A pendulum consists of a wooden bob of mass `M` and length `l`. A bullet of mass `m` is fired towards the pendulum with a speed `v`. The bullet emerges immediately out of the bob from the other side with a speed of `v//2` and the bob starts rising. Assume no loss of mass of bob takes place due to penetration.
If the bob stops where the string becomes horizontal then `v` is

Options

A. `(2M)/m sqrt(3gl)`
B. `(2M)/msqrt(5gl)`
C. `(2M)/msqrt(gl)`
D. `(2M)/msqrt(2gl)`

Explanation

To solve the problem, we will use the principles of conservation of momentum and conservation of energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Apply Conservation of Momentum
Initially, the bullet has momentum, and the bob is at rest. The initial momentum of the system can be expressed as:
\[ P_{\text{initial}} = m \cdot v \]
where \( m \) is the mass of the bullet and \( v \) is its initial velocity.

After the bullet passes through the bob, the bob moves with a velocity \( v' \) and the bullet continues with a velocity of \( \frac{v}{2} \). The final momentum can be expressed as:
\[ P_{\text{final}} = M \cdot v' + m \cdot \frac{v}{2} \]
where \( M \) is the mass of the bob.

By conservation of momentum:
\[ m \cdot v = M \cdot v' + m \cdot \frac{v}{2} \]

### Step 2: Rearranging the Momentum Equation
Rearranging the equation gives:
\[ M \cdot v' = m \cdot v - m \cdot \frac{v}{2} \]
\[ M \cdot v' = m \cdot \frac{v}{2} \]

Now, solve for \( v' \):
\[ v' = \frac{m \cdot \frac{v}{2}}{M} \]
\[ v' = \frac{mv}{2M} \] (Equation 1)

### Step 3: Apply Conservation of Energy
Next, we apply the conservation of energy. The kinetic energy of the bob just after the bullet passes through it will convert into potential energy when the bob reaches its maximum height \( h = l \).

The kinetic energy of the bob is:
\[ KE = \frac{1}{2} M (v')^2 \]

The potential energy at height \( l \) is:
\[ PE = M \cdot g \cdot l \]

By conservation of energy:
\[ \frac{1}{2} M (v')^2 = M \cdot g \cdot l \]

### Step 4: Substitute \( v' \) into the Energy Equation
Substituting \( v' \) from Equation 1 into the energy equation:
\[ \frac{1}{2} M \left(\frac{mv}{2M}\right)^2 = M \cdot g \cdot l \]
\[ \frac{1}{2} M \cdot \frac{m^2 v^2}{4M^2} = M \cdot g \cdot l \]
\[ \frac{m^2 v^2}{8M} = M \cdot g \cdot l \]

### Step 5: Solve for \( v^2 \)
Multiplying both sides by \( 8M \):
\[ m^2 v^2 = 8M^2 g l \]
Now, divide by \( m^2 \):
\[ v^2 = \frac{8M^2 g l}{m^2} \]

### Step 6: Take the Square Root
Taking the square root gives:
\[ v = \frac{2M}{m} \sqrt{2g l} \]

### Final Answer
Thus, the value of \( v \) is:
\[ v = \frac{2M}{m} \sqrt{2g l} \]