Question regarding inelastic/elastic collision.

Is this further evidence to show that the collision is inelastic?Yes, that is further evidence that this collision is not elastic. The fact that the angles do not add up to 90 degrees means that the total momentum of the two marbles is not in the direction of the initial motion. For an elastic collision, the marbles should "scatter" at right angles to the initial motion.Does it make sense for the collision to be elastic AT ALL? Like I did this experiment on the hardwood floor in my house, clearly the KE could not be conserved because energy left the system through the friction force and dissipated as heat. Therefore, energy was lost and not constant
  • #1
jfnn

Homework Statement


[/B]
Hi,

I have a quick question regarding elastic and inelastic collisions. I preformed an experiment in my house with two large, heavy marbles. I rolled one horizontally towards a stationary one, recorded a video of the collision. I uploaded the video to Tracker software, obtained motion graphs for each, and calculated the horizontal and vertical velocities. I then used pythagoreans theorem to find the scatter angle. I got 28 got marble one and 31 for marble two.

My question is, after plugging these values with the velocity obtained into the conservation of momentum equation, I see that momentum is conserved, which is a good thing.

However, how can I tell if the collision is elastic or not? I tried looking at the relationship between the kinetic energy and found that it was NOT constant. If it was not constant, would that be enough to say the collision was inelastic?

I also add those two angles, which do not add up to 90 degrees. For an elastic collision, the angles should add up to 90 degrees for two identical masses; however, it does not, it is not even close! Is this further evidence to show that the collision is inelastic?

Does it make sense for the collision to be elastic AT ALL? Like I did this experiment on the hardwood floor in my house, clearly the KE could not be conserved because energy left the system through the friction force and dissipated as heat. Therefore, energy was lost and not constant. Would it be possible to have an elastic collision at all in this situation?

Furthermore: Could there be enough experimental error that my values just deviate so much that the collision should be elastic instead of inelastic? Is my logic wrong?

Thank you so much for your help,
J

Homework Equations


[/B]
p=mv

KEi=KEf

The Attempt at a Solution



Mentioned above.[/B]
 
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  • #2
jfnn said:

Homework Statement


[/B]
Hi,

I have a quick question regarding elastic and inelastic collisions. I preformed an experiment in my house with two large, heavy marbles. I rolled one horizontally towards a stationary one, recorded a video of the collision. I uploaded the video to Tracker software, obtained motion graphs for each, and calculated the horizontal and vertical velocities. I then used pythagoreans theorem to find the scatter angle. I got 28 got marble one and 31 for marble two.

My question is, after plugging these values with the velocity obtained into the conservation of momentum equation, I see that momentum is conserved, which is a good thing.

However, how can I tell if the collision is elastic or not? I tried looking at the relationship between the kinetic energy and found that it was NOT constant. If it was not constant, would that be enough to say the collision was inelastic?

I also add those two angles, which do not add up to 90 degrees. For an inelastic collision, the angles should add up to 90 degrees for two identical masses; however, it does not, it is not even close! Is this further evidence to show that the collision is inelastic?

Does it make sense for the collision to be elastic AT ALL? Like I did this experiment on the hardwood floor in my house, clearly the KE could not be conserved because energy left the system through the friction force and dissipated as heat. Therefore, energy was lost and not constant. Would it be possible to have an elastic collision at all in this situation?

Thank you so much for your help,
J

Homework Equations


[/B]
p=mv

KEi=KEf

The Attempt at a Solution



Mentioned above.[/B]

QUICK EDIT: For an ELASTIC collision the angles should add up to 90 degrees!
 
  • #3
jfnn said:
I have a quick question regarding elastic and inelastic collisions. I preformed an experiment in my house with two large, heavy marbles. I rolled one horizontally towards a stationary one, recorded a video of the collision. I uploaded the video to Tracker software, obtained motion graphs for each, and calculated the horizontal and vertical velocities. I then used pythagoreans theorem to find the scatter angle. I got 28 got marble one and 31 for marble two.
OK. Sounds like fun!

My question is, after plugging these values with the velocity obtained into the conservation of momentum equation, I see that momentum is conserved, which is a good thing.
Yes. Momentum should be conserved if external forces can be neglected during the collision. The fact that the momentum is conserved is an indication that there was no significant impulse from external forces during the collision.

However, how can I tell if the collision is elastic or not? I tried looking at the relationship between the kinetic energy and found that it was NOT constant. If it was not constant, would that be enough to say the collision was inelastic?
Yes. By definition, total KE must be conserved in an elastic collision. However, you need to be looking at all forms of KE. The marbles are probably rolling without slipping. So, there is some rotational KE as well as translational KE. You can easily correct for that by using physics to determine how the total KE (translation plus rotation) depends on the speed. You should find that the total KE of one of the marbles is (7/10)mv2. The fact that the total KE is proportional to the square of the speed would still lead to

v1f2+ v2f2 = v1i2 (if the masses are equal ,the collision is elastic, and no extra spin is created in the collision [see below])

So, if your data is not obeying the above relation, then either the collision is not elastic or something else is going on. One thing that could be going on is that during the collision the marbles picked up some "spin" over and above the rolling rotational motion. This would also need to be included as part of the KE. But you probably didn't get any data on spin motion.

I also add those two angles, which do not add up to 90 degrees. For an elastic collision, the angles should add up to 90 degrees for two identical masses; however, it does not, it is not even close! Is this further evidence to show that the collision is inelastic?
If there is significant spinning motion generated in the collision, then I don't think the angles would need to add to 90 degrees even if there is no loss of total KE.

Does it make sense for the collision to be elastic AT ALL? Like I did this experiment on the hardwood floor in my house, clearly the KE could not be conserved because energy left the system through the friction force and dissipated as heat. Therefore, energy was lost and not constant.
There was probably not much KE lost due to friction with the floor. The fact that momentum was conserved indicates that the friction was not significant. But, even without any external forces acting on the system, it is still possible for a large percentage of the KE to be converted into heat (internal energy within the marbles). It depends on the material properties of the colliding objects. I would think that marbles would not lose a lot of KE to heat.

Would it be possible to have an elastic collision at all in this situation?
It's not possible to have a true elastic collision with real everyday objects making contact during a collision. Some heat will be generated. But, there are situations where the collisions are close to being elastic. Steel balls can have collisions that are pretty elastic. And I would think the marbles would be pretty elastic, too.



Furthermore: Could there be enough experimental error that my values just deviate so much that the collision should be elastic instead of inelastic? Is my logic wrong?
The "error" might be due to neglecting any extra spinning motion picked up in the collision. Or, the marbles just might not be very elastic. It's hard to say.
 
Last edited:
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1. What is the difference between an inelastic and elastic collision?

An inelastic collision is a type of collision where kinetic energy is not conserved, meaning that the total energy of the system before and after the collision is not equal. In an elastic collision, kinetic energy is conserved, meaning that the total energy of the system before and after the collision remains the same.

2. How is momentum conserved in an inelastic collision?

In an inelastic collision, momentum is still conserved even though kinetic energy is not. This means that the total momentum of the system before the collision is equal to the total momentum after the collision. However, some of the kinetic energy is transformed into other forms of energy, such as heat or sound.

3. What are some real-life examples of elastic collisions?

A classic example of an elastic collision is two billiard balls colliding on a pool table. Another example is a rubber ball bouncing off a hard surface. Both of these collisions exhibit the conservation of kinetic energy and momentum.

4. How do you calculate the velocity of two objects after a collision?

The calculation for the final velocity of two objects after a collision depends on whether it is an elastic or inelastic collision. In an elastic collision, the formula is vf = [(m1 - m2)/(m1 + m2)] * vi, where m1 and m2 are the masses of the two objects and vi is the initial velocity. In an inelastic collision, the formula is vf = [(m1 * v1) + (m2 * v2)] / (m1 + m2), where v1 and v2 are the initial velocities of the two objects.

5. How does the coefficient of restitution affect the outcome of a collision?

The coefficient of restitution is a measure of how much kinetic energy is conserved during a collision. In an elastic collision, the coefficient of restitution is equal to 1, meaning that all kinetic energy is conserved. In an inelastic collision, the coefficient of restitution is less than 1, meaning that some kinetic energy is lost. This affects the outcome of the collision by determining the final velocities of the objects involved.

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