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, we will follow these steps:

### Step 1: Understand the scenario
We have a simple pendulum with a wooden bob of mass \( m = 1 \, \text{kg} \) and length \( L = 1 \, \text{m} \). A bullet of mass \( m_b = 10^{-2} \, \text{kg} \) strikes the bob with a speed \( v_b = 2 \times 10^2 \, \text{m/s} \). We need to find the height \( h \) to which the bob rises after being struck by the bullet.

### Step 2: Apply conservation of momentum
When the bullet strikes the bob, we can use the conservation of momentum to find the velocity \( v \) of the combined system (bob + bullet) immediately after the collision.

The momentum before the collision is equal to the momentum after the collision:

\[ m_b \cdot v_b = (m + m_b) \cdot v \]

Substituting the values:

\[ 10^{-2} \cdot 2 \times 10^2 = (1 + 10^{-2}) \cdot v \]

Calculating the left side:

\[ 2 = (1.01) \cdot v \]

Now, solve for \( v \):

\[ v = \frac{2}{1.01} \approx 1.9802 \, \text{m/s} \]

### Step 3: Apply conservation of energy
Next, we apply the conservation of energy to find the height \( h \) to which the bob rises. Initially, the kinetic energy of the system right after the collision will convert into potential energy at the highest point of the swing.

The kinetic energy (KE) just after the collision is given by:

\[ KE = \frac{1}{2} (m + m_b) v^2 \]

Substituting the values:

\[ KE = \frac{1}{2} (1 + 10^{-2}) (1.9802)^2 \]

Calculating \( KE \):

\[ KE = \frac{1}{2} \cdot 1.01 \cdot (3.9208) \approx 1.979 \, \text{J} \]

At the highest point, all this kinetic energy will convert to potential energy (PE):

\[ PE = (m + m_b) g h \]

Setting \( KE = PE \):

\[ 1.979 = (1.01) \cdot 10 \cdot h \]

Solving for \( h \):

\[ h = \frac{1.979}{10.1} \approx 0.196 \, \text{m} \]

### Final Answer
The height to which the bob rises before swinging back is approximately \( 0.196 \, \text{m} \) or \( 0.2 \, \text{m} \).

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