A bob of mass `m` attached with a string of length `l` tied to a point on ceiling is released from a position when its string is horizontal. At the bottom most point of its motion, an identical mass `m` gently stuck to it. Find the maximum angle from the vertical to which it rotates.

Options

A. (a) `cos^-1(2/3)`
B. (b) `cos^-1(3/4)`
C. (c) `cos^-1(1/4)`
D. (d) `60^@`

Explanation

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

### Step 1: Determine the velocity of the bob just before the collision
When the bob is released from the horizontal position, it falls a vertical distance equal to the length of the string, \( l \). The potential energy at the top is converted to kinetic energy at the bottom.

Using the formula for kinetic energy:
\[ KE = PE \]
\[ \frac{1}{2} m v^2 = mgh \]
where \( h = l \) (the height fallen). Thus,
\[ \frac{1}{2} m v^2 = mg l \]
Solving for \( v \):
\[ v^2 = 2gl \quad \Rightarrow \quad v = \sqrt{2gl} \]

### Step 2: Apply conservation of momentum at the moment of collision
When the identical mass \( m \) strikes the bob, we can apply the conservation of linear momentum. The initial momentum before the collision is:
\[ p_{initial} = mv + 0 = m\sqrt{2gl} \]
After the collision, both masses stick together and move with a common velocity \( v' \):
\[ p_{final} = (m + m)v' = 2mv' \]
Setting the initial momentum equal to the final momentum:
\[ m\sqrt{2gl} = 2mv' \]
Cancelling \( m \) (assuming \( m \neq 0 \)):
\[ v' = \frac{\sqrt{2gl}}{2} = \frac{1}{2}\sqrt{2gl} \]

### Step 3: Determine the height reached after the collision
After the collision, the combined mass rises to a height \( h \) due to the kinetic energy it has. Using the work-energy principle:
\[ KE_{initial} = PE_{final} \]
The initial kinetic energy after the collision is:
\[ KE = \frac{1}{2} (2m) v'^2 = 2m \left(\frac{1}{2}\sqrt{2gl}\right)^2 = 2m \cdot \frac{1}{4} \cdot 2gl = mgl \]
The potential energy at height \( h \) is:
\[ PE = (2m)gh \]
Setting these equal gives:
\[ mgl = 2mgh \]
Cancelling \( m \) (assuming \( m \neq 0 \)):
\[ gl = 2gh \quad \Rightarrow \quad h = \frac{l}{2} \]

### Step 4: Relate height to the angle \( \theta \)
From the geometry of the pendulum, we know:
\[ h = l(1 - \cos \theta) \]
Substituting \( h = \frac{l}{2} \):
\[ \frac{l}{2} = l(1 - \cos \theta) \]
Dividing both sides by \( l \) (assuming \( l \neq 0 \)):
\[ \frac{1}{2} = 1 - \cos \theta \]
Rearranging gives:
\[ \cos \theta = \frac{1}{2} \]

### Step 5: Solve for the angle \( \theta \)
Taking the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]
This corresponds to:
\[ \theta = 60^\circ \]

### Final Answer
The maximum angle from the vertical to which the bob rotates is \( 60^\circ \).

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