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spratleyj
2009-Mar-25, 11:41 PM
In my physics class the other day we were taking about how when you bounce (drop) a ball off the floor it can never bounce higher than the point from which it was released at (unless of course you give it initial velocity). However, if you were to take a gallon of water and pour it into a sink you would most likely have water splash above the point from which you poured it. So my question is how is that possible? I posed that question to my teacher and he said something about the change (loss) of mass making it possible - but his explanation wasn't quite clear or complete. Anybody wish to enlighten me?

nauthiz
2009-Mar-25, 11:57 PM
If you hold a tennis ball on top of a basketball and drop them both, the tennis ball should bounce much higher than the point from which they were released. Long story short, on the bounce back up the tennis ball gets some extra kinetic energy from the basketball.

I imagine it's a similar effect with water - since you're not dealing with a discrete unit like with a single ball, it's possible for some of the droplets that are splashing back up to get more than their fair share of kinetic energy.

cjameshuff
2009-Mar-26, 01:29 AM
In my physics class the other day we were taking about how when you bounce (drop) a ball off the floor it can never bounce higher than the point from which it was released at (unless of course you give it initial velocity). However, if you were to take a gallon of water and pour it into a sink you would most likely have water splash above the point from which you poured it. So my question is how is that possible? I posed that question to my teacher and he said something about the change (loss) of mass making it possible - but his explanation wasn't quite clear or complete. Anybody wish to enlighten me?

Some of the water will splash higher, and some will splash lower. Momentum is transferred through the water unevenly...it doesn't bounce as a unit as a ball does, portions of it receive more upward momentum. If the ball shattered instead of bouncing as a unit, individual fragments could be thrown higher than the height it was dropped from.

If you consider the water as many small droplets, and add up the potential and kinetic energy of each droplet at any point in time, the total is the same as the potential energy the water had before being poured, minus losses to friction.

spratleyj
2009-Mar-26, 01:48 AM
Thanks for the answer cjameshuff - it appears my lack of knowledge about momentum is the problem here :) Is there a formula that describes how the momentum is split up and the result(s)?

WayneFrancis
2009-Mar-26, 04:15 AM
Yes but it would be nearly impossible within something like water. I don't know if fluid dynamics would cover this.

In the basket/tennis ball situation it is just the Newton’s 3 Laws of motion.

speedfreek
2009-Mar-26, 07:51 PM
In my physics class the other day we were taking about how when you bounce (drop) a ball off the floor it can never bounce higher than the point from which it was released at (unless of course you give it initial velocity). However, if you were to take a gallon of water and pour it into a sink you would most likely have water splash above the point from which you poured it.

Would the same thing happen if you poured your bucket of water onto a large flat surface instead of into a sink?

iquestor
2009-Mar-26, 08:05 PM
Isnt this just the Law of Conservation of Energy?

http://en.wikipedia.org/wiki/Conservation_of_energy

Basically, energy cannot be lost or gained in a system. In the case of the ball dropped, all of the energy in the system is stored as potential energy as it is held; This energy is the result of gravity. When the ball is released, it changes to kinetic energy. when it hits, some energy must be transferred to the surface it hits, and then the rest is released and causes the ball to bounce. Of course, a tennis ball will bounce better than a rock because of the mass and material, but if you measure everything accurately, the Law states that the sum of all these energies will remain constant.

With water, its also true, but it isnt neatly contained in one mass like a ball. It would be true is the surface were flat, convex, or concave. That just changes the energy absorbed by the surface on impact, etc.

George
2009-Mar-26, 08:17 PM
Is there a formula that describes how the momentum is split up and the result(s)?
m1v1 = m2v2

cjameshuff
2009-Mar-26, 09:15 PM
Thanks for the answer cjameshuff - it appears my lack of knowledge about momentum is the problem here :) Is there a formula that describes how the momentum is split up and the result(s)?

Depends on what you're asking. Fluid dynamics is extremely complicated and generally requires numerical simulations that can require enormous amounts of computation, especially when there's turbulence involved. The Navier–Stokes equations describe fluid flow with some simplifications, they're essentially Newtonian mechanics applied to fluids.
http://en.wikipedia.org/wiki/Navier–Stokes_equations

There's also much simpler ways of approximately solving similar problems dealing with water flowing through pipes or channels without having to compute the details of the flow...it's an entire field of engineering.

If you're asking something more general...energy is conserved. The sum of the potential and kinetic energy remains the same. Potential energy of a mass m at a height h is 9.8 m/s^2*m*h, kinetic energy is 0.5*m*v^2. If you split the mass into two masses, energy can be unevenly divided between the two, but the total remains the same:

m*(9.8m/s^2*h + 0.5*v^2) = m1*(9.8m/s^2*h1 + 0.5*v1^2) + m2*(9.8m/s^2*h2 + 0.5*v2^2)

If you want the absolute highest point that can be reached, that will be a droplet thrown straight up, and kinetic energy will be zero. Subtract the energy retained by the other portion from the total energy, and you have:
m*(9.8m/s^2*h + 0.5*v^2) - m2*(9.8m/s^2*h2 + 0.5*v2^2) = m1*9.8m/s^2*h1

Shuffle things around a bit, and you have the maximum possible height of one portion as a function of the initial height and velocity of the total mass, and the height and velocity of the remaining portion:
(m*(9.8m/s^2*h + 0.5*v^2) - m2*(9.8m/s^2*h2 + 0.5*v2^2))/(m1*9.8m/s^2) = h1

Momentum is conserved as well, but also exchanged with the planet...looking at the energy is a more useful approach for this problem.

spratleyj
2009-Mar-27, 01:48 AM
Yeah, I was looking for something along the lines of the Navier-Stokes equations you linked to, but even they don't give me the complete picture I'm looking for. By the looks of it the maths are way over my head, so it's not likely I'm going to get the comprehensive answer I was looking for, but thanks for the answers!

Noclevername
2009-Mar-30, 09:01 PM
Would the same thing happen if you poured your bucket of water onto a large flat surface instead of into a sink?

It can, but the sides of the sink contain the water and thus concentrate the momentum in a smaller area, resulting in an average of more splashbacks in a sink than on a flat surface. Smaller sinks also mean greater odds of high splashes than broad ones. But as with any randomized process, nothing is guaranteed for any single event.