1. Find the speed just after collision using conservation of momentum

Since the bullet gets embedded in the block, the collision is perfectly inelastic.

Given:

• Bullet mass $m = 0.1\,\text{kg}$
• Bullet speed $u = 400\,\text{m s}^{-1}$
• Block mass $M = 3.9\,\text{kg}$

Initial momentum: $p_i = mu = 0.1 \times 400 = 40\,\text{kg m s}^{-1}$

Let $v$ be the common speed just after collision. Then $(m+M)v = 40$ $(0.1+3.9)v = 40$ $4v = 40$ $v = 10\,\text{m s}^{-1}$

2. Use work-energy principle for motion after collision

After collision, the combined mass is $m+M = 4\,\text{kg}$

It moves $20\,\text{m}$ on a rough horizontal surface and comes to rest.

Initial kinetic energy of the combined body: $K = \frac{1}{2}(4)(10)^2 = 200\,\text{J}$

This entire kinetic energy is lost due to work done by friction.

Friction force: $f = \mu N = \mu (4g) = \mu (4 \times 10) = 40\mu$

Work done by friction over $20\,\text{m}$: $W = f s = 40\mu \times 20 = 800\mu$

Since this equals the initial kinetic energy: $800\mu = 200$ $\mu = \frac{200}{800} = 0.25$

3. Check options

• A: $0.65$ ❌
• B: $0.25$ ✅
• C: $0.50$ ❌
• D: $0.90$ ❌

Therefore, the correct answer is B.