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Homework Help: Need help with Conservation of E problem !

  1. Oct 6, 2005 #1
    Need help with Conservation of E problem ASAP! Please Help!

    Hey I have a test coming up and this is a problem that will be on the test. I think I have an idea of how it works but I can't really figure out how to do it. I know it's a conservation of energy problem but there is nothing like it anywhere I have looked. Any help would be appreciated but a walkthrough would really help.

    When a certain rubber ball is dropped froma height of 1.25 m onto a surface, it loses 18.0% of its mechanical energy on each bounce. a) how high will the ball bounce on the first bounce?

    I took 1.25(.18) and then subtracted this from 1.25 to get 1.03 which is the answer.

    b) how high will it bounce on the second bounce.

    I took 1.25(2(.18)) and subtracted this from 1.25 to get .8 but the book says .841 so I'm not sure what is going on there.

    c) with what speed would te ball have to be thrown downward to make it reach its original height on the first bounce?

    Here is my real problem. If I add 1.25(.18)+1.25+.841 I get 2.32 which is the answer but I know that's not how it is done. If there is a part I need to understand it is this one. Can anyone help me? Please?
  2. jcsd
  3. Oct 6, 2005 #2


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    Well for b) you must remember that when it loses 18% of its energy, it loses it on every bounce. After its first bounce, yes, it loses 18% and you get 1.0225. But now you have to figure out what 18% of that new energy amount is, not what another 18% of the original energy amount is.
  4. Oct 6, 2005 #3
    That makes sense. Simple oversight on my part. Thank you!

    Can anyone help on the third part?
  5. Oct 6, 2005 #4

    Any help at all would be great
  6. Oct 6, 2005 #5
    I am dying for some insight here. :frown:
  7. Oct 6, 2005 #6

    Physics Monkey

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    You want to throw the ball with enough kinetic energy so that after the energy loss at the bounce, the ball has just enough energy to reach the same height with no kinetic energy. That is the idea, try to translate that into an equation.
  8. Oct 6, 2005 #7
    How do I use this with the equation for Kinetic Energy if I don't know the mass??

    As far as I can see I take 1.25 + 1.25(.82) and get 2.275 but I really don't know where to go from here. I really need a more direct approach because We covered this section for about a millisecond and I understand the situation but not the mathematical mechancis....
    Last edited: Oct 6, 2005
  9. Oct 6, 2005 #8
    so you ask what power you add if it loses 18% at ground contact, and you start at 1.25m.

    haha i saw it. it's quite simple. i won't write how to do it, but what you do is figure out the height you must start from in order for it to bounce back to 1.25 and im sure you can figure out that it's basically working your equations backwards than you did on a) and b)

    after you found the height you will use it to find it's speed as it goes past the 1.25m marker...becasue that's what you want to know...how fast it moves at 1.25m in the air if it is to bounce back at 1.25m

    to do that...free fall or projectile motion...either way works for you.

    how to think this problem...i know that in college you need to know that.
    dropping the ball at some speed at 1.25 mark is the same as dropping the ball above..let's say 1.40m and letting it accelerate by itself to the speed that you would instantaneusly give it in the trow.

    in other words instead of you giving the ball the speed, you are analizing a similar problem, where the gravity gives the ball that speed. they're the same, becasue between 1.25m, the ground and the 1.25m the ball does the exact same thing.

    i hope it made sense...i can draw you a sketch if you like.

    edit: don't use no kinetic force. i'm not a college student, but i can tell you, kinetic can't help you AT ALL. just use what i told you...a "different scenario" with the same outcome and analize that. The problem doesn't ask where the speed comes from...so you can be smart and use the fastest way.
    Last edited: Oct 6, 2005
  10. Oct 6, 2005 #9
    I tried the free fall equations initially. They didn't give me the correct answers...

    What I take form what you said is find the height you woudl need to make the ball boucne back to the original height, which as I said above would be 1.25 + 1.25(.82)=2.275

    then you could take this and plug it into the free fall equation v^2=v^2-2gy which would give me v^s=2(9.8)(2.275) but this won't work. Did I do somethign wrong here? The answer is 2.32, but using the free fall equation gives an answer nowhere near that number by a variety of situations I have tried. If you spell it out for me what you mean a little mroe clearly, that would help.......
    Last edited: Oct 6, 2005
  11. Oct 6, 2005 #10

    Physics Monkey

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    You can calculate the energy it would take to have the ball bounce to the original height after the collisional loss in terms of the (unknown) mass, the gravitational acceleration, and the height. The difference between this energy and the potential energy at the height is the kinetic energy you must supply. You are correct that you don't know the mass, but the mass actually multiplies everything so it cancels out when you solve for the velocity. You don't actually need to know what it is!
  12. Oct 6, 2005 #11

    since I don't need to know the mass, I still don't see how this is put into the Kinetic Energy equation? Could I simply get a walkthrough on this?

    I tried 9.8(2.275) and took the square root of this and divided by two and get 2.36 which is pretty close. Is this the correct method.

    I am just in a fog on this question and need a more direct answer as it is the only one I can't get and each explanation winds me up further....
    Last edited: Oct 6, 2005
  13. Oct 6, 2005 #12

    Physics Monkey

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    The energy E you must start with to reach a height h after the bounce satisfies [tex] E - .18 E = mgh [/tex]. The kinetic energy [tex] K = 1/2 \,mv^2 [/tex] you start with is just [tex] K = E - mgh [/tex]. Solve for E in terms of mgh and then K in terms of mgh. You should end up with something like [tex] 1/2\, m v^2 = a mgh [/tex] where a is some number you get in the first steps. You can see how the mass cancels enabling you to solve for v without knowing m.
  14. Oct 6, 2005 #13
    so you say solve for E in terms of mgh and K in terms of mgh... but the equations seem to require m to solve so I end up with the same problem??

    honestly I am just in a fog with this one problem and I am completely lost now. I am not trying to take up anyone's time here and get a free answer for no reason, but I am just desperate and if you could show me your explanation a little clearer that would help. What you said above is convoluted to me. you say to solve for E to move on, but you say E-.18E=mgh... so how do I solve for E not knowing mass??

    If you could just help me step by step that would help me a ton. My teacher has not had office hours so I have had no assistance with this problem and I just need a simpler, step by step explanation...
  15. Oct 6, 2005 #14

    Physics Monkey

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    Ok, here goes:

    Step 1: Solve for energy in terms of mgh
    [tex] E-.18 E = mgh [/tex] implies that [tex] .82 E = mgh [/tex] or [tex] E = \frac{1}{.82}mgh [/tex].

    Step 2: Solve for kinetic energy in terms of mgh
    [tex] K = E - mgh = \frac{1}{.82}mgh - mgh = \frac{.18}{.82} mgh [/tex].

    Step 3: Solve for velocity
    [tex] \frac{1}{2} m v^2 = \frac{.18}{.82} mgh[/tex]. Cancelling the mass from both sides we find [tex] \frac{1}{2} v^2 = \frac{.18}{.82} gh[/tex].

    It should be pretty obvious now how to get the final answer.
  16. Oct 6, 2005 #15
    Oh wow.

    Honestly, that process was easier to understand than anything that has been written before it. I completely understand the process now. I'm sorry if I seemed rediculously stuck on this thing... but I really was. I am more of a learner by example so I truly thank you for this. This helped me more tahn any qualitative explanation could.
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