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Homework Help: 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.

    http://img174.imageshack.us/img174/2571/fig16vr6.png [Broken]

    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 by a moderator: May 3, 2017
  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:

    http://img142.imageshack.us/img142/4125/resolvingonefc5.png [Broken]

    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]
    Last edited by a moderator: May 3, 2017
  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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