3ΣFx=0
ΣFx=dm/dtv+dv/dtm=0
dm/dtv=-dv/dtm
ln(m)m0 to 0 =-lnv(v0 to v)
oops i am getting infinity
can anyone help me after this
A trolly full of sand is moving with velocity V0 there is a hole at the bottom with which sand drops out velocity of trolly will
a) increase
b) decrease
c) remain same
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18 Answers
3
33[3]
I never thought this one buzzed u all..
that was just a time pass question posting.. [3]
62note that the mass that is falling out also has the same momentum as the "main" mass of the sand
More numerically,
the total momentum at time 0 is MV
dm.V is the momentum of the sand (falling out!) (because it falls out)
M-dm is the mass of the remaining sand.. (let its velocity be u)
MV=dmV+(m-dm)u
thus, u=V
13is dat so...!!!
then final velocity of trolley will also be negligible... heisenberg's uncertainity principle...
u cannot make such odd assumptions sankara...
13@sankara... :
finally d mass is not zero dude... it is equal to mass of d trolley... dats d mistake in ur problem... hope u'll get it now...
[1]
13If the sand is dropping out wid negligible velocity, then the velocity remains same...
in this case, since velocity of sand dropping out is not given, it is assumed to be negligible... hence velocity of trolley remains same...
else, velocity will increase...
[1]
1remain same!
where does reduced mass come?
ye to aasan problem hai na? matlab x direction me momentum conservation.. simple?!
1yes... my duobt is similar as philip's.
mass is reducing, so shouldn velocity also reduce...
1there's no force in the horizontal direction , so no change in velocity is observed. as only a force can cause acceleration. and here only downward force and upward force exist.
1"so for momentum to remain conserved in the dir of motion of trollley, the trolley has to ..........................do nothing?"
plz someone explain this statement
for momentum to remain conserved the mass should also remain same is vel is constant [7][7]
so since mass decreases the vel should increase [7][2]
11IS it bcos the sand will drop downwards (a dir perpendicular to dir of motion of the trolley) and so for momentum to remain conserved in the dir of motion of trollley, the trolley has to ..........................do nothing? [3]
33Nice........
U understood the quetion... and not trapped by the title a mistake by me i meant variable mass in title....