Homework Help: Solving equation system

1. May 29, 2014

Maxo

1. The problem statement, all variables and given/known data
$$v_{f1}^2 = v_{01}^2 - m_{1}/m_{2} * v_{f2}^2\\ v_{f2} = m_{1}/m_{2} * (v_{01} - v_{f1})$$
Solve for Vf1 algebraically. How to solve this in the easiest way without having miles of calculations?

2. Relevant equations

3. The attempt at a solution
When I try to use the substitution method I get like miles of calculations where it's extremely easy to make careless misstakes. I only manage to get to some second order equation

$$v_{f1}^2 - 2*m_{1}*v_{01} / (m_{2}-m_{1}) * v_{f1} - v_{01}^2 = 0$$
but I doubt it's correct after all those hairy calculations, I probably made a careless misstake somewhere.

There must be some easier way to solve this?!

2. May 29, 2014

scurty

Edit: Nevermind, I misread the problem. Some problems are tedious and you just have to work through them.

3. May 29, 2014

Maxo

What is strange is that in my book they seem to just take all these calculations for granted. In many other cases they've shown the calculations, but those cases were much simpler math.

I have tried very much now but I don't manage to get to the simplification they get to here. I would appreciate some help in how to get to it.

Last edited: May 29, 2014
4. May 29, 2014

jbunniii

Assuming you are OK with the first line in the photo, let's declutter the problem by writing $x = v_{f1}$, $y = v_{01}$, and $c = -(m_2/m_1)(m_1/m_2)^2$. Then the first line becomes
$$x^2 = y^2 + c(x-y)^2$$
which is equivalent to
$$x^2 - y^2 = c(x-y)^2$$
or
$$(x-y)(x+y) = c(x-y)^2$$
Now clearly $x=y$ is one solution. We can obtain another by assuming $x \neq y$ and dividing by $x-y$ to obtain
$$x+y = c(x-y)$$
You should be able to solve this easily for $x$ (your $v_{f1}$).

Last edited: May 29, 2014
5. May 29, 2014

Ray Vickson

In these cases, always start by simplifying the notation before you do lots of algebra. Then, at the very end, you can put back the original notation. So, for example, you could use symbols $f_1, f_2$ instead of $v_{f1}, v_{f2}$, and use $r = m_1/m_2$, $w = v_{01}$. Then your system is
$$f_1^2 = w^2 - r f_2^2\\ f_2 = r(w-f_1)$$
It is pretty easy to get the two solutions for $f_1$ from these equations.

Here, I was assuming you meant
$$v_{f1}^2 = v_{01}^2 - \frac{m_1}{m_2} v_{f2}^2$$
and not
$$v_{f1}^2 = v_{01}^2 - \frac{m_1}{m_2 v_{f2}^2}$$
and not
$$v_{f1}^2 = \frac{v_{01}^2 - m_1}{m_2 v_{f2}^2}$$
Please clarify, and use parentheses or latex '\frac' commands to make things 100% clear.

6. May 29, 2014

Maxo

jbunniii: Thanks! That was a very good showing of how to do it. I understand perfectly what you are doing each step. I just wonder, how did you know that you were supposed to do these particular steps in order to solve this? I mean, suppose you would write an algorithm for how to solve these kind of equations generally, what are the actual general ideas behind each step?

To start with, and most importantly, how did you know/see that it would make everything look very much simpler if you choose the variable "C" the way you did? Instead of choosing C=m1 and D=m2 for example.

Secondly, how did you see that you were supposed to rearrange the terms in the way you did in order to simplify the equations? I mean what you did in order to see so "clearly" that x=y was a solution.

Ray Vickson: sorry if my original post was a bit unclear. You assumed correctly. Can you see the image I posted? There everything is written clearly. Thanks.

7. May 29, 2014

jbunniii

First general idea: identify what it is you're solving for, in this case $v_{f1}$. Everything else is just a constant for the purpose of solving the problem.

Second general idea: if there is a lot of clutter (subscripts, multiple constants, etc.) try to simplify the notation before proceeding. This makes the algebraic manipulations (and typesetting) easier, and more importantly, it can help you notice things you would otherwise miss.

Third general idea: identify what kind of equation it is. In this case, after decluttering, we have one variable, $x$, and some constants $y$ and $c$. Moreover we can see that $x$ only appears in polynomials, not something like $\sin(x)$, so the entire problem will reduce to solving a polynomial equation in $x$. We can also see that the highest power of $x$ that appears in the equation is 2, so it is a quadratic equation.

So far so good. We know how to solve quadratic equations - we use the quadratic formula. However, the quadratic formula is annoying - it has square roots and other fiddly things. We also know that any polynomial that has a root, say $r$, will have an $(x-r)$ factor. If we can find such a factor, we don't need to use the quadratic formula. So we look briefly to see if we can factor anything out. Fortunately we might recognize that $x^2 - y^2 = (x+y)(x-y)$ and that the only other term also has a $x-y$ factor. So we breathe a sigh of relief knowing that we can skip the quadratic formula.

Sure, you could do that, but it will only slightly improve the notation. If that's the only simplification in sight, it may still be worth doing. But if we look for a few more moments, we notice that all the $m$ junk appears in one big factor on the right hand side. So why not lump it all into one constant? It makes life easier if we can do this.

It's useful to remember a formula like $x^2 - y^2 = (x+y)(x-y)$ so well that it jumps to mind instantly when you see $x^2 - y^2$, or equivalently, $x^2$ on one side of the equation and $y^2$ on the other. You may not always use the formula even if you see $x^2 - y^2$, but certainly consider whether it will help if you do see it.

Anyway, hopefully these suggestions are helpful! In my mind, the key things you need in order to notice simplifications are (1) plenty of practice and (2) a desire to notice them so you don't have to work as hard.

Last edited: May 29, 2014
8. May 29, 2014

Ray Vickson

Yes, I can see the image, and it disagrees with what you posted. The image says, essentially, that
$$v_{f1}^2 = v_{01}^2 - \frac{m_2}{m_1}v_{f2}^2$$
but you wrote
$$v_{f1}^2 = v_{01}^2 - \frac{m_1}{m_2}v_{f2}^2$$
That makes all the difference in the world.

9. May 29, 2014

Maxo

I hope a lot of other members on this forum see how jbunniii is writing and adapt this style. It's extremely helpful and pedagogical, and even entertaining. Thanks again!