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I think this is an energy problem

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data
    A circus acrobat of mass M leaps straight up with an initial velocity of vo from a trampoline. As he rises, he takes a trained monkey of mass m off a perch of height h above the trampoline. What is the maximum height attained by the pair in terms of M, m, Vo, h, and g?


    2. Relevant equations



    3. The attempt at a solution
    I tried using kinetic energy and potential energy, but get stuck when trying to figure out the height after the mass increases at an unknown height
     
  2. jcsd
  3. Apr 20, 2008 #2

    alphysicist

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    Hi slammedtgs,

    There's a bit more to this problem than using conservation of energy. While he is taking the monkey, energy is not conserved. What do you think will be conserved?
     
  4. Apr 20, 2008 #3
    Well what i was thinking was that as he is rising hes slowing down, turning kinetic energy into potential energy. As soon as he grabs the monkey he is just added more gravitational potential energy to his system, thereby slowing him down.

    So if energy is not conserved we can look at the body's momentum as its rising, but how does that turn into max height?

    p=MVo initially, and once with the monkey p=(M+m)(v)?

    I know that max height will be when the systems velocity is zero.
     
  5. Apr 20, 2008 #4

    alphysicist

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    slammedtgs,

    We have to think of this motion in three separate parts. First the man rises to a height h; energy is conserved for this part. Then he grabs the monkey. Energy is not conserved here; grabbing the monkey is essentially a completely inelastic collision. After he has the monkey, he rises to a maximum height; during this part energy is conserved.

    So use conservation of energy to find his velocity at height h. Then use conservation of momentum to find his new speed after he grabs the monkey
     
  6. Apr 20, 2008 #5
    So this is what I was about to come up with, but im not sure if it is correct.

    (.5)(m)(v*v) v= sqrt(2*g*h) so. (M+m)(sqrt2*g*h). (m+M)(m+M)/(2*g) =H?

    Am i on the right track here?
     
  7. Apr 20, 2008 #6

    alphysicist

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    Hi slammedtgs,

    I don't think so. It looks like you have about four expressions on the same line and its a bit hard to tell where one begins and another ends.

    Look at the first part of the motion: the man starts with speed v0 moving upwards. Once he is at height h, what is his speed? Use conservation of energy here. What do you get for v at height h?


    Later in the problem:

    Now after he grabs the monkey, what is the speed of him+monkey (at the same height h)? Energy is not conserved during the grab, but momentum is.

    After that, the man+monkey rises to a maximum height H. As you said, energy is conserved during this process. You know their initial speed and height for this second part of the rising motion, so you can use conservation of energy to find the maximum height.
     
  8. Apr 20, 2008 #7
    The velocity at pt. H would be Sqrt(2gh)

    so P=mv at H would give (M+m)*(sqrt(2*g*h)

    Kinetic Energy would be (.5)*(m+M)(2*g*h)

    So when H is max V is zero

    0 +MGH = Etotal

    Etotal/MG= H? Is that another way of looking at it?
     
  9. Apr 20, 2008 #8

    alphysicist

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    slammedtgs,

    First we need to find the speed at point h (where the monkey is) not point H (the highest point.

    From conservation of energy, if we set the height to be zero at the starting point, the initial energy (which is all kinetic) equals the energy at height h (which is kinetic and potential). So we have:

    [tex]
    KE_i = KE_f + PE_f
    [/tex]

    or

    [tex]
    \frac{1}{2} M v_0^2= \frac{1}{2} M v^2 + M g h
    [/tex]

    Solving this for the speed v at height h gives:

    [tex]
    v = \sqrt{v_0^2 -2 g h}
    [/tex]

    So this is his speed right before he grabs the monkey. Multiplying this by his mass would give his momentum at that point.

    After he has grabbed the monkey, the momentum is equal to their total mass times their new speed. Since these two momentums have to be equal, what is that new speed?
     
  10. Apr 20, 2008 #9
    V at the point of the monkey is Sqrt(v0^2-2gh)

    So, (M+m)(v?) = M*Sqrt(v0^2-2gh)

    M*Sqrt(v0^2-2gh)/(M+m) will give (V?)

    So then take this new speed and use it in a new conservation of energy Equation?

    [tex]

    KE_i = KE_f + PE_f

    [/tex]

    (.5)(Mtotal)(v?) + MGH = Etotal

    (0)+MHG = Etotal

    Etotal/MG should give the max height right?
     
  11. Apr 20, 2008 #10

    alphysicist

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    Hi slammedtgs,

    For the new conservation of energy equation, you have the height H in the final potential energy, where H is the answer they are looking for. So that part is good.

    However, the initial potential energy right after the monkey is grabbed is not 0, because they are at height h. So you need another potential energy term.



    Remember that the speed in the kinetic energy is squared, and this equation is not quite right. You know that the final kinetic energy is zero (for the maximum height) so the energy equation will follow:

    [tex]PE_i + KE_i = PE_f[/tex]

    where the speed you found from the conservation of momentum equation does go in [itex]KE_i[/itex] as you said. The initial potential energy is at height h, and the final potential energy is at height H.
     
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