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Angle of inclination (using Newton's Laws)

  1. Apr 14, 2012 #1
    The problem statement, all variables and given/known data

    A physics student playing with an air hockey table (a frictionless surface) finds that if she gives the puck a velocity of 3.44 m/s along the length (2.73 m) of the table at one end, by the time it has reached the other end the puck has drifted 2.08 cm to the right but still has a velocity component along the length of 3.44 m/s. She correctly concludes that the table is not level and correctly calculates its inclination from the above information. What is the angle of inclination?


    Relevant equations

    Summing forces in x and y directions, according to the question.


    The attempt at a solution

    I could not visualize a way to solve this question using Newton's laws. The only path to a solution I could visualize was to assume that 2.08 cm was the hypotenuse of a triangle in which the height was the distance the puck had 'fallen', and then solve for theta. However, finding out the distance the puck had fallen has proved to be insurmountable and I may very well be trying to solve this question incorrectly.
     
  2. jcsd
  3. Apr 14, 2012 #2

    ehild

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    You can resolve the motion as one along the length of the table with constant velocity and the other along the slope with acceleration g sin(φ). The components of the displacements are given.

    ehild
     

    Attached Files:

  4. Apr 19, 2012 #3
    Thank you for your reply, ehild and I apologize for not replying much sooner with a result.

    That gif was quite helpful in visualizing the situation and I thought I understood the problem.
    Here is what I did:

    Find time taken to for the travel by Speed/distance.
    Use the kinematics equation S=ut+(1/2)at^2, where a=gsinθ, u(initial velocity in the direction of the slope)=0, S=distance drifted to the right (converted to m).

    I solve for θ using θ=arcsin(2S/g*t^2) but that answer seems to be incorrect. Could you point towards what I'm doing wrong?

    Thanks again.
     
  5. Apr 19, 2012 #4

    ehild

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    t^2 has to be in the denominator in the expression for sin θ.


    ehild
     
  6. Apr 20, 2012 #5
    That is indeed what I meant, sorry for my shoddy mathematical expression.

    sinθ=(2S)/(g*t^2)

    In this case, θ is 0.15° while the answer to the question is 0.39°.
     
  7. Apr 20, 2012 #6
    So, I may have misunderstood you...

    When I put the expression as as sinθ=(2S*t^2)/g I got the correct answer, which is a horrible way to go about things. Since our kinematics equation is S=0.5gsinθt^2, I don't understand how t^2 came to be in the numerator to get the correct answer.
     
  8. Apr 20, 2012 #7

    ehild

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    Your result is correct. The book might be wrong.


    ehild
     
  9. Apr 21, 2012 #8
    I suppose stranger things have happened.

    Thank you very much ehild.
     
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