# Homework Help: Circular elastic collision

1. Nov 11, 2012

### Lolagoeslala

1. The problem statement, all variables and given/known data
a 6 kg, originally held at rest in position 1 as shown in the diagram below, slides down a frictionless quadrant of a verticle circle of radius 15 m. It then collides elastically with another 16 kg mass at rest at position 2. They bounce off each other and move on the horizontal frictionless surface and up the frictionless quadrant of a verticle circle on each side. Calculate the verticle height "H" for each ball at which they would momentarily come to rest after elastic collision.

Diagram: http://s1176.beta.photobucket.com/user/LolaGoesLala/media/jkmlkm.jpg.html

3. The attempt at a solution

So i was thinking of starting with the kinetic energy formula btu then i realised i need to include the gravitational potential energy but i need the height for it... so can anyone help me out on starting with this problem?

2. Nov 11, 2012

### tiny-tim

Hi Lolagoeslala!
to start this problem, you need to find the speed of the first mass just before the collision, and then use conservation of energy and conservation of momentum to find the speeds just after the collision

3. Nov 11, 2012

### Lolagoeslala

so to start this problem...
i would use the equation...

1/2m1v1 + mgh = 1/2m1v2
This represents the mass at rest state with the height which equals to the velocity at no height

correct? but what would the height be then?

4. Nov 11, 2012

### tiny-tim

what is this ??

you can't add a momentum to an energy

5. Nov 11, 2012

### Lolagoeslala

OOPS sorry im so sorry...
the equation is all wrong...
its supposed to be like this:

1/2m1v1^2 = mgh + 1/2m1v1^2

this is for the conservation of energy just before it hits the collision for m1

6. Nov 11, 2012

### tiny-tim

ok, at least that makes sense now!

but v1' is 0

7. Nov 11, 2012

### Lolagoeslala

Wait isn;t the v1 zero.. that is the intially and that is the rest position....
so wouldn;t it be

mgh = 1/2m1v1^2 and we are trying to find the speed with which it hits the second ball?

8. Nov 11, 2012

### tiny-tim

yes, that's correct …

that gives you the initial speed (just before the collision) of the first ball

(btw, i would have used "u" rather than "v", for reasons which may shortly appear obvious )

now use conservation of energy and conservation of momentum for the elastic collision​

9. Nov 11, 2012

### Lolagoeslala

Wait ...

so i would use the equation
m1gh = 1/2m1v1^2

now whats the height??? is that the radius?

10. Nov 11, 2012

### tiny-tim

yes …
… it's a quadrant, so it goes from 3 oclock to 6 oclock … that's a height of one radius

11. Nov 11, 2012

### Lolagoeslala

so...

m1gh = 1/2m1v1^2
6kgx9.8m/s^2x15m = 1/2(6kg)v1^2
882 J / 3 kg = v1^2
√294 J = v1
17.14 m/s = v1

Is that correct? :D

12. Nov 11, 2012

### tiny-tim

(try using the X2 and X2 buttons just above the Reply box

yes

(but you could have deleted m1 from the equation completely, and so not used 6 kg at all )

13. Nov 11, 2012

### Lolagoeslala

Oh ok.. so now would i used the conservation of momentum

so like

m1v1 + m2v2 = m1v1 + m2v2
m1v1 = m1v1 + m2v2
but then i have two unknown variables v1 and v2

14. Nov 11, 2012

### tiny-tim

yes, so you also need to use conservation of energy (for the collision)

15. Nov 11, 2012

### Lolagoeslala

umm but wouldn't the energy be lost thenn ...
so they wouldn't be equal to one another would they...
lets see...

1/2m1v1^2 = 1/2m1v1^2 + 1/2m2v1^2

i still have two missing variables..

16. Nov 11, 2012

### tiny-tim

but the question says it's elastic (ie, no energy is lost) …
yes, you now have two equations and two unknowns …

so solve it​

17. Nov 11, 2012

### Lolagoeslala

v1=(m1-m2/m1+m2)v1
V1=(6 kg-16kg/22kg)17.14 m/s [E]
v1= 7.79 m/s [W]

v2 = (2m1/m1+m2)v1
v2= (12kg/22kg)17.14m/s[E]
v2`=9.34m/s[E]

18. Nov 11, 2012

### Lolagoeslala

YOU THERE?!?!?

19. Nov 12, 2012

### tiny-tim

(just got up :zzz:)
yes, those equations are correct

but how did you get them?

20. Nov 12, 2012

### Lolagoeslala

school lesson we had to derive them so i had them.. now what should i do ?? :O?

21. Nov 12, 2012

### Lolagoeslala

ARE YOU SURE IM ON THE RIGHT TRACK?!!? i asked this tutor about it, he was like no you don't need to find the 17.14 m/s, all you need is the energy .. and work with that. I am so CONFUSED

22. Nov 12, 2012

### haruspex

Your tutor is wrong. Without using conservation of momentum during the collision process you have no way of knowing how the kinetic energy is apportioned thereafter.
(Note that momentum is not conserved while the balls are travelling around the quadrants.)

23. Nov 13, 2012

### tiny-tim

Hi Lolagoeslala!

(just got up :zzz:)
You won't have them in the exam.

It would be good practice to solve the equations directly.

(and yes, you do need to use energy conservation during the collision)

Now use energy conservation for the arcs.

24. Nov 13, 2012

### Lolagoeslala

WHATS THAT?
sorry for the caps..
whats the energy conservation for the arcs
is there some new equation for that?

25. Nov 13, 2012

### tiny-tim

no, it's just the usual mgh stuff!

(i'm saying "energy conservation for the arcs" to distinguish it from "energy conservation for the collision" )