IIT JEE , AIEEE Forum - Physics , Mechanics

madman

and karthik i don think the acceleration is constant because the thrust force varies with velocity
since mgsin@ is going to accelerate the object the velocity is going to vary and hence the total acceleration

biki

@ waterdemon.......

does the sum require integration....
i think i m getting it

waterdemon

Yes..............there is integration in one step for sure.
go ahead if you solve it i'll solve it tomorrow.

biki

can i take the velocity constant for a short interval .......
or else waterdemon would not have told to take dt and all...

waterdemon

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!!!!!!!!!!!!!

biki

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

biki

thanx waterdemon ....... for the help of advising me on taking that change in momentum........

i tried conserving momentum at first..... but it did not work...

but why 1.23 s appears ???

biki

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....

rooney

@biki

Kyun tension le raha hai bhai. So many times we get junk values. Just reject them and goto sleep . lol

waterdemon

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!!!!!!!!!@@@!!!!!!!!!!

rhd92781

grt question n grt solution by biki
yaar i think its acceleration is always less than gsin30
so time taken should come >6.99 as it comes by taking a=gsin30
so 1.23 s can never be a soln
its only a biproduct of the quadratic

waterdemon

The reason is clear dude acceleration of (gSin@) is not enough
for the sledge to cover 120m in 1.23 sec.

Hope you got it.
Rate if useful.

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

ramyani

it is at a vertical height 120 sin 30 = 60 m
assuming free fall g = 10 m /s/s
time is greater than 1.23 s
so this is rejected.

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