My question is particularly about an oblique collision case. (For example a body having velocity along x axis approaching another with velocity along y axis) From what I know, in perfectly inelastic collisions the coefficient of restitution is equal to zero and there is maximum kinetic energy loss. If coefficient of restitution is zero it should mean that after collision the relative velocity of the bodies colliding should be zero along the plane normal to their surfaces (that is the plane along which they give impulse to each other). From my understanding this should not affect the velocities of the colliding bodies in the direction perpendicular to normal I.e. the common tangent. So if their net velocities are not same then the bodies should not stick together in case of oblique collision. But in most of the questions I have seen, it is assumed that their final speed and velocity will be same. So I wanted to ask if there is any fault in what I'm saying or any other reason why the bodies should always stick together?
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1$\begingroup$ Well, draw a force diagram to see what, if any, force is pushing them apart $\endgroup$– Carl WitthoftCommented Jun 21 at 18:05
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$\begingroup$ Keep in mind that even though the collision is completely inelastic losing kinetic energy, momentum is conserved $\endgroup$– Bob DCommented Jun 21 at 18:54
3 Answers
Do bodies stick together after an inelastic oblique collision?
If the collision is perfectly inelastic, yes.
If coefficient of restitution is zero it should mean that after collision the relative velocity of the bodies colliding should be zero along the plane normal to their surfaces (that is the plane along which they give impulse to each other)
If I understand what you’re saying correctly, you are asking if the bodies will have a velocity perpendicular to the $xy$ plane after the collision.
Assuming there are no external forces acting on the two bodies, the final velocity of the bodies after the collision will be determined by conservation of momentum. This applies regardless of the degree of elasticity of the collision.
Since prior to the collision there is no momentum of either body perpendicular to the $xy$ plane, if the bodies don’t stick together their final velocities should be such that there can be no net momentum perpendicular to that plane following the collision.
For a completely inelastic collision in which the two bodies stick together, this further means the final velocity of the combination cannot have a component of velocity perpendicular to that plane after the collision.
But suppose there was some prior motion of the bodies perpendicular to the x-y plane before the collision, in that case too they should they stick after a perfectly inelastic collision? (By stick here i just mean that they have same final velocity and move together)
It doesn't matter what the directions of the initial velocities are. The bodies will stick together. That's because a perfectly inelastic collision is by definition one in which the maximum amount of kinetic energy is lost, and that occurs when the bodies stick together.
The only difference if there is initial motion perpendicular to the x-y plane, such that there is initially a net momentum perpendicular to the plane, then the stuck together bodies will have have some component of velocity perpendicular to the x-y plane in order to conserve momentum perpendicular to the plane.
Hope this helps.
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$\begingroup$ But suppose there was some prior motion of the bodies perpendicular to the x-y plane before the collision, in that case too they should they stick after a perfectly inelastic collision? (By stick here i just mean that they have same final velocity and move together) $\endgroup$– UserCommented Jun 22 at 11:17
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$\begingroup$ @User I have updated my answer to respond. $\endgroup$– Bob DCommented Jun 22 at 11:56
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The case you are referring to is where the objects are like in this drawing:
where the surface of interaction (the red lines) is slippery and the objects are malleable enough to absorb all velocity perpendicular to the red lines. In that case they would indeed keep some final relative velocity as indicated in the drawing, but still a lot of kinetic energy would be lost.
This is not called "perfectly inelastic" simply because without the slippery surfaces (instead assuming that they have high friction and perhaps are even sticky) we would lose even more kinetic energy. They would then end up sticked together, or just touching each other which is kinematically the same.
Of course the collision in the drawing is still inelastic, just not perfectly so!
Some of this might be vocabulary. In the usual head on inelastic collision, the bodies wind up at the same velocity in contact. Typically they deform, turning kinetic energy into heat. I would say they stick together if they are sticky.
If you work in the center of mass frame, the two masses come straight at each other with no sideways component. So it is hard to see how there would be a sideways component left over.
If the collision was off center, there would be a torque. The final stuck together object would be rotating. If the masses weren't sticky, they would fly apart. Either way, that isn't an inelastic collision.
