Asked by Georgey

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
Answered by Georgey
If you could just tell me the steps and equations to use, without the numbers, that would be great. Thank you again in advance
Answered by Georgey
Nevermind. Disregard this post.
Answered by Georgey
Actually, can you do this anyway?
Answered by Damon
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
Answered by Georgey
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.
Answered by Damon
(1/2)(4.008) V^2 = (1/2)(2500)(.037)^2

V = .924 m/s

.008 Vb = (4.008)(.924 )
Vb = 463 m/s
Answered by damon
damon you the goat
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