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Potential/kinetic energy

  1. Feb 25, 2007 #1
    a ball of mass 522g starts at rest and slides down at a frictionless track. it leaves the track horizontally at 1.25m striking the ground 1m from the ramp.

    a. at what height above the ground does the ball start to move?
    b. what is the speed of the ball when it leaves the track?
    c. what is the speed of the ball when it hits the ground?

    is there even enough information to solve this equation??


    i tried setting the equation up as mgh= .5mv^2 but i end up with 2 variables...i dont know what to do from there. plz help!
     
  2. jcsd
  3. Feb 25, 2007 #2

    hage567

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    There sure is!

    I think the best way to start is to try to find out some things about the ball when it leaves the ramp and falls to the ground, by using equations of projectile motion. For instance, try to find the time it takes the ball to hit the ground. Since the ball leaves the track horizontally, it is fairly straight forward to do this. You will be using conservation of energy eventually, you just need to work a couple of things out first so you have enough info to use with it.
     
  4. Feb 27, 2009 #3
    using the same problem i cant seem to get it started any help?
     
  5. Feb 27, 2009 #4

    hage567

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    Try starting with finding the time the ball takes to reach the ground after leaving the track. You know the distance it falls during that time, and that's all you need.
     
  6. Feb 27, 2009 #5
    i found time using [delta]Dy=.5g[delta]t*t
    g=10 instead of 9.8

    time=.22

    to find the (C) i used v=d/t
    for d i input 1.25 time=.22

    so (C) is 5.68m/s but thats all i can seem to solve for all other equations leave me with more then one variable
     
  7. Feb 27, 2009 #6

    hage567

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    I don't get this answer. Are you using the 1m for Dy? It should be 1.25m. But even with that, I get a different number. 1m is the horizontal distance from the track to where the ball hits the ground. 1.25m is the vertical distance from the end of the track to the ground.
    Not quite. This will give you the answer for part (b). For (c) remember there are both vertical and horizontal components to the velocity when it strikes the ground. You also need to find the vertical component. You then add them together (remember they are vectors).
     
  8. Feb 27, 2009 #7
    i was doing change in distance so i took the only distances 1.25-1 so i did dy=.25
    after the correction i got 2.796m/s for b and am still working on c and a
    thanks
     
  9. Feb 27, 2009 #8

    hage567

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    I don't get the 2.796 m/s. What was your new value for the time after that correction?
     
  10. Feb 27, 2009 #9
    dy=.5g[delta]t*t
    1=.5(10)t*t
    1/5=t*t
    SQROOT1/5=t
    .447=t

    my teacher asks us to substitute gravity with 10 instead of 9.8 to simplify
    i feel confident with my anwsers
    a. 2.499m
    b. 2.796m/s
    c. 2.236m/s
     
    Last edited: Feb 27, 2009
  11. Feb 27, 2009 #10

    hage567

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    The g = 10 is fine. I see what you've done here though.
    This should be 1.25/0.5 = t*t

    1.25 is the vertical distance. You are finding the time it takes the ball to fall vertically, since there is no initial velocity in the vertical direction. You can't apply this equation to the horizontal direction because it contains acceleration and there isn't any acceleration in the horizontal direction. Gravity only acts down.

    Also, you left out a decimal place in the denominator. You need to divide through by 0.5 on both sides. That gives you 1/0.5 on the left side (so 2).

    Make sense?
     
  12. Feb 27, 2009 #11
    i did .5(10) then divded it i thought g still acts on it
    could you show the equation you are using i dont understand how you can omit g
     
  13. Feb 27, 2009 #12

    hage567

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    Sorry you are right about that part. For some reason I thought the g was still in there and you weren't dividing properly. Weird!

    But I still say you are using the wrong value for h! You need to use 1.25 here. That is the vertical distance the ball falls after leaving the track. You can find the time because you know there is no initial velocity in the vertical direction, and only gravity is acting on the ball. This part of the calculation has nothing to do with the horizontal direction AT ALL.

    Once you have the time, then use v = d/t to find the horizontal velocity. This is where you use the 1 meter!

    Are you sure the problem you're doing is exactly the same as the one stated in the very first post of this thread? The numbers aren't flipped around or something?
     
  14. Feb 27, 2009 #13
    sorry when it comes down to the basic parts of things i tend to be dyslexic i mixed them around
    so b is 2.23 finally
     
  15. Feb 27, 2009 #14

    hage567

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    According to the problem I'm doing, it's still mixed up. It should be like this for time

    1.25 = 0.5*g*t2
    1.25 = 0.5*10*t2
    1.25 = 5*t2
    t = sqrt(1.25/5)
    t = 0.5 s

    This does not give a velocity of 2.23 m/s.

    I would recommend drawing a diagram and labelling everything so you know what belongs where.
     
  16. Feb 27, 2009 #15
    i don't know how i got that but the velocity it 2m/s thanks for your assistance
     
  17. Feb 27, 2009 #16

    hage567

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    You're welcome. Are you OK with parts (a) and (c)?
     
  18. Feb 27, 2009 #17
    for c yes but i cant find c without a and i cant seem to find an equation that fits
     
  19. Feb 27, 2009 #18

    hage567

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    You don't need (c) to find (a), you need (b) to find (a), and you also need (b) to find (c).

    For (a), consider conservation of energy.

    For (c), you now need to find the vertical component of the velocity as the ball strikes the ground. Then you need to find the resultant velocity given that you know the x and y components of this velocity.
     
  20. Feb 28, 2009 #19
    using vf=Vo+g[delta]t
    x=2+10(.5)
    x=7
    c is 7m/s

    using conservation of energy for a i get ho = 2.45 height off the ground = 2.45+1.25

    final anwsers
    a.3.7m
    b. 2m/s
    c. 7m/s

    this has got to be right
     
  21. Feb 28, 2009 #20

    hage567

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    No. Why are you changing letters all over the place? That's very confusing. x is not the same as vf! Anyway, you can use this equation to find the vertical component of the velocity at landing. But you cannot use Vo=2 m/s because that is the horizontal velocity. You need the initial vertical velocity (which is zero). These things are not interchangeable.
    Instead of writing Vo for everything, use Vox for initial horizontal velocity, and Voy for initial vertical velocity.

    So the information summarizes like this:
    Vox = Vfx = 2 m/s (horizontal velocity) This does not change through the ball's motion.
    Voy = 0 m/s (initial vertical velocity) This is zero because the ball leaves the track in the horizontal direction only.
    Vfy = ? You need to find this (using the above equation).

    So Vfx and Vfy are vector components, so to get the final total velocity (which is what part (c) is asking for), you need to find the resultant of these two. Do you know about that? It should be in your textbook.

    I think you need to take more time to understand the characteristics of two dimensional motion. I've been saying the same things over and over and it's not helping. Maybe you need a visual description. This page may help http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tracon Take a good look at it and try apply it to your problem.
    I don't know how you came up with this, but I don't think it's right.
     
    Last edited: Feb 28, 2009
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