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Calculate the time it took the marble to travel from N to P?

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data

    A very small steel marble is shown rolling at a constant speed on a horizontal table. The marble leaves the table at L, falls, and hits the ground at P. This is illustrated in the diagram below which is drawn to scale. Calculate the time it took the marble to travel from N to P.

    Please look at the graph:

    http://i995.photobucket.com/albums/af79/huybinhs/plot.png



    2. Relevant equations

    p(t) = 0.5 * a * t^2 + v0 * t + p0
    where
    p(t) is position at time t
    a is acceleration
    v0 is initial velocity at time 0
    p0 is position at time 0



    3. The attempt at a solution

    p(t) = 0.5 * a * t^2 + v0 * t + p0
    where
    p(t) is position at time t
    a is acceleration
    v0 is initial velocity at time 0
    p0 is position at time 0

    We'll use the fall from L to P to compute v0.

    Vertical acceleration due to gravity = -9.8 m/s^2 = -980 cm/s^2
    Let t = 0 when the marble is at point L.
    horizonatal
    px(t) = 100 = 0.5 * 0 * t^2 + v0 * t + 50 = v0 * t + 50
    vertical
    py(t) = 0 = 0.5 * -980 * t^2 + 0 * t + 90 = -490 * t^2 + 90

    Two equations. Two Unknowns. Solve for t and v0.

    => t1 = 0.43; v0= 116.28

    To find the time need to travel from point N to L use v0 from above
    px(t) = 50 = 0.5 * 0 * t^2 + v0 * t + 10 = v0 * t + 10

    => t2 = 0.34

    Add the two times for the total time from L to P.

    => t = t1+t2= 0.774 which came out wrong. I dont know why???? Please help and let me know the result!!!

    Thanks so much!
     
  2. jcsd
  3. Feb 12, 2010 #2
    If you find the time it takes to fall to the ground you can find it's constant horizantal velocity.
    [tex] .7=\frac{1}{2}gt^2[/tex]
    [tex]t=.377[/tex]

    t1 obviously equals t2 since there is no x acceleration and the ball travels the same distance in both parts.
     
  4. Feb 12, 2010 #3
    So you mean final answer is 0.377^2 = 0.142129????
     
  5. Feb 12, 2010 #4
    Um no. Your final answer would be 2*.377 = .754
     
  6. Feb 12, 2010 #5
    I just tried to enter that number (.754). It came out wrong!!!!

    Please help!!!!!
     
  7. Feb 12, 2010 #6
    Oh, the marble starts at x=10, so the time is slightly less.
     
  8. Feb 12, 2010 #7
    ???? Could u be more specific and answer please!

    Thanks!
     
  9. Feb 12, 2010 #8
    You know the time it takes for the marble to reach the ground, .377. Using that time and the horizontal distance it travels you can find it's x velocity and therefore find out how long it takes to reach point L.
     
  10. Feb 13, 2010 #9
    Now, I have no clue. Please help!
     
  11. Feb 13, 2010 #10
    The time on the first part is
    [tex]t_1=x_1/v[/tex]
    You can easily find v since you know the time, t2, for the second part and the distance the marble travels.
    [tex] v = x_2/t_2[/tex]
    Then add up the times.
     
  12. Feb 13, 2010 #11
    Ok. Is this part correct?

    p(t) = 0.5 * a * t^2 + v0 * t + p0
    where
    p(t) is position at time t
    a is acceleration
    v0 is initial velocity at time 0
    p0 is position at time 0

    We'll use the fall from L to P to compute v0.

    Vertical acceleration due to gravity = -9.8 m/s^2 = -980 cm/s^2
    Let t = 0 when the marble is at point L.
    horizonatal
    px(t) = 100 = 0.5 * 0 * t^2 + v0 * t + 50 = v0 * t + 50
    vertical
    py(t) = 0 = 0.5 * -980 * t^2 + 0 * t + 90 = -490 * t^2 + 90

    Two equations. Two Unknowns. Solve for t and v0.

    => t1 = 0.43; v0= 116.28
     
  13. Feb 13, 2010 #12
    Anyone??? Please help!!!!
     
  14. Feb 13, 2010 #13

    rl.bhat

    User Avatar
    Homework Helper

    Velocity v at L is 50/0.377 cm/s.
    Time to move from N to L is 40cm/v.
     
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