A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m s$$^{-1}$$ in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m s$$^{-2}$$. Assume there is no rotational motion and loss of energy after the collision is negligible.]

We have a block of mass $$M = 1.9\ \text{kg}$$ initially at rest at the edge of a table of height $$h = 1\ \text{m}$$. A bullet of mass $$m = 0.1\ \text{kg}$$ comes horizontally with speed $$u = 20\ \text{m s}^{-1}$$ and sticks to the block. Because the bullet embeds itself, the collision is perfectly inelastic.

During the very short collision time, external horizontal forces are negligible, so we apply conservation of linear momentum in the horizontal direction. The formula is

$$\text{Initial momentum} = \text{Final momentum}.$$

Initially only the bullet moves, so

$$m u = (m + M)\,v,$$

where $$v$$ is the common horizontal velocity of the combined system immediately after the collision.

Substituting the numbers, we get

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

$$2 = 2.0\,v$$

$$v = 1\ \text{m s}^{-1}.$$

This horizontal speed remains unchanged during the subsequent flight because air resistance is neglected.

Now the block-bullet system moves off the table edge with

horizontal speed $$v_x = 1\ \text{m s}^{-1},$$

vertical speed $$v_y = 0$$ at the moment it leaves the table, and falls through a vertical distance $$h = 1\ \text{m}$$ under gravity.

For the vertical motion, we use the kinematic relation for a body starting from rest:

$$v_y^2 = u_y^2 + 2 g h,$$

where $$u_y = 0$$ (no initial vertical speed). Hence

$$v_y^2 = 0 + 2 \times 10 \times 1 = 20,$$

so

$$v_y = \sqrt{20}\ \text{m s}^{-1} \approx 4.47\ \text{m s}^{-1}.$$

Thus, just before striking the floor, the system has two perpendicular components of velocity:

$$v_x = 1\ \text{m s}^{-1}, \quad v_y = \sqrt{20}\ \text{m s}^{-1}.$$

The magnitude of the resultant speed is obtained from Pythagoras, but for kinetic energy we can add the squares directly. The total kinetic energy is

$$ K = \dfrac{1}{2}\,(m + M)\,\left(v_x^2 + v_y^2\right). $$

Substituting,

$$ K = \dfrac{1}{2}\,(0.1 + 1.9)\,\left(1^2 + (\sqrt{20})^2\right) = \dfrac{1}{2}\times 2.0 \times (1 + 20) = 1 \times 21 = 21\ \text{J}. $$

Hence, the correct answer is Option A.