# Billiard ball collision problem

1. Jan 21, 2007

### danield

1. The problem statement, all variables and given/known data
The question is the following:
A billiard ball at rest is truck by another ball of the same mass whose speed is 5.0m/s. After an elastic collision the striking ball goes off at an angle of 40 with respect to the original direction of motion and the struck ball goes off at an angle of 50 with respect to this direction. Find the velocity of both balls.

2. Relevant equations
Well i wasnt too sure about what to use but i was trying to use
KE= 1/2mv^2
m1v1 + m2v2 = m1v1f + m2v2f
also an equation for elastic colltions
v2f= 2m1v1
-------
(m1 + m2)

3. The attempt at a solution
I tryed using all those equations, but when i got to the point of solving by vectors..well by the angle. it was impossible for me to solve it...
any help or guiding on how to resolve this kind of problems with angles.. will be appreciated.. thx

Last edited: Jan 21, 2007
2. Jan 21, 2007

Draw a sketch, all the angles, directions, etc. and set up the equations once again. Further on, if it makes things easier for you, you can set up a vector equation of conservation of momentum, too. Then use conservation of kinetic energy as the second equation.

3. Jan 21, 2007

### danield

thats what im tyring to do... but thats what i dont know how to do... im not sure.. how the angles are actually laid out... im thinking something like

Code (Text):

/
/
/ 40
---------------------
\ 10
\
\

Last edited: Jan 21, 2007
4. Jan 21, 2007

I can't say I understand your sketch. What does '10' represent? Shouldn't the angle '50' be pointing 'left'? Btw, Set up a coordinate system, and everything should be just fine.

5. Jan 21, 2007

### danield

well i will appreaciate any help you could give me to see how to set up the coordinate system.. b/c i have no idea... ive been going throug this and the only thing i can get is
on the x
V1= V1(final) + V2(final)
other than that i cant get anything else

6. Jan 21, 2007

### cristo

Staff Emeritus
Your diagram looks correct. Set your coordinate system up so that postitive x goes from left to right, and postive y goes from the centre line upwards.

Now, you need to resolve the final velocities into x and y components, and then use conservation of linear momentum in each direction.

7. Jan 21, 2007

### danield

Well.. let me try to explain this better this is what i understand when i read the problem (see attached image)

Then i tried to solve by vectors on X
i said m1v1 + m2v2 = m1v1f + m2v2f
m1=m2, then masses cancel out
v1=v1f + v2f thats what i got for X
but because its on X i need to use Cosine of Angle...
so
v1= v1f cos 40 + v2f cos 10/50 deg
for y i tried
m1v1 + m2v2 = m1v1f + m2v2f
there is no initial momentum in Y so
m1v1f= -m2v2f
mases are equal then
v1f=-v2f...
then
v1f sin 40= -v2f sin 10/50 deg

other than that.. i dont know how to use the 5.0 m/s per second... or anything else.. im stuck..

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8. Jan 22, 2007

### danield

i kept on trying last night.. but i still cant get it... i dont know where to use the 5.0 m/s... and how does this divides amongst the two balls :( any help is appreciated

9. Jan 22, 2007

### cristo

Staff Emeritus
5m/s is initial velocity of the first ball. I'll rewrite your equations, since your notation seems to be a bit ambiguous (i.e. using v1f to mean both a vector, and each component of the vector!)

Let me define v1=(vx,vy)-- the vector corresponding to the final velocity of the first ball, with components vx,vy. Define v1 similarly: v1=(wx,wy).

Now, conservation of momentum gives us v=v1+v2, (since the masses cancel), where v is the initial velocity of the first ball.

Split this up into components, so we have conservation of momentum in the x and y directions, respectively:

v=vx+wx=> 5=v1cos40+v2cos10
0=vy+wy=> 0=v1sin40+v2sin10.

Where v1 and v2 denote the magnitude of the final velocity vectors for ball 1 and 2, respectively.

10. Jan 22, 2007

### danield

Well... still i dont knowhow to add that info that u gave me.. but i did the following and it worked
Code (Text):

5.0m/s
________________
\50                40/
\                   /
\               /
\           /
\      /
\  /

then i just did
sin 40 x 5
sin 50 x5
... but i would like to learn how to do it your way.. if u could gimme a brief explanation of how to add it using those vectors i would appreaciate it..knowing more is never bad

11. Jan 22, 2007

### cristo

Staff Emeritus
This new diagram is different to the one you posted before. I think you should use the first one.
I don't know what you mean here-- these are just numbers! Care to further explain how you did it using this diagram?
I'd be glad to, but we should make sure your solution is correct first.

12. Jan 22, 2007

### danield

well first of all the answers on the back of the book are 3.8 m/s and 3.2 m/s
the diagram i made.. its just a vector addition of the problem
just used the method.. head of the first with tail of the second... and the last one tail of the first and head of the last....
Code (Text):

^
/
/
/ 40
------------------------>
\ 50
\
\

then triangle is made out of 180deg.. so the other angle is a 90 deg...
making the use of sine possible to solve the ones needed....

13. Jan 22, 2007

### cristo

Staff Emeritus
Your angles seem rather ambiguous. I don't know how to write that code, so I can't draw what I think it is, but your angle 40 is from the horizontal, no? The angle 50 is from the other vector.. it should be 10 from the horizontal.

14. Jan 22, 2007

### danield

well im not sure about that... :(
but when i try ur method... if 0=v sin 40 - v sin 50
it works perfectly... but it doesnt work with the plus
its strange... :(

15. Jan 22, 2007

### cristo

Staff Emeritus
Ok, here's my solution. Let's take the situation as in your diagram in post #12. For this to be correct the question is very badly worded! Setting up the equations, using the same variables as above, yields
v=vx+wx=> 5=v1cos40+v2cos50
0=vy+wy=> 0=v1sin40-v2sin50.

Note that I've incorporated your suggestion about the negative sign in the second equation. If you think about it, it must be there, since the velocities in the y direction are opposite, whereas the velocities in the x direction are in the same direction.

Solving these equations gives the results in the back of the book. I hope this is the way in which the book wants the problem to be solved, since it gives those answers! Anyway, hope this helps.

(And a big thanks to Hoot, for amendments to my first attempt at this!)

Last edited: Jan 22, 2007
16. Jan 22, 2007

### danield

lol love you man.. that was what i was trying.. since the beggining.. but teh wording of the book confused me... and also that negative sign..
but thx for teh help love ya..