last day (186 days later) » 

John Rennie
John Rennie
6:50 AM
Hi :-)
 
PandaScientist
PandaScientist
Hello sir!
I was going to ping you, does the system ping automatically when a room is created?
 
John Rennie
John Rennie
It pinged me when you invited me to the new room.
 
PandaScientist
PandaScientist
I've this very small doubt:
Say I've to calculate the change in KE in a 2d collision
Can I find the difference in KE for each axis and then sum them to get the answer?
Inelastic collision
 
John Rennie
John Rennie
No, KEs don't sum because they depend on v² not v.
You need to find the final velocities then calculate the KE from those velocities.
 
PandaScientist
PandaScientist
So , I cannot find the change in KE for each dimension axis separately and add them right?
 
John Rennie
John Rennie
6:54 AM
No, that won't work.
 
PandaScientist
PandaScientist
I asked this, and now I realise , it worked for a very specific case :-)
It won't work because KE doesn't act like vectors right?
 
John Rennie
John Rennie
Yes
 
PandaScientist
PandaScientist
@JohnRennie , before coming here, i did a quick proof and it worked!
I can't spot the mistake here
 
John Rennie
John Rennie
Ah, I misunderstood what you were asking. I thought "2d collision" meant two balls colliding and bouncing off each other at an angle.
 
PandaScientist
PandaScientist
Won't this work in that case too?
 
John Rennie
John Rennie
7:13 AM
I'll have to work through a few examples to see what's going on ...
Ah, OK, it does work.
Suppose the initial velocity is u = (ux, uy) and the final velocity is v = (vx, vy)
Then initial KE is ¹⁄₂mu² = ¹⁄₂mux² + ¹⁄₂muy²
And likewise final KE is ¹⁄₂mux² + ¹⁄₂mvy²
So ΔKE = (¹⁄₂mvx² + ¹⁄₂mvy²) - (¹⁄₂mux² + ¹⁄₂muy²)
= (¹⁄₂mvx² - ¹⁄₂mux²) + (¹⁄₂mvy² - ¹⁄₂muy²)
= ΔKEx + ΔKEy
> And likewise final KE is ¹⁄₂mux² + ¹⁄₂mvy²
Typo: And likewise final KE is ¹⁄₂mvx² + ¹⁄₂mvy²
 
PandaScientist
PandaScientist
8:21 AM
Okk sir, :-)
@JohnRennie thank you :-)
 
John Rennie
John Rennie
Sorry I got it wrong at first :-(
 
PandaScientist
PandaScientist
Well, I conveyed it wrong :-)
@JohnRennie ngl I was expecting a "you're welcome" :-)
 
John Rennie
John Rennie
Well I feel like I didn't answer this particularly well!
 

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