# A and C collide with B inbetween; AB elastic, BC inelastic

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1. Homework Statement
Three bodies A,B,C on frictionless surface
masses= 1 kg each,
Positions at time 0: A is at x=0,. B is at x=1, C is at x=2 (unit is 1 meter).
velocities at time 0: A : 1m/s (to the right), B = 0 m/s, C = -1 m/s
Assume sizes are uniform, or just ignore the sizes in calculations
Collision A-B is elastic. Collision C-B is inelastic, and C sticks to B.
Question: what direction is the motion of the BC combination after collision, and why?
The relevant choices:
(a) to the right (i.e., positive velocity), due to the conservation of momentum
(d) to the left (i.e., negative velocity), due to difference of time force is applied by the respective bodies

2. Homework Equations
conservation of momentum, conservation of force , and (with v, F, p being vectors): p=mv, Fp/Δt (possibly KE=mv2/2, E= KE + friction, but I doubt it)

3. The Attempt at a Solution
My choice was (a) , due to the following: total momentum before and after, zero, so if there are two parts, they must be opposite signs. Since you cannot have A passing through BC, A then goes to the left (has a negative velocity), and BC has a positive. Hence BC goes to the right.
However, the answer is given as (d).
Analyzing it in terms of forces and time

The forces must also sum up to zero, that is, if after the collision the velocities are vA and vBC, then [mv-mvA]/ΔtA = [-(-Mv-2MvBC)]/ΔtBC, which has two variables, so assuming WLOG that the loss of energy due to friction is negligible, and that the velocities after the collision are both negative (as D would imply), I do not end up with a total zero force.
Is my answer correct, or is the answer given correct?
Thank you.

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#### haruspex

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My choice was (a) , due to the following: total momentum before and after, zero, so if there are two parts, they must be opposite signs. Since you cannot have A passing through BC, A then goes to the left (has a negative velocity), and BC has a positive. Hence BC goes to the right.
Yes.

Option d makes no sense anyway because you have no information on the durations of the collisions.

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Thanks, haruspex. This appears to be a mistake by the American College Board. Actually, a small apology: I wrote down A and D as the relevant solutions; that should have been A and B, but it appears to me that B is also incorrect. Here is the original question #### Attachments

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#### haruspex

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Thanks, haruspex. This appears to be a mistake by the American College Board. Actually, a small apology: I wrote down A and D as the relevant solutions; that should have been A and B, but it appears to me that B is also incorrect. Here is the original question
View attachment 225188
That is rather different.
As you originally posted, the glider moves to the right because momentum is conserved, but that is not the reason provided in A. So we know the answer is A or B, but which offers a valid explanation?
E.g. for A, is it true that:
1. The clay ball undergoes the greater change in momentum, and
2. That proves it moves to the right
?
Edit: I meant
1. The rubber ball undergoes the greater change in momentum, and

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Thanks again, haruspex. Precisely the apparent non-sequitur between the change of momenta and the direction made me pose the question. Answer A sounded better than answer B
But the connection between conservation of momentum and the way the answer is stated in A is not clear. However, to justify that at least the clay ball undergoes the greater change of momentum, I reasoned as follows:

The difference in the change of the magnitude of the momentum of the rubber ball versus the change of momentum of the class ball comes down to a comparison of the momenta after the collision, because before the collision the momenta (magnitudes) are equal.
Because after the collision, the momentum of the rubber ball (after) = the magnitude of the momentum of the (clay ball+ glider), that is,
Mvr= (M+m)vc > Mvc
So, the clay ball undergoes the greater change of momentum.

But I could see even less connection in the explanation given in answer B, because the whole system includes an inelastic collision, so the kinetic energy is not conserved, so the whole system is an inelastic collision, not an elastic one. Therefore to say that the collision of the rubber ball is an elastic collision is not really meaningful.

Therefore, I am still confused as to what answer would be the correct one (there is no "none of the above".)

#### PeroK

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These questions are poor, IMHO, because there is no one reason per se. The fact that the balls are of equal mass is important. If the clay ball were more massive, then everything would move to the left after the collision. And, as you pointed out, the fact that ball A is physically constrained to the left of ball B is also important.

Energy does not have a direction. Momentum is the vector quantity, so seems more directly related to the final direction of motion than energy.

My argument, however, would be that both energy and momentum are relevant to a collision so you ought to consider both in looking for a solution.

If I were learning the material I would be satisfied that I could analyse the problem using both energy and momentum and that's all that's important.

• #### haruspex

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Not sure if it matters for the question posed, but a full analysis would have to take into account that the order of events matters. Does the inelastic collision occur first or the elastic one? It is no good saying they're simultaneous because there is no method to analyse that as an instantaneous event. If they overlap then you get into the question of how long each contact lasts.

#### haruspex

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Mvr= (M+m)vc> Mvc
So, the clay ball undergoes the greater change of momentum.
I draw the opposite conclusion.

It seems to me that in both A and B the predicate and the conclusion are true, but the conclusion does not follow from the predicate. E.g. with the clay ball having the greater mass:
In A, we could have the greater momentum change in the rubber ball and yet both move off to the left.
In B, in an elastic collision with an initially stationary mass, the greater mass will continue in its original direction.

• #### ehild

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The problem says that both balls have the same mass m and same speed v. The momentum of the system is zero. If they hit the target at the same time the glider gets two equal and opposite momenta. Assume the mass of the glider M>>m. After the collision, the rubber ball will travel about the same speed but in the opposite direction (to the left), with momentum -mv. As the overall momentum is conserved, (still zero) what is the momentum of the glider and clay ball stuck to it? In what direction does the glider travel?

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First, thank you, PeroK, haruspex, and ehild. I think I have figured out what the examiners meant, but they just phrased it very, very sloppily. I think they just wanted to rule out the case where the rubber ball also sticks to the glider, so that you would have nothing moving after the collision -- that what they meant with B, even though technically what they said was not correct.
PeroK : I understand your case when M(glider)>>m (balls), that the rubber ball will travel with approximately the same speed (opposite velocity) after collision as before. But this is approximate only, and the approximation gets worse as M gets closer to m. This problem said nothing about such a condition, so this is a different problem.

#### PeroK

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PeroK : I understand your case when M(glider)>>m (balls), that the rubber ball will travel with approximately the same speed (opposite velocity) after collision as before. But this is approximate only, and the approximation gets worse as M gets closer to m. This problem said nothing about such a condition, so this is a different problem.
I'm not sure what "my case" is. There are numerous ways to explain the problem. Giving the underlying reason as momentum only (answer a)) seems poor to me, given the required assumption about energy and elastic collisions. The critical difference between the balls is the elasticity of their respective collisions, so, in a sense, must be the underlying reason! The different changes in their momentum is a result of the elasticity of the collisions.

In any case, there is nothing deep in this question that enhances our understanding of energy-momentum: it's really a question more of phraseology.

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oops
I'm not sure what "my case" is.
Sorry, PeroK, that was very poorly phrased (I was in a hurry when I typed it). My sincere apologies for that offensive phrasing. I meant the case posed by you: M>>m.
I agree that this is not a very deep problem: it is not meant to be -- a teacher of secondary school turned to me with this problem, and I found the phrasing so odd as to not point to any of the four answers, hence I thought to turn to the kind people on this forum.

#### PeroK

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oops

Sorry, PeroK, that was very poorly phrased (I was in a hurry when I typed it). My sincere apologies for that offensive phrasing. I meant the case posed by you: M>>m.
I agree that this is not a very deep problem: it is not meant to be -- a teacher of secondary school turned to me with this problem, and I found the phrasing so odd as to not point to any of the four answers, hence I thought to turn to the kind people on this forum.
It's okay. $M >>m$, I've just noticed, was part of @ehild 's analysis.

#### haruspex

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The problem says that both balls have the same mass m and same speed v. The momentum of the system is zero. If they hit the target at the same time the glider gets two equal and opposite momenta. Assume the mass of the glider M>>m. After the collision, the rubber ball will travel about the same speed but in the opposite direction (to the left), with momentum -mv. As the overall momentum is conserved, (still zero) what is the momentum of the glider and clay ball stuck to it? In what direction does the glider travel?
I think we all agree that by using conservation of momentum it is clear that the glider will move to the right. That was set out in post #1. (No need for M>>m, is there?)
But that does not answer the question posed. It requires us to consider some other lines of reasoning and say which of them leads also to that conclusion (validly).

That said, as I showed in post #8, none of the options achieves that.

I see that nomadreid offers another interpretation in post #10, but I am not persuaded. The original phrasing seems perfectly clear.

#### ehild

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Collision involving three bodies can not be solved in general without detailed knowledge of the collision process.
Here, option A is valid and clear if m<<M.
B is not, energy is not conserved in the process, and if conserved, it does not say anything about the direction of the velocities.
C speaks about greater inertia of the clay ball - has it sense?
D is about time of interactions, which can not be validated.

• #### haruspex

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option A is valid
How so? How does merely knowing that the change in momentum of the rubber ball has a greater magnitude than the change in momentum of the clay ball imply that the glider moves to the right?
Bear in mind that you cannot use the other given facts, like equal masses and speeds, in answering that. This is a question about logical inference.
E.g., as I wrote in post #8, we could increase the mass of the clay ball to such an extent that the glider continues to the left and yet the rubber ball still has the greater momentum change. Thus, the predicate does not imply the conclusion.
B is not, energy is not conserved in the process
It only claims that the rubber ball's collision conserves energy.

#### ehild

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Well, I think we have to solve the given problem, with equal masses and initial speeds.

#### haruspex

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Well, I think we have to solve the given problem, with equal masses and initial speeds.
The OP solved the question of which way it moves in post #1, using conservation of momentum, and making no assumptions about the mass of the glider. But as clearly laid out in post #3, that is only part of what is being asked. It says "and explains why it moves in that direction".
Option A says it is explained by the rubber ball having the greater change in momentum. Well, it does have the greater change in momentum, but does that constitute an explanation?

To assess that you cannot go combining it with all the other known facts and ask if that combination leads to the actual movement. For the reasoning to be valid, the conclusion must follow purely from the predicate; and it does not.

• #### neilparker62

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Whatever the change in momentum (impulse) experienced by A will be equally experienced by B (in the opposite direction). Similarly for C on B. But when considered separately the elastic AB collision impulse is twice that of the perfectly inelastic CB impulse. I would guess if the two happen simultaneously, the resultant impulse on B would simply be the sum of the afore-mentioned. Hence to the right.

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#### haruspex

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Hence to the right.
As noted in previous posts, everyone agrees that the motion will be to the right. The question is whether explanation A or B, if either, is the reason.
As I explained in post #8,I do not accept either as a sufficent reason in itself.

#### kuruman

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How so? How does merely knowing that the change in momentum of the rubber ball has a greater magnitude than the change in momentum of the clay ball imply that the glider moves to the right?
Consider the impulses delivered to the glider by the two balls. If the change in momentum of the rubber ball is greater (in magnitude) than that of the clay ball, then there will be a net impulse to the right. Isn't that sufficient? I think choice A does it with the tacit understanding that Newton's 3rd law comes into play.

#### neilparker62

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From my post I would say reason A. Change in momentum of ball A equates to the collision impulse. The impulse for collision AB is twice that for collision CB. For elastic collisions impulse Δp = 2μΔv and for perfectly inelastic it's just μΔv. But the result Δp = 2μΔv is applicable only to elastic collisions in which energy is conserved so reason B is equally applicable if a little less 'direct'.

#### haruspex

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Consider the impulses delivered to the glider by the two balls. If the change in momentum of the rubber ball is greater (in magnitude) than that of the clay ball, then there will be a net impulse to the right. Isn't that sufficient? I think choice A does it with the tacit understanding that Newton's 3rd law comes into play.
That works if we change A to read "the glider moves to the right because it started at rest and the magnitude.." etc. (Even then, formalising it is quite tricky.)
You may think I am being picky, but the whole basis for discriminating between A and B is that you must put aside all other information. For example, @neilparker62 in post #22 makes the same mistake as several others in this thread of reusing the information that the masses are equal.

#### kuruman

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That works if we change A to read "the glider moves to the right because it started at rest and the magnitude.." etc. (Even then, formalising it is quite tricky.)
You may think I am being picky, but the whole basis for discriminating between A and B is that you must put aside all other information. For example, @neilparker62 in post #22 makes the same mistake as several others in this thread of reusing the information that the masses are equal.
I think that both the equality of the masses and speeds of the two balls and the given that the glider is initially at rest are important considerations to the correct answer. If I were to write an explanation, I would say
"The glider moves to the right because the center of mass of the three-mass system is initially at rest."
The connective reasoning then is (a) because momentum is conserved, the CM must also be at rest after the collision, therefore the rubber ball and glider+clay ball must be moving in opposite directions or not at all; (b) the "not at all" option would be the case only if both balls stick to the glider (c) because the rubber ball cannot go through the glider, it must reverse direction, therefore the glider+clay ball must move to the right.

This reasoning also shows that the collisions need not be simultaneous. Regardless of the order of their occurrence, when all is said and done, the rubber ball will be moving to the left and the glider+clay ball to the right.

#### neilparker62

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For example, @neilparker62 in post #22 makes the same mistake as several others in this thread of reusing the information that the masses are equal.
Yes - perhaps I was being a little 'impulsive' in my assumptions :-)

Another point which occurs to me (in line with your assertion that the given reasons are insufficient) is that impulse is not force and we would need time information - for the respective changes in momentum (of balls A and C - assuming they have the same mass!) before we could decide which exerted the greater impact force.

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