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