Platform on Springs, how to find amplitude?

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Homework Help Overview

The problem involves a platform supported by springs and a chunk of modeling clay that collides with it, leading to oscillations. The participants are trying to determine the amplitude of oscillation after the collision, given the mass of the platform and the clay, the initial speed of the clay, and the spring constant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of energy and momentum during the collision, questioning whether kinetic energy is conserved. They explore the relationship between initial and final energies and how to calculate the final velocity after the collision.

Discussion Status

The discussion is ongoing, with participants providing insights into the energy conservation principles and the nature of the collision. Some participants have offered guidance on the correct energy equations to use, while others are still grappling with the calculations and assumptions involved.

Contextual Notes

There is a focus on the distinction between elastic and inelastic collisions, with participants noting that energy is not conserved in the latter. The problem also involves specific values for mass, velocity, and spring constant, which are critical for the calculations being discussed.

taveuni
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I have been working on this problem for an hour and can't seem to figure it out. I know there must be something fundamental that I must be doing wrong. Can someone give me a pointer?

A platform of mass 0.8 kg is supported on four springs. A chunk of modeling clay of mass 0.6 kg is held above the table and dropped so that it hits the table with a speed of v = 0.9 m/s. The clay sticks to the table so that the the table and clay oscillate up and down together. Finally, the table comes to rest 6 cm below its original position.

With what amplitude does the platform oscillate immediately after the collison?

In the first part of the problem I found that k=98.1 N/m.​


So, initially, the E is entirely Kinetic (KE = .5mv^2) for the intial velocity of 0.9 m/s and the mass of the clay (0.6 kg). This means that intial E = initial KE = 0.243.

Now, I don't know if the final E is kinetic and potential, or just potential? if I say it is kinetic too, then how do I find V right after the collision? I tried the equation v= (mvi+Mvi)/(m+M) and calculated (0.6*0.9+0)/(0.6+0.8)=0.3857 m/s. If I use this velocity for final KE, and solve for A in the final PE (PE=.5kA^2), I get the wrong answer (A=0.053m)

If I don't incorporate final KE and simply say that final E = final PE, then I get A=0.0703m, which is wrong too.

What am I doing wrong? Which one do I use? And why is it that both give me the wrong answer?
 
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taveuni said:
So, initially, the E is entirely Kinetic (KE = .5mv^2) for the intial velocity of 0.9 m/s and the mass of the clay (0.6 kg). This means that intial E = initial KE = 0.243.
I'm not quite sure why you calculated the kinetic energy of the clay, as energy is not conserved during the collision.

Now, I don't know if the final E is kinetic and potential, or just potential? if I say it is kinetic too, then how do I find V right after the collision? I tried the equation v= (mvi+Mvi)/(m+M) and calculated (0.6*0.9+0)/(0.6+0.8)=0.3857 m/s. If I use this velocity for final KE, and solve for A in the final PE (PE=.5kA^2), I get the wrong answer (A=0.053m)
How did you get that equation?
EDIT: Ah, you used vi for 2 different values.

Do notice that the system is not at the "new" equilibrium point right after the collision.
 
Last edited:
I don't understand this completely. Why isn't the energy of the clay conserved during the collision? Ei=KEi+PEi, right? And doesn't Ei=Ef?

v= (mvi+Mvi)/(m+M)
This equation gives the final velocity of 2 objects, of mass m and M with intial velocities vi1 and vi2, the collide and stick. It is more appropriately written vf= (mvi1+Mvi2)/(m+M) This is in my course text (Walker).
 
taveuni said:
I don't understand this completely. Why isn't the energy of the clay conserved during the collision? Ei=KEi+PEi, right? And doesn't Ei=Ef?

v= (mvi+Mvi)/(m+M)
This equation gives the final velocity of 2 objects, of mass m and M with intial velocities vi1 and vi2, the collide and stick. It is more appropriately written vf= (mvi1+Mvi2)/(m+M) This is in my course text (Walker).
If the masses stick together, that's called an inelastic collision. Energy is not conserved, but momentum is. You can easily derive the formula you have in your book from the conservation of momentum.

If I use this velocity for final KE, and solve for A in the final PE (PE=.5kA^2)
This I didn't quite understand. At the amplitude position, you're not supposed to have any kinetic energy.
 
Okay, so, correct me if I'm wrong:

The energy equation that should be used is KEi+PEi=PEf+KEf; However, there is no KE at the maximum amplitude (because it comes to rest), so the equation should thus be KEi+PEi=PEf.

Well, KEi+PEi = .5*m*v^2 + .5*k*x^2 right?
I assume I use the mass of both objects, m+M = 1.4 kg and the velocity calculated above through conservation of momentum (0.3857 m/s). The k=98.1 and x should be the distance the spring will compress, or, 0.06 m. That gives me 0.2807 on for KEi+PEi.

On the right side of the equation, we have PEf, which is .5*k*A^2.
Solving for A, I get 0.265 m, which is wrong.

What am I doing wrong?
 
taveuni said:
On the right side of the equation, we have PEf, which is .5*k*A^2.
Solving for A, I get 0.265 m, which is wrong.
Redo this last step. Everything else looks OK.
 
Wow. After a dozen sheets of scrap paper and forgetting to incorporate ^2 a many times - the answer is 7.57 cm! Thank you!
 

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