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A weird physics question...pls help !! watch

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    1) A truck with mass of 10000kg moves with 0.5ms^-1
    sand from above the truck is passed onto the truck with rates of 40kgs^-1
    how much additional horizontal force is needed for the truck to move at constant speeds of 0.50ms^-1 when weight is added to the truck constantly throughout the journey

    2) A parachutist falls from an airplane travelling at 67ms^-1.After traveling for 80m away from the plane,it opens up the parachute and its velocity at this time is approximately 40ms^-1.
    Determine the magnitude and direction of the resultant velocity of the parachuutist at this point

    3) How do you connect newton`s third law pair.Give 2 examples.

    i hope you guys can help me with this 2 questions
    thx a lot
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    (Original post by MalaysianDude)
    1) A truck with mass of 10000kg moves with 0.5ms^-1
    sand from above the truck is passed onto the truck with rates of 40kgs^-1
    how much additional horizontal force is needed for the truck to move at constant speeds of 0.50ms^-1

    2) A parachutist falls from an airplane travelling at 67ms^-1.After traveling for 80m away from the plane,it opens up the parachute and its velocity at this time is approximately 40ms^-1.
    Determine the magnitude and direction of the resultant velocity of the parachuutist at this point

    3) How do you connect newton`s third law pair.Give 2 examples.

    i hope you guys can help me with this 2 questions
    thx a lot
    1 is easy

    K.e=1/2mv^2
    initial K.e=1/2 10000 5^2
    =125000

    1 sec later
    K.e = 1/2 10040 5^2
    =125500

    change in Ke = 500
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    actually that looks weird..
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    v^2=u^2 +2as
    v^2= 2(9.8)80
    v=39.597979

    (horizontal velocity is same for plane and parachutst ignoring air resistan)

    40ms^-1 is horizontal or vertical?
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    (Original post by daydream)
    1 is easy

    K.e=1/2mv^2
    initial K.e=1/2 10000 5^2
    =125000

    1 sec later
    K.e = 1/2 10040 5^2
    =125500

    change in Ke = 500
    erm why are you giving the change of Ke
    they`re asking for the horizontal force
    and even if that was force
    the answer is wrong


    and 40ms^-1 is vertical
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    (Original post by MalaysianDude)
    erm why are you giving the change of Ke
    they`re asking for the horizontal force
    and even if that was force
    the answer is wrong


    and 40ms^-1 is vertical
    lol that question is weird.. i'm gonna give up
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    (Original post by daydream)
    lol that question is weird.. i'm gonna give up
    why dotn you try q3
    i dont think its a weird question
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    I think for q1 you'd have to use calculus for the mass in f=ma, dm/dt something or other.
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    (Original post by zoidberg)
    I think for q1 you'd have to use calculus for the mass in f=ma, dm/dt something or other.
    why dont you try
    put your answer here and ill see if its correct
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    dont know how, just an idea. plus im trying to get some revision done for p3
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    (Original post by MalaysianDude)
    1) A truck with mass of 10000kg moves with 0.5ms^-1
    sand from above the truck is passed onto the truck with rates of 40kgs^-1
    how much additional horizontal force is needed for the truck to move at constant speeds of 0.50ms^-1 when weight is added to the truck constantly throughout the journey

    2) A parachutist falls from an airplane travelling at 67ms^-1.After traveling for 80m away from the plane,it opens up the parachute and its velocity at this time is approximately 40ms^-1.
    Determine the magnitude and direction of the resultant velocity of the parachuutist at this point

    3) How do you connect newton`s third law pair.Give 2 examples.

    i hope you guys can help me with this 2 questions
    thx a lot
    is the first answer 19.9N every second
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    I think the answer to the first one is 10020 N.
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    (Original post by MalaysianDude)
    1) A truck with mass of 10000kg moves with 0.5ms^-1
    sand from above the truck is passed onto the truck with rates of 40kgs^-1
    how much additional horizontal force is needed for the truck to move at constant speeds of 0.50ms^-1 when weight is added to the truck constantly throughout the journey

    2) A parachutist falls from an airplane travelling at 67ms^-1.After traveling for 80m away from the plane,it opens up the parachute and its velocity at this time is approximately 40ms^-1.
    Determine the magnitude and direction of the resultant velocity of the parachuutist at this point

    3) How do you connect newton`s third law pair.Give 2 examples.

    i hope you guys can help me with this 2 questions
    thx a lot
    malaysian dude im in here
    ok
    after every second it gains 40kg, ok
    using conservation of momentum,
    10000(0.5) = (10000 + 40)v
    v=0.498m/s
    using f=ma
    f=10000(0.5-0.498/1second)
    f= 19.9N
    there u go
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    how bout the second one man?
    the questions are weird arent they

    oh and yeah
    the first answer is 20N
    which is correct,given by mathematician
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    (Original post by MalaysianDude)
    1) A truck with mass of 10000kg moves with 0.5ms^-1
    sand from above the truck is passed onto the truck with rates of 40kgs^-1
    how much additional horizontal force is needed for the truck to move at constant speeds of 0.50ms^-1 when weight is added to the truck constantly throughout the journey

    2) A parachutist falls from an airplane travelling at 67ms^-1.After traveling for 80m away from the plane,it opens up the parachute and its velocity at this time is approximately 40ms^-1.
    Determine the magnitude and direction of the resultant velocity of the parachuutist at this point

    3) How do you connect newton`s third law pair.Give 2 examples.

    i hope you guys can help me with this 2 questions
    thx a lot
    ok ive dun the second question
    horizontally the velocity of the parachuist is 67m/s(constant)(a=0)
    after 80m drop, g = 9.8, x=80, u=0, v^2=2gx, v=39.6m/s downwards.
    resolving components.using pythagoras. c^2=root of(39.6^2) + (67^2)
    c = 77m/s
    i dont have a calc on me at the moment but, the angle to the horizontal is
    acrtan 39.6/67
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    (Original post by mathematician)
    ok ive dun the second question
    horizontally the velocity of the parachuist is 67m/s(constant)(a=0)
    after 80m drop, g = 9.8, x=80, u=0, v^2=2gx, v=39.6m/s downwards.
    resolving components.using pythagoras. c^2=root of(39.6^2) + (67^2)
    c = 77m/s
    i dont have a calc on me at the moment but, the angle to the horizontal is
    acrtan 39.6/67
    the angle is below the horizontal
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    (Original post by MalaysianDude)
    1) A truck with mass of 10000kg moves with 0.5ms^-1
    sand from above the truck is passed onto the truck with rates of 40kgs^-1
    how much additional horizontal force is needed for the truck to move at constant speeds of 0.50ms^-1 when weight is added to the truck constantly throughout the journey

    2) A parachutist falls from an airplane travelling at 67ms^-1.After traveling for 80m away from the plane,it opens up the parachute and its velocity at this time is approximately 40ms^-1.
    Determine the magnitude and direction of the resultant velocity of the parachuutist at this point

    3) How do you connect newton`s third law pair.Give 2 examples.

    i hope you guys can help me with this 2 questions
    thx a lot
    as for the third part,
    connecting them.
    erm.
    equal and opposite forces acting on different bodies?
    e.g gravitational pull on prachutist from earth, and gravitational pull on earth from prachutist.
    contact force from the truck on the earth, and the contact force from earth on the truck
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    (Original post by mathematician)
    as for the third part,
    connecting them.
    erm.
    equal and opposite forces acting on different bodies?
    e.g gravitational pull on prachutist from earth, and gravitational pull on earth from prachutist.
    contact force from the truck on the earth, and the contact force from earth on the truck
    not too sure about last part
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    all right man
    you got the right answer
    now how did you do that?

    what q3 meant was
    describe newton`s third law pairs with a given example
    it does not neccesarily have to be from here
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    (Original post by MalaysianDude)
    all right man
    you got the right answer
    now how did you do that?

    what q3 meant was
    describe newton`s third law pairs with a given example
    it does not neccesarily have to be from here
    are u doing phy1 on the 14th june
 
 
 
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