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Change in linear momentum

  1. Sep 22, 2016 #1
    A 15 000kg loader traveling east at 20km/h turns south and travels at 25km/h. Calculate the change in the loader's
    a) kinetic energy
    b) linear momentum

    2. Relevant equations
    ΔKE= 1/2mvf^2 –1/2mvi^2
    Δp= mv2 –mv1
    m= 15000
    vi = 20km/h = 5.56m/s
    vf= 25km/h = 6.94m/s
    3. The attempt at a solution

    Part A I think I got.

    Part B is where I am confused.
    I am using Δp= mv2 –mv1
    I substitute the numbers in
    Δp= 15000*6.94 -15000*5.56
    The magnitude of Δp that I arrive at is equal to 20 700...is that correct?

    Also, when I am trying to find the change in direction when subtracting vectors you simply add the opposite direction. so rather than using east, i add my vector pointing south and add the initial vector but pointing west?

    I then use this to calculate the angle for the direction. Is this right?

    I tried to add an image to show you my diagram...but can't seem to figure out how!

    Thanks in advance!
  2. jcsd
  3. Sep 22, 2016 #2

    Simon Bridge

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    hint: momentum is a vector, and so is velocity
    ... so ##\Delta \vec p = \vec p_f - \vec p_i = m(\vec v_2-\vec v_1)##
    ... try subtracting the vectors by the heat-to-tail method.

    But notice: you are not asked to find the magnitude of the change in momentum.
    Last edited: Sep 22, 2016
  4. Sep 23, 2016 #3
    Thank you!

    I think I got it, it's essentially the hypotenuse formed by the triangle...my numbers weren't adding up because I was forgetting that I had to separate the vectors into components, N-S and E-W.
  5. Sep 23, 2016 #4


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    Sometimes it's usful to look at vectors graphically, but you can also look at vectors as pairs (x, y) or triplets (x, y, z). In this case, you have:

    Initial velocity ##= (u_x, u_y)##

    Final velocity ##= (v_x, v_y)##

    That might be an easier way to cacluate the change in velocity and momentum in this case.

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