Analyzing a Compressed Spring System

In summary: If you multiply the velocities by the distance between them, you can calculate the velocity of each block.
  • #1
Obliv
52
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 [tex] \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 [/tex] 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 [tex] a = v\frac{dv}{dx} = \frac{kx}{m} [/tex] [tex] \int v dv = \int \frac{kxdx}{m} [/tex] [tex] v = x(\frac{k}{m})^{\frac{1}{2}} [/tex] 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 [tex] \sum \vec P_i = \sum \vec P_f [/tex] [tex] m_1v_{1_i} + m_2v_{2_i} = (m_1+m_2)v_f [/tex] 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.
 
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  • #2
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?
 
  • #3
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.
 

1. How do you determine the spring constant of a compressed spring system?

To determine the spring constant, you can use Hooke's Law which states that the force exerted by a spring is directly proportional to the distance it is compressed or stretched. By measuring the force applied and the displacement of the spring, you can calculate the spring constant using the equation k = F/x, where k is the spring constant, F is the applied force, and x is the displacement.

2. What factors can affect the accuracy of analyzing a compressed spring system?

Some factors that can affect the accuracy of analyzing a compressed spring system include the material of the spring, the temperature, and the presence of any external forces or friction. It is important to control these factors as much as possible to obtain accurate results.

3. Can a compressed spring system store potential energy?

Yes, a compressed spring system can store potential energy. When a spring is compressed, it stores potential elastic energy which can be released when the spring is allowed to return to its original shape.

4. How does the mass of an object attached to a compressed spring affect its displacement?

The mass of an object attached to a compressed spring does not affect its displacement as long as the force applied to the spring remains constant. According to Hooke's Law, the displacement of a spring is directly proportional to the force applied, not the mass of the object.

5. Can the spring constant of a compressed spring system change over time?

Yes, the spring constant of a compressed spring system can change over time due to factors such as wear and tear, temperature changes, and material degradation. It is important to regularly check and recalibrate the spring system to ensure accurate results.

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