Question

A target with mass m=3.5 kg is attached to a spring constant k=200 N/m on a horizontal, frictionless surface. The target is struck by a bullet with mass of 20 g, traveling at a speed of v=550 m/s. Bullet embeds itself completely into target. Assuming that the target is initally at rest, what is maximum displacement of the tagret fromthe spring equilbrium position after collision? Assume instant collision occurs.

What percent of initial kinetic energy converted to non-mechanical forms of energy?

Answer 1

solution - uring from. me = w uring momentuan conseruation, velocity of mass +bullet system (M+m) uf V 20x10 kg x 550 m/s = 3.125 m/s (3.5+ 0.020)ky So, wsing energy conseruetion 1 kz² = (M+m) e ? /2X 17.1875 1² 200 N/m - 17. 1875 6.414m 15) x = 41.4em 0.02 Energy loss- -0.0057 Kf ko £ (M + m) Vt² im ve? Mtm 3.52 9943 - 99.43% loss | -0.0057 -