possible real solutions of the equations

Question

What are the possible real solutions of the equations $$1000=v_1^2+4v_2^2,100=v_1+4v_2$$ Its a physics question but I thought its not necessary to post here . Thank you.

Answer 1

I'd wanted to return to this question because there are some interesting things to be said about the graph for this set of equations.

This is a description for an "elastic" collision between two objects of different masses (which I presume is what the original physics problem was about), that is, a collision is which the total kinetic energy of the objects is unchanged (physicists say "is conserved"). I suspect the reason the posted problem has no solution is because one of the equations was not divided through consistently. The kinetic energy equation would be

$$ \frac{1}{2} \ m_1 \ v_1^2 \ + \ \frac{1}{2} \ m_2 \ v_2^2 \ \ = \ \ K \ \ , $$

and there is a second equation stating that the total linear momentum of the two objects also remains constant, or

$$ \ m_1 \ v_1 \ + \ m_2 \ v_2 \ \ = \ \ p_{total} \ \ . $$

The two equations given are consistent with the two masses being $ \ m_1 \ = \ 1 \ $ and $ \ m_2 \ = \ 4 \ $ , with a reasonable re-construction of the system being

$$ v_1^2 \ + \ 4 \ v_2^2 \ \ = \ \ 1000 \ \ , \ \ \ v_1 \ + \ 4 \ v_2 \ \ = \ \ \mathbf{50} \ \ , $$

for reasons that will hopefully become clear shortly.

Whether we solve this analytically or graphically, we obtain two pairs of solutions: $$ \ ( v_1, \ v_2 ) \ = \ (30, \ 5) \ \ , \ \ (-10, \ 15) \ \ . $$

These are interpreted as the velocities of the two objects before and after the collision: the object with mass $ \ 1 \ $ , moving at 30 velocity units, overtakes the object with mass $ \ 4 \ $ , traveling at 5 units. After the collision, the more massive object is now moving faster in the same direction as before, at 15 units, with the lighter object "recoiling" with a velocity of 10 units. On the graph, the new point is a sort of "mirror-reflection" about the line $ \ v_1 \ = \ v_2 \ $ (marked in light green).

(I don't think the original system of equations was intended to be $ v_1^2 \ + \ 4 \ v_2^2 \ \ = \ \ 2000 , $ $ \ v_1 \ + \ 4 \ v_2 \ \ = \ \ 100 \ \ , $ as this has the single solution $ \ (20, \ 20) \ $ [the line is tangent to the ellipse). This would be interpreted as the pair of objects simply moving together at 20 velocity units, which makes for a somewhat less interesting physics problem. The fact that the speeds in this reconstruction come out to be integers suggests that this may have been the intended solution.)

It might be worth mentioning that physicists often analyze such problems in mechanics using the "center-of-mass" or "center-of-momentum" reference frame, in which the total linear momentum of the objects is zero. On our graph, this would correspond to a line (marked in orange) parallel to the original momentum line (in red), but now passing through the origin ( $ v_1 \ + \ 4 \ v_2 \ \ = \ \ 0 $ ) . This alone does not give us solutions because those will not be on the original "kinetic energy ellipse": we are moving at a different speed relative to the pair of objects now, so their total kinetic energy is measured to be smaller.

We can locate these solutions by passing lines of slope $ \ 1 \ $ through our original solution points; these meet the new linear momentum line at $ \ (20, \ -5 ) \ $ and $ \ (5, \ -20) \ $ . These points are interpreted as the object of mass $ \ 1 \ $ moving to the right at 20 velocity units and the object of mass $ \ 4 \ $ moving to the left at 5 units before the collision, after which they rebound from one another with the same speeds but reversed directions. The use of the lines with slope $ \ 1 \ $ changes the reference frame speed by a specific amount, in this case by 10 velocity units to the right.

Note that doing so leaves the relative speeds of approach and separation of the two objects unchanged (closing on one another at 25 velocity units before the collision and separating at 25 units afterwards in either reference frame). The equal magnitude and opposite directions of the closing and separating velocities is a characteristic of elastic collisions.

Answer 2

Substitute $v_1=100-4v_2$ in the first equality, then $$ 20v_2^2-800v_2+9000=0, $$ reducing to $$ v_2^2-40v_2+450=0 $$ Then its quarter of discriminant is $20^2-450<0$, so no real solution exists. See also this picture.