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Ask iit jee aieee pet cbse icse state board experts Expert Question: conservation of mometum.
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[Post New]11 Feb 2007 12:12:26 IST
    Subject: conservation of mometum.
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dinesh_ddt (163)

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There are two trucks A and B of mass 40kg and 50 kg,moving with a velocity of 5m/s and 10m/s respectively in the same direction.There is a boy of mass 10kg standing on B.He First jumps from B to A then from A to B.What is the velocity of B when the boy comes to B.?The Velocity of Boy is 2m/s in both the jumps.
B.Tech Mechanical
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[Post New]11 Feb 2007 12:54:50 IST
    Subject: conservation of mometum.
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vineet (133)

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is it 9.7m/s?
Think different .........................think apple
ALBERT EINSTEIN
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[Post New]11 Feb 2007 15:58:34 IST
    Subject: Re:conservation of mometum.
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AAKRITI (228)

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i am not able to understand how the trucks are moving, one behind the other or are they parallel
you have to run faster and faster so as to remain where you are.. even if you are on the right track you will get run over if you just keep sitting there...
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[Post New]11 Feb 2007 19:03:33 IST
    Subject: conservation of mometum.
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krishna.gopal (2058)

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Another thing that will be important is how boy is jumping. Is he jumping in forward direction or in backward direction or in some oblique way. Because answer will vary according to that
Krishna Gopal Singh
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IIT Delhi 2002
Currently doing PhD from IIT Delhi
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[Post New]11 Feb 2007 19:30:56 IST
    Subject: conservation of mometum.
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dinesh_ddt (163)

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one behind the other and jumps in forward direction
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[Post New]11 Feb 2007 19:39:48 IST
    Subject: conservation of mometum.
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dinesh_ddt (163)

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vineet dude i too got the answer something like 9.67m/s!
plz can any1 clarify????????????????????
B.Tech Mechanical
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[Post New]19 Feb 2007 02:23:26 IST
    Subject: Re:conservation of mometum.
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krishna.gopal (2058)

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Mass of truck A = 40 kg
Velocity of truck A = 5 m/s
momentum of truck A = 200 kgm/s 
 
Mass of Truck B =50 kg
Mass of boy on it = 10 kg
Velocity of truck B = 10m/s
Momentum of truck B and boy system = 600 kgm/s
 
Just after the first jump
momentum of first truck = 200 kgm/s (same as earlier)
momentum of boy = 10*12= 120 kg m/s (Assuming 2m/s is velocity relative to truck B)
momentum of truck B - 600-120 = 480 kgm/s
Velocity of truck B= 9.6 m/s
 
After boy landing on truck A
Momentum of Truck A boy system = 200+120=320 kgm/s
velocity of truck A and boy system= 320/50=6.4 m/s
Momentum of truck B= 480 kgm/s (same as earlier)
velocity of truck B= 9.6 m/s
 
Just after jump of boy from truck A
momentum of boy = 10*(6.4+2) = 84kgm/s (2m/s is velocity relative to truck A)
Momentum of truck A = 320-84=236 kgm/s
Velocity of truck A = 236/40 = 5.9 m/s
Velocity of truck B= 9.6m/s (same as earlier)
 
After boy lands on truck B
Velocity of truck A = 5.9 m/s (same as earlier)
Momentum of truck B and boy system=480+84=566 kgm/s
Velocity of truck B and boy ystem = 564/60 = 9.4 m/s
 
So I think final velocity of B is 9.4 m/s
Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi
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