@JohnRennie , since the roller coaster is being accelerated greater than a speed of g, I expect the boxes in it will leave the car, when released and rise higher. This is similar to how if boxes are placed in a forward moving car on the road, and when they are released they will move backwards. A similar situation: If you have travelled in a forward moving bus, and you are standing and holding the handle; if you stop holding the handle you will be pushed backwards.
@JohnRennie, I made a mistake here, g is not a speed, its an acceleration due to gravity. So if boxes are accelerated downwards greater than the acceleration of g, when they are released they should go upwards, since the impending motion is upwards.
Do my ideas make sense? I'm willing to learn
I hope to learn something and leave this site with a better understanding of Physics
@ShashankVM Whenever any object accelerating it feel backward force due to property of inertia. Because objects don't want to change its velocity.
In order to keep boxes in roller coaster car when at some height it downward/invert it is necessary to create enough centrifugal force which is greater that gravity and forces acting on it downward. So, it feel its upward force.
I am trying to understand space groups in crystallography. In International tables for crystallography, for a nonsymmorphic space group, they list some symmetry operations. 8 of them are listed under the (0,0,0)+ set and 8 in the (1/2, 1/2, 1/2)+ set. What does this mean? Are there 16 operations ...
@ShashankVM suppose the roller coaster car and the boxes are all falling freely. So the observer in the ground frame sees all three of them accelerating downwards at g.
If you were sitting in the car then you, the car and the boxes would all be weightless and stationary with respect to each other.
If we now accelerate the car downwards, e.g. attach a rocket motor to it, then you and the boxes would still be weightless and falling freely but now you'd see the car accelerating away from you.
For the person watching from the ground you and the boxes would still be falling freely as before, so you and the boxes would only move apart if you and the boxes fell freely at different rates.
In a vacuum you and the boxes would stay together because you'd all be accelerating downwards at the same acceleration of g.
In air we need to include the drag due to air resistance, and that would make you and the boxes fall at different speeds so all three objects would separate from each other.
So the question comes down to what effect the air resistance has on the rate the boxes fall.
@JohnRennie In the question I asked, the boxes have the same size and shape. So they will experience the same air resistance. But when they are released, they will experience a force of friction, since the car faces downwards. The body of the car is perpendicular to the ground. Since these boxes have different masses, they will experience different frictional forces. So I feel the frictional force has more effect than air resistance
@ShashankVM Why will the boxes experience any friction? The car and the boxes are all falling freely so they are all weightless. There are no forces acing between them.
Initially the boxes are fastened to the car. They are release in the middle of the descent. When they are released the car accelerates away from the boxes. But when this happens the contact friction between the car and box comes into play
There are 2 boxes wooden boxes of the same size and shape. One is empty and another is filled with a heavy material, like Gold. These boxes are placed on top of an open car of a roller coaster ride and fastened to the car. The car goes over the top of the roller coaster and starts its descent ver...
I am not sure if I am overthinking but if you have an electrical wire loop, the even if you know $(x(t), p(t))$, you won't know the emf induced.
But I think this might not be the right question to ask because the whole system is a collection of particles. Maybe having individual $(x(t), p(t))$ does give all the information about the system.
@Yashas when people say that $(x(t), p(t))$ suffices, they mean a classical mechanical system, not a circuit
once electromagnetism becomes involved the electric and magnetic fields or rather their potentials become additional variables you need to know to know the entire state of the system
the trouble for ultrafinitism is you cannot use induction, so you do not have the luxury of the fundamental theorem of arithmetic to show it is impossible
@Yashas @NiharKarve so you brushed up against QM foundations/ interpretations, do encourage you to delve as deeply as possible, its a very deep rabbit hole, but myself think there is a (very elusive!) rabbit still yet waiting to be uncovered :)