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Homework Help: Impulsive Motion

  1. Apr 13, 2005 #1
    Problem:

    The initial velocity of block A is 9m/s. Knowing that uk=.3 Determine the time for the block to reach zero velocity. When (theta)=20 deg.

    Here is my work so far.

    mv_1+Imp_1,2=mv_2

    Imp_1,2=Ff(t)+sin(theta)W(t)
    Ff=Cos(theta)uk(mg)
    W=mg
    Imp=9.81cos(20).3m+9.81sin(20)m
    Imp=6.12mt

    9m+6.12mt=0
    t=1.47s


    This answer is Wrong It Should Be .963s

    What Have I Missed?
     
  2. jcsd
  3. Apr 13, 2005 #2
    Strange, i get the exact same answer 1.47

    marlon
     
  4. Apr 13, 2005 #3
    [tex]v_x = v_i -g(0.3cos(20)+sin(20))t[/tex]
    [tex]v_y = 0[/tex]

    X is along the incline and y perpendicular to it. I get the same answer using a different approach. Are you sure it's not 1.47 ?

    marlon
     
  5. Apr 13, 2005 #4
    Yes, I did place the positive x-axis in the direction of movement, but it really should not matter because I am using the Principle of Impulse and Momentum.
    (a) part was finding the time taken if (theta)=0deg. (3.06s I got this one right)
    So I know my method works. There is just something I am missing.

    The problem is from Beer and Johnston;
    VECTOR MECHANICS for ENGINEERS:STATICS and DYNAMICS 7th Ed.
    problem 13.124
    For anyone who knows the book. (They may have made a mistake but i doubt it.)
     
    Last edited: Apr 13, 2005
  6. Apr 13, 2005 #5
    The answer for part (a) is .306 or is is 3.06?
     
  7. Apr 13, 2005 #6
    My bad, 3.06s
     
  8. Apr 13, 2005 #7
    OK, so for 0 degrees I get 3.06 s

    and for 20 degrees I get 1.47 s

    I think the book is wrong
     
  9. Apr 13, 2005 #8
    If you take the angle to be zero, my method also gives 3.06, just plug in that value into the equation for v_x

    marlon

    ps : i think there must be a mistake here. Perhaps someone else can double check this problem ?

    Dexter, Doc Al, what do you get ??
     
  10. Apr 14, 2005 #9

    Doc Al

    User Avatar

    Staff: Mentor

    1.47s is correct. Using "impulse/momentum" is a perfectly OK way to solve this problem; it is equivalent to finding the acceleration and then using kinematics.
     
    Last edited: Apr 14, 2005
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