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Oblique collision of three disks

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Can you tell me how to solve this question. I have solved somewhat till the final velocity of the blocks and my final diagram looks like this :


But I have trouble calculating coeffecient of restitution. For starters do we calculate coefficient of restitution along the line of collision only, because impulse is imparted along this line so that means all velocities should be resolved along this line?

asked Aug 19, 2018 in Physics Problems by Harambe (130 points)
edited Aug 19, 2018 by sammy gerbil

1 Answer

2 votes
Best answer

The solution which you and Avnish Kabaj found (see diagram below) is almost certainly the one which the question-setter or examiner is looking for. However, I think the problem is more difficult than was intended, as you might see from considering what happens if the stationary disks were different sizes. It is not clear to me whether a satisfactory solution to this problem exists.


There are 2 related difficulties with this problem.

The 1st is that there is a simultaneous collision (ie simultaneous contact) between 3 rigid bodies. Such collisions are indeterminate : changing initial conditions infinitesimally such that one pair of disks collide first results in different outcomes depending on which pair collide first. See Multiple colliding balls and links therein. This is (probably) unrealistic.

All real bodies are deformable to some extent. Extending the object model to include laws of deformation (eg Hooke's Law) enables the outcome of collisions with multiple simultaneous contacts to be determined uniquely. However, if we wish to keep the rigid body model then the options are either (1) ensure that no more than 2 objects are in contact at any one time (which is always possible if you look on a small enough time scale because rigid body collisions are by definition instantaneous ), or (2) adopt an ad hoc or empirical rule to define how the initial momentum will be distributed among the 3 or more bodies which are in contact.

The 2nd (related) difficulty is that the Coefficient of Restitution may not be definable for such collisions, because there may be a different ratio of relative velocity of separation and approach for each pair of objects. For example, if the stationary disks have different sizes they will leave at different angles with different velocities, resulting in a different COR for each of the smaller disks which depends on the angle of incidence for each disk. The COR should depend only on the material, which is the same for both disks; it should not depend on angle of incidence. This difficulty is hidden here because the smaller disks have been chosen to have the same radius.

The alternative definition of COR, based on ratio of total kinetic energy after and before collision, is less ambiguous but does not necessarily give the same value, whereas it always does in two-body collisions.

In short, this problem may not be sufficiently well defined to have a single solution.

In my solution below I handle the collision by transforming it into sequential two-body collisions, by splitting the large disk into two equal parts, which collide with the smaller disks then collide with each other and coalesce. However, this method also fails to give a consistent value of COR for the two definitions, so it is not correct.

Yes you are correct : coefficient of restitution must be measured along the common normal. However, I think this is complicated by the above problem of having two simultaneous collisions, as I shall explain below.

Impulse is transferred along the common normal, which is the line joining the centres of the larger disk A and smaller disks B. This line lies at an angle of $\theta$ to the horizontal, where $\sin\theta=1/3$. See LH diagram below.

The smaller disks B are initially at rest, so their final velocities are along the common normal. Let this final velocity of disks B be $v$ and the initial velocity of the larger disk A be $U$. Disk A has the same mass is $m$ as disks B. Conservation of linear momentum in the horizontal direction gives $$mU=2mv\cos\theta$$ $$v=\frac{U}{2\cos\theta}$$.

The solution which you and Avnish Kabal proposed in the PSE Problem-Solving Strategies Chatroom resolves the vector velocity of A along and perpendicular to two different non-orthogonal directions at the same time. You cannot do this : the vector sum of the 4 components is greater than the original vector, which is clearly wrong. Your calculation assumes that the 2 collisions between disk A and each disk B are independent of each other by allowing disk A to approach each with velocity $U\cos\theta$ at the same time, but also inter-dependent by allowing momentum to be exchanged, otherwise momentum would not be conserved in each collision separately.

An alternative method of handling the simultaneous collision of the disk A with both disks B is to view A as two smaller disks C each of mass $\frac12m$ travelling together. Each disk C approaches disk B horizontally with speed $U$ the same as A, collides with it and rebounds so that linear momentum is conserved separately. See RHS of diagram above.

The disks C cannot be stationary after this collision with B because linear momentum would not be conserved. Instead the disks C must rebound vertically from B, then collide with each other and coalesce. This ensures that after they coalesce they have mass $m$ and are at rest again with no momentum, so they are equivalent to the larger disk A.

The vertical momentum with which each disk C rebounds from disk B must equal the final vertical momentum of disk B : $$\frac12mV=mv\sin\theta$$ $$V=2 v\sin\theta$$ where $V$ is the vertical velocity of the disks C.

Now we have enough information to calculate the Coefficient of Restitution, $e$ between the 2 disks C and B, which we assume is the same as between the 3 disks A and B.

The relative speed of approach between disks C and B along the common normal is $U\cos\theta$. The relative speed of separation along the same normal is $$v+V\sin\theta=v(1+2\sin^2\theta)=\frac{U(1+\frac12\sin^2\theta)}{2\cos\theta}$$ Therefore: $$e_V=\frac{U(1+2\sin^2\theta)}{2U\cos^2\theta}=\frac{1+\frac29}{2\times \frac89}=\frac{11}{16}$$

The total kinetic energies before and after collision are $$K_1=\frac12 mU^2, K_2=2\times \frac12 mv^2$$ An alternative way of defining the COR is related to the fraction of kinetic energy retained after collision : $$e_K=\sqrt{\frac{K_2}{K_1} }=\sqrt2 \frac{v}{U}=\sqrt2\frac{1}{2 \cos\theta}=\frac{3}{4}$$

This answer does not match any of the given options but is close to (b), which is $\frac{9}{16}$. The reason is clearly the way in which I have handled the simultaneous collision. It does not give a consistent method of calculating COR. KE is lost in the collision between the imaginary disks C when they coalesce. However, perhaps no method is consistent :

The solution which both you and Avnish Kabaj have obtained is that $$v=\frac{U}{2\cos\theta}=\frac{3}{4\sqrt2}U$$ which results in $$e_V=\frac{v}{U\cos\theta}=\frac{U}{2U\cos^2\theta}=\frac{9}{16}$$ which applies to each collision separately. The definition of $e$ using the ratio of kinetic energies is the same as above, viz. $e_K=\frac34$ which applies to both collisions collectively. However, note that $e_K=\sqrt{e_V}$. If we had obtained instead that $e_K=e_V^2$ then we might explain this by saying that the two contacts were sequential - ie a fraction $e_V$ of KE is lost in each contact.

If COR is defined consistently then these two methods should give the same result - ie $e_V=e_K$. Neither my method nor yours gives the a consistent result for COR.

Conclusion : it may not be possible to define the COR of a multi-body collision.

answered Aug 19, 2018 by sammy gerbil (28,466 points)
edited Aug 23, 2018 by sammy gerbil
@Sammy gerbil I applied conservation of momentum in this problem and also assumed $ u= ucos \theta$ was the initial velocity of the bigger disk like Avnish Kabal and Harambe did . But after reading your interesting reasoning about why this cannot be the case I want to redo it. Particularly I am thinking about the following: ‘ the vector velocity of A along and perpendicular to two different non-orthogonal directions at the same time: you cannot do this’ and then you argue than the sum of the vector components is larger that the vector itself. Did you mean velocity components involving the velocity of the two smaller disks after the collision? I have to say I’ve never thought about vector length argument in order to assert if momentum is conserved or not. Do you know of the existence of any paper/document related to this issue? I am pretty interested.
This is just as much an argument about conservation of momentum as it is about velocity. In the quote I am talking only about the velocity components of the larger disk.

In Avnish's solution, COR is calculated by resolving velocity of larger disk in the direction of each smaller disk independently and at the same time, so relative velocity of approach for each collision is $u\cos\theta$. However, the transfer of momentum is handled differently : it is assumed that each smaller disk exits along the line joining the centres **and** that total momentum is conserved (both of which are reasonable assumptions). It is not assumed that the momentum of the large disk along the line joining centres in each collision is $mu\cos\theta$. Different methods are being used for COR and momentum, so I think this solution is inconsistent.

No I haven't come across any document discussing this issue, or how to handle multiple simultaneous collisions. The only documents I have seen deal with experimental outcomes, and they did not draw any conclusions about theory.

It is one of those issues which get little or no attention. It would be useful if an expert addressed this issue in a journal article. Another issue is whether a string is elastic or inelastic. Both assumptions can be made in theory (separately of course). However this issue has received some attention in eg The Mathematical Gazette - see my answer to [Force Transfer Between two bodies linked by a rope](https://physics.stackexchange.com/q/274089).
Okey, thank you for the information. By the way, when you say 'the vector sum of the 4 components is greater than the original vector' you mean the sum of $ucos \theta + usin \theta$ corresponding to the triangle you drew + $ucos \theta + usin \theta$ corresponding to the symmetric triangle right? I know it may be trivial but I want to be sure of this argument.
Yes that is right. For example when $\theta \to 0$ the velocity of approach is approx $u$, so the initial momentum in each collision (upper and lower disks separately) is $mu$ and the total momentum (upper and lower disks together) is $2mu$. But this is bigger than the momentum of the incoming disk, so it cannot be correct.