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15 Sep 2007 14:44:54 IST
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A ball is dropped from some height on a heap of sand. The ball pireces into the heap and stops after travelling a certain distance.
(a) Only momentum is conserved
(b) Only Kinetic Energy is conserved
(c) Both are conserved
(d) None is conserved
I think answer should be (d) but in solution it is given (A)
Pls. explain........
Comments (45)
shashidhar sundareisan
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Joined: 15 Aug 2007
Posts: 36
17 Sep 2007 17:37:38 IST
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consider earth + ball system.hence momentum is conserved.
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17 Sep 2007 18:11:45 IST
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arrey guys & gals( if any
)
) why u people are considering only stone & heap as a system....?????
considering stone + sand + earth as a system i m gonna explain d answers...
starting from 1st option:-
1) momentum is conserved . purav says that stone is accelerating due to gravity.. its perfectly right . when u consider perfectly elastic colloision then why don't we consider gravity.?? because we talk in particular direction.......here also G force is internal force..............as soon as stone will get into the heap earth will also move. in the direction of stone.......( but as mass of earth is very very very very very large
its velocity will be very very small . so we can neglect it........
its velocity will be very very small . so we can neglect it........2) energy can be conserved here.. but as soon as stone will get into heap it will produce some heat hence energy is not conserved...
3& 4) cann;t be
hope its clear now
17 Sep 2007 19:55:07 IST
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@ Shubham,
even if we consider earth + stone + ball as a system, then too the frictional force provided by the heap of sand cannot be considered as internal force. It is a dissipative force. Also if it is the case of perfectly inelastic collision then what is the common velocity of com of heap and ball. I mean the com just shifts. Afterwards it will come to rest. So the momentum is finally becoming zero.
@ Swashata,
conservation of energy means only conservetion of KE (in this case) You can alternatively look that option as 'No Work is done'.
Now try to answer
even if we consider earth + stone + ball as a system, then too the frictional force provided by the heap of sand cannot be considered as internal force. It is a dissipative force. Also if it is the case of perfectly inelastic collision then what is the common velocity of com of heap and ball. I mean the com just shifts. Afterwards it will come to rest. So the momentum is finally becoming zero.
@ Swashata,
conservation of energy means only conservetion of KE (in this case) You can alternatively look that option as 'No Work is done'.
Now try to answer
23 Sep 2007 09:41:19 IST
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we cannot take...earth as a system because it is of infinite...mass...considering heap and ..stone as a system...as the impulse ....of gravity is ...very less and neglgible...( but i have a doubt...it is not necessarily ...negligible in all cases... so how do we check???) so we can conserve ...it in the vertical direction....but if we consider the system as only ball...then there will be impulse of normal contact force...which has to be taken to account ...so in that case we cant ....take the ...conservation of momentum...
23 Sep 2007 22:31:07 IST
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dats wat purav just said my dear.. dat there is force along its direction of motion(even wen in contact with the sand---gravity) which makes it impossible to conserve the momentum...
huge mass or small mass .. u can consider watever u want to b the system.... there is no mass limiti for systems...
huge mass or small mass .. u can consider watever u want to b the system.... there is no mass limiti for systems...
24 Sep 2007 10:43:58 IST
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ok thanx...but ...one more thing ...i think....if u considering sand and stone as one system its nothing but considering earth in the system rite??? anyways ..he is nt askin us to calculate rite??? so whats the grt deal in takin earth also in our system..??? if we take that there wud be no impulse rite??? so we can conserve ...momentum...rite???
24 Sep 2007 19:38:58 IST
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hey guys i think i got the answer .....
take the sand and stone as a system.... ..so there is no external force acting therefore ...momentum is conserved its as simple ...u see we got a looot confused.....and purav....friction .....force both action and reaction ....act on bodies of the system...so the net of it ..is zero....because it is an internal force...but if u consider any one particle as ur system....then it wud do some work...
if i am rigfht then pls rate me...!! hope its clear...if not then ...pls ask ill try my level best to answer it
take the sand and stone as a system.... ..so there is no external force acting therefore ...momentum is conserved its as simple ...u see we got a looot confused.....and purav....friction .....force both action and reaction ....act on bodies of the system...so the net of it ..is zero....because it is an internal force...but if u consider any one particle as ur system....then it wud do some work...
if i am rigfht then pls rate me...!! hope its clear...if not then ...pls ask ill try my level best to answer it
27 Sep 2007 22:38:47 IST
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If you have completed electrostatics then there is one more undiscussed (actually half discussed question) in Electricity section under the name 'System of Charges'. If you cant find it there then please visit my profile and search it there.
I am waiting for a satisfactory answer.
I am waiting for a satisfactory answer.
28 Sep 2007 07:26:59 IST
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clearly KE is not conserved
but momentum is conserved
because , the earth alone doesn't pull the ball.
the ball also pulls the earth with equal gravitational force
these two forces nullify and momentum will be conserved
MOMENTUM IS ALWAYS CONSERVED, or NEWTON'S THIRD LAW IS WRONG
28 Sep 2007 10:30:28 IST
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huh???? what are you talking about...ur answer is only valid if the system consists of both earth and stone....and for ur ...information....the action and reaction act on different bodies so they cant cancel each other unless and until they become internal forces...
28 Sep 2007 14:07:49 IST
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gravitational force acting on ball at any instant = GMm/R^2
gravitational force acting on earth at any instant = GMm/R^2
ie, the rate of change of momentum of both bodies is same but in opposite directions
so, momentum is conserved
as there was no reference to the system, we must take the entire thing ( universe) as the system
gravitational force acting on earth at any instant = GMm/R^2
ie, the rate of change of momentum of both bodies is same but in opposite directions
so, momentum is conserved
as there was no reference to the system, we must take the entire thing ( universe) as the system
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