An extremely difficult physics problem...please help?

Question

This originally had me completely befuddled, but I think I got it. Problem is, there is no way to "justify" my results, since I got to a certain point and simply used the "Guess and Check" method. I think I have a decent explanation and proof. Could someone check it for me? Here's the orignal problem:

One car has twice the mass of a second car, but only half as much KE. When both cars increase their speed by 6.0 m/s, they then have the same KE. What were the original speeds of the two cars?

Point where I got: (Initial Vi1 + 6)^2 = 2 (Initial Vi2 + 6)^2
This is what I tried from this point: I graphed each of them as
y = (x + 6)^2
y = 2(x+6)^2

I got to a point where at x = -6, y = 0 for both. Using this "explanation", I said V1 - V2 = 6.

Then, using guess and check with V1 - V2 = 6, I got approximately V1 Initial = 14.4775, V2 Initial= 8.4775

Is there a way to get the answers I got without using the guess and check method?

PLEASE HELP!!!

Answer 1

no, there is not a way to verify these numbers. this is mainly due to the fact that your answer is incorrect since you're going about it the wrong way thus far. don't make it so difficult on yourself... there's a reason the equation gives you two equalities and two variables... you've got to put them together like this:

1) 2MV(sub1)^2 = 0.5MV(sub2)^2 (the M's will cancel out and you can solve the equation for v(sub2) pretty easily)

2) 2M(V(sub1) + 6)^2 = M (V(sub2) + 6)^2 (now foiling out, this becomes...
2(V(sub1)^2 + 12 V(sub1) + 36) = V(sub2)^2 + 12V(sub2) + 36

now take the value for V(sub2) you found in the first equation and plug it into the second equation. this will give you one equation where the only variable is V(sub1). solve for V(sub1) and then plug this back into the first equation and find V(sub2). that's your initial velocities. i'm not going to do the work out for you, since it's your homework, but here's what i got (barring any algebra errors on my part)

V(sub1) = sqrt(18)
V(sub2) = sqrt(72)

Answer 2

Call their speeds u and v.

Then 2m u^2 = (1/2) m v^2
and so u^2 = (1/4) v^2, which means that
u = (1/2) v

Also 2m (u + 6)^2 = m (v + 6)^2
hence 2 (u + 6)^2 = (2u + 6)^2

This simplifies to
u^2 = 18
and so u = 3 sqrt(2), or about 4.242 m/s;
the other is 6 sqrt(2), about 8.484 m/s.

Answer 3

Your formation of equations is incomplete
There are 2 equations
status 1
2mv1^2/4 = mv2^2/2
v1^2=v2^2
This tells V1 = v2
This makes the formation wrong

Answer 4

I'm a chemist and I did physics years ago. Having 2 equations with 2 unknowns is not guess and check. Do you not know how to solve 2 equations with 2 unknowns without guess and check? Or perhaps I'm way off base.