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23 Nov 2007 19:13:35 IST
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63
5204
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.
Comments (63)
23 Nov 2007 22:42:55 IST
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Instead of that why don't you try Change in momentum for
a short interval.It will be easier.
Nudge me if not solvable bcoz many of nudged me to post the solution .......i am waiting for you.
But that does not mean that you lose your hope.............
I won't think anything negative about you even if you think
for 7 days for this sum and then say "I did not get it"
Still I am encouraging you all to ive your best shot at this
"BEST OF LUCK EVERYONE"
Cheers!!!!!!!!!!!!!
a short interval.It will be easier.
Nudge me if not solvable bcoz many of nudged me to post the solution .......i am waiting for you.
But that does not mean that you lose your hope.............
I won't think anything negative about you even if you think
for 7 days for this sum and then say "I did not get it"
Still I am encouraging you all to ive your best shot at this
"BEST OF LUCK EVERYONE"
Cheers!!!!!!!!!!!!!
23 Nov 2007 22:52:59 IST
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7 people liked this
having got the permission to integrate...... i ended up in this way
the answer comes.........
but i have to consider the velocity to be constant for some small interval of time dt.
i realised my previous mistake.... i took accl^n along the plane something other than gsin30
let v be the velocity over time interval t to t+dt.
we have to consider that the time t is that time by which the sledge goes 120m.
let M = mass of the sand box (sledge)
So (M + 40 - 2t)v - [M + 40 - 2(t + dt) ]v = change in momentum = Force x (time interval)
or 2vdt = Fdt
or 2v = F
F = force on the sledge at time t = mass at time t x gsin30
= (40 - 2t) x 5
So 5(40 - 2t) = 2v
or v = (5/2)(40 - 2t)
or dx/dt = (5/2)(40 - 2t)
or dx = (5/2)(40 - 2t)dt
integrating..
dx = 5/2{ 40dt - 2tdt }x = 5/2{ 40t - t2}
120 = 5/2. (40t - t2)
Now solving gives t = 1.23s and 38.8s
it took me two and a half hours ......... but still an imperfect answer with an impurity of 1.23s appears with 38.8s
23 Nov 2007 23:04:00 IST
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please friend....... temme why 1.23s appears...why should we reject it...???
today nite.... i'll sleep very well man for my ears turned red doing this sum....
i m feeling very relaxed now.... but make me fell more by telling why 1.23 s WHY WHY WHY....
today nite.... i'll sleep very well man for my ears turned red doing this sum....
i m feeling very relaxed now.... but make me fell more by telling why 1.23 s WHY WHY WHY....
23 Nov 2007 23:14:05 IST
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2 people liked this
Amazing answer by Biki !!!!!!!!!!! Congrats !!!!!!!!!!
Now let me show you how I got it.
Solution:
Let initially the mass of the sledge be 'm' and mass of
sand falling per second be say "
".
If V be the velocity of the sledge after "T" seconds,then
in a small interval "dT" the Change in Momentum is given by
(m -
T)V - [m-(T+dT)]V = F.dT....(1)
Here F is the force and
F = (m-T)gSin@ ....(2)
Substituting the value of "F" from Equation (2) in Eq.(1) &
Solving we wil get the following Expression:
V = (m-T)gSin@
.ds/dT = (m-T)gSin@
Now we integrate the following expression and we get:
.S = (mT - T2/2)gSin@
Substituting the values provded as:
S = 120m.
= 2 kg/sec
m = 40kg.
@ = 300.
2 x 120 = (40T - 2 x T2/2) x 10Sin(30)
240 = (40T - T2)5
48 = 40T - T2
T2 - 40T + 48 = 0
Now solve for T
T = 40 +
1408
2
T = 38.8sec.
Hope you all find it useful.
Rate if useful.
Biki you got your 5 salutes.Gr8 effort.
Cheers!!!!!!!!!@@@!!!!!!!!!!
Now let me show you how I got it.
Solution:
Let initially the mass of the sledge be 'm' and mass of
sand falling per second be say "
".If V be the velocity of the sledge after "T" seconds,then
in a small interval "dT" the Change in Momentum is given by
(m -
T)V - [m-(T+dT)]V = F.dT....(1)Here F is the force and
F = (m-
T)gSin@ ....(2)Substituting the value of "F" from Equation (2) in Eq.(1) &
Solving we wil get the following Expression:
V = (m-T)gSin@.ds/dT = (m-T)gSin@Now we integrate the following expression and we get:
.S = (mT - T2/2)gSin@Substituting the values provded as:
S = 120m.
= 2 kg/secm = 40kg.
@ = 300.
2 x 120 = (40T - 2 x T2/2) x 10Sin(30)
240 = (40T - T2)5
48 = 40T - T2
T2 - 40T + 48 = 0
Now solve for T
T = 40 +
1408 2
T = 38.8sec.
Hope you all find it useful.
Rate if useful.
Biki you got your 5 salutes.Gr8 effort.
Cheers!!!!!!!!!@@@!!!!!!!!!!
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