Springs, Simple Harmonic Motion

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SUMMARY

The discussion centers on a physics problem involving a 25.0g bullet striking a 0.600kg block attached to a spring with a spring constant of 6.70x103 N/m. The bullet sets the block into vibration with an amplitude of 21.5cm. The correct speed of the bullet before impact is determined to be 557m/s, calculated using conservation of energy principles, where the kinetic energy of the bullet equals the potential energy stored in the spring at maximum deflection. The participants also discuss the implications of momentum in the context of this inelastic collision.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with the concepts of kinetic and potential energy
  • Knowledge of simple harmonic motion and spring mechanics
  • Basic understanding of inelastic collisions and momentum
NEXT STEPS
  • Study the conservation of energy in mechanical systems
  • Learn about the equations governing simple harmonic motion
  • Explore inelastic collision dynamics and momentum conservation
  • Investigate the relationship between spring constant, mass, and amplitude in oscillatory motion
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Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of energy conservation and motion in spring systems.

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1. A 25.0g bullet strikes a .600kg block attached to a fixed horizontal spring whose spring constant is 6.70x10^3 and sets it into vibration with an amplitude of 21.5cm. What is the speed of the bullet before impact if the two objects move together after impact?2. V=A[tex]\sqrt{k/m}[/tex]
3. I used the equation above, adding the masses together (.625kg). I got the velocity to be 22.26m/s... but now i can't figure out how to get the speed of the bullet.


The answer in the book says that the speed of the bullet is 557m/s.
 
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One determines the maximum spring deflection from the amplitude, and from that the stored energy in the spring at max deflection, which by conservation of energy (and neglecting friction) should equal the initial kinetic energy of the bullet and block.

From the KE of the bullet and block, one can determine the velocity of the combination.

Since the bullet striking the block is an example of an inelastic collision, what about the momentum of the bullet and block?
 

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