Analyzing a Compressed Spring System

Click For Summary
SUMMARY

The discussion focuses on analyzing a compressed spring system using principles of conservation of energy and momentum. The user derived equations for kinetic and potential energy, specifically \(\frac{1}{2}m_1{v_{1_i}}^2 + \frac{1}{2}m_2v_{2_i} = \frac{1}{2}m_1{v_{1_f}}^2 + \frac{1}{2}m_2v_{2_f} + \frac{1}{2}kx^2\), and questioned the applicability of conservation of momentum in the absence of a collision between the spring and masses. Additionally, the user sought insights on analyzing a system where the second object's mass includes the spring, emphasizing the need for a framework that respects conservation laws.

PREREQUISITES
  • Understanding of conservation of momentum and energy principles
  • Familiarity with spring potential energy and kinetic energy equations
  • Knowledge of Newtonian mechanics and frame of reference concepts
  • Ability to analyze systems in the center of mass frame
NEXT STEPS
  • Study the application of conservation laws in non-collisional systems
  • Learn about the center of mass frame and its advantages in analyzing dynamics
  • Explore the relationship between acceleration, velocity, and displacement in spring systems
  • Investigate the effects of mass deformation on energy conservation in dynamic systems
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of spring systems and energy conservation principles.

Obliv
Messages
52
Reaction score
1
It's not a homework/coursework question but I did get the system from my textbook.
http://puu.sh/o03h7/32cdf7cffb.jpg

I solved the question by analyzing the system at different stages. Initially both objects are moving with a velocity and having some mass so their kinetic energies are the total energy of the system.
Eventually the spring is compressed fully and the masses are combined moving at some velocity ##v_f##. The system now has a kinetic energy as well as a potential energy due to the spring. By using the conservation of momentum, one can find the velocity ##v_f## of the system when the spring is completely compressed.

My first question is, how would the system be described at different points of compression? I gave it a go with energy conservation and got \frac{1}{2}m_1{v_{1_i}}^2 + \frac{1}{2}m_2v_{2_i} = \frac{1}{2}m_1{v_{1_f}}^2 + \frac{1}{2}m_2v_{2_f} + \frac{1}{2}kx^2 the velocities for ##v_{1_f}## and ##v_{2_f}## can be solved by relating the acceleration experienced by each mass to the compression by a = v\frac{dv}{dx} = \frac{kx}{m} \int v dv = \int \frac{kxdx}{m} v = x(\frac{k}{m})^{\frac{1}{2}} I'm guessing ##v_{1_f}## and ##v_{2_f}## can't be found by using conservation of momentum, right? Since we're not considering there to be any collision between the spring and the masses.
Are these equations/way of looking at this correct?

My second question is, if the 2nd object's mass included the spring, how would momentum/energy be described at different points of compression? I can't imagine Newtonian mechanics being able to describe this without violating some conservation laws or by utilizing other concepts aside from conservation of energy + momentum. The second mass can't be described as a point-like particle in equations \sum \vec P_i = \sum \vec P_f m_1v_{1_i} + m_2v_{2_i} = (m_1+m_2)v_f I don't expect an in-depth analysis of this new system but I would like some insight in how it would be analyzed (using what concepts, equations, etc.) I guess another way to look at it would be that the second mass is capable of deforming by some distance ##x## producing a potential energy, where the first mass cannot be deformed.
 
Last edited by a moderator:
Physics news on Phys.org
Darn, the velocity equation I derived doesn't make sense. But, I multiplied acceleration by distance shouldn't I be able to get ##v^2## and consequently ##v## from that?
 
Hi Obliv,

Have you considered analyzing the system in its center of mass frame? In that frame, the two blocks are moving towards each other, and their center of mass is fixed. This means that the sum of the momenta of the 2 blocks = 0, in other words, the velocity of one is completely linearly dependent on the velocity of the other (this is true in any frame, but harder to analyze if not in CM frame). You have only 1 variable here, 1 velocity.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
17
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 22 ·
Replies
22
Views
2K
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K