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 from point of firing) of the lighter part is:

Options

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

Answer

To solve the problem step by step, we will analyze the motion of the projectile before and after the explosion.

### Step 1: Understand the Initial Conditions
The projectile is fired with a horizontal range of 4 km. This means that the total horizontal distance covered by the projectile before it hits the ground is 4 km.

### Step 2: Determine the Velocity at the Highest Point
At the highest point of its trajectory, the vertical component of the velocity is zero, and the horizontal component remains constant. The horizontal velocity \( V_x \) at the highest point is given by:
\[ V_x = u \cos \theta \]
where \( u \) is the initial speed and \( \theta \) is the angle of projection.

### Step 3: Apply Conservation of Momentum at the Explosion
When the projectile explodes at the highest point, it splits into two parts with masses \( \frac{M}{4} \) and \( \frac{3M}{4} \). The heavier part (mass \( \frac{3M}{4} \)) falls vertically with zero initial speed. According to the conservation of momentum, the horizontal momentum before the explosion must equal the horizontal momentum after the explosion.

Before the explosion:
\[ \text{Initial momentum} = M \cdot V_x = M \cdot u \cos \theta \]

After the explosion:
\[ \text{Final momentum} = \frac{3M}{4} \cdot 0 + \frac{M}{4} \cdot V' \]
where \( V' \) is the horizontal velocity of the lighter part after the explosion.

Setting the initial momentum equal to the final momentum:
\[ M \cdot u \cos \theta = \frac{M}{4} \cdot V' \]

### Step 4: Solve for \( V' \)
Cancelling \( M \) from both sides and solving for \( V' \):
\[ u \cos \theta = \frac{1}{4} V' \]
\[ V' = 4u \cos \theta \]

### Step 5: Calculate the Time of Free Fall for the Heavier Part
The time taken for the heavier part to fall from the maximum height to the ground can be calculated using the formula for free fall:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
The time \( t \) to fall this height is given by:
\[ H = \frac{1}{2} g t^2 \]
Setting the two expressions for height equal gives:
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} g t^2 \]
Solving for \( t \):
\[ t = \frac{u \sin \theta}{g} \]

### Step 6: Calculate the Horizontal Range of the Lighter Part
The horizontal distance \( X' \) traveled by the lighter part after the explosion can be calculated as:
\[ X' = V' \cdot t \]
Substituting \( V' \) and \( t \):
\[ X' = (4u \cos \theta) \cdot \left(\frac{u \sin \theta}{g}\right) \]
\[ X' = \frac{4u^2 \sin \theta \cos \theta}{g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ X' = \frac{2u^2 \sin 2\theta}{g} \]

### Step 7: Relate to the Range of the Projectile
The horizontal range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Thus, we can express \( X' \) in terms of \( R \):
\[ X' = 2R \]

### Step 8: Total Horizontal Distance from the Firing Point
The total horizontal distance from the firing point for the lighter part is the sum of the distance traveled before the explosion (which is half the range) and the distance traveled after the explosion:
\[ \text{Total Distance} = \frac{R}{2} + X' = \frac{R}{2} + 2R = \frac{R}{2} + 2R = \frac{R}{2} + \frac{4R}{2} = \frac{5R}{2} \]
Given \( R = 4 \text{ km} \):
\[ \text{Total Distance} = \frac{5 \times 4}{2} = 10 \text{ km} \]

### Final Answer
The horizontal range of the lighter part is **10 km**.

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