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# EK 1001 Phys # 328 (Conservation of Energy)

Discussion in 'MCAT Study Question Q&A' started by member232, May 25, 2012.

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1. ### member232New Member

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A 2 kg ball and an 8 kg ball are placed on the same spring at the same time. The spring is compressed and released. Which of the following is true about the maximum height reached by the balls?

A. The 2 kg ball will go four times as high.
B. The 2 kg ball will go twice as high.
C. The balls will reach equal maximum heights.
D. The 8 kg ball will go four times as high.

Answer: C

I guess my intuition is skewing things a bit for me. Would someone please explain the problem conceptually and/or with the math? The way I'm looking at it is:

Since the spring is the same, both balls start off with the same potential energy of 1/2kx^2
By conservation of energy 1/2 kx2 = mgh , so shouldn't the less massive ball reach a greater height?

I know there's something wrong in the way I'm looking at this... any clarification? Thanks in advance!
2. ### bdc142New Member

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Your reasoning assumes that all the PE of the spring goes into each ball. The PE is split somehow between the two balls.
3. ### member232New Member

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How is the PE split?

If PE is 1/2 kx2 , the spring constant, k, should be the same for both balls, and the displacement due to compression, x, should also be the same. Right?
4. ### dollarbillNew Member

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the way i look at it conceptually is that energy is conserved so PE=KE, so when you set the formulas up you get mgh=1/2mv^2. mass cancels itself out and has no influence on the rest of the scenario. chew on that and let me know what you think.
5. ### member232New Member

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hmm... i thought of that... but as far as energy conservation goes isn't it going:

PE of spring --> KE of balls --> PE of balls?

1/2kx^2 turns into 1/2 mv^2 which turns into mgh ?

That's why I thought you cant neglect mass, because the velocity of the more massive ball should be less?

The explanation to the question says "The balls leave the spring at the same time, thus they have the same velocity." I guess I'm not seeing how their velocities are equal. I feel like the kinetic energies will be equal, but due to the different masses, they will have different velocities.

(again, I know there's something wrong in the way I'm looking at this, I can't figure out what)

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7. ### ikjadoonNew Member

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EDIT: should've refreshed the page...lol, same as bdc142.

http://forums.studentdoctor.net/showpost.php?p=12144303&postcount=6

The spring can't push two things at different speeds. And, then if two things leave at the same speed, they'll reach the same height, as here:

.5*m*v^2 = m*g*h

For the 2kg ball with velocity 4m/s (made up the velocity)

= .5*2*16 = 2*10*x
= 8 = 10x
x = 8/10

For the 8kg ball with velocity 4m/s (made up the velocity, same as 2kg ball)

= .5*8*16 = 8*10*x
= 8 = 10x
x = 8/10
8. ### member232New Member

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got it, thanks guys!!
9. ### sotto voceNew Member

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The simplest explanation is this:

-The balls must leave with the same velocity, because there is one spring.

-The height is dictated by v=sqrt(2gh). Thus for equal velocity, the maximum height will be equal.

-Mass plays no role.

-Note that they don't ask you what the heights actually are. The trick is to know whether you launch a car into the air at 5 m/s or a baseball, the max height will be the same.

-But I agree that not being able to use elastic PE is confusing.
10. ### Ibn Alnafis MDMS-0

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According to EK, for this type of questions, you could use the scenario of extremes. For example, if one object was a hair clip and the other was a piano, would the hair clip travel a million times in the air higher than the piano?
11. ### osprey099Established Member

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PE(spring) = (1/2)kx^2 and that's gg to this question

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