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 angle `theta` is

Options

A. `53^(@)`
B. `37^(@)`
C. `45^(@)`
D. `30^(@)`

Explanation

To solve the problem step by step, we will use the principles of conservation of momentum and conservation of energy.

### Step 1: Understanding the scenario
A bullet of mass \( m \) is fired with 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. After the collision, the combined system moves through an angle of \( 60^\circ \).

### Step 2: Analyze the collision
At the highest point of the bullet's trajectory, its vertical component of velocity is zero, and only the horizontal component contributes to its momentum. The horizontal component of the bullet's velocity is given by:
\[ v_{x} = 10 \cos \theta \]

The momentum of the bullet just before the collision is:
\[ p_{\text{bullet}} = m \cdot v_{x} = m \cdot (10 \cos \theta) \]

The bob is initially at rest, so its momentum is:
\[ p_{\text{bob}} = 0 \]

After the collision, the bullet and bob move together with a combined mass of \( 4m \) and a common velocity \( v \). By conservation of momentum:
\[ m \cdot (10 \cos \theta) = (4m) \cdot v \]
Cancelling \( m \) from both sides:
\[ 10 \cos \theta = 4v \quad \text{(Equation 1)} \]

### Step 3: Analyze the motion after the collision
After the collision, the bob with the bullet rises to a height corresponding to the angle of \( 60^\circ \). We can find the height \( h \) using the length of the string \( L = \frac{2}{5} \, \text{m} \):
\[ h = L - L \cos(60^\circ) = L(1 - \cos(60^\circ)) = \frac{2}{5} \left(1 - \frac{1}{2}\right) = \frac{2}{5} \cdot \frac{1}{2} = \frac{1}{5} \, \text{m} \]

### Step 4: Apply conservation of energy
The kinetic energy just after the collision is converted into potential energy at the highest point:
\[ \frac{1}{2} (4m) v^2 = (4m) g h \]
Cancelling \( 4m \) from both sides:
\[ \frac{1}{2} v^2 = g h \]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( h = \frac{1}{5} \, \text{m} \):
\[ \frac{1}{2} v^2 = 10 \cdot \frac{1}{5} = 2 \]
Thus,
\[ v^2 = 4 \quad \Rightarrow \quad v = 2 \, \text{m/s} \]

### Step 5: Substitute \( v \) back into Equation 1
Substituting \( v = 2 \, \text{m/s} \) into Equation 1:
\[ 10 \cos \theta = 4 \cdot 2 \]
\[ 10 \cos \theta = 8 \quad \Rightarrow \quad \cos \theta = \frac{8}{10} = \frac{4}{5} \]

### Step 6: Find \( \theta \)
Now, we find \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{4}{5}\right) \]
Calculating \( \theta \):
\[ \theta \approx 36.87^\circ \quad \text{(or approximately } 37^\circ\text{)} \]

### Final Answer
The angle \( \theta \) is approximately \( 37^\circ \).

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