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What is the change in momentum?

  1. Mar 30, 2006 #1
    Okay, I did this problem, and my answer disagrees with the book.

    A 15.0 g rubber bullet hits a wall with a speed of 150 m/s. It bounces straight back with a speed of 120 m/s. what is the change in momentum?

    I get:

    delta p=(m*v-i) - (m*v-f)
    delta p= (15.0g * 150 m/s) - (15.0 g - 120 m/s) = 15.0 g(150m/s - 120 m/s) = 15.0g * 30 m/s
    delta p = 450g * m/s = .45 kg * m/s
    the book says the answer is 4.05 kg * m/s

    Am I wrong?
  2. jcsd
  3. Mar 30, 2006 #2
    Yes. Momentum involves direction as well.
  4. Mar 30, 2006 #3
    But that doesn't actually answer my concern. It's the magnitude that disagrees.

    Oops, now I see why.
  5. Mar 30, 2006 #4
    Well, if you look at it vectorally... I usually assume that going away from the wall will be positive and going towards the wall is negative.

    So the final momentum minus the initial moment will be your answer.

    A shortcut (I prefer) is that the mass is 'constant' for this system so all we see is a change in the velocity.

    The change in the velocity is NOT 30 m/s, it is in fact 270 m/s.

    So 270 m/s * .015 kg = 4.05 N*s (newtons per second)

    Beware that you might also get this question in many forms with momentum... They try to trick you :-p
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