# Two carts are forced apart by a compressed spring

• paulimerci
In summary, when two carts are pulled apart, the bigger one takes longer to move than the smaller one due to their opposite velocities. This leads to the conclusion that momentum is conserved before and after the explosion. However, for energy to be conserved, there must be no external forces such as friction. The force applied to both carts may be different if the time taken for them to be pulled apart is different, resulting in different impulses. The contradiction between the problem statement and discussion is due to the carts being pushed apart by a compressed spring rather than pulled. The transfer of energy to each cart depends on their respective masses and velocities, with Newton's third law and second law playing a role in the process.
paulimerci
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.

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?

paulimerci
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.

paulimerci
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.

phinds
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.

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?

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.

Lnewqban
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 .

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.

Lnewqban, jbriggs444, SammyS and 1 other person
haruspex 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.

Lnewqban
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?

nasu
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.

paulimerci
Thank you everyone. I truly appreciate!

Lnewqban
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?

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?

haruspex said:
Yes, but how is that relevant?
It's not relevant. I'm just clarifying myself.

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!

paulimerci

## What is the principle behind two carts being forced apart by a compressed spring?

The principle behind this scenario is Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. When the spring is compressed and then released, it exerts equal and opposite forces on the two carts, causing them to move apart.

## How does the mass of the carts affect their motion when the spring is released?

The mass of the carts affects their acceleration when the spring is released. According to Newton's Second Law of Motion (F = ma), the force exerted by the spring will cause a greater acceleration in the cart with less mass and a smaller acceleration in the cart with more mass, assuming the force exerted by the spring is the same on both carts.

## What role does the spring constant play in this experiment?

The spring constant (k) determines the stiffness of the spring and the amount of force it can exert when compressed or stretched. A higher spring constant means the spring is stiffer and can exert a greater force, resulting in the carts being pushed apart with more force when the spring is released.

## How can the velocities of the carts be calculated after the spring is released?

The velocities of the carts can be calculated using the conservation of momentum and the energy stored in the compressed spring. Assuming no external forces and frictionless conditions, the total momentum before and after the release should be equal. The kinetic energy imparted to the carts can be determined from the potential energy stored in the spring (1/2 k x^2, where x is the compression distance) and then used to find the velocities.

## What happens if the surface on which the carts move has friction?

If the surface has friction, the motion of the carts will be affected by the frictional force, which will oppose their movement and cause them to eventually come to a stop. The kinetic energy imparted to the carts by the spring will be partially converted into heat due to friction, reducing the distances the carts travel compared to a frictionless surface.

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