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Help with this physics problem (acceleration in two dimensions).

  1. Feb 12, 2008 #1
    A hockey puck rebounds from a board as shown in Figure 16.

    Figure 16 looks like this.


    The puck is in contact with the board for 2.5ms. Determine the average acceleration of the puck over the interval.


    So I know you have to use vector components. The answer says the avg. acceleration is 7.3x10^3m/s^2 [7.5 N of W] (degrees).

    How do I solve it? =/

    I have made an attempt.

    I found that the vector components for v1x and v2x to add up to 43.57, and the components of v1y and v1x to add up to (-1.87). However, I'm not getting the correct average acceleration value in the end.
    Last edited: Feb 12, 2008
  2. jcsd
  3. Feb 12, 2008 #2
    For the second vector the vector in the ight of the picture), what angle did you use?
    The angle should be 180°-22° = 158°
  4. Feb 12, 2008 #3
    I just tried it using 158 as the angle and I didn't get the correct answer given again.

    I'm calculating the vector components using sin(theta)(Ay), sin(theta)By, cos(theta)Ax, and cos(theta)Bx of course.

    I can't seem to find out what I'm doing wrong.
  5. Feb 12, 2008 #4
    First you need to fix a coordinate system within which you'll work. Let the 'N' be our y-axis and 'E' our z-axis.

    Now try to find the vector of the two velocities. For the first one, it is simply

    [tex]\overrightarrow{v_i} = 26(\cos{(22)}\hat{i} - \sin{(22)}\hat{j})[/tex]

    it is important to note here that i have taken component on the 'y' axis to be negative. This is because, the direction of the 'y' component is downwards. Here is the diagram to resolve this vector:


    Now.. find the vector [itex]v_f[/itex] yourself. And then, remember that acceleration is given by:

    [tex]\overrightarrow{a} = \frac{\overrightarrow{v_f} - \overrightarrow{v_i}}{\Delta t}[/tex]
  6. Feb 12, 2008 #5
    |vf| = 21(sin(22)+cos(22))
    |vf| = 27.337

    |a| = (|vf|-|vi|)/(/\t)
    |a| = (27.337-14.37)/0.0025s
    |a|= 5186.8

    Hmmm, did I do something wrong?
  7. Feb 12, 2008 #6
    well.. for one.. you did many things wrong. First of all, you did not assign unit vectors to the [itex]\overrightarrow{v_f}[/itex]. Also, there is a difference between [itex]\overrightarrow{v_f}[/itex] and [itex]|v_f|[/itex]. Also, [itex]|v_f|[/itex] cannot be anything other than 21.

    When finding the acceleration, you need to subtract vectors first and then take modulus. I suggest u refer your textbook and understand vectors.
  8. Feb 12, 2008 #7
    I couldn't assign unit vectors to vf (with direction) as I don't know how to show that on this forum. =/

    Also, we haven't been taught "modulus" or is that modules? I don't know what they are.
  9. Feb 12, 2008 #8
  10. Feb 12, 2008 #9
    To learn how to write in LaTeX, check this: https://www.physicsforums.com/showthread.php?t=8997

    also.. u can show unit vectors using 'i' or 'j' like: vf = 21(cos(22)i + sin(22)j). Modulus of a vector is the 'magnitude' of the vector. If there is a vector:

    \overrightarrow{l} = a\hat{i} + b\hat{j}


    |\overrightarrow{l}| = \sqrt{a^2 + b^2}

    the modulus gives the length of the vector. For a velocity vector, the modulus is the magnitude of the vector, in the case of [itex]v_f[/itex], it is 21. For a vector in the form:

    \overrightarrow{l} = p(\cos{(\theta)}\hat{i} + \sin{(\theta)}\hat{j})

    the magnitude will always be 'p'. Also, even if u do it using impulse, it eventually will end up the same away. You'll assume a mass.. you'll find the change in momentum [which is nothing but the change in velocity multiplied by mass].. and then you'll divide it by time to find force and then reduce the 'm' to give u acceleration. Taking an extra 'm' will serve no purpose. So don't do it using impulse or collisions since only the acceleration is asked.
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