Mechanics, taut string
I don't understand the question 🤷 Isn't the string already taut? What am I supposed to do here
25 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.It's not going to be taut after they're released
Since the component of velocity of each along the string is not the same initially
which app?
The tension acting on the ball at rest tends to give it velocity while taking it away from the moving one.
Oh wait sorry misread the question.
ohk managed to figure that part out, but what causes the velocities to change?
because in my mental image the velocities of the balls dont end up changing and the answer just becomes
mv^2, which is wrong
yeh both of them are moving
but huh
tension changing velocity
lemme think about how that happens
allen, they gave us a bitsat test seriesAfter some time the string becomes taught, as a result the velocity changes due to impulse
ok i think you get the taut part, next think from the frame of the left ball
oh wow, how exactly
my brain broke a little trying to frame that numerically
ohhhh
okk
thnks
oke?
am doing
not sure what changing frame of reference does tho
After each ball covers l distance the string becomes taut once again, and impulse from tension distributes velocity among 2 balls.
You mean tension?
the velocity of ball 2 wrt ball1 is always at 45 degrees from ball1, so when ball2 is just above ball1, the distance between them is L and at that point string will apply impulsv
I got this, but how will the kinetic energy be 3mv²/4?
1/2mv^2+1/2m(v^2/4+v^2/4)=3/4mv^2
Oh nvm i forgot the formula for kinetic energy
I'm overworked today. I just did K = mv²
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
does it not add newlines? oh well
lemme try
Ig its \newline I'm not sure tho
For upwards ball, J=mv-mu and for the bottom ball its mu=J so mu=mv-mu right? (Also J is magnitude of impulse here.)
Nimboi [ping if answering]
nice
oh shit
i added
u in the startingYeah the impulse=mu not -mu (where u is final velocity)
wait initially its
v
then its v-u
ah yes
i did a brainfart
my bad
brilliant thank you
really nice problem
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