Project name idea

1. Jan 7, 2005

UrbanXrisis

Hi, i'm doing a project when I drop a tennis ball and a basketball at the same time they hit the floor and try to show conservation of momentum after the seperate. of course the V(initial) would be the same before the objects hit the ground and after they hit, the Tennis ball would recieve the force the basketball delivers and so the tennis ball would fly off and the basketball would lose velocity.

I need to find a good name to call this project. It's actaully a reseach project trying to analyze if Momentum is conserved outside the ideal lab.

Any good names would be helpful. I though of

"Basketball Tennis Ball Preservation of Inelastic Momentum"

any ideas?

Also, how could I prove Newton's 3 laws with this experiment? (I'm using a video analyzing software to graph the velocity of both the tennis ball and basketball before and after collision)

Last edited: Jan 7, 2005
2. Jan 7, 2005

quasar987

Here's my suggestion. You only need to know a little bit of calculus.

Newton's second law:
$$\vec{F}=\frac{d\vec{p}}{dt}$$

Newton's second law for the tennis ball:
$$\vec{F}_{\rightarrow tennis}=\frac{d\vec{p}_{tennis}}{dt}$$

Newton's second law for the basketball:
$$\vec{F}_{\rightarrow basket}=\frac{d\vec{p}_{basket}}{dt}$$

Proving the third law means proving that
$$\vec{F}_{\rightarrow tennis}=-\vec{F}_{\rightarrow basket}$$

If we assume that at all time during and right after the collision, the sum of the momentum of the tennis ball plus the momentum of the basketball is constant, then, if we ADD the two newton's law together, we have

$$\vec{F}_{\rightarrow tennis} + \vec{F}_{\rightarrow basket} = \frac{d\vec{p}_{tennis}}{dt} + \frac{d\vec{p}_{basket}}{dt} = \frac{d(\vec{p}_{tennis}+\vec{p}_{basket})}{dt}=\vec{0}$$

This last equality is what is implied by our assumption that at all time during and right after the collision, the sum of the momentum of the tennis ball plus the momentum of the basketball is constant. All that remains to do now in order to complete the demonstration is to transfer one of the force "on the other side":

$$\vec{F}_{\rightarrow tennis} + \vec{F}_{\rightarrow basket} = \vec{0} \Leftrightarrow \vec{F}_{\rightarrow tennis} = -\vec{F}_{\rightarrow basket} \ \ \blacksquare$$ :tongue2:

I'm not 100% sure this is correct either, and there may be a non-calculus based way of doing it too, but I don't see it.

3. Jan 7, 2005

UrbanXrisis

Thanks for the post quasar987. That's a good idea you have there.

I was wondering if anyone had a research name for this project. I dont like mine: "Basketball Tennis Ball Preservation of Inelastic Momentum"

and was wondering if there were any creative titles that your guys can think of.
thanks.

4. Jan 7, 2005

quasar987

It doesn't have to have basketball and tennis ball in it. And there is no such thing as Inelastic Momentum by the way. I would go for

"On the conversation of linear momentum in an inelastic collision"

or just

"Conversation of momentum in an inelastic collision"

5. Jan 7, 2005

UrbanXrisis

see, other people are doing that in my class too, so I need to specify the objects so my title does not get mixed with the others.

6. Jan 7, 2005

UrbanXrisis

Can Newton's second law be applied here? The second law never states momentum. It states F=ma, and not really $$\vec{F}=\frac{d\vec{p}}{dt}$$. Or am I wrong?

7. Jan 7, 2005

quasar987

These two ways of writting the second law are equivalent, as you will see. The acceleration is defined as the rate of change of speed. So

$$F=ma=m\frac{dv}{dt} \ \ (*)$$

On the other hand, p=mv, but since m is a constant, if you know the rules of differentiation, you know that the derivative of a constant times a variable, is the constant times de derivative of the variable. So
$$\frac{dp}{dt} = \frac{d}{dt}(mv)=m\frac{dv}{dt}=F$$

(according to (*))

So we see that F=ma and F=dp/dt are two ways of writing the same thing, only using different concepts (acceleration and momentum).

8. Jan 7, 2005

UrbanXrisis

how would i go about proving the 1st and 3rd law?

ex.
1st: Since the balls are accelerating, it means there's an outside force acting on the balls. This can be described as gravity. Newton's first law states that "Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it." Gravity is the external force that supplies the constant change in velocity. However, the balls will reach terminal velocity due to air drag.

3rd: Newton's third law "For every action there is an equal and opposite reaction." How could I relate this to momentum? ex. after the balls hit the floor, the floor pushes back with the same amount of force. That force is transfered through the basketball to the tennis ball. Then tennis ball flies away, pushing on the basketball. So, total momentum = momentum of basketball +momentum of tennis ball

9. Jan 7, 2005

quasar987

Note that we cannot prove Newton's laws. We can only verify that they apply in a certain situation.

For 1st law: You got the right idea. (However I doubt you lauched the balls from high enough so they reached terminal velocity in your experiment!)

For 3rd law: ...have you forgotten my post #2 already?

10. Jan 7, 2005

UrbanXrisis

ohhhh, momentum of the tennis ball should have equal magnitude as the momentum of the basketball after they hit. Right?

11. Jan 7, 2005

quasar987

No. That is not what the law of conservation of momentum says.

What is the law of conversation of momentum?

'If no external forces act of a system of particles (here the 2 balls), the sum of the momenta of the system of particles remains unchanged.' (Kind of like the law of inertia (1rst law) when you think about it!)

The idea of the proof I sketched in post #2 to show that Newton's third law apply, is to neglect the force of gravity during the time of the collision* so that there are no external forces acting on the system and so that we can apply the law of conservation of momentum.

So to complete the answer to your question, it is the SUM of the momenta $\vec{p}_{tennis}+\vec{p}_{basket}$ RIGHT BEFORE and RIGHT AFTER the collision that are of equal magnitudes (AND direction, by the way. It is the total momentum vector that is conserved).

*This is the assumption I have made in post #2 and it is very reasonable because the force of impact/collision is very big (in comparison to the force of gravity), as the resulting very rapid change in velocity/momentum of the balls attest.

P.S. I'm going to bed now.