These equations have an answer - how to get solve to find it?

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RobNik
RobNik on 29 Nov 2017
Commented: RobNik on 7 Dec 2017
I'm not sure where to ask this question. I solved some equations by hand, and I know the answer is right. I can't get `solve` to give me the same answer, and I'm wondering why.
It's a physics problem with two objects and conservation of momentum. Their final velocities are v_1f and v_2f. The question is "what is u when the two velocities are equal".
Why doesn't `solve` give me that value for `u`?
syms m m_1 m_2 v_1 v_2 v_1f v_2f u;
expression1 = (v_1*m_1 + v_1*m + u*m) / (m+m_1); % for final velocity 1
expression2 = (v_2*m_2 + m*(v_1f-u)) / (m_2 + m); % for final velocity 2
eq1 = v_1f == expression1
eq2 = v_2f == expression2
eq3 = v_1f == v_2f
solve([eq1 eq2 eq3], u) % fails
answerByHand = -(m_2*(m + m_1)*(v_1 - v_2))/(m*(m + m_1 + m_2)) % this is u, found manually
X = simplify(subs(expression1,u,answerByHand))
Y = simplify(subs(subs(expression2,v_1f,expression1),u,answerByHand))
isequaln(Y,X) % true
Update:
It might be useful or interesting to know more about the physics problem...
It's about 2 astronauts in zero-gravity throwing a mass between them.
Everything is 1-dimensional. v_1 and v_2 are the initial velocities of the two astronauts. They are both moving to the right. #2 is initially faster. #1 throws a mass (m) to #2, at relative speed u. v_1f and v_2f are their final velocities. So the question is: how fast must #1 through the mass m so that the two astronauts have the same speed after astronaut #2 catches it.
The equations come from the conservation of momentum (m*v).

Answers (2)

John D'Errico
John D'Errico on 29 Nov 2017
Edited: John D'Errico on 29 Nov 2017
Anyway, lets check the solution that you have posed. I'll write it a bit more simply.
v_1f = (v_1*m_1 + v_1*m + u*m) / (m+m_1);
E2 = v_2f == (v_2*m_2 + m*(v_1f-u)) / (m_2 + m);
But, you also have v1_f == v_2f. So we can eliminate v_2f completely, which has the nice feature that it also eliminates v_1f.
E3 = subs(E2,v_2f,v_1f)
E3 =
(m*u + m*v_1 + m_1*v_1)/(m + m_1) == (m_2*v_2 - m*(u - (m*u + m*v_1 + m_1*v_1)/(m + m_1)))/(m + m_2)
We can solve that for u.
u = solve(E3,u)
u =
-(m*m_2*v_1 - m*m_2*v_2 + m_1*m_2*v_1 - m_1*m_2*v_2)/(m*m_1 + m*m_2 + m^2)
u0 here is the solution that you posed.
u0 = -(m_2*(m + m_1)*(v_1 - v_2))/(m*(m + m_1 + m_2));
simplify(u-u0)
ans =
0
So in fact, the two solutions are the same.
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Stefan Wehmeier
Stefan Wehmeier on 30 Nov 2017
Why not just solve(expression1 == expression2, u) ?
What you did is to confuse assignemnts/assumptions with equations. You want to assume that v1_f and v2_f are given by certain expressions, and ask for a value of u that makes the two equal. What you typed is the question whether there exists u such that any two numbers v1_f and v2_f automatically become equal to the given expressions, and equal to each other.
  1 Comment
RobNik
RobNik on 7 Dec 2017
I guess I don't understand `solve`. Can you restate your answer using the following simpler example? If I were told to "solve these two equations for y":
  • x = y
  • x = 1
I'd use basic algebra and say that the answer is "y = 1". But this has no answer:
solve([x == y, x == 1], y) % Empty sym: 0-by-1
This one however, does what I was thinking:
solve([x == y, x == 1], [x,y])
and says "x=1, y=1".
What is the first one really saying/asking for?

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