A bullet of mass 0.1 kg moving horizontally with a speed of 100 m s ⁻¹ strikes and embeds itself into a block of mass 1.9 kg resting on a rough horizontal surface. The block is attached to one end of a relaxed horizontal spring of spring constant 160 N m ⁻¹ , the other end of which is fixed to a vertical wall. The coefficient of kinetic friction between the block and the surface is 0.5 . The maximum compression of th

Step-by-step solution

By conservation of linear momentum during the perfectly inelastic collision: m u = (m + M) v 0.1 100 = (0.1 + 1.9) v 10 = 2 v v = 5 m s ⁻¹ Initial kinetic energy of the combined mass just after the collision: K = 1 2 (m + M)v^2 = 1 2 (2)(5)^2 = 25 J Let the maximum compression of the spring be x . By the work-energy theorem, the initial kinetic energy is dissipated by friction and stored as elastic potential energy in the spring: K = 1 2 kx^2 + (m + M)gx 25 = 1 2 (160)x^2 + 0.5(2)(10)x 25 = 80x^2 + 10x Dividing by