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Homework Help: Puck on table problem

  1. Sep 4, 2007 #1


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    Gold Member

    Hello there,

    I'm trying to help someone with a physics problem. So this is what he described:

    "(01:01:49) ka0s1337the0ry: if she gives the puck a velocity of 4.6m/s along the length (1.75m) of the table at one end
    (01:01:53) RAMEN 3bun BR: are you going to be a freshman in high school?
    (01:02:06) ka0s1337the0ry: by the time it reached the other end the puck has drifted 3.60 cm to the right
    (01:02:13) B0x0Rr0X0r402: vectors
    (01:02:14) B0x0Rr0X0r402: etc
    (01:02:17) ka0s1337the0ry: but it still has a velocity component along the length of 4.60m/s.
    (01:02:32) ka0s1337the0ry: She conludes correctly that the table is not level and correctly calculates its inclination from the above information.


    Okay - so we have the formula [tex]v_f^2 = v_i^2 + 2 a (x_f - x_i)[/tex] and F=ma.

    He says that v_f = 0. Now we have [tex]\delta x = \sqrt{0.036^2+1.75^2}[/tex]

    We now have [tex]a = \frac{-4.6^2}{2 \sqrt{0.036^2+1.75^2}} = -6.16 m / s^2[/tex]

    Then we use [tex]F = ma[/tex]. Since gravity is acting against the object, [tex]mg \sin \theta = ma[/tex]. Since we want theta, we can have [tex]\theta = \arcsin{(\frac{ma}{mg})[/tex] We get [tex]\theta = -38.08[/tex], which is effectively 38.08 degrees. Is my reasoning correct? He said that most of his classmates got values around 2 degrees
  2. jcsd
  3. Sep 4, 2007 #2


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    Would you mind explaining the question more lucidly? For example, what did you mean by "t still has a velocity component along the length of 4.60m/s"? Did you mean that the overall magnitude of the velocity increased (since it still had the same component of velocity along the original direction and an added component along another direction) or did the magnitude of the velocity remained unchanged?
    Furthermore, how is [tex] v_f= 0[/tex]? I thought you just said "it still has a velocity component along the length of 4.60m/s"?
  4. Sep 5, 2007 #3


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    Homework Helper

    A more direct approach would be to calculate the time of travel

    [tex]t = \frac{v_x}{x}[/tex]

    and then substitute it in

    [tex]y = \frac{1}{2}at^2[/tex]

    to obtain the acceleration down the plane, which is given by


    as you correctly remarked. Note that this assumes that the puck was shot along the direction in which the table was not tilted.
    Last edited: Sep 5, 2007
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