Question

A 4.00 kg block sits at rest on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.275. Attached to the right side of the block is a spring which is also attached to a wall farther to the right. The spring has a spring constant of 220 N/m and is initially neither compressed nor stretched. A bullet of mass 50.0 grams is fired at the block from the left side. The bullet has a speed of 400 m/s when it hits the block. If the collision between the block and bullet is perfectly inelastic,

a. Determine the distance that the block travels after the collision (which is the same amount the spring is compressed).

b. Determine the percent loss of kinetic energy in the collision.

Answer 1

Mass of block, M=4kg,

Mass of bullet, m=0.05kg

(miu) friction coefficient =0.275

g=10m/s^2

Due perfectly inelastic collision both bullet and block will stick and move after collision with same velocity.

So, using conservation of momentum, we can write,

now when System(block + bullet) will compress the spring. There will be spring force which will opposing the motion of the system and friction force is also opposing the motion.

Net force at x distance =kx+(M+m)g*miu

M'=M+m=4.05 kg

Now net force we can write it as,

Now, integrate it from v=4.938 to v=0,

after integration,

x=0.621 m

part 2)

loss in K.E= Initial K.E -final K.E

  

  

=4000-49.4=3950.6 J

% loss =(3950.6/4000)*100=98.8%