physics
posted by Georgey on .
An 8 g bullet is shot into a 4.0 kg block, at rest on a frictionless horizontal surface. The
bullet remains lodged in the block. The block moves into a spring and compresses it by
3.7 cm. The force constant of the spring is 2500 N/m. In the figure, the initial velocity of
the bullet is closest to:
A) 460 m/s
B) 440 m/s
C) 480 m/s
D) 500 m/s
E) 520 m/s

If you could just tell me the steps and equations to use, without the numbers, that would be great. Thank you again in advance

Nevermind. Disregard this post.

Actually, can you do this anyway?

use conservation of momentum to get the velocity of the total mass (4.008 kg) just after collision in terms of bullet velocity Vb. Call that V.
(1/2) m V^2
= (1/2) k x^2
that gives you V
go back and get Vb 
That gives me an answer of 4625.93 m/s. The correct answer is 460 m/s. Can you tell me what I did wrong? I found V=.925 m/s, then did m1v1=m2v2.

(1/2)(4.008) V^2 = (1/2)(2500)(.037)^2
V = .924 m/s
.008 Vb = (4.008)(.924 )
Vb = 463 m/s