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23 Nov 2007 19:13:35 IST

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New Challenge (IIT'92)

None

5 Salutes to one who gets it right.

(No Diagram given)

Q)A sledge filled with sand slides down without friction down
a 30⁰ slope.Sand leakes out through a hole in the sledge at
a rate of 2 Kg per second.If the Sledge starts from rest
from the top of inclineand if total initial mass of Sledge is
40 Kg, How long does it take to go down 120 metre in the
plane?

Everyone who gives a good effort will be rated.
Solution will be posted tomorrow by 4.00 pm.

ALL THE BEST.....WISH YOU ALL GOOD LUCK.

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Karthik M

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23 Nov 2007 19:30:28 IST

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Is it 1.1 seconds?

Rahul

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23 Nov 2007 19:34:23 IST

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i may go wrong but i think when a body goes down an inclined plane without friction, its acceleration does not depend on its mass
so its acc should be gsin30
so 120=1/2gsin30t^2
but \\
this is a jee problem, may not be so easy,

Karthik M

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23 Nov 2007 19:37:33 IST

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I'm getting 1.105 and 6.9 seconds!!. The third value is negative

waterdemon

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23 Nov 2007 19:42:20 IST

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Well,Not close but atleast trying....... :)
Try keep trying.....

The answer is:
38.8 sec.

Cheers!!!!!!!!@@@!!!!!!!!!

It took me more then 45 mins to crack it when I did it.

waterdemon

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23 Nov 2007 19:45:40 IST

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It took me more then 45 mins to crack it when I did it.

Anyways Best of Luck to all.
Cheers!!!!!!!!@@@!!!!!!!!!

Rahul

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23 Nov 2007 19:45:59 IST

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blunder by me! trying again...

Karthik M

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23 Nov 2007 19:47:39 IST

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Mass of sledge is not given? If sledge is assumed to be massless, then t=38.8 gives negative mass

waterdemon

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23 Nov 2007 19:50:02 IST

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Hint:Make assumptions say "m" is the mass" and take some
"dt" time and then use some Brains !!!!!!!!!

Karthik M

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23 Nov 2007 19:52:02 IST

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Bleh... stupid mistake by me... I think i got the logic. Trying to find out the mass of the sledge now

Karthik M

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23 Nov 2007 20:11:57 IST

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If the sledge mass is found... the problem is over. But finding it is the whole problem :(

Karthik M

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23 Nov 2007 20:12:19 IST

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Still trying.... dont post the solution now

biki ....

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23 Nov 2007 21:00:13 IST

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force = (dm/dt)v + (dv/dt)m

the sledge suffers a vertical displacement of 120sin30 = 60m

So let us find the velocity of the body after t sec while it covers 120m.

initial velocity = 0, initial mass = 40kg

final velocity = v , final mass = (40 - 2t)kg

using energy conservation....

40x10x60 + 0 = (1/2)(40-2t)v²

or v = 40

30 /

(40 - 2t)

So let us find the force acting on the sledge at the final position.

F = (dm/dt)v + (dv/dt)m

= {2x40

30 /

(40 - 2t)} + {10sin30 x (40 - 2t)} N

and force acting on the body initially = F' = 40 x 10sin30 = 200 N

initial accl^n = 200/40 = 5 m/s^2 = a'

final accl^n =[ {2 x 40

30 /

(40 - 2t)} + {10sin30 x (40 - 2t)} ] / (40 - 2t) = a

here i am having a problem...

can i take avg. accl^n as ( a + a' ) / 2 = A (say)

and use the equation 120 = (1/2)At² to obtain an equ^n in t and solve for t ..??

biki ....

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23 Nov 2007 21:01:01 IST

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i don't think i am correct

Karthik M

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23 Nov 2007 21:04:40 IST

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@biki - after the body slides down after a time t, it still has potential energy yaar.

rini s

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23 Nov 2007 21:11:00 IST

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do u think this kinda ques may come in iit paper.. i meat its too much subjective type.......

Karthik M

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23 Nov 2007 21:11:18 IST

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It tests your concepts

Priyesh

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23 Nov 2007 21:21:08 IST

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Also biki energy cannot be conserved as

40x10x60 + 0 = (1/2)(40-2t)v²

because some K.E is also taken away by the flowing sand (as it moves from the hole with some velocity)

biki ....

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23 Nov 2007 21:24:20 IST

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i think you are correct.............

answer does not come this way....

ramyani chakrabarty

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23 Nov 2007 21:25:49 IST

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biki dada, the accln is always g sin

along the incline.

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