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Pendulum hitting static object

  1. Oct 24, 2004 #1
    Ok so i got this assignment to return tomorrow:

    "A particle P with mass m is fastened to the end of a string, that has the length L. The particle starts at rest in a position where the string is rigid, and the strings direction makes a 60° with the vertical axis. At it's lowest point of the pendulums swing, the particle P hits an object A, which is at rest on a frictionless horizontal table. The object A has a mass of n*m, where n is a positive number. The collision between P and A is completely elastic. Gravitational acceleration we call g and we ignore wind resistance. So the known variables are: n, m, l, and g.

    1. Just before the collision, the particle P has the kinetic energy [tex]T_0[/tex], find [tex]T_0[/tex].

    2. Find the velocity of the object A and the particle P right after the collision, and find out how big a part "q" of the kinetic energy [tex]T_0[/tex] has been transfered to A.

    3. Investigate q for [tex]n -> \infty[/tex]. Give a short comment about the observed effect.

    4. Find [tex]cos\theta_0[/tex], where [tex]cos\theta_0[/tex] is the angle between the string and the vertical axis, when the particle P is at it's highest point in the gravitational field.

    5. Inveistigate the expresion for [tex]cos\theta_0[/tex] for [tex]n -> \infty[/tex], give a short comment on the observed effect"

    Ok this was translated over to english so i just hope i got it written down right :)

    So ok, i think i got 1 and 2, and the most of 3.

    First we find the height of P and use conservation of energy to find it's velocity just before it hits A

    [tex]h=L-(cos(\theta)*L)[/tex]
    [tex]mgh=1/2*mv^2 => v=\sqrt{2g(L-(cos60°*L))}[/tex]

    And cos60 being0.5 we get [tex]v=\sqrt{g*L}[/tex]

    Then since it's totally elastic, no energy is lost and E before and E after is the same. That coupled with conservation of momentum i have two unknowns with two variables. It's rather tedious handworking but i got to

    [tex]1/2mgl=1/2mu_1^2 + 1/2mnu_2^2[/tex]
    [tex]mgl=mu_1 + mnu_2[/tex]
    [tex]u_1=\sqrt{gl}*((n-1)/(n+1))[/tex]
    [tex]u_2=2*\sqrt{gl}/(n+1)[/tex]

    and the to find the fractional quantity of q transfered kinetic energy to A i just took the kinetic energy of the object A right after being hit, and divided it with the kinetic energy of the particle P just before the collision which should give me the % of energy transfered, and i got:

    [tex]\frac{4n}{(n+1)^2}[/tex]

    Now for #3, when n goes towards infinity. Obviously (n+1)^2 grows exponantually and much faster then 4n, so when n goes towards infinity, the whole shebang goes towards zero, which i can see mathematicly, but i'm not sure why physicly this is the case. I can also see that this must be correct looking at the formula for kinetic energy, since if V gets very small, then V^2 must be even smaller and the kinetic energy lessens, but i just don't feel like i have an understanding of it :/ Even though i could propably give an answer by simply quoting the formulas that would be satisfactory for this assignment. If i could get some better explenation that would be great too.

    Now #4 i'm pretty lost on. First off of course i'll have a new v since it's an unknown variable, so [tex] v=\sqrt{2g(L-(cos\theta_0*L)}[/tex]. And i've tried playing a bit with that in the formulas, but i can't seem to get anything that makes good sense to me :/

    If anyone could give me some tips on #4 and perhaps a better explenation of #3 (and of course if you see something wrong with 1 and 2) that'd be great :)
     
  2. jcsd
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