Question on Conservation of Momentum

In summary, the conversation discusses a toy called Astro Blaster which consists of multiple balls that bounce, with the top ball bouncing the highest due to the combined mass exerted upon it. The question is whether it is possible to calculate the height of the bounce based on the potential energy of the top ball. The masses of the balls are provided, as well as the data for a test where the top ball was dropped from 5cm and bounced an average of 52cm. The conversation also mentions attempts at finding a solution through Google search and making assumptions about the behavior of the balls during collision. Ultimately, it is concluded that the top ball will bounce approximately 5 times the initial drop height.
  • #1
agentorange812
2
0

Homework Statement



So I have one of these trippy toys
http://www.spacehall.com/shop/images/AstroBlaster.JPG ,[/URL] and have a question.

Basically the way that it works, is that all the balls bounce, and the top one that isn't secured on rockets off very high while the others are secured in place on a rod. I know that this is because the combined mass is exerted upon the small top ball (Potential Energy = mass*gravity*height, thus the combined mass means that it has a huge potential energy).

Anyways, my question is that if I have recorded all the data about it, know the elasticity of the top ball, and have the potential energy calculated. Is there any way to calculate how high the ball will bounce if I know the potential energy?

(The potential energies for the entire thing and just the top ball are ~5000 and 200 J respectively)

Homework Equations



Potential Energy = MassGravityHeight , Kintetic Energy = 1/2 MassVelocity^2

The Attempt at a Solution



So far I have tried to divide the PE's and find a ratio, however the balls bounce at about 50cm and 3.8cm combined and just the top ball, but the ratios don't work out right.
 
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  • #2
Did you try a Google search on this. I'm sure I have seen the whole thing worked out more than once.

I'm not finding it now though. Do you know the masses of the balls?
 
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  • #3
the masses are --

3.88g for the top ball
9.96 for the next bigger
23.15 for the next bigger
64.11 for the next bigger

and 101.1g total

Dropped from 5cm, the top ball bounced an average of 52cm

Yeah, I tried googling it... haven't had any luck though. This SHOULD be really easy... I just can't find the right equation or am missing something!
 
  • #4
agentorange812 said:
the masses are --

3.88g for the top ball
9.96 for the next bigger
23.15 for the next bigger
64.11 for the next bigger

and 101.1g total

Dropped from 5cm, the top ball bounced an average of 52cm

Yeah, I tried googling it... haven't had any luck though. This SHOULD be really easy... I just can't find the right equation or am missing something!

Actually, I don't think it is easy. When you have that many objects involved in a process things get complicated in a hurry. I have come up with a solution based on an assumption that may not be completely valid, but it does lead to the conclusion that the top ball will rise to five times the initial height from which the blaster is dropped. The "five times" number is quoted in all the product descriptions I have seen.

The assumption is that each ball deforms in proportion to its radius, and that the balls remain in contact with one another while they are compressed. That means the top ball moves farther than the bottom ball during the collision process. Imagine the balls all being compressed by some fraction of their radius at the lowest point in the collision, with the center of mass being half way between the top and bottom of each ball. Let's number the balls 1 through 4 from top to bottom and call the distance the bottom ball CM moves d4 = kR4. The CM of the next ball up moves d3 = k(2R4 + R3). The next will be d2 = k(2R4 + 2R3 + R2) and finally d1 = k(2R4 + 2R3 + 2R2 +R1).

Assuming the balls all move together, the velocity of each ball at the end of collision will be proportional to the distance it moved up from the lowest point during the collision. To find those distances, you need the ratios of the radii. I calculated those assuming solid balls of the same density, which is not quite right because of the holes even if they are made of the same material. Using the velocity ratios from this calculation, I assumed all the mechanical energy was conserved and calculated the height each ball would reach if they were all free to move separately. (It looks to me from the picture that they can all separate a bit before the middle two get stopped by the stick) Then I took the total momentum of the bottom 3 balls and treated those as having a secondary collision that keeps them connected as one object.

The result of this calculation is that the top ball bounces just over 5 times the drop height, while the bottom three ball combination bounces to about 0.61 times the drop height.
 
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  • #5
Do you mind telling me where you got one of those trippy toys. Looks like a bit of fun
 

What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time, unless acted upon by an external force.

Why is conservation of momentum important?

Conservation of momentum is important because it helps us understand the behavior and interactions of objects in motion. It allows us to predict the outcomes of collisions and other interactions between objects.

How is conservation of momentum calculated?

Conservation of momentum is calculated by multiplying an object's mass by its velocity. In a closed system, the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

What is an example of conservation of momentum in action?

An example of conservation of momentum is a game of billiards. When the cue ball strikes the other balls, the total momentum before the collision is equal to the total momentum after the collision. This is why the cue ball stops after striking the other balls.

Are there any exceptions to conservation of momentum?

There are certain situations where conservation of momentum may not hold, such as in a system where external forces are present or in situations involving relativistic speeds. However, in most cases, conservation of momentum is a valid principle in physics.

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