Two carts are forced apart by a compressed spring

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Homework Help Overview

The discussion revolves around an explosion problem involving two carts being pushed apart by a compressed spring. Participants explore concepts of momentum, energy conservation, and the effects of varying masses on velocity and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of momentum conservation and energy conservation in the context of the problem. Questions arise regarding the interpretation of forces acting on the carts and the effects of mass differences on their motion.

Discussion Status

The discussion is active, with participants questioning assumptions and clarifying concepts related to the problem. Some guidance has been offered regarding the relationship between mass, acceleration, and energy transfer, but no consensus has been reached on specific interpretations of the problem.

Contextual Notes

There are contradictions noted in the problem statement regarding whether the carts are pushed or pulled apart. Additionally, participants are grappling with the implications of different kinetic energies for carts of varying masses and the interpretation of a related problem from external material.

paulimerci
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Homework Statement
Two carts are held together. Cart 1 is more massive than Cart 2. As they are forced apart by a compressed spring between them, which of the following will have the same magnitude for both carts.
(A) change of velocity (B) force (C) speed (D) velocity
Relevant Equations
conservation of momentum and Impulse
It's an explosion problem.
When two carts are pulled apart, the bigger one takes longer than the smaller one. So the velocity of the bigger one is small, and the velocity of the smaller one is large, and they are opposite each other. So the momentum before the explosion must be equal to the momentum after the explosion.
The energy is conserved if no energy is lost due to friction or any other external force. If the pulling force is an external force applied to the system, is energy conserved?
The force applied to both the carts should be different if the time taken for the carts to be pulled apart is different. And so the impulse is different for both carts.
I don't know how to interpret the following options given: I know some of the interpretations were wrong, and I would greatly appreciate it if anyone could explain in detail.
 
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Your problem statement says the carts are PUSHED apart by a compressed spring but your discussion says that they are PULLED apart, and apparently by different forces. Do you see the contradiction?
 
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And there was an explosion! Yay!
 
paulimerci said:
...
The energy is conserved if no energy is lost due to friction or any other external force. If the pulling force is an external force applied to the system, is energy conserved?
You can consider the energy-conservative system as cart 1 - spring - cart 2.
In that way, the elastic energy contained in the compressed spring remains within the system all the time.
paulimerci said:
The force applied to both the carts should be different if the time taken for the carts to be pulled apart is different. And so the impulse is different for both carts.
Reconsider this.
You can't push a heavy piece of furniture is you are standing on a slippery surface.
In order to effectively push in one direction, the spring must be supported or pushed at its other end by a reactive force.
 
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phinds said:
Your problem statement says the carts are PUSHED apart by a compressed spring but your discussion says that they are PULLED apart, and apparently by different forces. Do you see the contradiction?
Oh I read the question wrong. If it pushed by the spring then it should have exerted an equal and opposite force on the cart? Right?
 
paulimerci said:
Oh I read the question wrong. If it pushed by the spring then it should have exerted an equal and opposite force on the cart? Right?
Right.
 
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Lnewqban said:
You can consider the energy-conservative system as cart 1 - spring - cart 2.
In that way, the elastic energy contained in the compressed spring remains within the system all the time.

Reconsider this.
You can't push a heavy piece of furniture is you are standing on a slippery surface.
In order to effectively push in one direction, the spring must be supported or pushed at its other end by a reactive force.
Thank you. You mean stored EPE is transferred to a larger mass with less energy and a smaller mass with more energy, and the system's energy is conserved?
Since the objects have varying masses, they will move with different velocities, even though the carts experience the same force, so the options for c and d are wrong, right?
 
Last edited:
Consider two carts with masses m and 2m and their respective velocities as ##v## and ##\frac{v
}{2}##.
$$K.E_m = \frac{1}{2}mv^2$$
$$K.E_2m = \frac{mv^2}{4}$$
The two carts doesn’t acquire equal K.E's. right?
 
paulimerci said:
Thank you. You mean stored EPE is transferred to a larger mass with less energy and a smaller mass with more energy, and the system's energy is conserved?
Since the objects have varying masses, they will move with different velocities, even though the carts experience the same force, so the options for c and d are wrong, right?
If you are pushing two heavy objects simultaneously in opposite directions, how could you transfer more energy to one than to the other?

Use Newton's third law for action and reaction forces.
Use the second law for comparing the acceleration gained by each cart during the time the common spring is pushing on them.

The cart accelerating the most will be the first losing contact with the spring.
At that instant, the spring will lose its support to push the second cart any longer.
Therefore, the transfer of mechanical energy will beigin and end simultaneously for both carts.
 
  • #10
paulimerci said:
Consider two carts with masses m and 2m and their respective velocities as ##v## and ##\frac{v
}{2}##.
$$K.E_m = \frac{1}{2}mv^2$$
$$K.E_2m = \frac{mv^2}{4}$$
The two carts doesn’t acquire equal K.E's. right?
Right. So what is your answer to the question?
 
  • #11
Lnewqban said:
If you are pushing two heavy objects simultaneously in opposite directions, how could you transfer more energy to one than to the other?
I hope you are not suggesting it cannot be done.
 
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  • #12
Lnewqban said:
If you are pushing two heavy objects simultaneously in opposite directions, how could you transfer more energy to one than to the other?

Use Newton's third law for action and reaction forces.
Use the second law for comparing the acceleration gained by each cart during the time the common spring is pushing on them.

The cart accelerating the most will be the first losing contact with the spring.
At that instant, the spring will lose its support to push the second cart any longer.
Therefore, the transfer of mechanical energy will beigin and end simultaneously for both carts.
Your argument is not convincing .
 
  • #13
The magnitude of the forces are identical so the lighter cart experiences more acceleration. This means it travels a larger distance. Since the magnitude of the forces are identical, this means more work is performed on it.
 
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  • #14
haruspex said:
Right. So what is your answer to the question?
I see the answer is "B" for post #1.

I'm trying to solve mit problem that is similar to this. which you'll find below. I should have posted this in a separate homework statement, but since this looks similar, I thought I could try it here. My apologies!

Two toy cars with different masses originally at rest are pushed apart by a spring between them. Which TWO of the following statements be true? (A) both toy cars will acquire equal but opposite momenta (B) both toy cars will acquire equal kinetic energies (C) the more massive toy car will acquire the least speed (D) the smaller toy car will experience an acceleration of the greatest magnitude.
And the answer was found to be B, which contradicts post 8. I see that all the answers look correct except for B.
 
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  • #15
paulimerci said:
I see the answer is "B" for post #1.

I'm trying to solve mit problem that is similar to this. which you'll find below. I should have posted this in a separate homework statement, but since this looks similar, I thought I could try it here. My apologies!

Two toy cars with different masses originally at rest are pushed apart by a spring between them. Which TWO of the following statements be true? (A) both toy cars will acquire equal but opposite momenta (B) both toy cars will acquire equal kinetic energies (C) the more massive toy car will acquire the least speed (D) the smaller toy car will experience an acceleration of the greatest magnitude.
And the answer was found to be B, which contradicts post 8. I see that all the answers look correct except for B.
I agree with you. But if it asks which two, how come the answer it gives is B only?
Can you supply a link?
 
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  • #17
paulimerci said:
Okay,
https://web.mit.edu/~yczeng/Public/WORKBOOK 1 FULL.pdf
The question is on page 179 (Question No. 15), and the answer is on page 206.
The text that goes with the answer clearly states that the momenta are equal and opposite but the energies differ. So I would guess the question was supposed to ask which one is false.
 
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  • #18
Thank you everyone. I truly appreciate!
 
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  • #19
haruspex said:
The text that goes with the answer clearly states that the momenta are equal and opposite but the energies differ. So I would guess the question was supposed to ask which one is false.
Adding the kinetic energies of the two masses is equal to the EPE of the spring initial and thus energy is conserved in the system. Right?
 
  • #20
paulimerci said:
Adding the kinetic energies of the two masses is equal to the EPE of the spring initial and thus energy is conserved in the system. Right?
Yes, but how is that relevant?
 
  • #21
haruspex said:
Yes, but how is that relevant?
It's not relevant. I'm just clarifying myself.
 
  • #22
paulimerci said:
Adding the kinetic energies of the two masses is equal to the EPE of the spring initial and thus energy is conserved in the system. Right?
Correct!
 
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