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Conservation of momentum of a skateboard

  1. Nov 7, 2015 #1
    1. The problem statement, all variables and given/known data
    A 48.0-kg boy, riding a 1.60-kg skateboard at a velocity of 5.80 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is 6.00 m/s, 7.40° above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

    2. Relevant equations

    Pfinal = Pinitial

    3. The attempt at a solution

    I start with ##\vec{P}_0 = \vec{P}## then

    ##(m_b + m_s)\vec{v}_0 = m_b\vec{v}_b + m_s\vec{v}_s##
    then
    ##\displaystyle \vec{v}_s = \frac{(m_b + m_s)\vec{v}_0 - m_b\vec{v}_b}{m_s}##
    then
    ##\displaystyle v_{sx} = \frac{(m_b + m_s)v_0 - m_bv_b\cos7.4}{m_s}##
    then
    ##v_{sx} = 1.3~m/s##

    However, this is not the right answer. What am I doing wrong?
     
  2. jcsd
  3. Nov 7, 2015 #2

    gneill

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    I don't see a problem with your method. If it's an online quiz perhaps it's a stickler for significant figures?
     
  4. Nov 7, 2015 #3

    TSny

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    I agree with gneill that your answer looks correct except for the number of significant figures.

    I will add a nit-picky comment. The following two equations are not really correct for this problem:

    The above equations imply that both the x and y components of total momentum are conserved. But, in this problem, only the x component is conserved.
     
  5. Nov 8, 2015 #4
    Those vector equations are not correct because the earth applies a normal force in the y-direction, and that is an external force, which makes conservation of momentum invalid, right? I just want to make sure I understand you.

    Also, I get a positive number. Shouldn't I get a negative one since the boy is pushing off the skateboard in the opposite direction? Could one of you do the problem and see if you come out with the same number as I do? I don't think that it is a significant figure problem because the answer can be within +/- 2%, which would more than make up for significant figure difference.
     
  6. Nov 8, 2015 #5

    BvU

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    1.299191 -- if there are no typos (7.4##^\circ## isn't all that much)
     
  7. Nov 8, 2015 #6

    gneill

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    As far as significant figures is concerned, 1.3 is not equal to 1.30 . It's off by 1 in 3 :smile:
     
  8. Nov 8, 2015 #7

    TSny

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    Yes, that's exactly right.

    OK, I had not worked out the numbers. When I do, I get 1.30 m/s to 3 significant figures. So, 1.3 should have worked.
    There is no reason why the skateboard must end up going backwards just because the boy pushed backward on the skateboard. If the force was not very strong, it might just slow down the skateboard but leave it going in the same direction.
     
  9. Nov 8, 2015 #8
    This is the picture from the problem:
    cssd6vB.jpg

    However, it seems strongly misleading. Originally, I entered -1.3 m/s, because I thought that I had got a sign error, because the picture clearly shows that the skateboard should be going in the negative direction. I re-entered with what I had originally got, and what you guys got, 1.3 m/s, and it was correct. Perhaps the problem was written for different numbers in which the skateboard ended up having a negative velocity; with these, it came out positive. Thank you guys for your help!
     
  10. Nov 8, 2015 #9

    TSny

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    Great! Good work.
     
  11. Nov 9, 2015 #10

    BvU

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    I second that. Perhaps the exercise maker had a bad day. I am too old for a skateboard, but I suppose you want to jump up instead of forward when approaching a vertical barrier.
     
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