What is the motion of the cylinders after the collision?

[haruspex]

If one ball collides with another stationary ball completely in-elastically, then the deceleration of the first ball and the acceleration of the second would be slower due to the increased contact time. For example compare dropping a tennis ball and its collision with the ground with that of a golf ball. The collision time of the tennis ball is longer because of its inelasticity.

No, it doesn't work like that.

During any collision, there is a compression phase then an expansion phase. Take a spring as an easy model. In most questions we assume that there is a spring constant k such that the force at displacement x is kx. But real springs are not like that: they have hysteresis. In effect, the ratio between displacement and force depends whether it is moving towards or away from the relaxed state. That is, if you compress a spring some distance, and its resistant force has reached F, as soon as you start to let it expand the force will immediately drop to something less. This means you cannot get the same energy back as you put in, i.e. that it is not perfectly elastic. None f this has anything to do with the time taken.

If anobject impinges on the end of a spring,compressing it, the duration of the compression phase depends on the momentum of the object and the spring constant. The higher the constant, the less time taken to reach maximum compression, and the less time for the overall process.

In summary, the duration of each phase depends on the magnitude of the corresponding spring constant, while the elasticity depends on the ratio of the two constants.

 

[Game_Of_Physics]

No, it doesn't work like that.

During any collision, there is a compression phase then an expansion phase. Take a spring as an easy model. In most questions we assume that there is a spring constant k such that the force at displacement x is kx. But real springs are not like that: they have hysteresis. In effect, the ratio between displacement and force depends whether it is moving towards or away from the relaxed state. That is, if you compress a spring some distance, and its resistant force has reached F, as soon as you start to let it expand the force will immediately drop to something less. This means you cannot get the same energy back as you put in, i.e. that it is not perfectly elastic. None f this has anything to do with the time taken.
If anobject impinges on the end of a spring,compressing it, the duration of the compression phase depends on the momentum of the object and the spring constant. The higher the constant, the less time taken to reach maximum compression, and the less time for the overall process.

In summary, the duration of each phase depends on the magnitude of the corresponding spring constant, while the elasticity depends on the ratio of the two constants.

Ah...yes that's a good point! But then given that is the case, you could in theory have an impulse between an object colliding with another object, where the collision time is long due to a "small spring constant". In this case the force against time graph would be a line with a lower gradient, but as long as the area under it is the same as an equivalent graph that has a steeper gradient, then the impulses would be the same, but for the lower spring constant the collision time (the top limit of the integral) would be greater. My point is that the force that the first object feels because of the impulse from the second object is not necessarily as large as you make out in post 24, and therefore not necessarily great enough to overcome weight (in the case of the colliding cylinders).

 

[zwierz]

Spoiler: about impact problems

Nevertheless not all problems about collision are so trivial. Consider a wedge of mass $M$ that rests on a smooth horizontal floor. The slope of the wedge is also smooth. The wedge can not lose contact with the floor. It is an ideal constraint. Somebody drops a point mass $m$ on the wedge such that this point mass falls onto wedge with vertical velocity $u$. It then bounces from the wedge with velocity $v$ and the wedge begins to move with velocity $w$.
The velocity $u$ is known, the impact is perfectly elastic. Find $v,w$.

 

[haruspex]

Spoiler: about impact problems

Nevertheless not all problems about collision are so trivial. Consider a wedge of mass $M$ that rests on a smooth horizontal floor. The slope of the wedge is also smooth. The wedge can not lose contact with the floor. It is an ideal constraint. Somebody drops a point mass $m$ on the wedge such that this point mass falls onto wedge with vertical velocity $u$. It then bounces from the wedge with velocity $v$ and the wedge begins to move with velocity $w$.
The velocity $u$ is known, the impact is perfectly elastic. Find $v,w$.View attachment 196007

Spoiler: Impact on wedge

$w=\frac{mu\sin(2\theta)}{M+m\sin^2(\theta)}$

 

[haruspex]

the impulse from the second object is not necessarily as large as you make out in post 24

But then it would no longer qualify as an "impact problem", and you would need rather more data to be able to solve it.

Edit: and you mean the force, not the impulse.

 

[zwierz]

Yes. And I forgot to introduce angle :(