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longhorn991
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1. A 4kg block is suspended vertically from a spring with k = 500N/m. A 0.05kg bullet is fired into the block from directly below with a velocity of 150 m/s and becomes embedded in the block. Find the amplitude of the resulting simple harmonic motion.
V cm = (Bullet mass)(Bullet velocity)/(Bullet mass + block mass)
(1/2)mv^2 + (1/2)Kx^2 = MgX
This seems like a pretty straightforward problem, when I solve for the velocity after the the inelastic collision of the bullet and the block, I get 1.8519 m/s, which gives an initial kinetic energy of 6.9448 J. However, there is no solution to the above equation for any positive kinetic energy, so I must be setting up the work theorem equation wrong. Can anyone point out what mistake I am making?
Homework Equations
V cm = (Bullet mass)(Bullet velocity)/(Bullet mass + block mass)
(1/2)mv^2 + (1/2)Kx^2 = MgX
The Attempt at a Solution
This seems like a pretty straightforward problem, when I solve for the velocity after the the inelastic collision of the bullet and the block, I get 1.8519 m/s, which gives an initial kinetic energy of 6.9448 J. However, there is no solution to the above equation for any positive kinetic energy, so I must be setting up the work theorem equation wrong. Can anyone point out what mistake I am making?