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Kinetic energy / projectile motion question

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img690.imageshack.us/img690/5756/12281060.jpg [Broken]

    2. Relevant equations

    k = 0.5 mv^2
    projectile motion

    3. The attempt at a solution

    Ok so for part one i just did some algebra and got 1:1 as the ratio, no problems there
    For part two i treated it as a projectile after it leaves the roof with v = root (2g)
    with that i used y = v sin theta t - 0.5gt^2
    to find t
    so -3 = root (19.6) sin240 - 4.9 t^2
    root (19.6) sin240 = -3.834
    simplified and changed to a quadratic 4.9t^2 + 3.834t - 3 = 0
    found t with the quadratic formula, t= 0.4836
    then subbed into Vy = V sin theta - gt
    find Vy = -8.57

    So thats the velocity of the y component
    Do i have to add that to the velocity of the y component of the initial velocity?
    because i've tried it both ways, i'm a bit confused, my tutor told me to add both the inital and final y components, but he didnt treat it as projectile motion, he did it a different way
    But lets just say 8.57 is Vy

    Vx = v sin 60 = 3.834

    the velocity = the magnitude of 3.834i + 8.57j = 9.39

    then K = .5 * 5 * 9.39^2 = 220.43

    Kinetic rotational energy = 5g

    so 220.43/ 5g = 4.5

    so the ration is 4.5:1 ?

    when i find the ratio by adding the inital Vy and the final Vy i get 6.7:1 or somthing like that

    is this right?
    my tutor for 7.04:1, but i noticed afew errors so im not sure if its correct, it's sure a cleaner answer than 4.5 though
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 11, 2010 #2

    Doc Al

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    Staff: Mentor

    Since all you care about is the kinetic energy, not the speed, stick to energy methods and you'll get the answer cleanly with little calculation.

    You are correct that the ratio of KE/KEr is 1:1 while it's rolling. So what's the KE as it leaves the roof? How much energy does it pick up as it falls 3 m? What's the final KE?
     
  4. Apr 11, 2010 #3
    As is leaves the roof KE = 0.5 x 5 x 2g = 5g ?
    KEf = 0.5 x 5 x (2 x g x 3) v^2 =2gh h = 3
    = 15g
    15g:5g = 3:1?

    sheeeeet
    is that right?
     
    Last edited: Apr 11, 2010
  5. Apr 11, 2010 #4

    Doc Al

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    Write it as 0.5 x mgh1 (where h1 = 1 m).
    The amount of energy gained during the fall is mgh2 (where h2 = 3 m).

    So what's KEf? How does that compare to KEr?
     
  6. Apr 11, 2010 #5
    ohh
    sorry, so its 20g:5g 4:1

    mgh2 = 15 mgh1 = 5
    add them both because mgh2 is the ammount of energy gained, then compair with initial energy?

    aw man
    massive over think of that question
    if you saw the amount of work i did you would laugh
     
    Last edited: Apr 11, 2010
  7. Apr 11, 2010 #6
    oh wait sorry again

    196:49 4:1 ?
     
  8. Apr 11, 2010 #7

    Doc Al

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    No. Answer the questions in my last post, one by one.
     
  9. Apr 11, 2010 #8
    ok so, KEr = KEi and KEr doesnt change, so KErf is = 5g
    is the 5g part right? and the v^2= 2gh?

    KEf = KEi + KE

    KEf = 0.5mgh1 + mgh2 = 24.5 + 147 =171.5

    171.5:49
    3.5:1

    im kinda confused now, how come we are 0.5mgh1 instead of K = 0.5 m v^2
    and mgh2 is the potential energy of the fall right?
     
  10. Apr 11, 2010 #9
    i think it must be the v^2 = 2g that is stuffing me up
    i just did it like
    KE = 0.5mgh1 + mgh2 KEr = 0.5mgh1
    KEr:KEf = 24.5:(24.5 + 147) = 1:7
    which is what my tutor got kind of
    which is weird because i saw he made mistake with his trig early on


    v^2 = vi^2 + 2ah
    so before falling v^2 = 2ah = 2g
    after falling v^2 = 2a + 2ah = 8g

    so KEi = KEr = 0.5 x 5 x 2g = 5g
    KEf = (0.5 x 8g x 5) + (5 x 9.8 x 3) = 343

    KEf:KEr = 343:49 = 7:1
    that kind seems right to me
    im kinda confused why i had to add the gravitational potential energy to the kinetic energy
    is it because (0.5 x 5 x 8g) covers the diagonal KE and mgh2 the vertical? or is that just completely wrong?
     
    Last edited: Apr 11, 2010
  11. Apr 12, 2010 #10
    sorry to be rude and post again Doc Al, but was 1:7 right?
    this is from another question but if somthing is show like
    r>>R does that mean r is directly proportional to R?
     
  12. Apr 12, 2010 #11

    Doc Al

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    This is exactly right.

    KE = 0.5mgh1 + mgh2 = 3.5 mg (plugging in the values for h)
    KEr = 0.5mgh1 = 0.5 mg (plugging in the values for h)

    There's no need for any trig at all.


    The notation r >> R means that r is much greater than R.
     
  13. Apr 12, 2010 #12
    thanks a bunch Doc Al, now to stress over the final 3 questoins ;) and start some math

    was my second way of getting 7:1 just a fluke?
    when i did v^2 = vi^2 + 2ah and so on?
     
  14. Apr 12, 2010 #13

    Doc Al

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    Sorry, but I don't really understand what you did. If you explain it step by step, I'll take a look.

    To solve this kinematically, you'd have to find the speed as it leaves the roof. (How would you find the acceleration as its rolling?) Once it leaves the roof it's a projectile. Find the vertical and horizontal components of the speed. Then solve for the final vertical component of speed. Then combine them to find KEf. (Sounds like a lot of work!)
     
  15. Apr 12, 2010 #14
    haha yea
    good 4 pages
    with errors on the first 3
    butt with this other thing

    I have this formula in my physics text book in the kinetic energy part, it
    v^2 = vi^2 + 2ah
    where vi = initial velocity and in this case a = g
    so i applied it to this question and got
    before falling off the edge
    vi = 0 h = 1
    therefore v^2 = 2a = 2g

    after falling off the roof
    vi = 2g from what i found as the velocity before it falls
    h = 3 a = g
    v^2 = 2g + 2g3 = 8g

    so back to the ratio
    Before the fall:

    KEi = KEr = 0.5 x 5 x 2g = 5g
    KEr = 5g so KEr after the fall = 5g because it doesnt change as a it falls through space

    After the fall v^2= 8g what i found by putting into that equation before
    KEf = 0.5mv^2 + mgh where h = 3
    =(0.5 x 8g x 5) + (5 x 9.8 x 3) = 343

    so KEf = 343 and KEr = 5g = 49
    KEf:KEr = 343:49 = 7:1

    but looking at it the way you told me to makes much more sense and is alot easier
    but i dont get why its 0.5 mgh1 instead of 0.5mv^2
    is it because its sin30 mgh ?
     
    Last edited: Apr 12, 2010
  16. Apr 12, 2010 #15

    Doc Al

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    Careful: While it's rolling down the roof, a ≠ g. (a = g for free fall, but rolling isn't free fall.)

    Your starting v² isn't right, so your final won't be right either.

    If v^2= 8g, KEf would equal 0.5 x m x v^2 = (0.5 x 5 x 8g) = 20 g. (Note: This is incorrect, since v^2 does not equal 8g.)

    Seems like you made a few errors that canceled each other. :smile:

    KE = 0.5 mgh1 because the full change in potential energy equals mgh1 and half goes to KE and the other half goes to KEr. (This is conservation of energy.)

    Of course, KE also equals 0.5mv^2. The correct value of v^2 (after rolling down the roof) is gh1, not 2gh1.
     
  17. Apr 12, 2010 #16
    oh fair enough
    thanks heaps Doc Al
     
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