dmckee's comment is correct, this problem is hard (impossible?) to solve with pen and paper. But it's not too hard to set up, which hopefully will answer your question. There will advanced calculus by high school standards (including differential equations), no way around that, but I'll try to explain the equations at a high school level, I hope...
I'll skip the ballistic portion of the lab, which is fairly similar in setup but simultaneously more complicated (because the ball is not constrained to follow a particular path) and less complicated (because there is no rolling friction, only air drag). I'll include air drag, but you could get rid of this term if you wanted to simplify things (though at this point you'll be solving with a computer either way, so you may as well include it).
Start with conservation of energy. This means that at any point in time, the energy released from gravitational potential must be equal to the kinetic energy (translational & rotational) plus the work done (energy lost) so far by friction and drag.
$$U_{g}-U_{g,i} = K_t + K_r + W_f + W_d$$
We'll go through these term by term, but we need to set up a few things before that. First, what is our goal? Ultimately, we'll have two unknowns, a constant each for friction ($\mu$) and for drag ($C_d$, which I'll define later). There's also going to be a major intermediate result, which is to find $s(t)$: I'll call the distance travelled along the track $s$, so $s(t)$ is a function that describes how long it takes the ball to get to any given location on the track - clearly this is a useful quantity!
I'll point out that the speed of the ball is $\frac{{\rm d}s}{{\rm d}t}$. When we need to work with vectors, I'll use $\hat{x}$ & $\hat{y}$ coordinates, where gravity is in the $-\hat{y}$ direction. The other very important vector will be the normal vector of the track, $\hat{N}$. I'll write this vector as:
$$\hat{\bf N} = N_x(s)\hat{\bf x} + N_y(s)\hat{\bf y}$$
There is also the tangent vector, which we'll need at one point:
$$\hat{\bf T} = T_x(s)\hat{\bf x} + T_y(s)\hat{\bf y}$$
The two functions $N_x(s)$ and $N_y(s)$ depend on the shape of the track, but they are known: you can write them down using the known shape of the track and geometry.
Now we're ready to go:
- $U_g = mgy(s)$ and $U_{g,i} = mgy_0$, these are pretty easy, just the usual gravitational potential energy, expressed in terms of the height $y(s)$ at a given track position, and the initial height $y_0$.
- $K_t = \frac{1}{2}m\left(\frac{{\rm d}s}{{\rm d}t}\right)^2$, the usual translational kinetic energy, recalling the speed that I pointed out earlier.
- $K_r = \frac{1}{2}\frac{I}{r^2}\left(\frac{{\rm d}s}{{\rm d}t}\right)^2$, the usual rolling kinetic energy, assuming rolling without slipping. $I$ is the moment of inertia of the ball, which you should measure/calculate, and $r$ is the radius of the ball.
- $W_f$ starts to get tricky. You might know work is force times distance, but what if the force changes with distance? Then you need an integral: $W = \int_{s_0}^{s_1} F(s)ds$ (no need to worry about angles because friction always acts opposite the direction of travel). The friction force is the normal force times the coefficient of friction, so we need to know the normal force at any time. Luckily, this is not too hard to get. We know that the ball stays on the track, which tells us the force in the normal direction to the track must be exactly right to keep the ball on the track. On a circular track segment, this boils down to the centripetal acceleration $v^2/R$; on a more generic track we can pretend the piece of track we're interested in is part of a circular track, and ask what it's radius would be to get the centripetal acceleration. That radius is:
$$R(s) = \left|\frac{{\rm d}\hat{\bf T}}{{\rm d}s}\right|^{-1}$$
Now we know that the acceleration in the $\hat{\bf N}$ direction needs to be $\left(\frac{{\rm d}s}{{\rm d}t}\right)^2/R(s)$ (that's "$v^2/r$" centripetal acceleration for our problem). Now we can use ${\bf F}=m{\bf a}$ in the $\hat{\bf N}$ direction to get friction force. The only forces that ever act in the $\hat{\bf N}$ direction are gravity and the normal force; friction and drag are exclusively in the $\hat{\bf T}$ direction. So:
$$mg\hat{\bf y}\cdot\hat{\bf N} + F_N = m\frac{\left(\frac{{\rm d}s}{{\rm d}t}\right)^2}{R(s)}$$
Solving for the normal force $F_N$ (and evaluating that dot product):
$$F_N = m\frac{\left(\frac{{\rm d}s}{{\rm d}t}\right)^2}{R(s)} - mgN_y(s)$$
and the friction force is:
$$F_f = \mu\left[m\frac{\left(\frac{{\rm d}s}{{\rm d}t}\right)^2}{R(s)} - mgN_y(s)\right]$$
So finally we can write down an expression for $W_f$:
$$W_f = \mu m\int_0^s\left[\frac{\left(\frac{{\rm d}s'}{{\rm d}t}\right)^2}{R(s')} - gN_y(s')\right]{\rm d}s'$$
- $W_d$, the work done by drag, is fortunately a lot easier. The same idea, integrating the force along the path, applies. Again, no need to worry about angles, because drag opposes the direction of motion exactly. A simple expression for the drag force that should be applicable here is $F_d = \rho v^2 C_d \pi r^2$ - the density of air is $\rho$, the speed of the ball is $v$, the cross sectional area of the ball is there as $\pi r^2$ and there's a constant $C_d$, a "drag coefficient", this is similar to $\mu$ for drag. Writing this consistently with the notation above,
$$F_d = \pi r^2 \rho C_d \left(\frac{{\rm d}s}{{\rm d}t}\right)^2$$
Then just putting this into an integral to get the work:
$$W_d = \pi r^2 \rho C_d \int_0^s\left[\left(\frac{{\rm d}s'}{{\rm d}t}\right)^2\right]{\rm d}s'$$
So that's all the terms, now I'll just stuff them all into the first equation at the top of this answer (and put everything on one side)...
$$0 = mg\left(y(s) - y_0\right) + \frac{1}{2}m\left(\frac{{\rm d}s}{{\rm d}t}\right)^2 + \frac{1}{2}\frac{I}{r^2}\left(\frac{{\rm d}s}{{\rm d}t}\right)^2 +\\ \mu m\int_0^s\left[\frac{\left(\frac{{\rm d}s'}{{\rm d}t}\right)^2}{R(s')} - gN_y(s')\right]{\rm d}s' + \pi r^2 \rho C_d \int_0^s\left[\left(\frac{{\rm d}s'}{{\rm d}t}\right)^2\right]{\rm d}s'$$
This is quite a mess, but it is an integro-differential equation for $s(t)$. The idea is to find a function $s(t)$, when substituted into this equation, makes the right hand side equal to $0$. In this case, I'm almost certain that this cannot be done by hand (pen and paper) and instead you'd need to find an approximate solution using a computer. There are a couple of last complications before you can solve this. First, equations like this need what are called initial and boundary conditions. These give some extra constraints on the function $s(t)$. Here they're pretty straightforward, and just express that the length of the track is $L$, that the initial speed is $0$ and that the speed at launch time $t_L$ (which you would measure in the lab) is $v_L$ (which you would also measure in the lab). Here are the conditions:
- $s(t=0) = 0$
- $s(t=t_L) = L$
- $\left.\frac{{\rm d}s}{{\rm d}t}\right|_{t=0} = 0$
- $\left.\frac{{\rm d}s}{{\rm d}t}\right|_{t=t_L} = v_L$
Last but not least, there's the matter of the two unknowns we are trying to measure, $\mu$ and $C_d$. Unless my intuition is failing me, I think you'll end up with a system of equations with too many unknowns to solve, given your input. This is easy enough to remedy by making some additional timing and/or velocity measurements in the lab. Things are pretty dicey since $s(t)$ depends on $\mu$ and $C_d$, and you need to know the constants before finding $s(t)$ because you're finding an approximate solution. This ends up meaning you need to find solutions $s(t)$ for many combinations of $\mu$ and $C_d$ (probably in some clever way that helps find the right one quickly), then pick the one that works with your measurements.
So uh... I guess you can see why your teacher suggested a simple assumption to keep the problem from getting out of control!
Best Answer
tok3rat0r probably has the right of it.
You have not told us exactly how your timing data was acquired. If it was done by some sort of stopwatch (mechanical or electrical) you should assume an uncertainty of at least 0.1 seconds, and perhaps more for a situation where you have to push a button twice in 1/2 second.
If you assume a worst-case error of 0.1 seconds, your timing measurement becomes 0.54 seconds, and if you redo your calculations you'll get a final energy of 0.15 J. While this does not conclusively prove that your timing technique was the problem, it suggests that you should look very closely at it.
One approach would be to increase your ramp height so that your duration gets much longer - several seconds at a minimum. If you are using an electronic timer, you could replace your timing switch with two switches - one at the top of the ramp and one at the bottom, with a mechanical block release such as burning through a thread.