In regards to the momentum principle ( SPECIFICALLY ) updating your momentum after you have calculated your net force.


- Why couldn't use just calculate the acceleration to multiply with mass of the object in the system over your change in time.

INSTEAD OF

- Using velocity and mass.

Notice correct SI conversions * -- specifically Cm - m and g = kg.


For example, Note* my net force was calculated by multiplying a negative k ( spring constant ) by the difference between a stretched (24 cm.) and relaxed (21.2 cm.) spring. = 3.2 cm* ( - 7.5 k x .032* m ) = - 0.24 N )

acceleration
My acceleration ( fnet / mass ) = - 0.24 N / *95 g ( * .095 kg ) = - 2.526 m/s ^2


Updating momentum using acceleration ( 0.095 kg ) ( - 2.526 m/s ) ( 0.1 sec = my delta t / change in t ) = -.024 N

or updating momentum using velocity and mass MISTAKE FOUND THIS IS FOR VELOCITY UPDATE NOT MOMENTUM
Velocity and mass route ( 0 The initial velocity ) + ( - .24 N / 0.095 kg ) x 0.1 ( change in t ) also = -.2526

HOLD UP.....

Using the acceleration to find the momentum would not work because the initial velocity might be different each time especially since this is a iterative process and a looped process....

im glad I just looked this up in some coding....

The best way to update momentum is to just use the velocity and mass.... just like a computer does really...

momentum update = v.init + ( fnet / m) * dt

initial velocity or v.init = ( zero or non zero value that changes )

fnet = self explanatory ( derived first in update problem anyway )

....divided by mass multiplied by the change in position for me was 0.1


Glad I caught this while typing it out....




This post was edited by lilzombiehoe on Feb 13 2019 11:37pm

wut

This post was edited by marioo1182 on Feb 15 2019 12:01am

Which part confused you? Maybe I can help... Did you not know that computers are able to perform pretty much every kind of calculation you can think of ?
A lot of times, ( that input / code ) can help you to better understand the process...

It’s very cut and dry and you are only focusing on the meat and potatoes of the problem without the added jargon; without the bs.

looks good as far as i see
are you coding in python?

Yes sir! Exactly

Basic while loops. Making sure position and momentum is updated and that all constant etc... are defined.


All that kind of basic tasks

This post was edited by lilzombiehoe on Feb 15 2019 12:30pm

you're gonna have to store a new momentum each time so you have one you're updating (each timestep is gonna call for a new v, rather than v initial)
add some linear air resistance if you want a fucking challenge haha
computational physics i got a C but i learned to code and learned a lot of phsyics xD

well in math

velocity = dy / dx

where y is acceleration

velocity is a derivative of the acceleration.

principles of momentum quantify a tangential result in the function. Force quantifies the function itself.

momentum =/= force

So you can't simply use the acceleration over a change in time to find momentum, that is working backwards, that would find you the snap of the acceleration.

I thought Velocity is the integration of acceleration and acceleration is the derivative of Velocity. not the other way around.

wait ur right

The derivative is the slope of the function. So if the function is f(x)=5x−3, then f′(x)=5, because the derivative is the slope of the function. Velocity is the change in position, so it's the slope of the position. Acceleration is the change in velocity, so it is the change in velocity. Since derivatives are about slope, that is how the derivative of position is velocity, and the derivative of velocity is acceleration. So if the position can be expressed with the function f(x)=x2−3x+7, then the derivative would be f′(x)=2x−3 since that is the slope of the function at any given point, and since it is the slope of the position function, it is velocity. Same for acceleration; f"(x)=2, which is the derivative of velocity, which makes it slope. The slope of velocity is acceleration.

This post was edited by Phaggot on Feb 16 2019 12:45pm

Yes. That’s correct

1st derivative of position in respect to time is velocity

Derivative of velocity is acceleration

Derivative of acceleration is “the jerk” —not talked about that much for us..
Unless you need to know higher order derivatives in Calculus...

Besides Calc. 1 there is very little emphasis on differentiation compared to integration..

Just a simple mix up on words really...

But if you were working out any type of iterative process - you would right off the bat catch such a mix up...