A bag of sand of mass $$9.8 \text{ kg}$$ is suspended by a rope. A bullet of $$200 \text{ g}$$ travelling with speed $$10 \text{ m s}^{-1}$$ gets embedded in it, then loss of kinetic energy will be

A bullet of mass $$m = 0.2 \text{ kg}$$ moving at $$v = 10 \text{ m s}^{-1}$$ embeds in a sandbag of mass $$M = 9.8 \text{ kg}$$. This is a perfectly inelastic collision.

Conservation of momentum yields:

$$mv = (m + M)V$$ $$0.2 \times 10 = (0.2 + 9.8) \times V$$ $$2 = 10V$$ $$V = 0.2 \text{ m s}^{-1}$$

We now calculate the kinetic energies before and after the collision.

Initial kinetic energy (only the bullet is moving):

$$KE_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.2 \times 10^2 = 10 \text{ J}$$

Final kinetic energy (combined system):

$$KE_f = \frac{1}{2}(m + M)V^2 = \frac{1}{2} \times 10 \times (0.2)^2 = \frac{1}{2} \times 10 \times 0.04 = 0.2 \text{ J}$$

The loss in kinetic energy is therefore:

$$\Delta KE = KE_i - KE_f = 10 - 0.2 = 9.8 \text{ J}$$

Hence, the correct answer is Option B: $$9.8 \text{ J}$$.