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Newton's Third Law Problem

  1. Mar 31, 2005 #1
    So I have a homework question that says:
    A 100 kg hockey player a 112kg hockey player collide with each other each travelling with a force of 50N.

    The text book answer says that the acceleration of the 112kg is 0.89m/s^2
    and the acceleration of the 100kg player is -1.0 m/s^2.

    I'm confused because according to Newton's third Law there will be equal and opposite reaction forces so wouldn't they not move? Is the text book wrong or am I?
    Last edited: Mar 31, 2005
  2. jcsd
  3. Mar 31, 2005 #2
    Remember by the second law:

    [tex] F = m a [/tex]

    They both impart a force of 50N on each other, but the acceleration is also scaled by their masses. 50N on a 100kg person will push them farther than 50N on a 112kg person.

    Imagine pushing a shopping cart with 50N. An empty shopping cart will go really far, whereas a full shopping cart will go a much shorter distance. In both instances the force is the same, but:

    [tex] a = \frac{F}{m} [/tex]
  4. Mar 31, 2005 #3
    calculating acceleration

    So you do 100/100 and 100/112 even though the total of the forces equals 0 because its 50 + -50??
  5. Mar 31, 2005 #4
    The net force of the entire system is 0, but the force on each player is of magnitude 100N. You can find the acceleration of each player using

    [tex] a = \frac{F}{m} [/tex]. Knowing F = 100N and m = 100kg for person 1,

    [tex] a = \frac{100N}{100kg} = 1.0m/s^2 [/tex]

    F = 100N and m = 112kg for person 2:

    [tex] a = \frac{100N}{112kg} = 0.89m/s^2 [/tex]
  6. Mar 31, 2005 #5
    Thank You!!

    Thank-you you are a saviour!
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