A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass `10^(-2)` kg moving with a speed of `2 xx 10^(2) m s^(-1)`. The height to which the bob rises before swinging back is (Take `g = 10 m s^(-2)`)

Options

A. 0.2 m
B. 0.6 m
C. 8 m
D. 1 m

Explanation

To solve the problem of how high the bob of a simple pendulum rises after being struck by a bullet, we can follow these steps:

### Step 1: Calculate the momentum of the bullet
The momentum \( p \) of the bullet can be calculated using the formula:
\[ p = m \times v \]
where:
- \( m \) is the mass of the bullet, \( 10^{-2} \) kg
- \( v \) is the velocity of the bullet, \( 2 \times 10^{2} \) m/s

Calculating:
\[ p = 10^{-2} \, \text{kg} \times 2 \times 10^{2} \, \text{m/s} = 2 \, \text{kg m/s} \]

### Step 2: Apply conservation of momentum
Before the collision, only the bullet is moving. After the collision, both the bullet and the bob move together. Let \( V \) be the velocity of the bob and bullet after the collision.

Using conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
\[ 10^{-2} \, \text{kg} \times 2 \times 10^{2} \, \text{m/s} = (10^{-2} \, \text{kg} + 1 \, \text{kg}) \times V \]
\[ 2 = 1.01 \, \text{kg} \times V \]

### Step 3: Solve for \( V \)
Rearranging the equation gives:
\[ V = \frac{2}{1.01} \approx 1.9802 \, \text{m/s} \]

### Step 4: Use conservation of energy to find height
The kinetic energy of the bob and bullet right after the collision will convert into potential energy at the highest point of the swing. The kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} (m + M) V^2 \]
where \( m \) is the mass of the bullet and \( M \) is the mass of the bob.

The potential energy (PE) at the highest point is given by:
\[ PE = (m + M) g h \]

Setting KE equal to PE:
\[ \frac{1}{2} (10^{-2} + 1) V^2 = (10^{-2} + 1) g h \]

### Step 5: Cancel out the common terms
Since \( (10^{-2} + 1) \) appears on both sides, we can cancel it out:
\[ \frac{1}{2} V^2 = g h \]

### Step 6: Solve for height \( h \)
Rearranging gives:
\[ h = \frac{V^2}{2g} \]

Substituting \( V \approx 1.9802 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ h = \frac{(1.9802)^2}{2 \times 10} = \frac{3.9208}{20} \approx 0.196 \, \text{m} \]

### Step 7: Final answer
Thus, the height to which the bob rises before swinging back is approximately:
\[ h \approx 0.2 \, \text{m} \]