Conservation of Momentum (Elastic Collisions)

Jimmy87

Problem Statement
Part (iii)
Relevant Equations
conservation of momentum and conservation of kinetic energy Part (iii) is the part I am stuck on and is a 5 mark question. I have some idea of how to attempt it shown below

momentum is conserved so mux = mvy + mvz

(where ux is the initial velocity before the collision of ball x, vy is the velocity after the collision of ball y and vz is the velocity after the collision of ball z)

They are all indentical masses so they all cancel to give ux = vy + vz

Since kinetic energy is conserved then we have:

½(mux)2 = ½(mvy)2 + ½(mvz)2

The halves and masses cancel to give:

ux2 = vy2 + vz2

Since ux = vy + vz from earlier we have:

(vy + vz)2 = vy2 + vz2 which gives:

vy2 + vz2 + 2vyvz = vy2 + vz2

2vyvz = 0

Now I am stuck as I was going to say that vy must be zero because vz isn't (it is 1.5 m/s). Then I realised the question asks to prove vz is not zero so I can't really use this assumption?

edit: I was going to go on to say that since ball y must be zero then we can go back to the original equation which is ux = vy + vz to say that ux = vz and the question tells us that ux is 1.5 m/s. However, I have had to assume vy is zero but it doesn't give me this information in the question so I can't really do that can I?

Any help is appreciated

Last edited:
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haruspex

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Problem Statement: Part (iii)
Relevant Equations: conservation of momentum and conservation of kinetic energy

View attachment 243907

Part (iii) is the part I am stuck on and is a 5 mark question. I have some idea of how to attempt it shown below

momentum is conserved so mux = mvy + mvz

(where ux is the initial velocity before the collision of ball x, vy is the velocity after the collision of ball y and vz is the velocity after the collision of ball z)

They are all indentical masses so they all cancel to give ux = vy + vz

Since kinetic energy is conserved then we have:

½(mux)2 = ½(mvy)2 + ½(mvz)2

The halves and masses cancel to give:

ux2 = vy2 + vz2

Since ux = vy + vz from earlier we have:

(vy + vz)2 = vy2 + vz2 which gives:

vy2 + vz2 + 2vyvz = vy2 + vz2

2vyvz = 0

Now I am stuck as I was going to say that vy must be zero because vz isn't (it is 1.5 m/s). Then I realised the question asks to prove vz is not zero so I can't really use this assumption?

edit: I was going to go on to say that since ball y must be zero then we can go back to the original equation which is ux = vy + vz to say that ux = vz and the question tells us that ux is 1.5 m/s. However, I have had to assume vy is zero but it doesn't give me this information in the question so I can't really do that can I?

Any help is appreciated
You know that vy and vz cannot both be zero.
What is happening if vz is zero?

• Jimmy87

Jimmy87

You know that vy and vz cannot both be zero.
What is happening if vz is zero?
If vz is zero then vy has to be moving which isn't possible if they are next to each other. The question stills says to show that ball x is stationary after the collision and I don't get how to show that mathematically either.

Jimmy87

You know that vy and vz cannot both be zero.
What is happening if vz is zero?
Ok hang on I have over complicated this. Forget my vy term and replace it with vx (velocity of ball x after the collision). I can do this since the question before shows that vy has no resultant force and hence must be at rest anyway. This gives me:

2vxvz = 0

So can I now say vx is zero so vz has to be 1.5 m/s. I am still uncomfortable with this as I know vx is zero but I am not actually still proving it am I?

edit: or can I just say vx must be zero since the last question shows ball y is stationary so it is impossible for vx to carry the 1.5m/s and have y stationary? Although this is a qualitative answer rather than a quantitative one.

kuruman

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Can you show that when two identical masses collide elastically they exchange velocities? Once you have shown this, then you can apply it here. See Newton's cradle.

haruspex

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Can you show that when two identical masses collide elastically they exchange velocities? Once you have shown this, then you can apply it here. See Newton's cradle.
The trouble with that approach is that it takes the balls to be not quite touching. Indeed, sufficiently separated that the first two balls lose contact before the second makes contact with the third.
The question here specifies that they are touching. Since a collision always involves some compression, however slight, things are more complicated.
I attempted a general solution assuming the balls have identical "spring constants", and found Newton's Cradle doesn't work! So I looked it up online and found that, indeed, a small separation and correspondingly high spring constants are required.
The question as given in this thread is flawed.

kuruman

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The question as given in this thread is flawed.
That is true. However it is a problem that is intended to be answered using introductory level physics, in this case momentum and energy conservation. Solutions that involve pressure waves, normal modes of vibration etc. are beyond the scope of the problem. One could question the exact meaning of "touching". The masses cannot come any closer than the repulsive forces between atoms will allow so there is a gap in between after all. As with other collision problems. it is necessary to make certain assumptions in order to be able to proceed with the solution. For example, with the ballistic pendulum one assumes that the bullet is completely embedded in the pendulum mass before the combined mass moves appreciably. Here, the assumption is that all the momentum and energy of mass $x$ are transferred to mass $y$ before any momentum and energy from mass $y$ is transferred to mass $z$.

haruspex

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it is a problem that is intended to be answered using introductory level physics, in this case momentum and energy conservation.
The solution I found obeys those, therefore the problem is not solvable without making additional assumptions. To get the expected answer must require an unrealistic assumption, but there is no way to be sure which one the question setter made.
One could question the exact meaning of "touching".
In regards to getting the Newton's Cradle result, not touching means that there is no instant at which both neighbours of a ball are interacting with it to any significant extent.
Here, the assumption is that all the momentum and energy of mass x are transferred to mass y before any momentum and energy from mass y is transferred to mass z.
I.e., not touching.

neilparker62

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Alternative solution via transfer of momentum - elastic collision:

a) From ball X to ball Y
$$Δp=2μΔv=2\frac{m^2}{m+m}Δv=mΔv$$
Since the transfer of momentum is equal and opposite, ball X comes to a standstill and ball Y takes off (momentarily) with velocity Δv (1.5 m/s in this case).

b) From ball Y to ball Z
$$Δp=2μΔv=2\frac{m^2}{m+m}Δv=mΔv$$
Since the transfer of momentum is equal and opposite, ball Y comes to a standstill and ball Z takes off with velocity Δv (1.5 m/s).

All but trivial when looked at from this perspective! Note that in principle, ball Y experiences two successive (almost instantaneous) impulses which are equal and opposite in magnitude. Therefore no net change in momentum and no net force.

haruspex

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Alternative solution via transfer of momentum - elastic collision:

a) From ball X to ball Y
$$Δp=2μΔv=2\frac{m^2}{m+m}Δv=mΔv$$
Since the transfer of momentum is equal and opposite, ball X comes to a standstill and ball Y takes off (momentarily) with velocity Δv (1.5 m/s in this case).

b) From ball Y to ball Z
$$Δp=2μΔv=2\frac{m^2}{m+m}Δv=mΔv$$
Since the transfer of momentum is equal and opposite, ball Y comes to a standstill and ball Z takes off with velocity Δv (1.5 m/s).

All but trivial when looked at from this perspective! Note that in principle, ball Y experiences two successive (almost instantaneous) impulses which are equal and opposite in magnitude. Therefore no net change in momentum and no net force.
Yes, that's all fine if there is never a time that the central ball is subject to forces from both its neighbours.
But in this thread we are told that all the balls are initially touching their neighbours. Not only does that invalidate your method, it really does change the outcome.

haruspex

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I modeled the three ball case and obtained these displacement curves Where the red and blue lines cross, the first and second ball lose contact; similarly the second and third lose contact where the red and yellow lines cross.
Finding those points analytically involves solving equations like $\sin(\sqrt 3\theta)=\sqrt 3\sin(\theta)$.

This model omits that it takes time for the compression wave to transit a ball. So in practice the behaviour may be closer to a "perfect" Newton's Cradle, i.e. as though no simultaneous contact occurs.
However, that does not excuse the question setter in this thread. The analysis required is surely beyond a reasonable level for the student. (Well, it's beyond me anyway.)

neilparker62

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Yes, that's all fine if there is never a time that the central ball is subject to forces from both its neighbours.
But in this thread we are told that all the balls are initially touching their neighbours. Not only does that invalidate your method, it really does change the outcome.
Isn't this a more general problem with collision theory: namely that we consider the changes in momentum to occur instantaneously and do not consider at all the large transient forces which actually lead to those changes? (I think that part iii of the OP's problem merely highlights the fact that there must be such transients since the transfer of momentum from ball X to ball Z occurs in 2 distinct steps)

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haruspex

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Isn't this a more general problem with collision theory: namely that we consider the changes in momentum to occur instantaneously and do not consider at all the large transient forces which actually lead to those changes? (I think that part iii of the OP's problem merely highlights the fact that there must be such transients since the transfer of momentum from ball X to ball Z occurs in 2 distinct steps)
No, it becomes a problem when there are multiple bodies and the order of events is uncertain. As long as you can reduce it to a known sequence of two body interactions it is reasonably straightforward.

neilparker62

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I modeled the three ball case and obtained these displacement curves
Could you perhaps show a graph of corresponding velocity curves ?

kuruman

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I modeled the three ball case and obtained these displacement curves
View attachment 244041
Where the red and blue lines cross, the first and second ball lose contact; similarly the second and third lose contact where the red and yellow lines cross.
Finding those points analytically involves solving equations like $\sin(\sqrt 3\theta)=\sqrt 3\sin(\theta)$.
I got very similar plots for the displacements (see below) in the lab frame. I have also appended a graph of the velocities. Why is x1 = x2 the criterion for separation and not v2 - v1 > 0? The two do not occur at the same time.  haruspex

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Why is x1 = x2 the criterion for separation
If x1>x2 then 1 and 2 are in a state of mutual compression.
I got very similar plots for the displacements
Except that you have modelled them as stuck together, going into tension rather than separating.

haruspex

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Could you perhaps show a graph of corresponding velocity curves ? Visually, it looks like rounding errors have led to a net gain in KE, but I checked for that and it is negligible. Just goes to show the quadratic dependence.

• neilparker62

kuruman

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If x1>x2 then 1 and 2 are in a state of mutual compression.
Perhaps I do not understand what x1 and x2 stand for. I thought they were the displacements of the center of each mass. In my model the masses can be considered as point masses for the purposes of writing down the equations, however the condition for separation would be $x_2 - x_1 \geq D$ where $D$ is the diameter of the disks.
Except that you have modelled them as stuck together, going into tension rather than separating
That is true and I apologize for not making it clear that I did not modify the model to take into account the first and second separations. That's because I wanted to be sure I understood the meaning of separation before I proceeded.

haruspex

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I thought they were the displacements of the center of each mass.
Quite so... displacements from their respective positions when the balls are in contact but uncompressed. Thus, x1-x2 is a reduction in the distance between the first and second mass centres, so is the extent of compression.

kuruman

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I think this question can be understood independently of the model simply on the basis of energy and momentum conservation and the indistinguishability of the colliding masses. First consider the collision between two arbitrary masses. The momentum conservation equation results in the equality $m_1 (v_{1f}-v_{1i})=-m_2( v_{2f}-v_{2i})$. This allows dividing the energy conservation equation by each of the terms to reduce it from quadratic to linear. The operation is justified by observing that there is no division by zero as the final velocities must be different from the initial velocities if there is to be a collision. In other words, we eliminate the case that the masses move through each other. This is not because it is physically impossible for this to happen; the masses can move past each other on parallel tracks, which is certainly a situation that conserves both energy and momentum.

Now consider a thought experiment in which two masses are moving on a single track, say towards each other. Part of the track is hidden from view and we are told that behind the screen the track divides into parallel tracks which merge again so that a collision may or may not occur depending on whether one of the balls is switched to the other track. The question is, "Can we tell if there has been a collision behind the screen?" In general the answer is "yes". All we have to do is see if the outgoing velocities are different from the incoming velocities which will eliminate the "move through" case. When the masses are equal however, we cannot tell whether one of the balls was switched to the other track because the collision, if it occurs, simply exchanges velocities and all we see in either case is the same incoming and outgoing velocities.

Suppose now we are told that that that there is an identical ball at rest behind the screen. We see a ball go in and another come out. Can we tell whether there has been a collision? Of course not. Suppose instead we are told that there two balls at rest behind the screen. We send a ball in. Can we tell from what comes out whether there has been a collision or not? More specifically, if we can ascertain that there has been a collision, can we tell whether the incoming ball collided on a track bearing both balls? If only one ball comes out, then it has to have the same velocity as the incoming ball and we cannot tell whether there has been a collision.

Can two balls come out? Energy and momentum conservation require that
$$v_0^2=v_1^2+v_2^2$$
$$v_0=v_1+v_2$$
The first equation is the Pythagorean theorem for a triangle of hypotenuse $v_0$ and right sides $v_1$ and $v_2$. The second equation asserts that the sum of the two right sides is equal to the hypotenuse. The two equations can be simultaneously true only if one of the outgoing velocities is zero and the other is equal to $v_0$.

It is easy to show in general that, if $C^2=\sum_i x_i^2$ and $C=\sum_i x_i$, ($C > 0$), then only one of the $x_i$ is non-zero and equal to $C$. It follows that the rule "one mass in, one mass out" can be generalized to any number of balls on any number of tracks behind the screen and that the indistinguishable "move through" case must result from the conservation equations. Furthermore, the rule can be extended to $N$ masses in $N$ masses out with any number of tracks, balls and switches behind the screen. All we have to do is split the incoming masses to one per track behind the screen and bring them back together before they exit.
(I shall need a different argument in view of $21. Last edited: haruspex Science Advisor Homework Helper Gold Member 2018 Award It is easy to show in general that, if$C^2=\sum_i x_i^2$and$C=\sum_i x_i$, ($C > 0$), then only one of the$x_i$is non-zero and equal to$C$. C=1, xi=2/3, 2/3, -1/3. kuruman Science Advisor Homework Helper Gold Member C=1, xi=2/3, 2/3, -1/3. Thanks, it's not as easy as I deluded myself into believing by not considering values of$x_i$less than 1, yet another blind spot. I need to rephrase the argument, if there is a way and I can find it. How does Nature pick the solution? (Rhetorical question). haruspex Science Advisor Homework Helper Gold Member 2018 Award Thanks, it's not as easy as I deluded myself into believing by not considering values of$x_i$less than 1, yet another blind spot. I need to rephrase the argument, if there is a way and I can find it. How does Nature pick the solution? (Rhetorical question). As I posted, I think the transit time of the shock wave through each ball is important. To make this easier to model, could consider instead cylinders placed end to end, forming a rod. The result may be a wave that travels through the rod, whipping off the last cylinder. neilparker62 Homework Helper View attachment 244081 Visually, it looks like rounding errors have led to a net gain in KE, but I checked for that and it is negligible. Just goes to show the quadratic dependence. Many thanks - will need to take some time to digest what is going on here. Very different dynamics from that envisaged by the question setter (as you have pointed out). kuruman Science Advisor Homework Helper Gold Member As I posted, I think the transit time of the shock wave through each ball is important. To make this easier to model, could consider instead cylinders placed end to end, forming a rod. The result may be a wave that travels through the rod, whipping off the last cylinder. Yes, the time is important. I am not knowledgeable enough to model the shock wave however whatever model works with the collision of two masses should work with any number of masses. I reached the point of asking myself what it means to have an "elastic collision". The standard answer, "one that conserves mechanical energy" is insufficient. I think the answer ought to be "one that conserves kinetic energy" or one with a coefficient of restitution (CoR) equal to 1. In collisions where only two masses at a time are involved there is no distinction, but in collisions where three (or more) masses may be in contact at the same time, there is a difference when, as in the present case, making contact is simultaneous but breaking contact is not. Here, when contact is broken between masses 1 and 2, the "spring" between masses 2 and 3 is still compressed. If, at that moment, a Maxwell's demon were to latch the spring permanently in the compressed state, we would conclude that the collision is inelastic because some of the kinetic energy has been converted into internal energy amd is therefore mot conserved. I have been seeking ways to make the collision time very short, but "very short" relative to what? Two periods are involved in the three-masses-with-springs model but both are scale with$\sqrt{m/k}$and have a constant ratio of$\sqrt{3}$. Thus, increasing$k$to make the springs stiffer doesn't buy much. A second time scale related to the collision is needed to compare$\sqrt{m/k}$against it. A possible model might be to have all collisions take place and be over simultaneously, perhaps in zero time, and have a CoR = 1. This is equivalent to starting with the masses separated by$\delta$and then let$\delta\rightarrow 0.##

Many thanks - will need to take some time to digest what is going on here. Very different dynamics from that envisaged by the question setter (as you have pointed out).
I agree.

"Conservation of Momentum (Elastic Collisions)"

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