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.
What is the momentum transferred to the bob by the bullet?

Options

A. `mv`
B. `(mv)/2`
C. `(Mv)/2`
D. Mv

Explanation

To solve the problem of the momentum transferred to the bob by the bullet, we will follow these steps:

### Step 1: Understand the initial and final momentum of the bullet
The bullet has an initial momentum before it hits the bob, which can be calculated using the formula:
\[ \text{Initial momentum of bullet} = m \cdot v \]
where \( m \) is the mass of the bullet and \( v \) is its initial speed.

### Step 2: Calculate the final momentum of the bullet after it exits the bob
After passing through the bob, the bullet exits with a speed of \( \frac{v}{2} \). Thus, the final momentum of the bullet is:
\[ \text{Final momentum of bullet} = m \cdot \frac{v}{2} \]

### Step 3: Determine the change in momentum of the bullet
The change in momentum of the bullet, which is the momentum transferred to the bob, can be calculated as:
\[ \text{Change in momentum} = \text{Final momentum} - \text{Initial momentum} \]
Substituting the values we calculated:
\[ \text{Change in momentum} = m \cdot \frac{v}{2} - m \cdot v \]
\[ = m \cdot \frac{v}{2} - m \cdot \frac{2v}{2} \]
\[ = m \cdot \left(\frac{v}{2} - \frac{2v}{2}\right) \]
\[ = m \cdot \left(-\frac{v}{2}\right) \]
\[ = -\frac{mv}{2} \]

### Step 4: Calculate the momentum transferred to the bob
The momentum transferred to the bob is the negative of the change in momentum of the bullet, which is:
\[ \text{Momentum transferred to bob} = \frac{mv}{2} \]

### Final Answer
Thus, the momentum transferred to the bob by the bullet is:
\[ \frac{mv}{2} \]