Confused by answer to mechanics question 
I am confused by the answer to this question. Well, I am trying to explain my thought process about this. I used net energy in the system to calculate the maximum speed. So I assume at max velocity, net change of energy is totally converted to kinetic energy. So, net change in energy after the trolley has been moved 1.5 cm = (0.5)(1600)(0.06^2-0.03^2)=2.16J. So I assume that at max speed, KE=2.16J hence the velocity can be calculated. But apparently it is the wrong method to use. I understand that the total elastic potential cannot be zero because the strings are bound to the wall hence there will always be extension but if I consider directions, at equidistance, wouldn't the energies on the left spring and right spring cancel out? That was my thought process in answering this question.
The mark scheme used this method:
KE gained = 0.5(k)(0.060^2-0.045^2)+0.5(k)(0.030^2-0.045^2)=0.36J hence leading to a different max velocity, which kinds of make sense too since that is the energy you add to the system, so at max velocity, it will be totally converted to KE back to equilibrium again. 
I am just want to know the flaw in my method and I hope I can get a better understanding about this topic after this! :D Thx in advance!
 A: The energy is a scalar, so you add it just like numbers. There is no such thing as "energy in that direction". The way to calculate the total energy is to add the energy of each component regardless of other components.
So in this particular question - 
the energy at is (left -start, right - equilibrium)- 
$$ \frac{1}{2}k (x_0+d)^2+\frac{1}{2}k (x_0-d)^2 = \frac{1}{2}m v_{max}^2 +2\frac{1}{2}k (x_0)^2  $$
$ x_0 $ - the spring extention when the trolley at equilibrium
$ d $  - dispacement of trolley from equilibrium
A: Both energies are positive. You should add, not subtract.
Compression and elongation will both store positive potential energy. The $x^2$-part will remove any minus sign.
$$U_{spring}=\frac{1}{2}kx^2$$
Your error is that the expression:

(0.5)(1600)(0.06^2-0.03^2)=2.16J

should have been

(0.5)(1600)(0.06^2+0.03^2)=2.16J

A: Your mistake is that you didn't take into account the original position of the trolley, you'll notice that you don't have 4.5cm appear anywhere in your answer.
If you wrote down the equation describing the words you wrote for your answer before your equation, you'd actually write down the equation for the correct answer.
I'll go into more details about your answer and what is wrong with it and the correct answer and where it comes from.
Your equations only know what you write down, and the world of things you can right down is vast indeed.  For instance you could have a trolley made of Lego bricks that could be rebuilt to be shorter or longer.  It seems like you are writing equations for that.  And there is no reference to the original energy in your answer (all the talk about directions of energy make no sense, and can't make sense, energy is a scalar), so you can't be measuring a change in energy, a change in energy needs to have an energy before and an energy after.  You have nothing that refers to the energy before, nothing referring to the 4.5 cm.  The entire next two paragraphs will only talk about the situation before you move the trolley.  Seriously.
So now lets look at the energy before, the energy of a spring depends on how much it is stretched or compressed.  So we need to quantify that.  So we need to figure out the natural length (the length it is when it is neither compressed nor stretched).  Each spring naturally wants to be a certain length, let's call them $x_1$ and $x_2$, when they are that length, they exert no force and we will also make that the zero of our potential energy.  When we hook up the first spring we either have to compress it (if $x_1 > 4.5cm$ or we have to stretch it (if $x_1 < 4.5cm$).  Either way, afterwards it will exert a force of $\pm k(4.5cm-x_1)$ on the trolley.  Same thing happens when we hook up the second spring, we either have to compress it (if $x_2 > 4.5cm$ or we have to stretch it (if $x_2 < 4.5cm$).  Either way, afterwards it will exert a force of $\pm k(4.5cm-x_2)$ on the trolley.  But at 4.5cm, the trolley is in equilibrium so those two forces (when we take the directions into account) must cancel.  We know that when $4.5cm-x_1>0$ we had to stretch the spring, so the spring pulls it towards it, so $k(4.5cm-x_1)$ is the force towards the first spring.  So the net force is the sum of the forces, so if you pick the direction towards the first spring to be positive, the total force due to both springs is $k(4.5cm-x_1)-k(4.5cm-x_2)=k(x_2-x_1)$, so since originally that force is zero (the trolley started out in equilibrium, we know that $x_2-x_1=0$ or $x_2-x_1$.  I breezed over that before because I thought you'd seen that before, and I just called them both $x_0$ since they were equal.
So that's just us noting that if the trolley was at equilibrium originally when both springs were at 4.5cm then the equilibrium extension ($x_0$) of each spring is pretty much unknown, but they are equal for the two springs ($x_1=x_2$ so let's call them both $x_0$).  Now lets talk about potential energy.  The potential energy of a spring (with spring constant $k$) of length $\ell$ that is in equilibrium when it is length $x_0$ is $\frac{1}{2}k(\ell-x_0)^2+C$, and the constant $C$ doesn't matter in Newtonian Mechanics, and is usually set to zero.  If you've studied calculus you can get this from integrating the force law.  Or you can take it as given and differentiate it to get the force law, you have to start somewhere.  So springs either have that potential energy and you get the force from it, or else springs have a force law and you integrate the force to get the potential energy.  Either way you get $\frac{1}{2}k(\ell-x_0)^2$ as the potential energy.  So in particular, the springs have no potential energy when they are each length $x_0$ but we don't know what that length is.  Hopefully this unknown thing drops out in our final answer! It does. But we do know the unknown $x_0$ is equal for the two springs since there is an equilibrium when they are both at 4.5cm. So the energy before we move the trolley is $\frac{1}{2}k(4.5cm-x_0)^2+\frac{1}{2}k(4.5cm-x_0)^2$, since the length of each spring was 4.5cm before we moved the trolley.
We are now done talking about the energy before we move the trolley, it is time to move the trolley.  Nothing changes except the lengths of the springs. They both used to be 4.5cm and now one is 6cm and the other is 3.0cm.  But they are the same springs so $x_0$ (the length of the free spring) is the same (and still unknown),and the formula for the potential energy $\frac{1}{2}k(\ell-x_0)^2$ is the same, but this time each spring has a different length, so we get a total potential energy of $\frac{1}{2}k(6.0cm-x_0)^2+\frac{1}{2}k(3.0cm-x_0)^2$.
Before we moved the trolley and after we moved the trolley there was no kinetic energy so all the energy was potential, so you are totally correct that the energy added by moving the trolley was just the difference in the potential energies.  So we take the energy after $\frac{1}{2}k(6.0cm-x_0)^2+\frac{1}{2}k(3.0cm-x_0)^2$ and subtract from it the energy before $\frac{1}{2}k(4.5cm-x_0)^2+\frac{1}{2}k(4.5cm-x_0)^2$.  So the energy added is:
$$\left(\frac{1}{2}k(6.0cm-x_0)^2+\frac{1}{2}k(3.0cm-x_0)^2\right)-\left(\frac{1}{2}k(4.5cm-x_0)^2+\frac{1}{2}k(4.5cm-x_0)^2\right).$$
This added energy will be the kinetic energy at the position where both are at 4.5cm. Because back there the potential energy was the energy before.  And I breezed over that because you seemed to suggest that so I though you understood.  But we can do more steps.
Energy at 6.0cm,3.0cm is $$\left(\frac{1}{2}k(6.0cm-x_0)^2+\frac{1}{2}k(3.0cm-x_0)^2\right)+\frac{1}{2}m_{trolley}0^2.$$
Energy at 4.5cm,4.5cm is $$\left(\frac{1}{2}k(4.5cm-x_0)^2+\frac{1}{2}k(4.5cm-x_0)^2\right)+\frac{1}{2}m_{trolley}v^2.$$
And conservation of energy says the two are equal. So we can solve for the kinetic energy (it is the difference between the two potential energies).
If you do a little bit of math, the unknown $x_0$ drops out and you'll see that it looks like the formula for the correct answer.
