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Acceleration in two dimensions. Hockey puck question

  1. Dec 30, 2012 #1
    Problem: A hockey puck rebounds from a board as shown in figure 16. The puck is in contact with the board for 2.5 ms. Determine the average acceleration of the puck over the interval.

    t = 0.0025 s |vi|= 26 m/s |vf| = 21 m/s aav = ? *v and a are vectors*

    I have made several attempts to solve this question and have seen the previous thread about this question and I am still unable to get the correct answer. The correct answer is
    7.3 x 10^3 m/s^2[7.5 N of W]. In my opinion my closest attempt was that I broke the initial vector and final vector into components, found the change in vx to equal 4.6 m/s and the change in vy to equal -1.9 and then found the change in v to equal 5. my final answer and direction were completely wrong. Can someone help me with this awful question?
     

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  2. jcsd
  3. Dec 30, 2012 #2

    cepheid

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    Welcome to PF TimelessTime!

    Don't worry, this question is quite straightforward, and you already have the right method! The average acceleration is given by change in velocity divided by change in time:

    a = Δv/Δt

    The thing that needs to be done is finding the change in velocity Δv, in each direction, using those to calculate separate horizontal and vertical components of the acceleration, and then combining these into the overall vector. Could you please post your work (the actual calculations you used to find the change in horizontal and vertical speeds), so that we can track down where you went wrong?
     
  4. Dec 30, 2012 #3

    cepheid

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    Also: working through some of the numbers, here is another hint: pay very close attention to the signs of your components. Are they in the same direction both before and after the rebound? This affects your computation of the Δv's
     
  5. Dec 30, 2012 #4
    These were my calculations:
    |Vx| = |Vfx| + (-|Vix|)
    = 21 m/s(sin68) - 26 m/s(cos22)
    = -4.6 m/2 oops, i got -4.6 but on the next page i left out the negative, not sure if that is the real problem though

    |Vy| = |Vfy| + (-|Viy|)
    = 21 m/s(cos68) - 26 m/s(sin22)
    = -1.9
    |V| = sqrt((-1.9)^2 + (-4.6)^2)
    = 5.0 m/s
    |aav| = 5/0.0025 != 7.3 x 10^3
     
  6. Dec 30, 2012 #5

    cepheid

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    One of your two components of Δv is wrong here. I'll reiterate my point from post #3. Is the puck going in the same direction (along the east-west axis) after the rebound? What about along the north-south axis? Same direction as before, or opposite?
     
  7. Dec 30, 2012 #6
    so what the problem is is that for the x components the should both be positive and for the y components it is a subtraction?
     
  8. Dec 30, 2012 #7

    cepheid

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    There's another problem as well. Don't compute the final acceleration from the magnitude of the velocity. Remember, acceleration is a VECTOR. So you should calculate the individual components of a first, and then combine them:

    ax = Δvx/Δt

    ay = Δvy/Δt

    a = (ax2 + ay2)1/2
     
  9. Dec 30, 2012 #8
    alright, ill try solving it again and post if theres anymore problems. Thanks for the help.
     
  10. Dec 30, 2012 #9

    cepheid

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    You tell me. Just be systematic about it. Let's call north the positive vertical direction, and south the negative vertical direction. So if the vertical component of the vector is northward, it should be positive. If it is southward, it is negative.

    Likewise for the horizontal components. Call east + and west -

    Initially, is the puck going east or west? What about finally? So what is the sign of the horizontal component before and the one after?

    Initially, is the puck going north or south? What about finally? So what is the sign of the vertical component before, and the one after?

    These questions I've been asking you are not rhetorical, I intended for you to actually answer them. I'm trying to walk you through the reasoning.
     
  11. Dec 30, 2012 #10
    Ok, so I found the average acceleration to be 7.3 x 10^3 however, the direction I got was 75degrees N of W. Here's what I did:

    |Vx| = 21 m/s(cos22) - 26 m/s(cos22)
    = -4.64 m/s

    |Vy| = 21 m/s(sin22) + 26 m/s(sin22)
    = 17.6 m/s

    aav,x = -4.64/0.0025 = -1856 m/s^2

    aav,y = 17.6/0.0025 = 7040 m/s^2

    Then, aav = sqrt(7040^2 + (-1856)^2) = 7.3 x 10^3

    for the direction I found theta in the pic I attached and it ended up being 75NofW instead of 7.5NofW
     

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  12. Dec 30, 2012 #11

    haruspex

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    7.5 degrees N of W must be a misprint. 75 is right. It must be fairly close to perpendicular to the wall.
     
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