A bullet hits a lock kept at rest on a smooth horizontal surface andgets embedded into it. Which of the following does not change?

Options

A. linear momentum of the block
B. kinetic energy of the block
C. gravitational potential energy of the block
D. temperature of the block

Explanation

To solve the problem, we need to analyze the situation where a bullet strikes a block at rest on a smooth horizontal surface and becomes embedded in it. We will determine which physical quantity does not change during this interaction.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The bullet is moving with some initial velocity (let's call it \( v_b \)).
- The block is at rest, so its initial velocity is \( 0 \).
- The bullet has a certain mass \( m_b \), and the block has a mass \( m_{block} \).

2. **Conservation of Linear Momentum**:
- When the bullet hits the block and they stick together, we can apply the principle of conservation of linear momentum.
- The total momentum before the collision is equal to the total momentum after the collision.
- Mathematically, this can be expressed as:
\[ m_b \cdot v_b + m_{block} \cdot 0 = (m_b + m_{block}) \cdot v_f \]
- Here, \( v_f \) is the final velocity of the bullet-block system after the collision.

3. **Change in Kinetic Energy**:
- Before the collision, the kinetic energy of the bullet is given by:
\[ KE_{initial} = \frac{1}{2} m_b v_b^2 \]
- After the collision, the kinetic energy of the combined system (bullet + block) is:
\[ KE_{final} = \frac{1}{2} (m_b + m_{block}) v_f^2 \]
- Since some kinetic energy is transformed into other forms of energy (like heat due to deformation), the kinetic energy after the collision will be less than before. Therefore, the kinetic energy changes.

4. **Change in Temperature**:
- As the bullet embeds into the block, some kinetic energy is converted into heat, which can increase the temperature of the block. Thus, the temperature of the block changes.

5. **Gravitational Potential Energy**:
- The gravitational potential energy (GPE) is given by:
\[ PE = m \cdot g \cdot h \]
- In this scenario, since the block is on a smooth horizontal surface and there is no vertical displacement, the height \( h \) remains constant. Therefore, the gravitational potential energy does not change.

### Conclusion:
The quantity that does not change when the bullet hits the block and gets embedded into it is the **gravitational potential energy** of the block.

### Final Answer:
The gravitational potential energy does not change.

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