Simple Spring Problem with something that I'm sure im not seeing

  • Thread starter wildcatdude90
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In summary, the energy in the compressed spring is evenly distributed between the particles, and the kinetic energies of the particles are equal when the spring is released.
  • #1
wildcatdude90
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Particle A and particle B are held together with a compressed spring between
them. When they are released, the spring pushes them apart, and they then
go in opposite directions, free of the spring. The mass of A is 2.00 times the mass
of B, and the energy stored in the spring was 60 J. Assume that the spring has
negligible mass and that all its stored energy is transferred to the particles. Once
that transfer is complete, what are the kinetic energies of particle A and
particle B?




Homework Equations


It seems so simple, but I am sure I am just missig something (this problem is under the "conservation of momentum" section, so I am sure it has to do with that, but i just can't figure out any equation(s) that i can use)
 
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  • #2
first you must assume that it pushes out (with a force) equally towards particles A & B, this is a reasonable assumption if you think about it.

Next, you know that force is equal to change in momentum per second, right? d(mv)/d(t). next you should assume that the contact time (t) for the interaction is equal for each mass, which again, is a reasonable assumption if you were to model it (the spring is pushing them both out, not hitting). so you know that it pushes mass A & B away with equal (and opposite) momentums.

do you see why? if not ask again

if so, write an equation equalling the two momentums to see the speed of them relative to one another.

Now! you know that the spring stored 60J of energy, where does that energy go to when the spring isn't tense? try to work it out using the equation you created earlier.
 
  • #3
so, mv of a = mv of b? and since a is twice as heavy it has to be going only half the speed. and if we reduce the velocity by half, the kinetic energy is reduced to 1/4. for b, its mass is reduced by half, so its kinetic energy is reduced to 1/2. so the ratio of kinetic energy of a to b is 1:2? So kinetic energy of a = 20J, and kinetic energy of b = 40J?
 
  • #4
wildcatdude90 said:
so, mv of a = mv of b? and since a is twice as heavy it has to be going only half the speed. and if we reduce the velocity by half, the kinetic energy is reduced to 1/4. for b, its mass is reduced by half, so its kinetic energy is reduced to 1/2. so the ratio of kinetic energy of a to b is 1:2? So kinetic energy of a = 20J, and kinetic energy of b = 40J?

That sort of logic is always hard to follow, since there's ample room for errors.

Try writing out the conservation laws and solving directly.
 

1. What is a simple spring problem?

A simple spring problem refers to a physics problem that involves a spring and its behavior under certain conditions. The spring can be compressed, stretched, or released, and the problem typically involves finding the displacement, velocity, or acceleration of the spring.

2. What are the key factors that affect a spring's behavior?

The key factors that affect a spring's behavior are the spring constant, mass of the object attached to the spring, and the displacement of the spring from its equilibrium position. Other factors such as friction and air resistance may also play a role.

3. How do you calculate the spring constant?

The spring constant, denoted by k, can be calculated by dividing the force applied to the spring by the displacement of the spring. This can be represented by the equation k = F/x, where F is the force and x is the displacement.

4. What is Hooke's Law and how does it apply to a simple spring problem?

Hooke's Law states that the force exerted by a spring is directly proportional to the amount of displacement from its equilibrium position. This means that as the spring is stretched or compressed, the force applied to the spring increases or decreases, respectively. In a simple spring problem, Hooke's Law can be used to calculate the force exerted by the spring at a given displacement.

5. What are some real-life applications of simple spring problems?

Simple spring problems have many real-life applications, such as in the design of suspension systems for vehicles, shock absorbers, and even in musical instruments like guitars and pianos. They are also used in engineering and construction to absorb and distribute forces, such as in the construction of buildings and bridges.

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