A bullet of mass `20 g` and moving with `600 m/s` collides with a block of mass `4 kg` hanging with the string. What is the velocity of bullet when it comes out of block, if block rises to height `0.2 m` after collision?

Options

A. `200 ms^(-1)`
B. `150 ms^(-1)`
C. `400 ms^(-1)`
D. `300 ms^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial momentum of the bullet.
The initial momentum of the bullet can be calculated using the formula:
\[ P_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}} \]

Where:
- \( m_{\text{bullet}} = 20 \, \text{g} = 0.02 \, \text{kg} \) (converted to kg)
- \( v_{\text{bullet}} = 600 \, \text{m/s} \)

Calculating:
\[ P_{\text{initial}} = 0.02 \, \text{kg} \times 600 \, \text{m/s} = 12 \, \text{kg m/s} \]

### Step 2: Use conservation of momentum.
According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]

The final momentum can be expressed as:
\[ P_{\text{final}} = m_{\text{bullet}} \times v + m_{\text{block}} \times v' \]

Where:
- \( v \) is the velocity of the bullet after it exits the block.
- \( v' \) is the velocity of the block after the collision.

Given:
- \( m_{\text{block}} = 4 \, \text{kg} \)

Thus, we can write:
\[ 12 = 0.02v + 4v' \]
This is our **Equation 1**.

### Step 3: Calculate the velocity of the block after the collision using conservation of energy.
The block rises to a height of \( h = 0.2 \, \text{m} \). The potential energy gained by the block is equal to the kinetic energy it had just after the collision.

Using the conservation of energy:
\[ \frac{1}{2} m v'^2 = mgh \]

Where:
- \( m = 4 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \) (approximate value for gravitational acceleration)
- \( h = 0.2 \, \text{m} \)

Cancelling \( m \) from both sides:
\[ \frac{1}{2} v'^2 = gh \]
\[ v'^2 = 2gh \]
\[ v' = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2 \, \text{m/s} \]

### Step 4: Substitute \( v' \) back into Equation 1.
Now, substitute \( v' = 2 \, \text{m/s} \) into Equation 1:
\[ 12 = 0.02v + 4(2) \]
\[ 12 = 0.02v + 8 \]
\[ 0.02v = 12 - 8 \]
\[ 0.02v = 4 \]
\[ v = \frac{4}{0.02} = 200 \, \text{m/s} \]

### Final Answer
The velocity of the bullet when it comes out of the block is \( 200 \, \text{m/s} \).

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