Question

A 5.0-g bullet traveling horizontally at an unknown speed hits and embeds itself in a 0.195-kg block resting on a frictionless table. The block slides into and compresses a 180-N/m spring a distance of 0.15m before stopping the block and bullet.

Determine the initial speed of the bullet.

Answer 1

Find Us=1/2k(deltax)^2
=1/2(180)(.15)^2
=2.025
This is spring potential energy. This must have been conserved from kinetic energy. Therefore Us=K=1/2mv^2
2.025=1/2(.195+.005)(v^2)
v=4.5 m/s
This is explains the time that the bullet is imbedded in the wood. To find the speed of the bullet before it is imbedded in the wood, use momentum.
p=mv
p(f)=4.5*(.195+.005)
p(f)=0.9
p(i)=p(f)
m*v=p(f)
.005*v=0.9
v=180 m/s