A bullet of mass 0.1 kg moving horizontally with speed 400 m s$$^{-1}$$ hits a wooden block of mass 3.9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is

We have a bullet of mass $$m = 0.1$$ kg travelling at $$u = 400$$ m/s that gets embedded in a block of mass $$M = 3.9$$ kg. The combined system then slides a distance $$s = 20$$ m before coming to rest.

Since the bullet gets embedded in the block, this is a perfectly inelastic collision. By conservation of momentum:

$$mu = (m + M)v$$

$$0.1 \times 400 = (0.1 + 3.9) \times v$$

$$40 = 4v$$

$$v = 10 \text{ m/s}$$

Now, after the collision, the block-bullet system decelerates due to friction and comes to rest. Using the kinematic equation $$v^2 = u^2 - 2as$$ with final velocity zero:

$$0 = (10)^2 - 2a(20)$$

$$a = \frac{100}{40} = 2.5 \text{ m/s}^2$$

Since friction is the only horizontal force causing this deceleration, we have $$a = \mu g$$. So:

$$2.5 = \mu \times 10$$

$$\mu = 0.25$$

Hence, the correct answer is Option 4.