Dropping two balls.

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myanmar
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Dropping two balls.

Postby myanmar » Wed Feb 11, 2009 5:35 pm UTC

This was a problem I couldn't understand in my Physics class today- please help!

You drop two balls, vertically aligned, from a height h. What is the best mass ratio such that

1) the top ball's final position is as high as possible

2) the bottom ball's final position is as high as possible

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theoreticallyKat
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Re: Dropping two balls.

Postby theoreticallyKat » Wed Feb 11, 2009 5:55 pm UTC

OK maybe I'm being a bit slow in not really understanding what this question is asking...
Also i don't know what level you're learning at, so sorry if you know what i'm about to write.

Basic principles of falling objects: the speed up at the same rate no matter what they weigh (air resistance comes into it if your falling far, but we can ignore if its just from the table height etc.) This is because the acceleration of the objects is just due to gravity ~9.81 m/s

What do you mean by the final position? Their very final position will both be on the floor?!
Are they bouncing, are they touching when you drop them?

I'm sure somebody will come along and help alot more than I can.....
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Re: Dropping two balls.

Postby idobox » Wed Feb 11, 2009 6:00 pm UTC

Since the balls are aligned, I guess it is a problem of elastic collision.
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Re: Dropping two balls.

Postby wisnij » Wed Feb 11, 2009 6:26 pm UTC

I'm interpreting the problem as being shaped something like this:

Code: Select all

  O      | g
  O      |
         v
-----

and asking what is the greatest height each ball will bounce after it hits the surface.

If the top sphere is smaller than the bottom one, it should bounce back up at a greater speed because of conservation of momentum. There's a toy made from rubber spheres aligned on a rod which uses that effect -- the sphere at the top is smallest and free to move along the rod, and it flies off with great enthusiasm when the device is dropped on the ground from a height. So if the ratio [imath]m_\mathrm{top}/m_\mathrm{bottom}[/imath] is small, the height of the top sphere should be maximized.

As for the bottom sphere... intuitively, at least, it seems that simply reversing the situation won't help, because a large upper sphere will just block the lower one from bouncing up. It might be the case that [imath]m_\mathrm{top}/m_\mathrm{bottom} = 1[/imath] will split the difference best and maximise the height of the bottom sphere. That isn't based on experiment or a rigorous calculation, though, so take it with a grain of salt.
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Re: Dropping two balls.

Postby gmalivuk » Fri Feb 13, 2009 3:00 pm UTC

Yeah, I'm assuming this is like when you hold a tennis ball on a basketball and drop them both. The basketball then mostly stays put while the tennis ball flies up much higher than it was originally dropped from.

Assuming perfect elasticity, though, I don't see how the bottom ball would bounce up at all, since it would pretty much transfer all its energy to the top ball, wouldn't it?
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Re: Dropping two balls.

Postby wst » Fri Feb 13, 2009 7:36 pm UTC

gmalivuk wrote:transfer all its energy to the top ball, wouldn't it?
Pretty nifty thing I saw on Blue Peter one day ages ago was to get 10 bouncy balls, drill holes completely through 9 of them and (iirc) glue a dowel into a hole that went halfway-through the biggest ball. Then slide the balls onto the dowel in descending order of size, and drop the contraption big-ball-first (the one with the dowel glued into it) to the ground from about 2 metres up. The top ball ended up about 10 metres in the air, and the others just feebly rolled around on the floor... it was quite amazing to my 7 year old mind.
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Re: Dropping two balls.

Postby Atre » Fri Feb 13, 2009 8:32 pm UTC

Heya, In the blurb below I've put inverted commas "" around the technical terms that get taught in school at some point - no worries if you haven't come across them yet.

1)The answer is that the smaller ball bounces back to "9h" maximum, and it does so in the limit where the lower ball is infinitely heavier than the upper ball. (this is assuming a "perfectly elastic" collision - ie. no energy is spent on deforming the colliding object, it all stays as kinetic energy)

The answer can be obtained by conservation of momentum.
-Big(lower) ball hits the ground first and bounces back upwards at the same speed it impacted the ground
- Big ball then collides with the smaller, ***this is where you calculate the effects of the collisions using "Newtons law of restitution" and "conservation of momentum"***. The trick comes from knowing that

a) In a perfectly elastic collision the "speed of separation" of the objects is the same as the "speed of approach" (conserving energy - the balls cannot separate at a higher speed than they came together - if they did, where would that extra energy have come from?)
b) If the big ball is REALLY MUCH bigger than the small ball, the collision will (effectively) not slow the big ball at all. Therefore we have

[imath]Big ball velocity before = v (upwards) .................... Small ball velocity before = v (downwards)[/imath]

And therefore they are approaching each other at 2v

[imath]Big ball velocity afterwards = v (upwards) ................... Small ball velocity after = v(up) + (speed of approach)
= 3v[/imath]

from here we just use
[imath]KE = (mv^2)/2[/imath]
Therefore 3v implies 9 times as much kinetic energy.
And with 9x KE the ball will bounce 9x higher than the height it was dropped from. (Bit of inference here, the KE came from the GPE. So 9x KE implies 9x more energy to store as GPE - this is done by attaining 9x the original height)




2) Same mass ratio as the first question. You need the lower ball to be slowed as little as possible by its collision with the upper ball, that means VERY BIG lower ball.

(However,the lower ball would reach a higher SPEED if it were infinitely smaller than the upper ball. It just wouldn't be able to bounce high because it would thunder into the upper ball continuously.)

PS. Lower ball>>upper ball and upper>>lower provide the same (optimal) answer for part 2... Can you see how/why?

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Re: Dropping two balls.

Postby sigsfried » Fri Feb 13, 2009 10:36 pm UTC

So
Taking a stack of balls.
Each with the larger very much bigger than the former we would get that the top ball leaves with
v(initial)*3^x
Escape velocity is 11.2*10^3
So assuming they fall for 1 second before bouncing they have an initial velocity of 9.8
3^x=1142.9
x=6.4

So a stack of 7 such balls has the top one reaching escape velocity.
Seems a tad dramatic.

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Kow
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Re: Dropping two balls.

Postby Kow » Sat Feb 14, 2009 6:49 pm UTC

sigsfried wrote:So
Taking a stack of balls.
Each with the larger very much bigger than the former we would get that the top ball leaves with
v(initial)*3^x
Escape velocity is 11.2*10^3
So assuming they fall for 1 second before bouncing they have an initial velocity of 9.8
3^x=1142.9
x=6.4

So a stack of 7 such balls has the top one reaching escape velocity.
Seems a tad dramatic.

At such high velocities, you'd have to incorporate wind resistance in as well. It still seems a bit too easy though. How big would the biggest ball need to be?
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Re: Dropping two balls.

Postby Sir_Elderberry » Sat Feb 14, 2009 7:07 pm UTC

sigsfried wrote:So
Taking a stack of balls.
Each with the larger very much bigger than the former we would get that the top ball leaves with
v(initial)*3^x
Escape velocity is 11.2*10^3
So assuming they fall for 1 second before bouncing they have an initial velocity of 9.8
3^x=1142.9
x=6.4

So a stack of 7 such balls has the top one reaching escape velocity.
Seems a tad dramatic.


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Re: Dropping two balls.

Postby Atre » Sat Feb 14, 2009 8:27 pm UTC

sigsfried wrote: So a stack of 7 such balls has the top one reaching escape velocity.
Seems a tad dramatic.


There are 3 potential error sources extending the model to such limits.

1) Balls would stop behaving elastically in such high energy limits (in addition to air resistance and other "big KE" effects coming out)

2) "Very much bigger" as an approximation would see far more dramatic scaling than the max KE over those 7 balls. If we take "much bigger" as 1000x mass, then starting with a 1 gram ball we end up with a 10^15 monster at the bottom. (mass Earth approx 6.02 x 10^24 kg)

3) The hidden approximation is that the "big" ball loses such a small amount of it's energy that we don't need to worry about it. Plug in some energy calculations for some multiball masses and you'll find out the only way to obtain the escape velocity is to use so much mass that we'll have to factor in the balls bumping the earth a significant amount.

An interesting exercise in where approximations break down and having to worry about extra effects as you scale a problem... Cheers.


Quick list of things that crop up in extreme limits for this problem: Air resistance, Elastic limit, Balls possessing a grav. field, Earth not being infinitely larger than the ball system, size of the balls require to get v. high mass (lift the top ball up by 100km and it's a different problem).


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