A 10 gm bullet is fired from a rifle horizontally into a 5 kg block of wood suspended by a string and the bullet gets embedded in the block. The impact causes the block to swing to a height of `2.5 cm` above its initial level. The velocity of the bullet is

Options

A. `286.8 m/sec`
B. `350.7 m/sec`
C. `1000 m/sec`
D. `523 m/sec`

Explanation

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

### Step 1: Understand the Problem
We have a bullet of mass \( m_b = 10 \, \text{g} = 0.01 \, \text{kg} \) that is fired horizontally into a block of wood with mass \( m_B = 5 \, \text{kg} \). The bullet gets embedded in the block, and together they swing to a height of \( h = 2.5 \, \text{cm} = 0.025 \, \text{m} \).

### Step 2: Apply Conservation of Momentum
Before the collision, the block is at rest, so its initial momentum is zero. The momentum of the bullet before the collision is:
\[ p_{\text{initial}} = m_b \cdot u \]
where \( u \) is the velocity of the bullet.

After the collision, the bullet and block move together with a common velocity \( V \):
\[ p_{\text{final}} = (m_b + m_B) \cdot V \]

By conservation of momentum:
\[ m_b \cdot u = (m_b + m_B) \cdot V \]

### Step 3: Calculate the Potential Energy at Maximum Height
When the block swings to a height \( h \), all the kinetic energy is converted into potential energy:
\[ PE = (m_b + m_B) \cdot g \cdot h \]
where \( g = 9.8 \, \text{m/s}^2 \).

### Step 4: Set Up the Energy Conservation Equation
The kinetic energy just after the collision is equal to the potential energy at the maximum height:
\[ \frac{1}{2} (m_b + m_B) V^2 = (m_b + m_B) \cdot g \cdot h \]

### Step 5: Simplify the Equation
We can cancel \( (m_b + m_B) \) from both sides (as long as it's not zero):
\[ \frac{1}{2} V^2 = g \cdot h \]
Thus,
\[ V^2 = 2gh \]

### Step 6: Calculate \( V \)
Substituting the values:
\[ V^2 = 2 \cdot 9.8 \cdot 0.025 \]
\[ V^2 = 0.49 \implies V = \sqrt{0.49} = 0.7 \, \text{m/s} \]

### Step 7: Substitute \( V \) Back to Find \( u \)
Now we substitute \( V \) back into the momentum equation:
\[ m_b \cdot u = (m_b + m_B) \cdot V \]
\[ 0.01 \cdot u = (0.01 + 5) \cdot 0.7 \]
\[ 0.01 \cdot u = 5.01 \cdot 0.7 \]
\[ 0.01 \cdot u = 3.507 \]
\[ u = \frac{3.507}{0.01} = 350.7 \, \text{m/s} \]

### Final Answer
The velocity of the bullet is \( u = 350.7 \, \text{m/s} \).

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