A bullet is fired horizontally and gets embedded in a block kept on a table. If table is frictionless, then

Options

A. kinetic energy is conserved
B. momentum is conserved
C. both a and b
D. none of these

Explanation

To solve the problem of a bullet being fired horizontally and embedding itself in a block on a frictionless table, we can follow these steps:

### Step 1: Understand the Scenario
A bullet is fired horizontally towards a block resting on a frictionless table. When the bullet embeds itself in the block, they move together as a single system.

**Hint:** Visualize the scenario by drawing a diagram of the bullet and block on a frictionless table.

### Step 2: Identify the Type of Collision
Since the bullet becomes embedded in the block, this is an example of a perfectly inelastic collision. In such collisions, the two objects stick together after the impact.

**Hint:** Remember that inelastic collisions involve objects sticking together, while elastic collisions do not.

### Step 3: Apply the Law of Conservation of Momentum
In the absence of external forces (which is the case here since the table is frictionless), the total momentum before the collision is equal to the total momentum after the collision.

- Let the mass of the bullet be \( m_b \) and its initial velocity be \( U \).
- Let the mass of the block be \( m_{block} \) and its initial velocity be \( 0 \) (since it is at rest).

The momentum before the collision is:
\[ P_{initial} = m_b \cdot U + m_{block} \cdot 0 = m_b \cdot U \]

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

According to the conservation of momentum:
\[ m_b \cdot U = (m_b + m_{block}) \cdot V \]

**Hint:** Make sure to set up the momentum equation correctly, considering the masses and velocities before and after the collision.

### Step 4: Solve for the Common Velocity \( V \)
From the momentum conservation equation, we can solve for \( V \):
\[ V = \frac{m_b \cdot U}{m_b + m_{block}} \]

**Hint:** This equation shows how the initial momentum of the bullet is distributed between the bullet and block after the collision.

### Step 5: Analyze Kinetic Energy
In a perfectly inelastic collision, kinetic energy is not conserved. Some kinetic energy is transformed into other forms of energy (like heat and sound).

- Initial kinetic energy of the bullet:
\[ KE_{initial} = \frac{1}{2} m_b U^2 \]

- Final kinetic energy after the collision:
\[ KE_{final} = \frac{1}{2} (m_b + m_{block}) V^2 \]

Since \( V \) is less than \( U \), it follows that:
\[ KE_{final} < KE_{initial} \]

**Hint:** Remember that kinetic energy is only conserved in elastic collisions.

### Conclusion
- Momentum is conserved in this scenario.
- Kinetic energy is not conserved due to the nature of the collision.

Thus, the correct answer to the question is that momentum is conserved, while kinetic energy is not conserved.

**Final Answer:** Momentum is conserved; kinetic energy is not conserved.