A bullet ofmass m is fired with a velocity `10m//s` at angle `theta` with horizontal. At the hightest point of its trajectory, it collsides head-on with a bob of mass `3m` suspended by a massless string of length `2//5m` and gets embedded in the bob. After th collision the string moves through an angle of `60^(@)`.
The horizontal coordinate of the intital position of the bob w.r.t. the point of firing of the bullet is n

Options

A. `(9)/(5)m`
B. `(24)/(5)m`
C. `(9)/(4)m`
D. None of these

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Understand the Problem
We have a bullet of mass \( m \) fired at a velocity of \( 10 \, \text{m/s} \) at an angle \( \theta \) with the horizontal. At the highest point of its trajectory, it collides with a bob of mass \( 3m \) suspended by a massless string of length \( \frac{2}{5} \, \text{m} \). After the collision, the string makes an angle of \( 60^\circ \) with the vertical. We need to find the horizontal coordinate of the initial position of the bob with respect to the point of firing of the bullet.

### Step 2: Draw the Diagram
Draw a diagram showing the bullet's trajectory, the highest point of the bullet's path, and the bob hanging from the string. Mark the angle \( \theta \) and the angle \( 60^\circ \) after the collision.

### Step 3: Find the Horizontal Distance at the Highest Point
At the highest point of its trajectory, the horizontal distance \( x \) from the firing point to the bob can be expressed in terms of the range \( R \) of the bullet:
\[ x = \frac{R}{2} \]
where \( R \) is the total range of the bullet.

### Step 4: Calculate the Range of the Bullet
The range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u = 10 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \).

### Step 5: Use Conservation of Momentum for Collision
Before the collision, the momentum of the bullet is:
\[ p_{\text{initial}} = m \cdot (10 \cos \theta) \]
After the collision, the total mass is \( 4m \) (bullet + bob), and they move together with speed \( v \):
\[ p_{\text{final}} = 4m \cdot v \]
Using conservation of momentum:
\[ m \cdot (10 \cos \theta) = 4m \cdot v \]
This simplifies to:
\[ v = \frac{10 \cos \theta}{4} = \frac{5 \cos \theta}{2} \]

### Step 6: Apply Conservation of Energy
The change in kinetic energy equals the change in potential energy:
\[ \Delta KE = \Delta PE \]
At the highest point, the kinetic energy is:
\[ KE = \frac{1}{2} \cdot 4m \cdot v^2 \]
The potential energy change as the bob rises to height \( h \) is:
\[ PE = 4mgh \]
Setting these equal gives:
\[ \frac{1}{2} \cdot 4m \cdot v^2 = 4mg \cdot h \]
This simplifies to:
\[ \frac{1}{2} v^2 = gh \]

### Step 7: Determine Height \( h \)
The height \( h \) can be expressed in terms of the string length \( l \) and the angle \( 60^\circ \):
\[ h = l - l \cos(60^\circ) = l(1 - \frac{1}{2}) = \frac{l}{2} \]
Substituting \( l = \frac{2}{5} \):
\[ h = \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{5} \, \text{m} \]

### Step 8: Substitute \( h \) into Energy Equation
Substituting \( h \) into the energy equation:
\[ \frac{1}{2} v^2 = g \cdot \frac{1}{5} \]
Substituting \( g = 10 \):
\[ \frac{1}{2} v^2 = 2 \Rightarrow v^2 = 4 \Rightarrow v = 2 \, \text{m/s} \]

### Step 9: Find \( \cos \theta \) and \( \sin \theta \)
Using the momentum equation:
\[ v = \frac{5 \cos \theta}{2} \Rightarrow 2 = \frac{5 \cos \theta}{2} \Rightarrow \cos \theta = \frac{4}{5} \]
Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \Rightarrow \sin^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = \frac{9}{25} \Rightarrow \sin \theta = \frac{3}{5} \]

### Step 10: Calculate the Range \( R \)
Now we can calculate the range \( R \):
\[ R = \frac{u^2 \sin(2\theta)}{g} = \frac{100 \cdot 2 \cdot \frac{3}{5} \cdot \frac{4}{5}}{10} = \frac{100 \cdot \frac{24}{25}}{10} = 96 \, \text{m} \]

### Step 11: Find \( x \)
Finally, we find \( x \):
\[ x = \frac{R}{2} = \frac{96}{2} = 48 \, \text{m} \]

### Final Answer
The horizontal coordinate of the initial position of the bob with respect to the point of firing of the bullet is \( 48 \, \text{m} \).

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