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In the collision between the red disk (mass $m_1$) and the green disk (mass $m_2$) as shown bellow, why the force that mass 1 makes on mass 2, $\mathbf{F}_{12}$, is along the line connecting the point of contact and the center of the ball (green arrow)?

Is this always true? Is there a simple explanation for this?

Collision between two disks

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    $\begingroup$ This answer might be of interest to you at this point: physics.stackexchange.com/a/220776/392 $\endgroup$ Commented Nov 23, 2020 at 0:05
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    $\begingroup$ @John Alexiou I looked to your answer in the post you recommend. I have a question. In point 2 you say to draw a circle in the momentum plane and you say that momentum is conserved if the two final vectors are diagonals in this circle. I understand that. But there are solution that conserve total momentum with both vectors outside the circle. Why are those solutions not important? Thanks. $\endgroup$
    – Blue
    Commented Nov 23, 2020 at 15:11
  • $\begingroup$ Actually, you do consider those for elasto-plastic collisions, but it turns out for elastic collision all possible solutions lie on the circle. And those solutions inside the circle have a coefficient of restitution less than one, and those outside the circle a coefficient of restitution larger than one (if possible). That was a very insightful observation @Blue. Good job. $\endgroup$ Commented Nov 23, 2020 at 20:17

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Why? Because that is the only direction where relative motion is not allowed. The centers cannot get any closer to each other.

If force was along a direction where motion was allowed, then the contact force would do work and either add or remove energy to the system violating the conservation of energy. The only time the contact force is at an angle to the contact normal is when friction is involved.

This rule can be generalized as follows

The line of action of the contact force must pass through the instant center of rotation between the two bodies.

At the instant center, the two bodies have zero relative velocity (only relative rotation) and thus the power transmitted through the contact is also zero.

fig1

The direction of the contact force is always such that

$$ \vec{F}_1 \cdot (\vec{v}_1 - \vec{v}_2) = \vec{F}_2 \cdot (\vec{v}_2 - \vec{v}_1) = 0 $$

when you work out the relative instance center of rotation (point C below), it will always fall on the line along the contact normal.

fig2

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  • $\begingroup$ I need some time do study all this. Thanks! When you say "Because that is the only direction where relative motion is not allowed", do you mean that at the exact moment they collide if I had the possibility to freeze the whole setup and move mass 1 freely I would not be able to move it in the direction of mass 2. Is this what you mean by relative motion? $\endgroup$
    – Blue
    Commented Nov 23, 2020 at 0:21
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    $\begingroup$ Yes, at the moment of collision, the only allowed motion is sliding in a direction perpendicular to the contact normal. $\endgroup$ Commented Nov 23, 2020 at 5:32
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In an idealised case: yes. You can decompose the force that is acting at the boundary in two parts: the normal force and the tangential force. The normal forces prevents objects from moving through eachother so this always appears when you try to move to objects into one another. The tangential force often occurs as a result of friction when two surfaces move with respect to eachother. You are probably familiar with this picture for the more simple case of the normal force:

enter image description here

It is easiest to reason from the center of mass frame. If the collision is head-on (the paths and velocities are parallel) then it is easy to see the tangential force is zero. So the force is normal to the surface so it is along the line connecting the point of contact and center of mass.

When the collision is off-centre, like in the picture below, it gets more complicated. When we assume friction is zero the tangential force is still zero. This is still an approximation since in real life the normal force is a result from the balls deforming and this doesn't happen instanteneously. But we only care about the ideal balls for now. When friction isn't zero we can have a tangential force. In the picture the left ball and right ball are moving in opposite directions so the surfaces have some relative velocity. If there is friction the balls get a nonzero torque which makes the ball spin and it can change the direction in which they leave the collision.

enter image description here

source of the first picture: Why is normal force perpendicular?

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  • $\begingroup$ Just want to clarify my understanding about your answer. Thank you. So, if i change to center of mass frame of the two balls the collision is heads on. In fact, in that situation the force is normal (tangential force is zero). So if I get back to the lab frame I shown in the picture I get forces in that direction. Is this right? $\endgroup$
    – Blue
    Commented Nov 22, 2020 at 23:23
  • $\begingroup$ Just to clarify, so that the disks move in a straight line in the center of mass frame, its seems that the frame is a rotating frame. In this case we have non-inertial forces, right? Sorry if I am making a confusion here. $\endgroup$
    – Blue
    Commented Nov 22, 2020 at 23:41
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    $\begingroup$ @Blue In the center of mass frame the collision need not be head on. The picture is an example of a center of mass frame collision that is off centre. If you imagine the center of mass it stays exactly still in the middle. The center of mass frame is just an intertial frame. There are no non-intertial forces. When you change between two intertial frames of reference all the forces stay the same. $\endgroup$ Commented Nov 23, 2020 at 10:17
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It means that there is no friction between the disks. Otherwise the forces would not be in the same line, and the resulting torque would cause a rotation of the disks after the collision.

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  • $\begingroup$ Ok. There is no friction. This is a 2D situation and I am neglecting rotation. I cannot understand from your answer why force is in that direction. This problem is solved here en.wikipedia.org/wiki/Elastic_collision#Two-dimensional and rotation is not even considered. But the forces in that directions as shown are assumed. $\endgroup$
    – Blue
    Commented Nov 22, 2020 at 23:03
  • $\begingroup$ I made an edit in my answer. $\endgroup$ Commented Nov 22, 2020 at 23:55
  • $\begingroup$ Thanks. I am so sorry. This seams a simple thought but I can't seem to figure it out. From my point of view, what you say corresponds to having mass 1 fixed and mass 2 coming from an angle. So, the resulting picture is simpler because one mass is fixed, but my question still remains. I guess! Please, let me know if I am making some reasoning mistake. Thanks! $\endgroup$
    – Blue
    Commented Nov 23, 2020 at 0:27
  • $\begingroup$ @blue I think that unfortunately my idea for simplification is wrong. I put the answer as it was before. $\endgroup$ Commented Nov 23, 2020 at 0:37
  • $\begingroup$ Thanks. Your thoughts have been useful. $\endgroup$
    – Blue
    Commented Nov 23, 2020 at 10:18

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