Relationship between height and velocity in conservation of mechnical energy

Question

I'm a high school physics student, and we recently did a lab on the conservation of energy where we measured the speed of a marble at varying heights on a rollercoaster track. We were supposed to graph the height and speed, and I ended up with something resembling a square root graph

We were required to linearize the data (so that we could have a single slope), so I squared the heights and got a slope of about $-0.11\,cm^{-1}\,s^{-1}$:

I don't understand what the slope is supposed to signify. I tried to take the derivative of velocity with respect to height in the equation $E = mgh + \frac{1}{2}mv^2$, where E is constant, and got a cryptic answer of $\frac{dv}{dh} = \frac{-g}{v}$. Can someone help me understand what's going on here?

Answer 1

Look at the equation you gave for the energy, $E=mgh + \frac12 m v^2$, and find two variables that are linearly related. It can help to make a new variable to show the linear relationship. For the attempt you show (plotting $h^2$ vs $v$), set $H=h^2$ so that the equation for the energy is $E=mg\sqrt{H} + \frac12 mv^2$ which does not suggest that there is a linear relationship between $H$ and $v$.

Answer 2

Are you sure that $h^2$ vs $v$ would be linear?

Consider $E = mgh + 1/2mv^2$. To linearize this, think about what variable is causing the "non-linearity," i.e. which variable could you transform in order to create a new, linear equation.

Answer 3

I get the same answer as you, $dv/dh = -g/v$ or $dh/dv = -v/g$. Here's what I think it means.

For the same energy $E$, there is a tradeoff between height and velocity. More height, less velocity. More velocity, less height.

So at any point in the fall, the rate at which it is trading height for velocity is equal to velocity divided by the gravitational constant. So at the top, when its velocity is small, it loses little height for each increment of velocity. At the bottom, when its velocity is large, it loses a lot of height for each increment of velocity.

This may be easier to see if you just consider that an increment of velocity is just $g$ times an increment of time.

EDIT: to put it another way, $-v/g = t$, the length of time it's been falling. So $dh/dv = -t$, and since $dv = -g dt$ (because velocity is in the downward direction), $dh/dt = -g t$ or $v = -g t$
:-)