# Simple Maths/Physics Simultaneous Equation.

• May 13th 2013, 04:41 AM
TwentyThree
Simple Maths/Physics Simultaneous Equation.
Given the following two equations:

m1*u = m1*v1 + m2*v2

m1*u^2 = m1*v1^2 + m2*v2^2

Taking this as a simultaneous equation, and taking x as v1 and y as v2. How would you find x and y? I.e. How would you isolate v1 in terms of the other variables without using v2.

I'm not sure whether my algebra is lacking or what, but I started off finding a v2 = equation and substituting to cancel out the v2. However I keep getting a v1 and v1^2 and I am unable to group them together.

• May 13th 2013, 08:30 AM
ebaines
Re: Simple Maths/Physics Simultaneous Equation.
You have two equations, but it's clear whether you have two unknowns or three. If m1, m2 and u are all known values then you have two equations in two unknowns (v1 and v2) so can solve for v1 and v2. Apparently you have arrived at a quadratic equation in v1, and so you can solve for v1 by using the quadratic formula. But if u is an unknown as well then this is indeterminate since you need three independent equations in order to solve for three unknowns.
• May 13th 2013, 09:54 AM
HallsofIvy
Re: Simple Maths/Physics Simultaneous Equation.
I assume you mean you want to solve for v1 and v2 in terms of m1, m2, and u.

First square both sides of the first equation: m1^2u^2= (m1v1+ m2v2)^2 gives m1^2u^2= m1^2v1^2+ 2m1m2v1v2+ m2^2v2^2.
Multiply the second equation by m1: m1^2u^2= m1^2v1^2+ m1m2v2^2

Now subtract the second equation from the first: 0= m1(m1+ m2)v2^2+ 2m1m2v1v2= v2(m1(m1+ m2)v2+ 2m1m2v1)= 0 so either v2= 0 or (m1+ m2)v2+ 2m1m2v1= 0.

If v2= 0 then m1u= m1v1 sothat v1= u.

If (m1+ m2)v2+ 2m1m2v1= 0 the v2= -(2m1m2/(m1+ m2))v1 so that m1u= m1v1- (2m1m2^2/(m1+m2))v1= [(m1^2+m1m2- 2m1m2^2)/(m1+ m2)]v1 so that
v1= [(m1^2+ m1m2)/(m1^2+ m1m2- 2m1m2^2)]u.
• May 13th 2013, 12:24 PM
TwentyThree
Re: Simple Maths/Physics Simultaneous Equation.
@ebaines

This is one of those exercise book 'prove' questions where you have to assume variables = variables as opposed to variables = numbers.

@HallsofIvy

Your method at the start was quite amazing, I never would've thought about squaring the first equation. However I think you might've made a mistake (or just simplified it differently) at this point:

Now subtract the second equation from the first: 0= m1(m1+ m2)v2^2+ 2m1m2v1v2= v2(m1(m1+ m2)v2+ 2m1m2v1)= 0 so either v2= 0 or (m1+ m2)v2+ 2m1m2v1= 0.

Subtracting the second equation from the first gives me 0 = m2(m2-m1)v2^2 - 2m1m2v1v2 instead. Additionally my final computation of v1 using the aftorementioned expression gives me the proof I wanted which is:

v1 = (m1-m2)u/m1+m2
v2 = 2m1m2/m1+m2