Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: I just wanna check my work on this problem

  1. Oct 3, 2004 #1
    [tex] \textrm{Hello, folks. I just wanna check my work on this problem. Thanks.} [/tex] :cool:

    [tex] \textrm{A certain ball has the property that each time it falls from a height} [/tex] [tex] h [/tex] [tex] \textrm{onto a hard, level surface, it rebounds to a height}[/tex] [tex] rh [/tex] [tex]\textrm{, where}[/tex] [tex]0<r<1[/tex]. [tex] \textrm{Suppose that the ball is dropped from an initial height of}[/tex] [tex] H [/tex] [tex] \textrm{meters.} [/tex]

    [tex] \textrm{(a) Assuming that the ball continues to bounce indefinitely, find the total distance that
    it travels.} [/tex]

    [tex] H + 2rH + 2r^{2}H + 2r^{3}H + \cdots = H + 2H \sum _{n=1} ^{\infty} \left( r \right) r^{n-1} = H + 2H \left( \frac{r}{1-r} \right) = H \left( \frac{1+r}{1-r} \right) [/tex]

    [tex] \textrm{(b) Calculate the total time that the ball travels.} [/tex]

    [tex] t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + \sqrt{\frac{2rH}{g}} + \sqrt{\frac{2r^2H}{g}} + \sqrt{\frac{2r^3 H}{g}} + \cdots [/tex]

    [tex] t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + \sqrt{\frac{2rH}{g}} \left( 1 + \sqrt{r} + \sqrt{r^2} + \sqrt{r^3} + \cdots \right) [/tex]

    [tex] t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + \sqrt{\frac{2rH}{g}} \left( \frac{1}{1-\sqrt{r}} \right) [/tex]

    [tex] \textrm{(c) Suppose that that each time the ball strikes the surface with velocity}[/tex] [tex]v[/tex] [tex]\textrm{it rebounds with velocity}[/tex] [tex] -kv [/tex][tex] \textrm{, where}[/tex] [tex] 0<k<1 [/tex]. [tex] \textrm{How long will it take for the ball to come
    to rest?} [/tex]

    [tex] v_{\textrm{REST}} = v + kv + k^2 v + k^3 v + \cdots [/tex]

    [tex] v_{\textrm{REST}} = v + \sum _{n=1} ^{\infty} \left( k v \right) k^{n-1} [/tex]

    [tex] v_{\textrm{REST}} = v + \left( \frac{kv}{1-k} \right) [/tex]

    [tex] \textrm{If } K=U, \textrm{we find} [/tex]

    [tex] \frac{1}{2}mv_{\textrm{REST}} ^2= mgH [/tex]

    [tex] \frac{1}{2}m\left[ v^2 + 2v^2 \left( \frac{k}{1-k} \right) + v^2 \left( \frac{k}{1-k} \right)^2 \right] = mgH [/tex]

    [tex] H = \frac{1}{2g}\left[ v^2 + 2v^2 \left( \frac{k}{1-k} \right) + v^2 \left( \frac{k}{1-k} \right)^2 \right] [/tex]

    [tex] \textrm{which gives} [/tex]

    [tex] t_{\textrm{REST}} = - \frac{2\left| \frac{v}{g\left( k-1 \right)} \right|}{\sqrt{r}-1} - \left| \frac{v}{g\left( k-1 \right)} \right| [/tex]
     
  2. jcsd
  3. Oct 4, 2004 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    At b)
    The time it takes to fall from a height h is [itex]\sqrt{\frac{2h}{g}}[/itex], so the time it takes to rebound to a height h and fall down to the ground again is twice as long.

    [tex]t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + 2\sqrt{\frac{2rH}{g}} + 2\sqrt{\frac{2r^2H}{g}} + 2\sqrt{\frac{2r^3 H}{g}} + \cdots [/tex]
     
  4. Oct 4, 2004 #3
    [tex] \textrm{Yes, indeed. I should have written}[/tex]

    [tex]t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + 2\sqrt{\frac{2rH}{g}} + 2\sqrt{\frac{2r^2H}{g}} + 2\sqrt{\frac{2r^3 H}{g}} + \cdots [/tex]

    [tex] t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + 2\sqrt{\frac{2rH}{g}} \left( 1 + \sqrt{r} + \sqrt{r^2} + \sqrt{r^3} + \cdots \right) [/tex]

    [tex] t_{\textrm{TOTAL}} = \sqrt{\frac{2H}{g}} + 2\sqrt{\frac{2rH}{g}} \left( \frac{1}{1-\sqrt{r}} \right) [/tex]

    [tex] \textrm{Thanks.}[/tex]
     
  5. Oct 12, 2004 #4
    Guys, I'm not so sure about what I found for part (c). Did I get it right?

    Thank you. ​
     
  6. Oct 12, 2004 #5

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Admin note: It is preferable to keep the text parts of your posts in plain text, and use LaTeX only to render pieces of math notation.

    - Warren
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook