# Elastic Collision of two pucks

1. Feb 19, 2005

### mrjeffy321

Here is the problem as it reads out of the book.

In the figure, puck 1 of mass .2 kg is sent sliding across the table [frictionless] to undergo a one-dimentional elastic collision with stationary puck 2. Puck 2 then slides off the edge and lands a distance d from the base of the table. Puck 1 rebounds from the collision and slides off the opposite edge of the table, landing a distance of 2d. What is the mass of puck 2?

The figure mentioned in the question is fairly well described in the problem, it is a leven, frictionless table with 2 pucks on it.

Here is what I make of the question.
since puck 2 slides a distance "d" from the table while falling and puck 1 slides "2d", I conclude that puck 1 has twice the horizontal velocity, although negative) as puck 2 when it left the table (and also just after the collision since the table is frictionless). Since puck 1 is sliding one direction at the start of the problem, stops and then reverses direction and has a larger speed, its mass must be smaller than that of puck 2. Since this is an elastic collision, conservation of kinetic enery and conservation of momentum apply.
I will make a guess that puck two has twice the mass (.4 kg) of puck 1, thus casing it to have half the kinetic energy aftter the collision. Although I dont know how to prove this, mathmatically, how do I show it?

2. Feb 19, 2005

### dextercioby

Could you please attach a drawing...?

Daniel.

3. Feb 19, 2005

### mrjeffy321

Here is the diagram given.

[it currently just needs approval]

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4. Feb 20, 2005

### mrjeffy321

5. Feb 20, 2005

### Staff: Mentor

Right!
Exactly right.
You were doing great up to this point. Why did you decide to guess? Write the equations describing conservation of energy and momentum and solve for the mass!

6. Feb 20, 2005

### mrjeffy321

Well I only know 1 thing about this problem, the mass of the first puck. I know that the final speed of puck 1 is two times that of puck 2, but I dont have any real numbers to use for them.

equations I can use:
conservcation of linear momentum,
m1*v1i = m1*v1f + m2*v2f

conservation of kinetic energy,
1/2 m1*v1i^2 = 1/2 m2*v1f^2 + 1/2 m2*v2f^2

but I only know the ratio of the final velocities and one of the masses? how can I get the other mass out of this?

mathmatically, I can solve for mass 2,
m2 = (m1*v1i - m1*v1f)/v2f
and
m2 = (1/2 m1*v1i^2 - 1/2 m2*v1f^2)/(1/2*v2f^2)

7. Feb 20, 2005

### Staff: Mentor

Right! Now add an equation relating the final speeds (v1f = - 2v2f).

You have three equations; give it a shot.

Keep playing around with the equations. Don't give up.

8. Feb 20, 2005

### mrjeffy321

I still am not getting it.

you can just pick any two final velocities you want, aslong as the are opposite directions and velocity one is double the other, then just pic some value for mass 2. then using conservation of momentum, solve for the initial velocity of mass one. then using conservation of kinetic energy, check this solution, and you find it works.
I tried this using values I just made up, and it worked in both cases, using diferent masses and velocities, so I know I am wrong somewhere.

should I be solving each variable for the things (one thing I know, numerically) I know and then substituing in? I dont know where to go from here, there too many unknows that I will need to know.

9. Feb 21, 2005

### Staff: Mentor

So you are telling me that any value of mass 2 will satisfy the conservation equations? I don't think so! :yuck:

$$\rmtex{(#1) }v_{1f} = -2v_{2f}$$
$$\rmtex{(#2) }m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}$$
$$\rmtex{(#3) }m_1 v_{1i}^2 = m_1 v_{1f}^2 + m_2 v_{2f}^2$$

Here's what to do: First, use equation 1 to eliminate $v_{1f}$ in the other equations. Rewrite equations 2 & 3 as simply as possible. Then square equation 2 (both sides). Then you can eliminate $v_{1i}^2$ by combining equation 2 (squared) with equation 3. Take that final equation and solve for $m_2$ in terms of $m_1$. You'll find that things will drop out nicely.

Once you do this I will show you another "trick" to use for elastic collisions.

10. Feb 21, 2005

### mrjeffy321

well, here are the simplified equations that you were talking about, I think,
--m1v1i=m1v1f+m1v2f --> m1(v1i-v1f)=m2v2f --> m1(v1i-2v2f)=m2v2f
--m1v1i^2=m1v1f^2+m2v2f^2 --> m1(v1i-v1f)(v1i+v1f)=m2v2f^2 -->
m1(v1i-2v2f)(v1i+2v2f)=m2v2f^2

now I square the momentum equation, it becomes
m1v1i^2-4m1v2f+2v2f^2 ?
now how do I combine this with the other one, add? that wont cancel anything, subtract, ...that will cancel some stuff but not everything.

what is that other trick that can be done, is it much simplier than this way , I hope.

11. Feb 21, 2005

### Staff: Mentor

Do it again, but this time don't move things from one side to another. Just replace v1f with -2v2f, then simplify. (Don't try solving for m2 at this point.)

12. Feb 21, 2005

### mrjeffy321

m1v1i=m1v1f+m2v2f -->
m1v1i=-2m1v2f+m2v2f -->
m1v1i=-v2f(2m1+m2)

13. Feb 21, 2005

### Staff: Mentor

Now do the same thing for equation 3.

14. Feb 21, 2005

### mrjeffy321

I now have:
m1v1=v2f(-2m+m2)
then square that and you get,
m1v1i^2 = v2f^2*4m1^2-v2f^2*2m1m2+v2f^2*m2^2

and

m1v1i^2 = v2f^2(-2m1+m2) -->
m1v1^2 = -2m1v2f^2 + m2v2f^2

Last edited: Feb 21, 2005
15. Feb 21, 2005

### Staff: Mentor

OK.
OK, write it like this:
m1^2v1^2 = v2f^2 (m2^2 - 4m1m2 + 4m1^2) [ = new equation 2]

Now write equation 3:
m1v1^2 = m1(-2v2f)^2 + m2v2f^2 => simplify [ = new equation 3]

Now combine the new equations 2 and 3.

16. Feb 21, 2005

### mrjeffy321

I dont know what you mean by "combine the equations"

17. Feb 21, 2005

### Staff: Mentor

Multiply (the new) equation 3 by m1. Then compare the left sides of both equations.

18. Feb 21, 2005

### mrjeffy321

so multilpy
m1v1^2 = m1(-2v2f)^2 + m2v2f^2 by m1 to get
m1^2v1^2 = m1^2(-2v2f)^2 + m1*m2v2f^2

no it would seem that I have 2 equations for the same thing,
m1^2*v1^2 = .....

I still dont know where to go from here, sorry for being so slow to pick this up.

19. Feb 22, 2005

### Staff: Mentor

Right. I'll rewrite it:
$$m_1 v_1^2 = m_1 (-2v_{2f})^2 + m_2 v_{2f}^2$$
$$m_1 v_1^2 = v_{2f}^2 (4m_1 + m_2)$$
Now multiply by m1:
$$m_1^2 v_1^2 = v_{2f}^2 (4m_1 + m_2)m_1$$
Right... and if two things are equal to the same thing...?

Please write a new equation: set the right side of both equations equal to each other and see if you can now solve for m2. (The $v_{2f}^2$ factors will cancel.)

20. Feb 22, 2005

### mrjeffy321

Oh!! reveleation!
I get it now.

t this point the acutall answer has become rather trivial to me, now that I have gotten the process down.

is that trick worth mentioning?