| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 00:14:46 IST
|
|
|
same confusion i have also got after 20 sec the box becomes massless then how come ans is comin as 38.8 sec coz remember no net force can be applied on a masssless body
|
"Imagination is more important than knowledge."
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 11:05:44 IST
|
|
|
Amazing sum man. Do post more. Hats off biki!
|
Will nip in at times to solve problems :)
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 11:13:19 IST
|
|
|
The thing to be remembered here is that the sledge has some mass. So the mass of the sand will be zero before 20 seconds, and after that, the sledge will move with its own mass.
|
Will nip in at times to solve problems :)
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 11:44:49 IST
|
|
|
even if the sledge is considered massless (as assumed in the problem) : the problem uses change in momentum to calculate force after 20 s there is no change in momentum so no force applied as we know no force acts on a massless body but it will have its velocity so it should be covering the remaining distance with a constant velocity taking a total of 38 sec
|
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 14:16:01 IST
|
|
|
wow that was something the solutioin and the question were brilliant and yes hats of to both waterdemon and biki
|
science-
the most fundamental
the most eternal
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 14:46:24 IST
|
|
|
congrats to both waterdemon and biki for solving it.i was just three steps away from the answer myself but being weak in calculus,couldn't find the solution till the end
|
there are numerous options besides I.I.T
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
  
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
there should be no confusion............
consider the mass of the box = M
in the equ^n for change in momentum...
( M + 40 - 2t ) v - { M + 40 - 2( t + dt ) } v = Fdt => 2vdt = Fdt...gives the same result...
i m modifying my solution....
|
salman khan |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 23:24:00 IST
|
|
|
i think the answer is appx 3.0985sec
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 23:35:31 IST
|
|
|
Hey Dude, The question has been already solved and we got the answer. No prblems Anyways............. Better luck next time Cheers!!!!!!!!!!!!!!
|
Always available for help !
But Remember Don't hesitate to ask a good Question but
Be damn serious for Questioning a weak one.
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Dec 2007 19:02:20 IST
|
|
|
hows tat -2t cming (M + 40 - 2t)v in this eq
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Dec 2007 19:16:10 IST
|
|
|
Dude, The sand is leaking out from the sledge at the rate of 2kg/s as it has been given in th Question. So the change in mass every sacond will be: (M+40-2t) Here M=mass of sledge as assumed by :"BIKI" 40 kg = weight of sand in the sledge. So totla initial mass=M+40 kg. And then he has given the conservation of momentum eq: (M+40-2t)V - M+40-2(t+dt)V = F.dT Here t+dt as assuming that the mass of sand that has leaked out in small interval "dt" to be very small. Hope you find it useful. Rate if useful. Cheers!!!!!!!!@@@!!!!!!!!!
|
Always available for help !
But Remember Don't hesitate to ask a good Question but
Be damn serious for Questioning a weak one.
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Dec 2007 20:24:58 IST
|
|
|
if u dont mind can u just gimme the source..i mean..on net.any link to dis question.. i couldnt find in IIT 92 paper[:o]
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Dec 2007 09:43:56 IST
|
|
|
well i got it less than 10 minutes
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Dec 2007 22:06:46 IST
|
|
|
Won't velocity also change over the time interval dt? Shouldn't that also be taken into consideration?
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Dec 2007 10:23:30 IST
|