# physics

posted by .

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

• physics -

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

• physics -

Nevermind. Disregard this post.

• physics -

Actually, can you do this anyway?

• physics -

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

• physics -

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.

• physics -

(1/2)(4.008) V^2 = (1/2)(2500)(.037)^2

V = .924 m/s

.008 Vb = (4.008)(.924 )
Vb = 463 m/s