1. Given data

• Mass of block: $m_1 = 1.9\,\text{kg}$
• Mass of bullet: $m_2 = 0.1\,\text{kg}$
• Speed of bullet before collision: $u = 20\,\text{m/s}$
• Height of table: $h = 1\,\text{m}$
• Acceleration due to gravity: $g = 10\,\text{m/s}^2$

The bullet sticks to the block, so this is a perfectly inelastic collision.

2. Find velocity just after collision

Since the collision happens horizontally, conserve horizontal momentum:

$m_2 u = (m_1 + m_2) v$

Substitute values:

$0.1 \times 20 = (1.9 + 0.1)v$

$2 = 2v$

$v = 1\,\text{m/s}$

So just after collision, the combined mass $2\,\text{kg}$ moves horizontally with speed $1\,\text{m/s}$.

3. Kinetic energy just after collision

$K_1 = \frac{1}{2}(m_1+m_2)v^2$

$K_1 = \frac{1}{2}(2)(1)^2 = 1\,\text{J}$

4. Gain in kinetic energy during fall

As the combined system falls through height $h=1\,\text{m}$, gravitational potential energy converts into kinetic energy:

$\Delta K = (m_1+m_2)gh$

$\Delta K = 2 \times 10 \times 1 = 20\,\text{J}$

5. Kinetic energy just before striking the floor

$K_{\text{final}} = K_1 + \Delta K$

$K_{\text{final}} = 1 + 20 = 21\,\text{J}$

6. Check options

• A: $23\,\text{J}$
• B: $21\,\text{J}$
• C: $20\,\text{J}$
• D: $19\,\text{J}$

Hence the correct option is B.