I have masses 2kg, 1kg and 1kg connected as shown...assume all threads to be really very long.
The mass 2kg is released from rest at t=0.
Whats its accn. at t=0??
Whats its velocity after a very very long time?
And one question..which of the following topics are you done with (Lets go for a majority): electrostatics, light (waves waala), SHM...
acc at t=0 is g....
ReplyDeleteafter it has fallen through h, v=sqrt(gh)...(here h is very large)
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ReplyDeletethis should be fun...
ReplyDeletei got v^2 = 4g (root{1+h^2} -h )/ 3
ReplyDeletehey..i don't remember mentioning any 'h' in the question...
ReplyDeleteso the ans is zero ?
ReplyDeletehmm i m also getting v=0
ReplyDeletetaking cosQ=1
and energy conervation
but is it actually possible?
a body stops motion without any reason
and fr ur second question....we (JEE 2011 batch)have only done shm out of the three
ReplyDeletedoes the 2kg mass carry out shm???
ReplyDeletei'll give you a few hints:
ReplyDelete1) the velocity saturates, ie it approaches a fixed finite value..and this is what i ask for.
2) Ankush, have you done Limits and Continuity..you can't actually put cosQ=1; you have to take limit of CosQ approaching 1.
3) Akash, there was something wrong with your result in terms of h..try it again.
but you people have done some limits, haven't you..i mean at least in skul..they have it in class 11???
ReplyDeleteis it sqrt of g?
ReplyDeletefinally..the ans. is sqrt(g)..but only one??
ReplyDeletehow sir please explain somebody?????????
ReplyDeletearey bhaiya i m so sry....pehle vale main maine sqrt(gh) h galti se daal dia tha....it is simple energy conservation
ReplyDeleteplease post your solution sambhav
ReplyDeleteokk..i got it
ReplyDeletelet the side bodies come up by h
conserve energy
gh+gh + 1/2 v^2 + 1/2 v^2 + v ^2 = 2g(h+1)cos Q
take cos Q = 1...as Q is almost 0
when i said v=0 ,i had put h instead of h+1
ur answer is right.....but solution is wrong...
ReplyDeleteu forgot about the length constraints....
please give your solution sambhav
ReplyDeletewhat length constraints?
ReplyDeletepost ur solution sambhav please
ReplyDeletei had logged on to say the same..ankush, your answer is correct but soln is wrong.
ReplyDeleteLemme give you a few hints:
you don't need any hints here...simply let the angle subtended by the two threads holding 2m be A...
apply cons. of energy..but to do that, you need to express the 'm's velocity in terms of 2m's velocity...
express the drop in height of all blocks in terms of A, and the given 1m length..
i did exactly the same...
ReplyDeletetry it out guys, if anyone still wants, i'll post the exact equation by midnight
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ReplyDelete[loss in GPE of mass 2m]=[gain in GPE of mass (m+m)] + [gain in KE of mass (m+m)] + [gain in KE of mass 2m]
ReplyDeletenow draw a rough figure, with some length of string moved to the other side...let this length be x, and the angle made by the string connected to the 2m mass with the vertical be Q...
at any instant, total heigth gained by each mass is x, and height fallen by mass 2m is [{1+x)cosQ]
now, let v1 and v2 be the respective velocities of mass m and 2m...so by length constraints, v1=v2cosQ
substitute all values in the first equation....put cosQ=1, x will cancel out....v2 will come out to be equal to sqrt(g)
had the separation been 'a' instead of 1m....the answer wud have been sqrt[ag]
ReplyDeletegot it thnx sambhav
ReplyDeleteThanks for pointing out that "ag" thing.. I was wondering why it doesn't match dimensionally(before seeing the solution)
ReplyDelete"now, let v1 and v2 be the respective velocities of mass m and 2m...so by length constraints, v1=v2cosQ"
ReplyDeletesambhav shouldnt it be
v1cosQ=v2
no sambhav..that was a very fundamental mistake from you.
ReplyDeletei was actually having a doubt about that...
ReplyDeletev1=v2cosQ...
ReplyDeletesahi to hai yr..!