Bullet fired from below into block suspended by spring

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SUMMARY

The discussion focuses on a physics problem involving a 4 kg block suspended from a spring with a spring constant of 500 N/m, into which a 0.05 kg bullet is fired at a speed of 150 m/s. The participants analyze the inelastic collision using momentum conservation, resulting in a combined velocity of 1.85 m/s for the block and bullet. The kinetic energy of the combined mass is calculated to be 113.9 J, which is then equated to the potential energy at maximum spring compression to find the amplitude of the oscillation. The discussion emphasizes that gravity is implicitly accounted for in the equilibrium position of the spring, negating the need for explicit inclusion in the equations.

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  • Understanding of momentum conservation in inelastic collisions
  • Familiarity with simple harmonic motion and spring mechanics
  • Knowledge of kinetic and potential energy equations
  • Ability to apply energy conservation principles in mechanical systems
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Homework Statement


A 4 kg block is suspended from a spring with k=500N/m. A .05kg bullet is fired from below into with a speed of 150m/s and comes to rest in the block.
a.)Find the Amplitude of the resulting simple harmonic motion.
b.)What fraction of the original KE of the bullet appears as mechanical energy in the harmonic oscillator?
^^^This is the problem word for word^^^

Homework Equations


KE=1/2mv^2
F=-ky
U=1/2ky^2
T=2pi(m/k)^1/2
P cons.
E cons.
etc...

The Attempt at a Solution


Where I am atm is, the problem basically has two parts, the collision which is completely inelastic so Pcons applies and then the compression of the spring and oscillation of the combined masses.

setting up the equations for the collision I apply Pcons.
MblkVblk+MbtVbt=MblkV`blk+MbltV`blt --> Vbb=(MblkVblk+MbtVbt)/(Mblk+Mblt)
-->Vbb=1.85m/s

after the collision having Mbb=4.05 and Vbb=1.85m/s the combined block and bullet mass has
KEbb=113.9J

now I know that at the point of maximum compression of the spring, the mass will be at Vbb=0. So, Applying E cons I set KEbb=Us

113.9J=1/2ky^2 -->227.8=ky^2 --> 227.8/k=y^2 --> (227.8/k)^1/2=y...

Now this is where confusion sets in for me. If I take the problem as I have been, and continue to ignore the hell out of gravity, then there is nothing to dampen the oscillations over time, and I've basically got amplitude since I just solved for maximum compression of the spring. However if I let gravity in the door, things get more complex. This problem is part of a huge packet of problems I'm doing to prep for senior exams so it covers everything from intro on up and this particular problem is located smack in the middle of the medium-difficult ranked intro problems so I don't think the intention here is from me to go charging off with a diff-eq into classical mechanics territory. Since they gave me just masses, velocities and the spring constant I'm kinda thinking this is supposed to be a fairly straight forward energy/momentum problem...

If anyone would look this over and see if there is some thing I'm missing here I'd appreciate it very much as it's been almost 4 years since I took intro. Thanks.
 
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No, don't use gravity. Gravity is already implicitly playing a role. The string has already stretched to a certain spot because of gravity. The only thing that this bullet does is give an initial velocity to the simple harmonic spring oscillator.

[tex]x(t) = c_1 sin(\sqrt{k/m}t) + c_2 cos(\sqrt{k/m}t[/tex]

Here is a more detailed explanation to leave out gravity (I copied it out of a paper I wrote a while back):
The reason behind gravity not explicitly appearing in equation (1) has to due with building the equation around the idea of the equilibrium point. When the spring and mass are at rest the sum of the forces must all equal zero, and since there is no velocity or acceleration at this point then ks = mg where s is the distance from the ceiling to the mass. The only time a restoring force will be applied is when the spring has been stretched away from this equilibrium point, but if the spring is lifted up from the equilibrium point the force of gravity on the mass will be equal to the restoring force that would occur if it were stretched, precisely because ks = mg. In other words, whether the spring is lifted up or stretched down the forces will be equal; thus, the equation does not need to explicitly use a term for the force of gravity because it is implicit in the restoring force.
 
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Thanks I figured it was something along those lines but wanted to run it by someone just because it's been so long...every time I do these problems it's like I know how to do it, but the right way is just out of reach.
 
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