Law of Gravity

Yes, but only so far as friction is concerned. If friction was removed as a component in the experiment, then mass would not play a part.

In your experiment, the wheels and their bearings are the largest source of friction. Mass will help overcome the static frictional force much more than the dynamic friction, which as you know, is proportional to the mass forces applied to the bearings.

Aerodynamic drag is another form of friction, but that has more to do with cross sectional area. If the cross section area is the same, more mass will render that less and less of a factor. Think of a feather falling. If you made the feather out of brass or a heavy material it would drop much faster.

I have to agree but would add that surface friction can play a large part in this equation. A higher density mass will depress the surface greater causing a pushing effect of upper surface thereby increasing rolling friction and thus a change in the results. Increase density high enough and it will just sink. Devils in the details.

For all practical purposes, gravity is constant on earth, so given that, F=ma applies. Bodies will free-fall on earth with a constant acceleration, which is about 9.8 meters/second squared, regardless of their masses. When on an inclined plane, bodies will be subjected to a vector force in the plane of the incline which is a function of the slope of the incline and the earth's gravity. Given that the wheels on the cart are frictionless, the cart will accelerate down an incline at the same rate for all masses. Now, for rolling bodies, such as wheels or balls, the same is true only if bodies of different mass have the same mass moment of inertia. I'll leave it to you to figure out.

I left out something . . . In the real world, no, the more massive object won't be moving faster unless there's other factors that have a significan affect on rolling, such as wheel friction, aerodynamics, etc. The more massive object will, however, have something more than the less massive object, and that's "momentum," which is 1/2(mass X velocity squared).

well first of all there can't be any "rolling" without friction. There will be sliding and all objects will have same speed.

It depends a lot upon what forces we are willing to take into account in the real world. I guess all the forces present in the real world on an ordinary road like bumps and air resistance etc :) If there's only the road friction then mass wouldn't affect the velocity. True, there's more frictional force acting on the body, but the acceleration (and hence the velocity) is not affected by the mass. So as far as the road friction is concerned, mass of object doesn't change velocity. But air resistance would make a difference here. I would think that the air drag force would be the same on heavier and lighter object (if they have the same surface area, for the same velocity) so the loss of acceleration on heavier object would be less. My answer: heavier goes faster in real world.

Indeed tonight on Myth Busters there was a demonstration of the effect. The little blocky lead car significantly beat out the little aerodynamic aluminum car because, up to a point of course, the effect of the mass increase is proportionately greater than the slightly greater friction (air, wheels, et al) due to greater weight.

All the previous opinions are very close... but there is a lack... Although we may assume gravity is almost constant all over the world and it is no influenced by height (if height difference is a few hundreds meters we may really assume that "g" is constant, but really "g" decreases with height), there will always be a difference when object with different mass will roll downhill. The reason is not gravity law but energy conservation. At the top of the hill (assuming zero initial speed) all the energy is due the height difference "h" between the top and the basis of the hill. Total initial energy of the cart and the body (at the top) will be "(mc + mb)g h", where "mc" is the cart mass, "mb" is the body mass, "g" is the gravity acceleration and "h" the difference in height between top and basis of the hill. When the cart rolls down it loses height but gains speed (kinetical energy); assuming there is no energy lose due friction of mechanical components and air, when the cart reaches the basis it will only have kinetical energy "1/2 (mc + mb) v2" (there is no way to write down "v squared" and I adopted "v2" as "v squared"). If we disregard loses due friction, all the potential energy at the top will change to only kinetical energy downhill: "(mc+mb)g h = 1/2 m v2". In this case, when we solve the equation for "v", we will get "v = squared root (2gh)", meaning weight of body has no effect. But the above situation is not true in the real world as there are energy losses due to friction and aerodynamics behavior of the body and the cart: all those losses of energy are independent of the mass (both the body and cart) and the equation for energy conservation will then become: "mgh = 1/2mv2 + EL", where "m" is the total mass (m = mc + mb) and "EL" is the total amount of energy lost. Now, when we solve for "v", we get "v = squared root [2(mgh - EL)/m]" or "v = squared root [2(gh-EL/m)]". The last expression let us better understand that the higher the mass the higher the speed... except when the mass of the body is so high that the cart collapses, and then speed is zero :-)

(There is no way to write down "v squared" and I adopted "v2" as "v squared".) Would this help? (v^2) It is v (Shift+6 key) 2. The carrot symbol is assumed to mean to the power of. Since v to the power of 2 = v squared then v^2 should work. Your way works just as well too once you explained it.

Well I would prefer to solve this using momentum conservation ideally. Anyway, in your expression "v" also depends on "EL". Are you assuming "EL" is the same for objects with different masses. I would think loss of energy will be higher for heavier objects. If that is the case then we can't arrive at a conclusion from your equation.

Having read the golf ball theory, Dimples instead of smooth surfaces to reduce drag I am contemplating wheather the cart would have less (EL) where the cart to be made with it's surfaces arrigated in some specific manor.

or would the lift from specific incorrect aggigations create lift and add to the (EL) energy loss their by significatly reducing the expected potential kinetic energy? (v or/and v^2)

the cart obviously has more velocity when heaver although, the kinetic energy is potential, (from the height and mass) the mass, disregarding friction would have more potential energy in both forms v an m from a heavier weight.

I further contemplate back to the fact that an earwig can lift 1760 times it's own weight, that's like a human lifing a three bedroom house.