Does Conservation of Momentum Apply to an Object Moving on a Rigid Surface?

[e2m2a]

Suppose there is a black box resting on a flat surface, which surface is rigidly attached to the earth. Suppose inside the black box there is a process initiated which causes the black box to lurch to the right 3 centimeters and let's say after one second the box comes to a complete stop. (No observable ejectile exits the black box, nor any energy or radiation. Also, there is no external force acting on the black box that causes it to lurch in one direction and the other direction)

Now the only external forces acting on the black box after the internal process is triggered are friction forces, acting on the bottom of the box opposite to the direction of the motion of the box. Conservation of momentum would require that after the box comes to a stop that it should move to the left for the same amount of time before coming to a stop, if we assume the magnitude of the friction force remains constant. And this must be true even if the distance the black box travels to the left may be less than 3 centimeters. That is, the work done on the surface of the Earth in the right direction may not be the same as the work down on the surface of the Earth in the left direction. But the magnitude of the impulse must be the same.

Put another way, the Earth receives an impulse, friction force x time (1 second) in the right direction. In order for the Earth to not have a net impulse, there must be an equal and opposite impulse on the Earth in the opposite direction, friction force x time (1 second).

Bottom line. If the box moves to the right for 1 second, it must move to the left for 1 second. Is this conclusion correct? Again, assume the coefficient of static and kinetic friction remain the same such as the case with polytetraflouroethelene(Teflon).

 

[oz93666]

Imagine you were put in a box , large enough to stand up in ...

You could jump in the air at an angle to hit the side wall , the force from this jump would not be enough to overcome static friction , and so would not move the box , but when you hit the wall this sharper impulse would move the box . Repeat this and you can shift the box and yourself across the floor.

Or you could jump inside the box , while in mid air punch the box sideways , you would come back down close to the opposite side wall , then you walk back to the center of the box and repeat . this would move you and the box as far as you want across the floor.

 

[A.T.]

Conservation of momentum would require that after the box comes to a stop that it should move to the left

No it doesn't require that.

 

[e2m2a]

No it doesn't require that.

ok. Let me be more specific. Let's say in the black box there is a spring attached to one wall. The spring is compressed and a ball lies at rest at the end of the spring. Let's assume there is no friction acting on the box. (The box is in space, away from all gravitational effects.) Thus, the initial momentum of the center of mass of the box-ball system is zero with respect to an inertial reference frame.

At some point in time the spring is released. The box moves to the left and the ball moves to the right, eventually making an inelastic collision with the opposite wall. Now, since there were no external forces acting on the box-ball system, the box-ball system must come to a complete stop. Or, the momentum of the center of mass of the system must be zero because the initial momentum of the center of mass of the system was zero. Also, with respect to the inertial reference frame, the center of mass of the system did not move.

Now, let's place the box-ball system on the surface of the earth. We assume the bottom of the box is made of Teflon and the surface likewise. Thus, the coefficient of static and kinetic friction is the same, equal to .04, which would be the case for Teflon. And asssume this coefficient remains constant.
We repeat the same scenario. Also, assume the rolling friction of the ball remains constant.

Again, the compressed spring is released, the box moves to the left and the ball moves to the right, making an inelastic collison with the wall opposite where the spring is attached.

Now in this situation, there is a third body in this system-- the earth. So, we are now dealing with a 3 body system: the box, the ball, and the earth. The initial momentum of the center of mass of this 3 body system is zero, and therefore, the final momentum of the this 3 body system must be zero. That is, if we define the box,ball, and Earth as our system, then the friction forces are internal forces, not external forces to the system. Thus, by Euler's first law, these forces cannot change the momentum of the center of mass of the 3 body system. (We can ignore the rolling friction of the ball because this friction is internal to the box-ball system, and therefore, will not affect the momentum of the center of mass of the ball-box system).

I maintain because of this, any impulse on the Earth must be countered by an equal and opposite impulse on the earth. Else, after the box comes to a rest, the Earth would have acquired a net impulse in the left direction, and thus, the 3 body system as a whole would have acquired a net increase in momentum in the left direction. But this cannot be because the friction forces are internal forces to the 3 body system.

Hence, let's focus on the box first. As the box moves to the left, a friction impulse acts on the box to the right, slowing its momentum so that when the ball makes its inelastic collision with the wall, the box will have a mangitude of momentum less than the magnitude of the ball before it makes its collision. (Remember, the rolling friction between the ball and box do not have an impact on the momentum of the center of mass of the ball-box system). Thus, right before the collision of the ball, there will be a net momentum in the right direction of the center of mass of the ball-box system, and thus, the ball-box system must move to the right after the collision, unlike the situation in space where the ball-box system comes to a complete halt after the collision. It comes to a complete halt because in space, where there are no friction forces, there was no change in the momentum of the center of mass of the ball-box system.

Now, let's focus on the earth, the third member of the 3 body system. As the box moves to the left, there will be an impulse acting on the Earth in the left direction. If this impulse is not countered by an opposite impulse, the Earth would have acquired a net impulse or change in momentum in the left direction after the box comes to a stop. Again, this is impossible because the friction forces experienced by the box and the Earth are internal forces to the 3 body system. Alas, it's because the box moves to the right after the collision of the ball that the Earth does experience an impulse in the right direction which must cancel out the impulse it experienced in the left direction. Hence, since the friction force is constant and since the impulses must be equal in magnitude and opposite in direction to cancel to zero, the time the box moves to the right must be equal to the time the box moved to the left.

 

[malemdk]

Earth receives equal impulse as the ball does and the earth-ball system conserve the momentum

 

[jbriggs444]

Now, let's place the box-ball system on the surface of the earth. We assume the bottom of the box is made of Teflon and the surface likewise. Thus, the coefficient of static and kinetic friction is the same, equal to .04, which would be the case for Teflon. And asssume this coefficient remains constant.
We repeat the same scenario. Also, assume the rolling friction of the ball remains constant.

Again, the compressed spring is released, the box moves to the left and the ball moves to the right, making an inelastic collison with the wall opposite where the spring is attached.

We are not told that the spring is strong enough and the box-ball are light enough to result in slippage against the coefficient of friction of Teflon on Teflon. Accordingly, it is not certain that the box moves to the left.

Accordingly, it is not certain that the box moves to the left while it imparts a leftward net impulse to the Earth.

Accordingly, it is not certain that the impulse imparted while the box is moving left is equal and opposite to the impulse imparted while the box is moving right.

Any imbalance is accounted for by the fact that a net impulse can be applied while the box is moving neither left nor right.

 

[Cutter Ketch]

Why argue that the scenario isn't necessarily the only possibility when A) it is possible and B) the conclusions are flawed and need correcting? Let's start with the premise as correct and then discuss what is wrong with the conclusions.

So a box contains a mass on a compressed spring. The box sits on the floor with some coefficient of static and kinetic friction. At time 0 the spring is released. The force of the spring is large enough that the impulse of releasing the spring causes the box to slide on the floor. Nothing wrong with any of that. I'm going to add one more thing to make the end state easier to understand. The spring is long enough that it traps the mass against the far wall of the box so very quickly the inside of the box is static again.

This is a perfectly reasonable and physically possible scenario. Practically everything the OP said about the result of this scenario is wrong. Conservation of momentum does NOT mean that the objects within a closed system cannot interact and PERMANENTLY move relative to each other. In fact that is how we almost always use conservation of momentum. Cars colliding masses separated by springs etc., in most of these kinds of problems the objects interact and change and move permanently relative to each other. With no external forces we invoke conservation of momentum and that puts constraints on what can happen, but the cars are absolutely not required to uncollide to conserve momentum!

So let's take your box. First suppose there is NO friction. The box is the system. The spring goes off. The ball is pushed right, the box is pushed left AND THEN THEY BOTH STAY IN THE NEW POSITIONS. How is momentum conserved? It IS conservation of momentum that required that the box move left! The mass moved right. The mass of the box had to move left for the center of mass to stay put. Put another way through the spring the box pushed on the mass. By Newton's law of action and reaction the mass pushed on the box. The actions have to be equal. Why did Newton conclude they had to be equal? No external forces equals no acceleration equals CONSERVATION OF MOMENTUM.

Ok, let's add the Earth and the friction. Now the box doesn't move quite as far left. Why? Because it pushed on the Earth and the Earth moved a tiny bit left. Now the closed system includes the earth. There are no forces external to THAT system, but there is no reason objects within the system can't move relative to each other. The mass moved right. The box AND the Earth moved left. Afterwards they all stayed in their new positions, but the center of mass did not move and momentum is conserved.

 

[jbriggs444]

Why argue that the scenario isn't necessarily the only possibility when A) it is possible and B) the conclusions are flawed and need correcting? Let's start with the premise as correct and then discuss what is wrong with the conclusions.

Because A) an alternative is also possible and B) because the conclusion follows in some cases but not in the alternative case.

The conclusion that I have in mind is, paraphrased:,

"If the box starts and ends at rest [and has no nasty vertical components to its motion, is finite and settles down to a fixed internal configuration, etc, etc] then the total duration over which the box moves left must be equal to the total duration over which the box moves right.

That conclusion is almost correct. If the box is moving leftward, it applies a leftward force of fixed magnitude to the ground and by Newton's third law is subject to a rightward force of equal magnitude. Similarly if the box is moving rightward. The net impulse applied to the box is the difference between the leftward and rightward impulses. But both of those are proportional to the durations of the leftward and rightward movement [and with the same constant of proportionality]. The net impulse will be zero if and only if the box spends the same amount of time moving left as it does moving right.

That argument actually holds... If one ignores the possibility of a non-zero force being delivered by a stationary box. [Or the possibility of a time-varying normal force, or of a coefficient of friction that varies depending on something other than normal force or of an infinite box or of a box that never settles down]

A way to poke holes in the argument is to use a scenario where a non-zero force is delivered while the box is stationary.

 

[Dale]

I maintain because of this, any impulse on the Earth must be countered by an equal and opposite impulse on the earth

This is correct. If follows for any interaction, due to Newton's 3rd law, regardless if the interaction is an internal or external force.

Hence, since the friction force is constant and since the impulses must be equal in magnitude and opposite in direction to cancel to zero, the time the box moves to the right must be equal to the time the box moved to the left.

The conclusion is certainly wrong as worded. If the impulses are equal and opposite and the magnitude of the force is constant then the duration of the force must be equal, not the motion. Indeed, the box may only move in one direction.

 

[e2m2a]

This is correct. If follows for any interaction, due to Newton's 3rd law, regardless if the interaction is an internal or external force.
The conclusion is certainly wrong as worded. If the impulses are equal and opposite and the magnitude of the force is constant then the duration of the force must be equal, not the motion. Indeed, the box may only move in one direction.

No. I am not asserting that the magnitude of the motions or the displacements of the box must be equal, I am only asserting that the time that the box moves to the left and to the right must be equal. As you said, "the duration of the force must be equal." This is what I am saying.

The box might move to the left, let's say 1 centimeter, before coming to a stop after, let's say .5 seconds. But then the box must move to the right for .5 seconds and it doesn't matter what distance it travels: it could be only .3 centimeters or .1 centimeters. It doesn't matter how far it moves. It only matters that the friction force acts for .5 seconds. This would be possible because the ball-box system after the ball collides with the box will be going slower to the right. Thus, it will still take it .5 seconds to travel the .3 centimeters to the right whereas it took .5 seconds for the box to move 1 centimeter to the left because it was going faster.

The box cannot move in "one direction" as you said because then the Earth would only receive an impulse in one direction. The box must move in the right direction to cancel out the intial impulse on the Earth in the left direction. If the box only moved to the left and came to a stop, where would be the equal and opposite impulse on the Earth to cancel out this initial impulse to the left on the earth? If there was no cancellation, the center of mass of the Earth would acquire a net momentum.

 

[Dale]

I am only asserting that the time that the box moves to the left and to the right must be equal.

No, one may be zero and the other non zero.

As you said, "the duration of the force must be equal." This is what I am saying.

And I am pointing out that the duration of the force and the duration of the movement are completely different things.

The box cannot move in "one direction" as you said because then the Earth would only receive an impulse in one direction

This is not correct. Again, force and movement are different things. For a constant mass the force is the derivative of the movement.

 

[Cutter Ketch]

The box might move to the left, let's say 1 centimeter, before coming to a stop after, let's say .5 seconds. But then the box must move to the right for .5 seconds and it doesn't matter what distance it travels: it could be only .3 centimeters or .1 centimeters. It doesn't matter

I think I'm beginning to catch on here. I think you are confusing impulse with movement. impulse is the change in momentum. It is proportional to the derivative of the displacement, not the displacement. If you accelerate to the right and then stop the impulse is toward the right when you accelerate, but toward the left when you stop. The impulse is both directions. The net impulse is zero. But the motion is only toward the right and the displacement is not zero.

That is what happens with the mass in the box. The mass is accelerated toward the right by the spring and then it hits the wall and is decelerated. (I'm still sticking it to the wall for simplicity). The force is in both directions first right and then left. The impulse is in both directions first right and then left. The net impulse is zero. But the motion, the displacement, is only toward the right. Likewise the box and the Earth are accelerated toward the left. Then when the ball hits the wall and stops they are decelerated and stopped by a force, an impulse to the right. However the movement, the displacement of the box and Earth is only to the left.

Mass moved right. Box and Earth moved left. All impulses were balanced and cancelled. The center of mass never moved.

 

[e2m2a]

Ok. Dale, maybe we are not understanding each other. I would be very happy to have my thinking cleared up if I am in error, and I would be the first to admit it if I am wrong. So be patient with my questions and my assertions.

Lets just look at the Earth for a moment in a free body diagram. When the box slides to the left, there will be an impulse on the Earth to the left equal to the normal force of the weight of the ball and box times the coeficient of friction times the duration that box slides to the left on the surface of the earth, relative to an inertial reference frame not attached to the earth. Do you agree with that?

Now what would the net impulse on the Earth be if the box did not slide to the right? Would it be zero?

 

[Cutter Ketch]

Ok. Dale, maybe we are not understanding each other. I would be very happy to have my thinking cleared up if I am in error, and I would be the first to admit it if I am wrong. So be patient with my questions and my assertions.

Lets just look at the Earth for a moment in a free body diagram. When the box slides to the left, there will be an impulse on the Earth to the left equal to the normal force of the weight of the ball and box times the coeficient of friction times the duration that box slides to the left on the surface of the earth, relative to an inertial reference frame not attached to the earth. Do you agree with that?

Now what would the net impulse on the Earth be if the box did not slide to the right? Would it be zero?

I'm sorry to interrupt this question to Dale, and I'm sure he'll answer. I just wanted to point something out. What you say here is true. Have you considered that the impulse you describe on the Earth actually causes the Earth to start moving? When the mass hits the wall and stops the box the friction with the Earth is going to have to stop the earth. There will be an impulse on the Earth to the right that would be easy to miss if you forget the Earth was accelerated.

 

[Dale]

When the box slides to the left, there will be an impulse on the Earth to the left equal to the normal force of the weight of the ball and box times the coeficient of friction times the duration that box slides to the left on the surface of the earth, relative to an inertial reference frame not attached to the earth. Do you agree with that?

Yes, where the coefficient of friction you refer to is the coefficient of kinetic friction and the time arbitrary.

For clarity, let's use the term "the system" to refer to the box and the ball together. That way we can distinguish between the movement of the box and the movement of the (center of mass) of the system. The whole purpose of the system arrangement is to decouple the motion of the box and the motion of the system.

Now what would the net impulse on the Earth be if the box did not slide to the right? Would it be zero?

No, it could be anything, with the maximum force being limited by the coefficient of static friction and the time being arbitrary.

 

[e2m2a]

I'm sorry to interrupt this question to Dale, and I'm sure he'll answer. I just wanted to point something out. What you say here is true. Have you considered that the impulse you describe on the Earth actually causes the Earth to start moving? When the mass hits the wall and stops the box the friction with the Earth is going to have to stop the earth. There will be an impulse on the Earth to the right that would be easy to miss if you forget the Earth was accelerated.

Yes. And there can only be an impulse on the Earth to the right if the box momentarily moves to the right after the ball collides. The box cannot come to a dead stop after the ball collides with the inner right wall. If it did, there would be no friction impulse on the Earth to the right.

 

[A.T.]

Yes. And there can only be an impulse on the Earth to the right if the box momentarily moves to the right after the ball collides. The box cannot come to a dead stop after the ball collides with the inner right wall. If it did, there would be no friction impulse on the Earth to the right.

Impulse can be transferred without relative movement, through static friction.

 

[e2m2a]

Yes, where the coefficient of friction you refer to is the coefficient of kinetic friction and the time arbitrary.

For clarity, let's use the term "the system" to refer to the box and the ball together. That way we can distinguish between the movement of the box and the movement of the (center of mass) of the system. The whole purpose of the system arrangement is to decouple the motion of the box and the motion of the system.

No, it could be anything, with the maximum force being limited by the coefficient of static friction and the time being arbitrary.

Ok. Now think about this. The box has come to a stop. There is no relative motion between the box and the earth. Now, the Earth and box will move at the same velocity to the left relative to an inertial frame because the Earth acquired a net impulse greater than zero to the left. This means the center of mass of the earth-box-ball system will be moving to the left, which means there will be momentum to the left of the 3 body system. This cannot happen unless we can account for an "external" force that acted on the 3 body system per Euler's first law.

 

[jbriggs444]

Ok. Now think about this. The box has come to a stop. There is no relative motion between the box and the earth. Now, the Earth and box will move at the same velocity to the left relative to an inertial frame because the Earth acquired a net impulse greater than zero to the left. This means the center of mass of the earth-box-ball system will be moving to the left,

The "box" and the "center of mass of the box and ball" are two different things that can be in relative motion.

 

[Dale]

the Earth and box will move at the same velocity to the left ... This means the center of mass of the earth-box-ball system will be moving to the left

No it doesn't. Remember, by design you have decoupled the motion of the box from the motion of the center of mass of the system. They are not the same thing, by design. So the fact that the box is moving left with the Earth does not imply that the CoM of the box+ball system is moving left.

 

[Cutter Ketch]

Yes. And there can only be an impulse on the Earth to the right if the box momentarily moves to the right after the ball collides. The box cannot come to a dead stop after the ball collides with the inner right wall. If it did, there would be no friction impulse on the Earth to the right.

No, that's my point. The Earth is moving left. Stopped box puts an impulse on the Earth to the right without the box moving right because the Earth is moving left.

I think the Earth may be confusing the issue. The temptation is to switch from it being part of the moving system to being the immobile reference. We might ought to put the box on a small cart that has no friction with the floor and see if the three body system of the internal mass the box and the cart makes sense.

 

[e2m2a]

No it doesn't. Remember, by design you have decoupled the motion of the box from the motion of the center of mass of the system. They are not the same thing, by design. So the fact that the box is moving left with the Earth does not imply that the CoM of the box+ball system is moving left.

Here is a mathematical proof of my assertion that the time the box slides to the left must be equal to the time the box slides to the right.

We define the left direction as the negative x- direction and the right direction as the positive x- direction.

In this derivation we will not use a ball, but just a small cube designated as m1. The box will be designated as m2.

The coefficient of friction between the surface of the Earth and the bottom of the box we designate as mu. We assume mu is constant for both sliding and static friction as in the case of Teflon. We also assume the surface m2 slides on is perfectly flat and level.

To simplify, we assume there is no friction between the bottom of m1 and the “floor” of the box m2. We can do this because any friction force pair between the bottom of m1 and the floor of m2 by Newton’s Third Law would cancel out any effect on the center of mass of the m1-m2 system. That is, these friction forces are “internal” to the m1-m2 system, and cannot impact the center of mass of the m1-m2 system by Euler’s first law.

A spring is rigidly attached to the inside of the left wall of m2. The mass of this spring is included as part of the mass of m2.

Initially, m1 is compressed against the spring. The spring is compressed by a light fish line. We ignore the mass of the fish line because we can make this mass insignificant by making m1 and m2 sufficiently large.

At some point in time, the line breaks, and immediately m2 accelerates in the negative x-direction, and m1 accelerates in the positive x-direction. We define the time interval that the spring force acts on m1 and m2 when there is physical contact between the free end of the spring and m1 as delta_ti, where delta_ti stands for the initial time interval.

During the expansion of the spring, a friction force acts on the bottom of the box in the positive x-direction as the box accelerates to the left. We define this force as ff, for friction force, where:

ff = mt x g x mu (1)

Where, mt is the total mass of the m1-m2 system and g is the acceleration due to gravity.

During the expansion of the spring it is axiomatic by Newton’s third law that the magnitude of the forces on m2 and m1 due to this spring are always equal, but opposite in direction.

Since this force is not constant, we must integrate this force. We designate this force obtained by integration as If or integrated force. Thus, the impulses I experienced by m1 and m2 during the expansion of the spring we denote as I = If x delta_ti.

I1 is the impulse on m1 and I2 is the impulse on m2 due to the expansion of the spring. Note that I1 plus I2 must equal zero. We are now ready to set up some equations.

During the expansion of the spring, the net impulse acting on m2 is:

-I2 + ff x delta_ti (2)

The net impulse on m1 is:

I1 (3)

At the end of the initial time interval delta_ti, both m1 and m2 have acquired a velocity or momentum derived as follows:

Using the definition of impulse,

I = mvf – mvi

Where I is impulse, m is mass, and vf and vi are the final and initial velocity of the mass, we have for the momentum of m2, after rearranging terms:

m2v2f = Inet2 + m2v2i (4)

Where, m2v2f is the final momentum of m2, m2v2i is the initial momentum of m2, and Inet2 is the net impulse on m2 during the expansion of the spring.

The net impulse on m2 during the expansion of the spring is defined as:

Inet2 = -I2 + ff x delta_ti (5)

The initial momentum of m2 is zero. Thus, the final momentum of m2 after the expansion of the spring is:

m2v2f = -I2 + ff x delta_ti (6)

By the same analysis, the final momentum of m1 after the expansion of the spring is:

m1v1f = I1 (7)

Remember, there are no friction forces between m1 and m2. (If we included these friction forces, they would cancel out in the final derivation).

In this derivation we let m1 collide with the inside of the right wall of m2 with an inelastic collision immediately after m1 has no contact with the spring. (After the spring has reached its non-compressed state).

We now derive the final momentum of the m1-m2 system after the collision as follows:

m1v1i + m2v2i = mtvf (8)

Where, mt is the total mass of m1 and m2, m1v1i and m2v2i are the initial velocities of m1 and m2, and vf is the final velocity of the m1-m2 system.

Substituting equations(6) and (7) into (8), we have:

mtvf = I1 + -I2 + ff x delta_ti (9)

This simplifies to:

mtvf = ff x delta_ti (10).

Solving for the final velocity of the m1-m2 system, we have:

vf = (ff x delta_ti)/mt (11).Note that vf is positive, or the m1-m2 system has acquired a velocity in the positive x-direction after the collision of m1 with m2.

Now we are interested in finding the time interval it takes for the m1-m2 system to come to a complete stop. We can use the kinematic equations for constant acceleration to do this since the only force acting on the m1-m2 system is friction.

We use this general kinematic equation for velocity:

vf = vi + a x delta_tf

Where, vf is the final velocity, vi is the

initial velocity, a is the acceleration due to friction and delta_tf is the final time interval.

We set vf to zero and solve for delta_t, giving us:

delta_tf =vi/a (12)

Here, a = g x mu, where g is the acceleration due to gravity and mu the coefficient of friction we previously defined.

Now we can substitute equation (11) for vi in equation(12), giving us:

delta_tf = (ff x delta_ti)/(mt x a) (13)

And we can substitute equation(1) for ff and g x mu for a, giving us:

delta_tf = (mt x g x mu x delta_ti)/(mt x g x mu) (14).

After dividing out like terms, we are left with:

delta_tf = delta_ti (15).

QED

 

[jbriggs444]

During the expansion of the spring, a friction force acts on the bottom of the box in the positive x-direction as the box accelerates to the left. We define this force as ff, for friction force, where:

ff = mt x g x mu (1)

Where, mt is the total mass of the m1-m2 system and g is the acceleration due to gravity.

If box m2 does not move relative to the Earth, ff is bounded above by mt x g * mu ( 1 ) but need not attain exactly this value.

 

[sophiecentaur]

The OP seems to be thinking that some basic principle is being violated here. But the model is not being strictly adhered to, I think, and that is producing confusion. Energy is (or can be) expended from some source inside the box, which will allow it to 'hotch' along by using the difference between static and dynamic friction.

 

[Dale]

Here is a mathematical proof

Seriously, if you are going to post stuff like this you really need to use LaTeX. It is illegible otherwise. I am simply going to assume that your algebra is correct. However, when proving something you want to use general cases, as much as possible. Otherwise you run the risk that you have proved something for your specific case, but not for the general case. That is what you have done here (assuming the math is otherwise correct). Although you may have shown one example where it works, there are other examples that show where it doesn't work.

For clarity let $m_1$ be the mass of the cube, $m_2$ be the mass of the box, $m=m_1 + m_2$ be the mass of the cube+box system, $f_1 (t)$ be the force on the cube, $f(t)$ be the external (friction) force on the box, and $f_2(t)=f(t) - f_1(t)$ be the net force acting on the box. We are neglecting vertical forces and motion. So by Newton's 2nd law we can write $f_2(t)=m_2 \dot v_2(t)$ for the motion of the box, $f_1(t)=m_1 \dot v_1(t)$ for the motion of the cube, and $f(t)=m \dot v(t)$ for the motion of the CoM of the system.

Now, let's suppose that we have the kind of friction force that you were describing where the maximum static friction is equal to the kinetic friction $F=\mu m g$, so if the box is moving at time $t$ then $f(t)= \pm F$ and if the box is stationary at time $t$ then $-F<f(t)<F$. So then as a counter example to your proof, let's consider
$$f(t) = \begin{cases}
F/2 & 0<t<2 \\
-F & 2<t<3 \\
0 & otherwise
\end{cases}$$
$$f_2(t) = \begin{cases}
F & 2<t<2.5 \\
-F & 2.5<t<3 \\
0 & otherwise
\end{cases}$$
$$f_1(t)=f(t)-f_2(t)$$
If you integrate these functions out then you will find that $v(0)=v_1(0)=v_2(0)=0$ and $v(3)=v_1(3)=v_2(3)=0$ so the system starts and stops at rest. You will also find that $v_2(t)>0$ for $2<t<3$ and that it is never less than 0. So the time that the box moved to the right is 1 and the time that it moved to the left is 0, so this is a counter example to your proof, obtained by considering more general force profiles than the one you considered.

 

[e2m2a]

Seriously, if you are going to post stuff like this you really need to use LaTeX. It is illegible otherwise. I am simply going to assume that your algebra is correct. However, when proving something you want to use general cases, as much as possible. Otherwise you run the risk that you have proved something for your specific case, but not for the general case. That is what you have done here (assuming the math is otherwise correct). Although you may have shown one example where it works, there are other examples that show where it doesn't work.

For clarity let $m_1$ be the mass of the cube, $m_2$ be the mass of the box, $m=m_1 + m_2$ be the mass of the cube+box system, $f_1 (t)$ be the force on the cube, $f(t)$ be the external (friction) force on the box, and $f_2(t)=f(t) - f_1(t)$ be the net force acting on the box. We are neglecting vertical forces and motion. So by Newton's 2nd law we can write $f_2(t)=m_2 \dot v_2(t)$ for the motion of the box, $f_1(t)=m_1 \dot v_1(t)$ for the motion of the cube, and $f(t)=m \dot v(t)$ for the motion of the CoM of the system.

Now, let's suppose that we have the kind of friction force that you were describing where the maximum static friction is equal to the kinetic friction $F=\mu m g$, so if the box is moving at time $t$ then $f(t)= \pm F$ and if the box is stationary at time $t$ then $-F<f(t)<F$. So then as a counter example to your proof, let's consider
$$f(t) = \begin{cases}
F/2 & 0<t<2 \\
-F & 2<t<3 \\
0 & otherwise
\end{cases}$$
$$f_2(t) = \begin{cases}
F & 2<t<2.5 \\
-F & 2.5<t<3 \\
0 & otherwise
\end{cases}$$
$$f_1(t)=f(t)-f_2(t)$$
If you integrate these functions out then you will find that $v(0)=v_1(0)=v_2(0)=0$ and $v(3)=v_1(3)=v_2(3)=0$ so the system starts and stops at rest. You will also find that $v_2(t)>0$ for $2<t<3$ and that it is never less than 0. So the time that the box moved to the right is 1 and the time that it moved to the left is 0, so this is a counter example to your proof, obtained by considering more general force profiles than the one you considered.

You are right that I should learn LaTex, but I have attempted it before and I could not find a place to practice it.

And you are right that my derivation was for only a special case and not a general expression. But I wasn't trying to prove a general case. Just the special case with the given initial conditions.

Thank you for looking at my derivation and for your suggestions.

 

[Dale]

But I wasn't trying to prove a general case.

Then you should have stated your claim as a specific claim rather than a general claim. I doubt that anyone would object to a narrow claim like "in this scenario the box moves right for the same amount of time as to the left" as opposed to the broad claim "the box must move to the right for the same amount of time as to the left" (which is not generally true).

One other point I noticed in working through this that the Earth is not relevant. The net impulse is zero, so it winds up at rest regardless of its mass.

You are right that I should learn LaTex, but I have attempted it before and I could not find a place to practice it.

You can practice here. When you are composing a post there is a button marked "preview" which allows you to see your message without posting it. Checking my LaTeX is the main thing that I use it for.

Thank you for looking at my derivation and for your suggestions.

Note, I didn't check your derivation carefully, it was too difficult to read without LaTeX. So I am not saying it is correct, just assuming that it was correct for the sake of argument. I have no objection to the specific claim that there exists some scenario where the times happen to be equal. I only object to the general claim that the times must be equal.