Messing up simplifying formula for head on collision problem

[slambert56]

Ok I have been going crazy and I think I must be making a simple algebra mistake. Basically in my book I read over an example dealing with a head on collision between 2 balls and the collision is elastic so the kinetic energy intial= final KE. The momentum is also conserved so i can use momentum final=momentum initial. I tried combining these formulas but I always get quadratic. My work is in the image.
The two formulas I am combining are the 1/2m1*vf1^2+ 1/2 m2*vf2^2=1/2m1*vo1^2
and m1*vf1+m2*vf2=m1*vo1
http://img825.imageshack.us/img825/2450/001copys.jpg

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the problem just gave 2 balls first one has inital velocity and second one is initially at rest. the book gave the simple formula vf1^2=v01(m1-m2/m1+m2)
THANKS YOU FOR ANY HELP! i have done this wayy too many times and gotten wrong answers.
The answers from book is vf1=-2.62m/s and vf2=2.38m/s

 

[grzz]

You can simplify the notation by assuming one mass is M and the other is m and the initial velocity is u and the final two velocities are v and w.

 

[slambert56]

I did that but I still couldn't get the final velocity of one to be just vf squared. I always seemed to come up with a vf^2 and a vf which led to the quadratic equation again.

 

[grzz]

The two masses and the initial velocity of one of them are given.And the other is initially at rest.Right?

 

[LawrenceC]

For one thing, you are stating what the book provided as an answer incorrectly. The mass ratio is dimensionless so you have a velocity squared having the same dimensions as velocity to the first power. Something is obviously wrong there. Both sides of an equation MUST have the same dimensions.

I have not gone through your algebra but, instead, did it myself. Use the momentum equation to get an expression for the velocity of the ball that was hit (initial velocity was zero). Plug that into the kinetic energy equation and simplify. It'll require a little factoring but I am sure you can do that.

Keep at it.

 

[slambert56]

Thanks for the help guys. Yes I am given the two masses and 1 ball is moving at 5m/s and the other is at rest. For some reason I keep on getting over 3m/s. I simplified the masses in this one to make it faster because I am all about finding the easiest way.
http://img845.imageshack.us/img845/375/002raw.jpg

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[grzz]

First choose the most simple notation as possible, eg m,M,u,v,and w.
Then do as LaurenceC adviced you and write down the conservation of momentum equation and the conservation of KE equation and then enter the known values straight away.
Simplify the two equations to obtain a quadratic in only one unknown.

 

[LawrenceC]

You have a sign error in the right column of equations near the top...your last post.

 

[gneill]

If you're looking for an easy approach, it can proven that for an elastic collision the relative velocity of the two objects after collision is equal to the negative of their relative velocity before collision. This little gem can replace the conservation of KE equation as the second equation in the simultaneous equation system for solving for the velocities, and avoids the dreaded velocity-squared terms.

 

[slambert56]

ok thanks for that error but I still get 5m/s

http://img831.imageshack.us/img831/4113/003lzn.jpg

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gneill so i could just say that -(v01+v02)=vf1+vf2 ? I think I am a confused on that one. Isnt what you said pretty much the momentum conservation?

 

[gneill]

Relative velocities: (v1_o - v2_o) = -(v1_f - v2_f)

 

[slambert56]

gneill I feel like I have found the holy grail. That so makes it easier. Thanks. would you happen to know of a website explaining the theory and all. If i use it on a test I will want to explain it.

 

[slambert56]

ahh i saw another post from you

It's really a very simple way to get the required two equations in two unknowns that are required to solve for the two final velocities. It relies on a (provable) characteristic of elastic collisions whereby the relative velocities of the two objects, before and after collision, are of equal magnitude but opposite sign.

Another way to solve the problem is to use conservation of kinetic energy as the second equation (KE is conserved in elastic collisions). But this introduces the squares of the velocities into the mix, which turn into square roots of expressions, and are a bit harder to work with.

If you want you can prove the relative velocity relationship by solving the general case elastic collision using the conservation of energy approach to yield expressions for the final velocities, and then use them to find the final relative velocity. You will then find that the relationship is true in general.

ok so I can just basically use the KE final=KE initial and solve for vo1-vo2 ?

 

[gneill]

gneill I feel like I have found the holy grail. That so makes it easier. Thanks. would you happen to know of a website explaining the theory and all. If i use it on a test I will want to explain it.

Offhand I don't know of a website that deals with this little theorem directly. I did prove it myself a long ways back. It's not too hard. It's just a matter of slogging through the math using the KE (!) as one of the constraints, then comparing the relative velocities that result.

So you only have to the the KE method once to prove the theorem, then never look back!

 

[gneill]

ahh i saw another post from you


ok so I can just basically use the KE final=KE initial and solve for vo1-vo2 ?

You can try

There's not enough information in the KE equation alone to obtain the relative velocities.

I just checked Wikipedia. Will wonders never cease! -- they have a derivation of the relative velocity formula.

http://en.wikipedia.org/wiki/Elastic_collision

 

[nasu]

It's not an independent theorem. It follows from combining the energy and momentum conservation equations. It's just a matter of a few lines.
You need to re-arrange the terms in the energy equation so that you have differences of the velocities squared.
For example,
m1(v1i^2-v1f^2)=m2(v2i^2-v2f^2)
where 1 and 2 mean the first and second body and i and f mean initial and respectively final.
Once you do this you expand each parenthesis in a product of sum and difference.
Using the momentum conservation the sum factors (and the mass) will simplify and you are left with the conservation of the relative velocity.

Note:Sorry, I started to write before the previous message was posted.

 

[slambert56]

yes! now I will surprise my teacher with a curveball. :)

 

[slambert56]

http://img535.imageshack.us/img535/8316/004plx.jpg

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I am going to frame this on my wall haha

 

[LawrenceC]

Since there seems to be so much difficulty with the algebra...

Subscript 11 denotes ball 1, initial velocity
Subscript 12 denotes ball 1, final velocity

Subscript 22 denotes ball 2, final velocity

 

[slambert56]

Thanks for the help. I usually make little mistakes so my plan is to derive the relative velocity because it is faster and applies to all elastic collisions. I can always do both to check my work though.

 

[grzz]

It is good that the notation be simple and LaurenceC did just that.

May I take the method of LaurenceC and simplify it further.

Let small mass be m with initial velocy u and final velocity v and big mass be M with initial velocity 0 and final velocity w.

KE

(1/2)mu^{2} = (1/2)mv^{2} + (1/2)Mw^{2}

multiplying by 2: mu^{2} = mv^{2} + Mw^{2}




Momentum

mu = mv + Mw
hence m(u - v)/M = w

therefore

mu^{2} = mv^{2} + M{m(u - v)/M}^{2}

u^{2} = v^{2} + (Mm){(u - v)/M}^{2}


(u - v)(u + v) = (m/M)(u - v)^{2}

(u - v)(u + v) = (m/M)(u - v)(u - v)

But v cannot be equal to u because otherwise KE is not conserved. So one can divide by (u - v).

therefore u + v = (m/M)(u - v)

Mu + Mv = mu - mv
Mv + mv = mu - Mu
v = u(m - M)/(M + m)