Counter-intuitive 2D collision

In summary, the conversation discusses the calculation of final velocities for a ball dropped onto a slope with a restitution coefficient e. The problem involves two collisions - the first between the ball and the slope, and the second between the slope and the Earth. The meaning of the restitution coefficient in this problem is also discussed, as well as its dependence on the colliding materials. It is noted that the restitution coefficient, like the coefficient of friction, is an engineering approximation and can be compared to real-world simulations using 2D physics engines.
  • #1
jaumzaum
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Consider a slope with mass, that can move in the horizontal plane without friction. A ball is dropped and hits the slope with restitution coefficient e. How to calculate the final velocities?

How can I solve something like this? Note that it's not a simple 2D collision, it has a restriction that the slope can only move in the horizontal plane. There are actually 2 collisions happening, the collision of the ball with the slope (partially inelastic) and the collision of the slope with the Earth (that is completely inelastic). What is the meaning of the restitution coefficient in this problem? Will it be the restitution coefficient for the first collision or for the overall collision? I know the answer could be "well, it could be for both, you need to specify", but practically, do I really need to specify this? Isn't anything implied?

I have another problem when I try to think about the energy dissipated in the problem. The first collision will drain energy from the ball and convert in another type of non-translacional kinetic energy. If the second collision happened after the first (the ball hits the slope and then the slope hits the ground), the second collision would drain energy only from the slope. That way we could say there would be 2 different restitution coefficients. But both things are occurring at the same time, so I don't know if this "second collision" will drain energy only from the slope anymore, it could drain energy from the ball, and the final velocity of the ball when both collisions happens simultaneously could actually be different from those calculated if the collisions happened one another.
 
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  • #2
You need to make an assumption about the nature of the collision, about the velocity of the ball immediately after bouncing off. The assumption I would make is that the component of the ball's velocity parallel to the incline is the same before and after the collision. WIth this assumption you can find the impulse delivered to the inclined plane and hence its horizontal velocity. I would use the coefficient of restitution as it is meant to be used. Find expressions of the relative speeds before and after the collision and take the ratio. The collision of the incline with the Earth just negates the vertical component of the impulse while the horizontal component gets the incline sliding.

Disclaimer: I have not solved this problem, but this is how I would proceed, at least initially, to solve it.
 
  • #3
You have a coefficient of restitution for each impact. At each impact you can construct a basis consisting of unit vectors normal and tangential to the interface ##\{\hat{n}, \hat{t}\}##, for a 2D space. If the velocities of the colliding parts of the two bodies before the collision are ##\vec{u}_A## and ##\vec{u}_B##, and those after the collision are ##\vec{v}_A## and ##\vec{v}_B##, then the coefficient of restitution is defined with$$(\vec{v}_B - \vec{v}_A) \cdot \hat{n} = -e(\vec{u}_B - \vec{u}_A) \cdot \hat{n}$$i.e. proportional to the ratio of the normal components of the relative velocities before and after the collision. If the wedge is always in contact with the Earth, I don't think you need to worry about that interface. For the collision between the wedge and the ball, you will need to account for the final velocity of the wedge in your restitution calculation!
 
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  • #4
etotheipi said:
##\dots## then the coefficient of restitution is defined with$$(\vec{v}_B - \vec{v}_A) \cdot \hat{n} = -e(\vec{v}_B - \vec{v}_A) \cdot \hat{n}$$
Please edit the equation and put ##\vec u_A## and ##\vec u_B## where they belong. :oldsmile:
 
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  • #5
kuruman said:
Please edit the equation and put ##\vec u_A## and ##\vec u_B## where they belong. :oldsmile:

Thanks for the assist 🙌... I think that's a sign I should pack up shop for today :wink:
 
  • #6
Thanks guys! So I think both of you would agree with me that if the ground was not there (if there was no restriction that the slope can only move in the horizontal plane) the restitution coefficient would be different. That is because the ground exerts an impuls in the vertical direction, and the vertical direction is not perpendicular to the collision. So if the same ball was dropped from the same height to the same slope, the restitution coefficient would be different in these two different scenarios. That way, if we assume the restitution coefficient is calculated in the old-fashioned way as mentioned, we are indirectly implying that the Earth plays a hole in the coefficient. This was what I don't understand properly. What does the restitution coefficient depend on? I know they could depend on the velocities, and now I found it depends on the restrictions. It seems a bit useless to me to calculate something that varies all the time and does not depend only on the surface material (like friction coefficient, in almost all the time). So why do we use it?
 
  • #7
jaumzaum said:
So why do we use it?
Like the coefficient of friction, it is not a physical principle, but an engineering approximation.
 
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  • #8
To add to what @jbriggs444 noted, the coefficient of restitution depends on the colliding materials much like the coefficient of friction depends on the surfaces in contact with each other. Take a look at this short video. Same table, different balls.
 
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  • #9
jbriggs444 said:
Like the coefficient of friction, it is not a physical principle, but an engineering approximation.
Or like coefficient of rolling resistance, which accounts for various phenomena lumped together for practical reasons.
 
  • #10
For this kind of problems, it's good to learn to use some simple Java or C# 2D physics engine realistic enough to simulate what happens and compare it to calculations. The documentations of those will tell if it's sophisticated enough to include the coefficient of restitution, friction and so on.
 

Related to Counter-intuitive 2D collision

1. What is counter-intuitive 2D collision?

Counter-intuitive 2D collision refers to the phenomenon where the outcome of a collision between two objects in a 2D space does not match our intuition or expectations. This can occur due to various factors such as the shape, size, and mass of the objects, as well as the angle and speed of the collision.

2. Why is understanding counter-intuitive 2D collision important?

Understanding counter-intuitive 2D collision is important for scientists and engineers working in fields such as physics, computer graphics, and game development. It allows them to accurately predict the behavior of objects in a 2D space and create realistic simulations and animations.

3. How does counter-intuitive 2D collision differ from 3D collision?

Counter-intuitive 2D collision differs from 3D collision in that it only takes place in a two-dimensional space, whereas 3D collision occurs in a three-dimensional space. This means that the objects involved in a 2D collision have limited movement and can only collide along a single plane.

4. What are some common misconceptions about counter-intuitive 2D collision?

One common misconception about counter-intuitive 2D collision is that the objects involved will always bounce off each other at the same angle they collided at. In reality, this only occurs in perfectly elastic collisions, and most real-world collisions involve some loss of energy and a change in direction.

5. How can we accurately predict the outcome of a counter-intuitive 2D collision?

To accurately predict the outcome of a counter-intuitive 2D collision, we can use mathematical equations such as the conservation of momentum and energy. These equations take into account factors such as mass, velocity, and angle of collision to determine the final outcome of the collision.

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