roller coaster with weight requirement for a car My kids went on a roller coaster type ride on the weekend.  The staff loading individual (separate) cars on the ride said each had a minimum and maximum requirement to clear a hill - too little, it wouldn't accelerate enough down one hill to clear the next up, too much it wouldn't go up that hill either.  Can someone briefly sketch the equations here?  (I did high school physics decades ago - can't remember much of it now.)
 A: 
Consider the diagram above.
Roller coasters mostly rely on two simple principles: conversion of Potential Energy to Kinetic Energy and vice versa, and conservation of energy.
Suppose the ride starts where the car is drawn at a height $h_1$ above a common reference line (like the horizon).
We say that the car at that point has a Potential Energy $U$ according to: 
$U=mgh_1$, where $g=9.81\:\mathrm{ms^{-2}}$ the Earth's acceleration constant, $m$ is the mass of the car (and riders) and $h_1$ that height above the reference line.
Now we give the car a nudge and it starts accelerating down the slope and starts acquiring Kinetic Energy $K$:
$K=\frac{mv^2}{2}$, where $m$ is the mass of the car (and riders) and $v$ is the velocity of the car.
The Law Energy Conservation tells us that:
$T=K+U$, with $T$ the Total Energy of the car (and riders).
This means that as the car accelerates down the slope, $U$ is being converted to $K$.
At the start of the ride, when $v=0$, the Total Energy $T$ was:
$T = mgh_1$ and when the car reaches the lowest point $h_3$, then:
$T=mgh_3+\frac{mv^2}{2}$.
A little rearranging gives:
$\large{v=\sqrt{2g(h_1-h_3)}}$.
Note something important here: the expression for $v$ does not contain the mass $m$ of the car (and riders).
Now as the car starts moving up the hill towards $h_2$ the acquired Kinetic Energy $K=\frac{mv^2}{2}$ is being converted back to Potential Energy $U$ and as long as:
$\large{h_1-h_3 > h_2-h_3}$, then the car will be able to negotiate the hill $h_3,h_2$ without problems. This is the principal mode of operation of roller coasters (without any propulsion systems installed in the cars) which allows them to negotiate a series of slopes and hills without running out of energy.
There is however one complication: my ideal scenario doesn't account for any energy losses due to friction between car and track or air drag losses. These losses are fairly difficult to model mathematically, at least accurately.
A responsible roller coaster designer would allow for a considerable safety margin, by starting the ride from a sufficient height $h_1$ to prevent any stalling of the cars on inclines.
For that reason I'm very surprised that a roller coaster operator would specify any loads, instead of relying on a fail safe design.
In addition to being able to negotiate ups and downs safely, the car must also be able to safely negotiate bends. When the car takes a turn, the track must be able to supply a force $F_c$ known as the centripetal force in order to keep the car on the track:
$\large{F_c=\frac{mv^2}{R}}$, where $R$ is the radius of the bend.
As you can see, the mass of car (and riders) $m$ does play part here but again it seems folly to rely on loading specifications rather than safe design.
