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Homework Help: Impulsive force

  1. Apr 30, 2007 #1
    1. The problem statement, all variables and given/known data
    A 200g puck sliding on ice strikes a barrier at an angle of 53degree and bounces off at an angle of 45degree. The speed of the puck before the bounce was 15m/s and after the bounce it is 12m/s. The time of contact during the bounce is .05seconds

    A) write these velocities in rectangular components relative to the barrier
    B) Determine the x and y components of the impulsive force. Which one can be identified as a normal force and which one can be identified as frictional force

    2. Relevant equations
    We were just introduced to impulse last class so i really don't know if I'm doing this right, but i wiki'ed it and see
    Impulse = Change in Momentum

    3. The attempt at a solution

    A) Pretty sure i did this part right
    V_x=15cos53 = 9.02723
    V_y=15sin53 = 11.9795

    V_x=12cos45 = 8.48528
    V_y=12sin45 = 8.48528

    B)This is were I'm kinda lost
    Formula i used: I=M(v_i - V_f) (M=.2, v_i= velocity initial, v_f= velocity final)

    I_x=.2(15cos53-12cos45)=.541944(.2) = .108389

    I_y=.2(15sin53-12sin45)=3.49425(.2) = .69885

    Guessing the I_y is caused from the normal force of the wall and the I_x is caused from friction.

    Any help would be appreciated
    -Thanks Tom
    Last edited: Apr 30, 2007
  2. jcsd
  3. Apr 30, 2007 #2

    Doc Al

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    Staff: Mentor

    You'll need to know how impulse is defined in terms of force and time. Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html" [Broken]
    Careful. Signs matter! If the puck has a positive y-component of velocity when it approaches the barrier, what sign must it have after it bounces off?

    Two problems: (1) You need to fix your signs, as already mentioned; (2) You've calculated the components of the impulse, but not the force. (Check out the link I gave you.)

    You're on the right track here.
    Last edited by a moderator: May 2, 2017
  4. Apr 30, 2007 #3
    Ok now lets see if i get it

    V_x=15cos53 = 9.02723 (going to the right)
    V_y=15sin53 = -11.9795 (going down)

    V_x=12cos45 = 8.48528 (going to the right)
    V_y=12sin45 = 8.48528 (going up)

    For part B I'm still a little confused because if F_average = M(change in v/change in t)
    Then Impulse = F_average*Change in t then doesn't that cancel out the t completely? (I'm still working on trying to understand the whole concept, sorry its a little confusing)
  5. Apr 30, 2007 #4

    Doc Al

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    That looks good.

    Realize that you're just writing the same equation twice, since Impulse = F_average*Change in t (by definition) and also Impulse = M * change in v (this can be derived from Newton's laws). Pick one version and solve for F!
  6. Apr 30, 2007 #5
    So this is the same thing i did before but updated with the new signs
    I_x=.2(8.48528-9.02723)= -.10839

    I_y=.2(8.48528-(-11.9795))= 4.092996

    and to solve for the F_average i just do divide by .05
    F_x=-.10839/.05 = -2.1678 N

    F_y=4.092996/.05 = 81.8591 N
    Last edited: Apr 30, 2007
  7. Apr 30, 2007 #6

    Doc Al

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    Staff: Mentor

    Good. I didn't check your arithmetic, but it looks reasonable. Now find the force components. Hint: Use the time that was given.
  8. Apr 30, 2007 #7
    just updated it with the time

    now i got to work on

    C)what is the direction of the impulsive force with respect to the barrier?

    I can solve for this by taking the arctan of the two impulse components ?

    Tan x = opposite / adjacent = 81.8591/ -2.1678
    so this means almost all the force is bouncing straight up and a little bit of friction is pushing it back, which seems right to me

    D) how much energy was lost to heat in the bounce

    would this be equal to the total impulse ?

    square root ( -.10839^2 + 4.092996^2) = 4.09443
  9. Apr 30, 2007 #8

    Doc Al

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    Looks good.

    Sounds good.

    No. Compare the puck's energy before and after the collision.
  10. Apr 30, 2007 #9
    so for part D i can just use the difference in Kinetic energy?

    (1/2)(.2)(15)^2 - (1/2)(.2)(12)^2 = Energy lost due to heat
    (.1)(225) - (.1)(144)= 8.1 N

    Well that seems simple if i did it right.

    And i just want to make sure that on part B when it says "find the components of the impulsive force"
    Impulsive force is the same thing as F_average?
  11. Apr 30, 2007 #10

    Doc Al

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    Staff: Mentor


    Careful with units. Newton's (N) are a measure of force, not energy. (Otherwise: Good!)

  12. Apr 30, 2007 #11
    Thanks for all your help
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