phsyics
posted by joyce on .
an 8g bullet is shot into a 4kg block at rest on a frictionless horizontal surface. the bullet remains lodged in the block. the block movies into a spring and compresses it by 3cm. the force constant of the spring is 1500N/m. the initial speed of the bullet is closest to?

1/2 kX^2 is the kinetic energy of the bullet and block after the bullet gets stuck inside. That tells you the velocity V' after impact.
(1/2) k X^2 = (1/2)(M+m)V'^2
V' = sqrt[k/(m+M)]* X
Once you know V', apply conservation of momentum to the process of the bullet lodging in the block, to get the initial bullet velocity V.
m V = (m+M) V' 
thank you so much. i was struggling through this problem for so long. but now everything makes perfect sense! =D