Conservation of Energy Problem - Need Help

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Homework Help Overview

The problem involves a conservation of energy scenario with a cart and a spring, where one cart collides with another stationary cart. The participants are exploring the maximum compression of the spring during the collision and the subsequent speeds of both carts after the collision.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using conservation of momentum and energy to solve for the maximum compression of the spring and the final velocities of the carts. There is an exploration of different methods to approach the second part of the problem, including substitution into equations and the use of the quadratic formula.

Discussion Status

Some participants have provided guidance on comparing initial and final momentum and energy, suggesting that this approach may simplify the problem. There is acknowledgment of potential errors in calculations, particularly regarding the signs of velocities.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also a focus on ensuring the assumptions made during the calculations are valid.

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[SOLVED] Conservation of Energy Problem - Need Help

Homework Statement



A 2.0 kg cart has a sping with k= 5000 N/m attached to its front, parallel to the ground. This cart rolls at 4.0 m/s toward a stationary 1.0 kg cart.

a) What is the maximum compression of the spring during the collision?

b) What is the speed of each cart after the collision?


Homework Equations



KE=(1/2)mv^2 Us=(1/2)kx^2

3. Attempt at a solution

I got the first part:

Using first conservation of momentum and assuming the cars stick together momentarily
then using conservation of energy

(2kg)(4m/s)=(3kg)v1
v1= 2.67 m/s
(.5)(2kg)(4m/s)^2=(.5)(5000N/m)(x)^2 + (.5)(3kg)(2.67m/s)^2
x= .046 m

But I'm not sure what to do with the second part. Any suggestions?
 
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Apply the conservation laws. What's conserved? (The difference here is that there's no spring PE before and after the collision.)
 
Ok, I see.

So I would use conservation of momentum first:

(3kg)(2.67 m/s) = (2kg)v1 + (1kg)v2 Solving to get an expression for v1
v1= 4.005 - .5v2

Then I would substitute that into conservation of energy:

(.5)(3kg)(2.67 m/s)^2 + (.5)(5000 N/m)(.046 m)^2 = (.5)(2kg)(4.005-.5v2)^2 + (.5)(1kg)(v2)^2

Then I would solve for v2 using the quadratic equation. I come up with values for v2:
.016 & 5.3

Plugging them into v1= 4.005- .5v2 I see that 5.3 m/s must be the velocity for the cart that was initially at rest because the velocity of the cart that was intially moving cannot be greater than the velocity of the cart initially at rest.

So: v1= -1.3 m/s & v2= 5.3 m/s

Does this look right to you?
 
Looks OK, except for v1 being negative. (Double check that.)

I would have simply compared the initial momentum and energy (prior to collision) to the final momentum and energy (after the collision). That way you don't need the results from the first part and you don't have to worry about spring compression. (You get the same answer either way, of course.)
 
Ok, v1 shouldn't be negative. Not sure why I put that there.
I also tried it the way you suggested, and it is much easier.

Thanks for your help!
 

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