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 _.

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(2) 0.25. Idea: two stages, and they need different conservation laws. The embedding is a perfectly inelastic collision — momentum is conserved, energy is not. The slide afterwards is pure friction. Stage 1, the collision. The bullet buries itself, so the combined mass is 0.1+3.9=4.0 kg: 0.1×400=4.0× v, so v=10 m s⁻¹. Stage 2, the slide. Friction decelerates the block at μ g over 20 m: v²=2(μ g)s 100=2μ(10)(20)=400μ μ=0.25. Using the bullet's 400 m s⁻¹ in stage 2 is the trap: almost all of that speed is lost in the collision itself, along with 97.5% of the kinetic energy.

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 _____.

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