Nimboi [ping if answering]
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
+solved @SirLancelotDuLac
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
really nice problem
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
brilliant thank you
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
my bad
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
i did a brainfart
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
ah yes
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
then its
v-u51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
wait initially its
v51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
i added
u in the starting51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
oh shit
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
nice
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
lemme try
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
does it not add newlines? oh well
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
so here's what i did $\newline$
For the top ball, velocity upwards is $v$ and then a downward velocity by the tension impulse $u$ acts on it $\newline$
So $J = \delta p = \int{Tdt} = (mv-mu) - (mv) \newline$
Which means $-mu = \int{Tdt} \newline$
For the bottom ball in the $y$ direction, there's $v$ towards the right and $u$ upward. $\newline$
Conserving momentum in $y$, we get $J = mu - 0 = \int{Tdt} \newline$
Therefore $mu = -mu$ and u = 0 $\newline$
What did I do wrong
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
not sure what changing frame of reference does tho
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
am doing
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
oke?
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
my brain broke a little trying to frame that numerically
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
oh wow, how exactly
51 replies
IiTeachChem
•Created by Nimboi [ping if answering] on 4/26/2025 in #💭│doubts
Mechanics, taut string
allen, they gave us a bitsat test series
51 replies