Conservation of Linear Momentum (spring)

In summary: The center of mass frame has a velocity of v0 and its own kinetic energy. The original energy is 3Mv0. The center of mass energy is then 2Mv0 - 3Mv0.
  • #1
songsteel
4
0

Homework Statement



A mass 3M moving horizontally with velocity v[itex]_{0}[/itex] on a frictionless surface, strikes head-on and sticks to a horizontal spring system of natural length l and spring constant k with masses M at each end. The spring has negligible mass.

Determine the maximum compression of the spring.

Homework Equations



Conservation of Linear Momentum:
Initial Linear Momentum = Final Linear Momentum
Conservation of Energy:
Initial Kinetic Energy + Initial Elastic Potential Energy = Final Kinetic Energy + Final Elastic Potential Energy

The Attempt at a Solution



3Mv[itex]_{0}[/itex] = (3M + M)Vfinal
Vfinal = [itex]\frac{3}{4}[/itex]v[itex]_{0}[/itex]

At Maximum compression: all the Kinetic Energy would have been converted to Elastic Potential Energy; let x be the maximum compression.

[itex]\frac{1}{2}[/itex](4M)([itex]\frac{3}{4}[/itex]v[itex]_{0}[/itex])[itex]^{2}[/itex] = [itex]\frac{1}{2}[/itex](k)(x)[itex]^{2}[/itex]

[itex]\frac{9}{8}[/itex]Mv[itex]_{0}[/itex][itex]^{2}[/itex] = [itex]\frac{kx^{2}}{2}[/itex]

x = √([itex]\frac{9Mv_{0}^{2}}{4k}[/itex])

According to the answersheet I was given, the final answer should be x = √([itex]\frac{9Mv_{0}^{2}}{20k}[/itex]) instead.

Would really appreciate it if anyone could help me out. Thanks!
 
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  • #2
Your assumption that at max compression KE = 0 has no basis. In fact, it is not compatible with conservation of momentum.
 
  • #3
I see, would it be correct to write

[itex]\frac{1}{2}[/itex](4M)([itex]\frac{3}{4}[/itex]v[itex]_{0}[/itex])[itex]^{2}[/itex] = [itex]\frac{1}{2}[/itex](k)(x)[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex](5M)(V1[itex]^{2}[/itex])

If it is then what method should I use to find v1?
 
  • #4
The latter equation is meaningless, because 5M is the mass of the entire system, but V1, even if it was the velocity of the center of mass, is not the only contribution to kinetic energy.

Perhaps the easiest approach here would to use the center of mass frame. Find its velocity. Find the kinetic energy of the center of mass. Then, the internal energy in the center of mass frame is the original energy minus the center of mass energy. In the center of mass frame, the system undergoes simple harmonic motion, whose energy is known.
 
  • #5
I think I got it. Thanks!
 

What is the definition of Conservation of Linear Momentum?

The conservation of linear momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time, regardless of any external forces acting on the system.

How does the spring affect the conservation of linear momentum?

The spring is a crucial element in the conservation of linear momentum. When a spring is compressed or stretched, it stores potential energy which can be converted into kinetic energy when the spring is released. This change in energy can affect the momentum of the system.

What is the equation for calculating the conservation of linear momentum in a spring system?

The equation for calculating the conservation of linear momentum in a spring system is:
Pinitial = Pfinal
Where P represents the total momentum of the system, and the subscripts "initial" and "final" indicate the initial and final states of the system, respectively.

What are some real-life applications of the conservation of linear momentum in springs?

Some examples of real-life applications of the conservation of linear momentum in springs include car suspension systems, pogo sticks, and trampolines. In all of these cases, the spring allows for the conversion of potential energy into kinetic energy, which helps to maintain the momentum of the system.

How does the mass of an object affect the conservation of linear momentum in a spring system?

The mass of an object does not directly affect the conservation of linear momentum in a spring system. However, the momentum of an object is directly proportional to its mass, so a heavier object will have a greater momentum and therefore a greater impact on the conservation of linear momentum in the system.

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