A bullet of mass 10 g is fired horizontally with a speed of 400 m/s into a stationary wooden block of mass 390 g resting on a smooth horizontal surface. The bullet penetrates 20 cm into the block and comes to rest inside it as the block begins to slide. The average resistive force exerted by the wooden block on the bullet is:

Step-by-step solution

By conservation of linear momentum along the horizontal direction, the common velocity v of the block and bullet after penetration is: mu = (m + M)v v = mu m + M Given m = 0.01 kg, u = 400 m/s, and M = 0.39 kg: v = 0.01 400 0.01 + 0.39 = 4 0.4 = 10 m/s The initial kinetic energy of the system is: K_i = 1 2 mu^2 = 1 2 0.01 (400)^2 = 800 J The final kinetic energy of the system is: K_f = 1 2 (m + M)v^2 = 1 2 0.4 (10)^2 = 20 J The loss in kinetic energy is equal to the work done by the internal resistive force F over