Room for RyderRude and 123

Ryder Rude

07:26

@user430580 hi

123

Hi

Ryder Rude

hi

123

First If you like to share your education background and university. Because your knowledge is good

In my country education is very very worst

Ryder Rude

i just do physics as hobby. I'm not professional..

i self study

123

Nice.. Same as me

Ryder Rude

07:29

yes

123

Your country?

Ryder Rude

i wanna stay private..

123

Ookay... What about education system in your country? out of 10

Ryder Rude

idk.. 5 maybe. It depends on the teacher

123

Okay...

means average. In school time you also had difficulty in physics

Ryder Rude

07:30

yes

123

Ookay.

Ryder Rude

i started learning qm 3 years ago

and I'm doing physics on and off since then

123

Is English mother language in your country?

@RyderRude Oh nice

Ryder Rude

pls don't ask personal questions

123

Ookay... NP

Ryder Rude

07:31

I'm strict about privacy

123

Lets begin..

Ryder Rude

ok

123

We know from yesterday experiment three experiments between three objects, in each experiment we used two objects in different combination.

Ryder Rude

we should finish this fast... discussions keep going in circles

@123 yes

123

We found that no matter how fast they were before collision. But ratio between velocities change always stays the same.

If then we multiply all three constants which is always = 1

Ryder Rude

07:35

we multiply the negative of the ratios

that multiplication is 1

123

$K_{12}K_{23}K_{31} = 1$

Ryder Rude

the multiplication of the ratios themselves is -1

@123 yes k12 is - a1/a2, which is the negative of the ratio a1/a2

123

Yes

Using this way we can separate any one constant which can be write between the ratio of other two.

Ryder Rude

yes, like k23=1/k12k31

123

$\Large{K_{23} = \frac{K_{13}}{K_{12}}}$

Ryder Rude

07:38

@123 yes

123

Now my observation starts

I found that the object 1 which is common in denominator and numerator gives the value in terms of object 1

Ryder Rude

i dont understand...

123

Like $K_{23} = 3$ , $\Large{K_{23} = \frac{3/2}{1/2}}$

Ryder Rude

yeah

pls write ur whole problem....

123

Here value $2$ is common in the ratio but which is value of change in velocity object 3 from object 1 for above

Let me write the whole equation , then it will be easier to ask

Let me write on paper. It will be quicker.

We used here the data of experiment 1 and experiment 3 to write the value of experiment 2.

Ryder Rude

07:55

@123 1/2 x 1/4 x 2 is not equal to 1

123

Means. experiment-1 give the change in velocity of object2 in terms of object1. And experiment-3 gives the change in velocity of object3 in terms of object1. Both in terms of object-1 so object1 is common.

@RyderRude It was typo. pls correct it to 1/2

Ryder Rude

1/2 x 1/4 x 1/2 is not 1 either

let's say 1/2 x 1/4 x 8

123

Sorry correct one more value. $K_{23} = 4/1$

Ryder Rude

ok

1/2, 4 and 1/2

123

Let me share picture again/ By correcting it

Ryder Rude

08:01

@123 yes. Object 1 is common

@123 it is fine

123

Now it is Okay

Ryder Rude

yes

123

Above value $K_{13} = 2/1$ it means value of object3 is 1 in terms of object1 from experiment-3

Value $K_{12} = 1/2$ means value of object2 is 2 in terms of object1 from experiment-1.

So we found value of object2 and object3 of experiment2 in terms of object1 from other two experiments.

Ryder Rude

what is the problem??

123

How can we interpret this results. because we can see that. ratio of their masses between mass 1 and 3 are 1:2

and same for object1 and object2 their ratios also 2:1

Ryder Rude

08:10

ok i will summarise experimental findings. 1. In inertial frames, when u collide any two objects, a1/a2 is always the same 2. we defined $k_{ij}= -\frac{a_i}{a_j}$ and found that, for any three objects, k12k23k31=1

123

Problem is that in actual there are three different masses we used. But the above experiment shows object1 taken to mass = 1kg

Ryder Rude

to define mass, u have to choose a reference, right?

123

Yes

We use $m_1 = 4kg$ but using above ratios we are insist to take it $m_1 = 1kg$

How can we interpret this result?

Ryder Rude

u r doing it backwards... the actual way to get these results is to collide two objects in experiment

u didn't do that. u used conservation of momentum on paper

if u collide objects in experiment, u don't know the mass of anything initially

and u can use this experiment to define mass

123

Yes i used this thought experiment as m_1 = 1 lumps. And we created m2 and m3 using identical lumps by joining or dividing

Ryder Rude

08:15

m1= 1 is a definition here...

123

No...

Ryder Rude

we r using m1 as a reference to define the mass of other objects

123

Pls see the results of experiment1 and experiment3 in terms of change in velocities.

which gave us different values of $\Large{V_1 - U_1}$

Ryder Rude

they're different. What's the problem

i told u that the same object can have different accelerations in different experiments.this is y i used a1 and a1'

123

It automatically made m1 = 1 magnitude. How. I can see the calculations

But i don't know how to interpret this

Ryder Rude

08:19

what automatically made??

123

Let me explain this way.

$m_1 = 4kg$ , $m_2 = 2kg$ and $m_3 = 8kg$

The above unit is settled by anyone in the other planet. We know from them.

Ryder Rude

ok...

but we can define new units, can't we?

123

But $\Large{K_{23} = \frac{K_{13}}{K_{12}} = \frac{2/1}{1/2}}$

@RyderRude Yes this is what i am doing here

Ryder Rude

i dont know what's the problem.....

i cant say anything.... Sorry

123

Look at the result $\Large{K_{13} = \frac{2}{1}}$

I am explaining. Pls give me time

here in $\Large{K_{13}}$ , $\Large{\Delta{V_1} = 2}$ and $\Large{\Delta{V_3} = 1}$

In $\Large{K_{12} = \frac{1}{2}}$

Here, in $\Large{K_{12}}$ , $\Large{\Delta{V_1} = 1}$ and $\Large{\Delta{V_2} = 2}$

How the above $K_{12}$ and $K_{13}$ gave the constant $K_{23}$ in terms of object1

idk how to explain it further. But i can clearly see. what ever the ratio of $K_{12}$ and $K_{13}$. we need to make $m_1 = 1$ magnitude. Despite the fact $\Delta{V_1}$ is different in both $K_{12}$ and $K_{13}$

it automatically defined $m_1 = 1$

I think i confused you. By not explaining problem clearly.

Ryder Rude

08:36

@123 i think u r overthinking it. k12k23k31=1 holds. so we can get k23 in terms of k12 and k31. there is nothing more to this

123

No i am not overthinking my friend. I will explain the main key point or idea of getting mass m1 = 1 automatically.

Pls give me 1 hour. I need to pick my kids from school.

Thank you.

Ryder Rude

ok..

123

: )

Transcript for

Room for RyderRude and 123