2 Variables and 2 Unknowns Returning Incorrect Results

I am trying to solve a problem involving collisions, oblique collisions to be exact.

First attempt:

VA1=2
VB1=3
m1=0.5
m2=0.4
ThetaA1=30
ThetaB1=50
VA1x = m1*VA1*cosd(ThetaA1);
VB1x = m2*VB1*cosd(ThetaB1);
syms VA2x VB2x
equ1 = m1*VA1x + m2*VB1x == m1*VA2x + m2*VB2x;
equ2 = -0.5*(VB1x - VA1x) == (VB2x - VA2x);
sol = solve ([equ1,equ2],[VA2x,VB2x]);
solVA2x = sol.VA2x;
fprintf('solution is %i \n', solVA2x)

For my second attempt:

VA1=2
VB1=3
m1=0.5
m2=0.4
ThetaA1=30
ThetaB1=50

VA1x = m1*VA1*cosd(ThetaA1);
VB1x = m2*VB1*cosd(ThetaB1);

syms VA2x VB2x
sol=solve([m1*VA1x + m2*VB1x == m1*VA2x + m2*VB2x , 0.5 == -(VB2x - VA2x)/(VB1x - VA1x)],[VA2x,VB2x]);
C=sol.VA2x;
E=sol.VB2x;

fprintf('Value for VA2x %i \n',C);
fprintf('Value for VB2x %i \n',E);

I am pretty unexperienced and willing to learn so any additional references would be appreciated. I have read that I should add vpasolve however I have already tried that and the results are the same.

[Rik]

Removed text restored from Google cache. Please don't remove it again.

http://web.archive.org/web/20210309114706/https://www.mathworks.com/matlabcentral/answers/766876-2-variables-and-2-unknowns-returning-incorrect-results

Answer 1

[Walter Roberson]

fprintf('solution is %g \n', solVA2x)

The %i format is only for integers.

Answer 2

[John D'Errico]

This is a linear system of two equations in two unknowns. The solution is unique, since the sysem is full rank. If we look at your first attempt, we see:

VA1=2;
VB1=3;
m1=0.5;
m2=0.4;
ThetaA1=30;
ThetaB1=50;
VA1x = m1*VA1*cosd(ThetaA1);
VB1x = m2*VB1*cosd(ThetaB1);
syms VA2x VB2x
equ1 = m1*VA1x + m2*VB1x == m1*VA2x + m2*VB2x;
equ2 = -0.5*(VB1x - VA1x) == (VB2x - VA2x);
sol = solve ([equ1,equ2],[VA2x,VB2x]);
sol.VA2x

ans = 

sol.VB2x

ans = 

format long g
[double(sol.VA2x),double(sol.VB2x)]

ans = 1×2

0.802905222344044 0.85024535842434

Your second attempt is no different, you did nothing significant but manipulate one of the equations.

Is this a solution? We can verify that.

simplify(subs(equ1,sol))

ans = 

symtrue

simplify(subs(equ2,sol))

ans = 

symtrue

I could also have solved it as:

[A,b] = equationsToMatrix(equ1,equ2,[VA2x,VB2x])

A = 

b = 

As we see there, the problem is clearly non-singular,, but we can verify that as:

rank(A)

ans = 
     2

And now the solution is simple.

sol2 = double(A\b)

sol2 = 2×1

0.802905222344044 0.85024535842434