Conservation of Energy Problem - Need Help

In summary, after the cart collision, the first cart has a velocity of 5.3 m/s and the second cart has a velocity of -1.3 m/s.
  • #1
bcjochim07
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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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  • #2
Apply the conservation laws. What's conserved? (The difference here is that there's no spring PE before and after the collision.)
 
  • #3
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?
 
  • #4
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.)
 
  • #5
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!
 

1. What is the conservation of energy problem?

The conservation of energy problem refers to the principle that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant.

2. Why is the conservation of energy important?

The conservation of energy is important because it is a fundamental principle in physics. It helps us understand and predict the behavior of energy in various systems, and is crucial for many technological advancements such as renewable energy sources and efficient machines.

3. How is the conservation of energy applied in real life?

The conservation of energy is applied in various ways in real life, such as in the design and operation of energy-efficient buildings, vehicles and appliances. It also plays a role in understanding natural phenomena, such as weather patterns and the movement of celestial bodies.

4. What are some common examples of the conservation of energy?

Some common examples of the conservation of energy include a pendulum swinging, a ball rolling down a hill, and a car accelerating. In each of these cases, energy is transformed from one form to another, but the total amount remains constant.

5. Are there any exceptions to the conservation of energy?

The law of conservation of energy is considered to be a fundamental principle in physics and has been observed to hold true in all known cases. However, in extreme situations such as near black holes or during the early stages of the universe, some exceptions may be observed due to the effects of gravity and quantum mechanics.

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