# One-dimension collision question

1. Jan 8, 2009

### physics_geek

1. The problem statement, all variables and given/known data
A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are released from rest at the same moment, to fall through a distance of 1.08 m,

(a) Find the magnitude of the downward velocity with which the basketball reaches the ground. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down.

(b) Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound?

2. Relevant equations
p = mv
KE = .5mv^2
gravitational KE = mgy

3. The attempt at a solution

so i know its an elastic equation, thus meaning energy is convserved. i know that means i'll have at least two equations to solve for.

2. Jan 8, 2009

### NBAJam100

(a)Well if you have to find the magnitude of the downward velocity of the basketball, why dont you start by finding the v of the basketball just before it hits the ground and it is reversed?

As you said, its elastic so why dont you try using PE(i)+KE(i)=KE(f)+PE(i) to solve for the v of the basketball.

(b) once you know the speed of the basketball on the way up, you can then use momentum considerations to find the velocity of the tennis ball, which can then be used to determine how far up it goes. One thing though, it doesnt tell you exactly how far down the tennis ball falls (because it was still falling when the b-ball hit the ground), so it couldnt have went the whole distance of 1.08m... hmmmmm

3. Jan 8, 2009

### physics_geek

o ok
sounds a little confusing..but i'll try it
for the PE are we talking about gravitational potential = mgy ??

4. Jan 8, 2009

### Delphi51

Yes. It might be helpful to think that, as the basketball falls, its PE is being converted into KE - when it hits the bottom all of the PE it had at the top has been converted into KE. Start by writing "PE at the top = KE at the bottom", then fill in the details (which should leave just one unknown that you can solve for).

The question is a bit confusing. I'm picturing the tennis ball practically touching the basketball initially, so both balls actually fall the full 1.08 m. In the absence of the separation distance number, I would treat it as too small to matter - perhaps less than 1 mm.

5. Jan 9, 2009

### physics_geek

ok so i got the velocity of the basketball as it falls
now i just need to know how to go about solving the second part

no idea

6. Jan 9, 2009

### physics girl phd

An elastic collision is defined as a collision where energy as well as momentum is conserved... energy isn't lost to heat or used to deform the objects. In these cases you generally will get two equations, with two unknowns (usually the speeds of the objects right after collision).

In this particular case, they tell you the collision is between the basketball (moving upwards) and the tennis ball (still moving down). If you can find the speed of the tennis ball right after the collision, you will be able to know what height it reaches later.

PS. you should do this experiment... it's fun. (You can also find videos online.)

7. Jan 9, 2009

### physics_geek

thanks a lot
i understand what you are saying

i just have no idea how to set up the equation

8. Jan 9, 2009

You said it yourself....

So if KE is conserved, then, using subscript A=baseball and B=basketball and subscript 1=initial and 2=final

$$(mv^2_1)_A+(mv^2_1)_B=(mv^2_2)_A+(mv^2_2)_B$$

You should be able to set up a very similar equation for Linear Momentum.

Casey

9. Jan 9, 2009

### physics_geek

im sorry i still just dont understand

im trying to read my book and it doesnt explain well

10. Jan 9, 2009

### Delphi51

I find that long line of symbols rather disheartening, too!
I like to start the solution in words and then fill in the details with symbols as needed.

kinetic energy before collision = kinetic energy after, momentum before = momentum after

Under energy, write a .5mv2 for each moving object, before and after. Maybe use a little m for the little ball and a big M for the big ball. Similarly, use the momentum formula for each moving ball, before and after. The velocities before and after will be different, so perhaps you need 4 symbols - or you could just rely on the words "before" and "after" in the line I wrote.

On the next line, substitute numbers for all the symbols you know, such as m and M and the velocities you found in the first parts. If you are lucky, there will only be one unknown in one or both of the equations, and you will be able to solve the equation to find it. More likely, you will be faced with two unknowns - the velocities of the balls after the collision. When you have two equations, you can use them together to get the two unknown answers. We call it "solving a system of equations" in Canada. No doubt you have learned several methods for doing this. Look them up if you don't remember.

11. Jan 9, 2009

Well, whether the equation is something that is pretty to look at is rather trivial

Why don't you be a little more specific? Which part do you not understand? Have you read your text or looked at your notes? We do not provide solutions here, but we can walk you through the problem.

So let's see what you have written so far so we can see where you are having trouble

12. Jan 11, 2009

### physics_geek

well
i solved for the velocity of the basketball on its way down

so i put mgh + .5mv^2 = 0
which gave me 4.6 m/s

now i know that i have to get some other equation right?
my teacher isnt the best english speaker so im trying to improvise here haha

so do i do some kind of momentum equation like mv1 + mv2 = something
i have no idea?

13. Jan 11, 2009

Did you even look at post number 8?

14. Jan 11, 2009

### physics_geek

yes i did
to me that doesnt mean anything. i feel so dumb

i guess i'll try asking one of my friends
thanks for the help..sorry to be an annoyance