A 30-g bullet traveling horizontally at 300 m/s strikes a 1.0-kg block which is attached to a horizontal spring with a force constant of 2000 N/m and rests on a frictionless horizontal surface. The spring is on the far side of the block and aligned with the direction of travel of the bullet. The bullet becomes embedded in the block and, as a result of the impact, and the block slides against the spring. How far is the spring compressed before it reverses its direction of travel?
A)0.20 m
B)0.58 m
C)1.2 m
D)0.40 m
E)0.039 m

Question

Quizplus · Accepted Answer

Momentum is conserved in the collision, so the block and bullet move off together with a speed of 27 m/s. The kinetic energy of the system is then:$$K=\frac{1}{2}(1.03	ext{ kg})(27	ext{ m/s})^2=365	ext{ J}$$This energy is stored in the spring when it is compressed a distance x:$$U=\frac{1}{2}kx^2=365	ext{ J}$$Solving for x:$$x=\sqrt{\frac{2U}{k}}=\sqrt{\frac{2(365	ext{ J})}{2000	ext{ N/m}}}=0.20	ext{ m}$$

A 30-G Bullet Traveling Horizontally at 300 M/s Strikes a 1.0-Kg