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Homework Help: Change in momentum

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    An object of mass m is launched at speed v from point P on horizontal ground at an angle of elevation of 45 degrees, as shown in the picture. When the object reaches point Q the magnitude of the change in momentum of the object is ?

    2. Relevant equations
    Change in momentum of the object
    mv - mu

    3. The attempt at a solution
    Honestly, I really don't know how to start
    All i know is use the formula of the change of momentum
    But then the angle 45 degrees I really dont know where to put..
    Can somebody help me

    Sorry for not attempting
    Is because I am a bit blur..
    Lastly the options for this question is

    A : Zero
    B : 1/2 mv
    C : 2mv
    D : 4mv

    Thanks so much..

    Attached Files:

  2. jcsd
  3. Feb 10, 2009 #2
    Okay i think I might be able to solve it..
    Finally I manage to attempt this question..

    Initial velocity, v = ( vcos45 )^2 + ( vsin45 )^2 ( using pythagorean theorum ) ( to find the resultant velocity
    Therefore initial velocity = v
    Hence, initial momentum, p =mv

    Final velocity, v is the same in magnitude but opposite direction,
    Hence final momentum, p =-mv

    Hence, change in momentum,
    is -mv -mv


    But the question just wants the magnitude,
    Hence = 2mv

    Can anybody tell me whether my working is correct??
  4. Feb 10, 2009 #3


    User Avatar
    Homework Helper
    Gold Member

    Really? Why do you think that?

    Also, momentum is a vector:


    What force is causing the momentum to change? Which direction does this force act in? Hence, which component(s) of the momentum change?
  5. Feb 10, 2009 #4
    The final velocity of the object will be pointing downwards..
    Hence, the it will be the same magnitude as the initial velocity but pointing downwards..

    That is what I think..

    Is my answer correct?
    Or are there any other answers..

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