Solving Simple Harmonic Motion Problems| Physics HW

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Homework Help Overview

The discussion revolves around a problem related to simple harmonic motion, specifically involving a collision and the conservation of energy and momentum. The original poster expresses uncertainty about how to approach the problem and references relevant equations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of a perfectly inelastic collision and question the conservation of energy in this context. There are discussions about using conservation of momentum and the role of the spring in the system.

Discussion Status

Participants have provided guidance on the relationship between momentum and energy, with some suggesting that energy considerations may not be necessary until later in the problem. The conversation indicates a productive exploration of the concepts involved, although the original poster's confusion remains evident.

Contextual Notes

There are assumptions being discussed regarding the nature of the collision and the initial conditions of the system, including the state of the spring and the velocities involved.

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Homework Statement



http://www.physics.umd.edu/courses/Phys122/Noyes/HW/hw1.pdf (pg 3)

Homework Equations



Im guessing that E = K + U has something to do with it


The Attempt at a Solution



Not really sure where to start..I tried a few approaches but none of them got me anywheres E = 1/2mv^2 + 1/2 kx^2..but I am missing too many things..

Any help?
 
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when the bullet colides, it sticks. This means that energy is not conserved but something else is.
 
by the way, the fact that the clay sticks and that the collision is therefore perfectly inelastic is one of those "assumptions"
 
Last edited:
conservation of linear momentum?
 
yes. now you can use that in combination with energy.
 
so I have [tex]\frac{mv +Mv_2}{m+M} = v_f[/tex] where [tex]Mv_2[/tex] is going to be 0 since it is at rest?

and [tex]E=1/2mv^2 + 1/2kx^2[/tex]

Do i let m = m+M? and will 1/2kx^2 be 0 since the block can be considered to be at point x=0?...set it to zero and solve for v?
 
you actually don't need to use energy until you are ready to find the amplitude. You already have enough information to find the velocity.
 
So imediately after the collision, the spring has no effect on the velocity?
 
that's correct. two ways to understand that
1) imediately after colision, spring is not compressed and therefore applies no force.

2) even if it were compressed, a force causes an acceleration which only changes velocity of some period of time.
 
  • #10
Solved it..Thanks for the help
 
  • #11
good job, np
 

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