ h_{\text{max}} = \frac{v^2 \sin^2 \theta}{2g} \]
The time taken to fall from this height to the ground can be calculated using the equation of motion:
\[ s = ut + \frac{1}{2} g t^2 \]
Here, \( s = h_{\text{max}} \), \( u = 0 \), and \( a = g \):
\[ \frac{v^2 \sin^2 \theta}{2g} = \frac{1}{2} g t^2 \]
Solving for \( t \):
\[ t^2 = \frac{v^2 \sin^2 \theta}{g^2} \]
\[ t = \frac{v \sin \theta}{g} \]

### Step 5: Calculate the Horizontal Distance Traveled by Part \( Q \)
The horizontal distance \( d \) traveled by part \( Q \) during the time \( t \) is given by:
\[ d = v_1 \cdot t \]
Substituting \( v_1 \) and \( t \):
\[ d = (2 v \cos \theta) \left( \frac{v \sin \theta}{g} \right) \]
\[ d = \frac{2 v^2 \sin \theta \cos \theta}{g} \]

### Step 6: Relate the Distance to the Range \( R \)
We know that the range \( R \) of the projectile is given by:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can express \( d \) in terms of \( R \):
\[ d = \frac{2 v^2 \sin \theta \cos \theta}{g} = \frac{R}{2} \]

### Step 7: Total Horizontal Distance from the Origin
At the highest point, the particle has traveled a horizontal distance of \( \frac{R}{2} \). After the explosion, part \( Q \) travels an additional distance of \( d \):
\[ \text{Total distance} = \frac{R}{2} + d = \frac{R}{2} + \frac{R}{2} = R \]

### Final Result
The total horizontal distance from the point of projection where part \( Q \) lands is:
\[ \text{Distance from origin} = R + \frac{R}{2} = \frac{3R}{2} \]

### Conclusion
Thus, the horizontal distance of point \( Q \) from the point of projection is \( \frac{3R}{2} \).

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