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Speed Question: Two bodies are hit/ final momentums?

  1. Dec 18, 2016 #1
    1. The problem statement, all variables and given/known data

    Two bodies are standing still upon a surface without friction. The mass of body 1 is bigger than the mass of body 2.

    i) A constant force is applied to body 1, so it accelerates in a segment to a distance d. The force stops being applied to body 1, and is applied to body 2. When body 2 has done the distance d, which of the following are valid?

    ii) When a force is applied to body 1, it accelerates for a time Δt. The force stops being applied to body 1, and is then applied to body 2. Which are valid if body 2 accelerates for Δt?

    a) p1 < p2
    b) p1 = p2
    c) p1 > p2
    d) K1 < K2
    e) K1 = K2
    f) K1 > K2

    2. Relevant equations

    I = ΣFΔt
    Δpf = Δpi

    3. The attempt at a solution

    I think I'm done with (ii), but here's a look:

    ii) I = FΔt => I1 = I2 => Δp1 = Δp2 => pf1 - pi1 = pf2 -pi2 => pf1 = pf2 (1)

    m1 > m2 (2)

    (1) & (2): V1 < V2 (3)

    K = 1/2*m*V2 (4)

    (2) & (3) & (4): K1 < K2

    The book has b & d as the answers for (ii), and c & e as the answers for (i).

    Now, it's one (i) I have a problem with. I just don't know how to tackle it. ΔP = 0 is for whole isolated systems, and for a very small time-frame. Here, from what I gather, I hit 1, it accelerates, it reaches d, and then the force stops being applied to it (so technically it keeps going on and on since there's no friction). After that, I do the same to 2.

    I'm probably missing something (I just got into the chapter, I'm five pages in so I haven't really gotten the hang of it yet), so I'd be grateful for any help!
     
    Last edited: Dec 18, 2016
  2. jcsd
  3. Dec 18, 2016 #2

    Simon Bridge

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    If there is an applied force - the system is not isolated.
    There is a relationship between force and change in momentum you learned a while ago... you need to relate that to the distance d.
    Shortcut: you can relate work to momentum via change in kinetic energy.
     
  4. Dec 18, 2016 #3
    Yeah, since the F is external, the system is not isolated, but so far, the only thing the book has talked about are forces that one body exerts to the other, and so it desribes the system as isolated to the momentum. In my case, it's different. The only formula I've got at my disposal atm is ΣF = dp/dt. That, plus the usual ΣF = ma and whatnot.

    So far, the only exercise I've seen dealt with internal forces, and the usual pf = pi. So I don't really know how to tackle and continue this one.
     
  5. Dec 18, 2016 #4
    All I can do is say again what Simon said. With an external force momentum is not conserved: pf does not equal pi. I feel very confident if you look back in your notes or book that they have told you what force x distance is.
     
  6. Dec 18, 2016 #5
    Yeah, I got that, I'm just saying what I've gone through with the book thus far. If you're refering to work, then I guess you're talking about this:

    W = Fdcos0 = Fd = ΔK1 = ΔK2 <=> Kf1 -Ki1 = Kf2 -Ki1 <=> Kf1 = Kf2 [So that's one question down]

    1/2m1Vf12 = 1/m2Vf22 <=> m1Vf12 = m2Vf22 (1)
    &
    m1 > m2

    so, Vf12 < Vf22 <=> |Vf1| < |Vf2|

    (1) p1*Vf1 = p2*Vf2
    &
    Vf1 < Vf2 (I guess they're both going towards the same direction, otherwise it'd needlessly complicate things)

    so, p1 > p2 [and that's the second one down]

    Is that what you mean? I did think about that in the beginning, but I had just gone into the momentum chapter, so I figured it'd be something more to do with that info.

    PS: Is my solution for (ii) correct? It's at my first post (attempt at solution).
     
  7. Dec 18, 2016 #6

    haruspex

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    Yes, that looks good.
    Your final step does not quite work, in that it does not follow from the three equations you cite.
    You do have the equation you need to use instead.
     
  8. Dec 18, 2016 #7
    Is there an equation for that part? It's 00:00 here so I'm probably blanking out, but isn't it more of a "logic problem" at the final step?

    m1 > m2
    V1 < V2
    K = 1/2*m*V2
    p1 = p2

    So, K1 = 1/2m1V12 = 1/2p1V1
    K2 = 1/2m2V22 = 1/2p2V2

    And so, since p1 = p2, the only deciding factors are the velocities, and with V1 < V2, K1 < K2
     
  9. Dec 18, 2016 #8

    haruspex

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    Yes, that's fine. Originally you cited eqns 2, 3 and 4, which omitted the crucial p1=p2.
     
  10. Dec 18, 2016 #9
    Oh yeah, my bad. Thanks for the confirmation, and thanks for the help everyone!
     
  11. Dec 18, 2016 #10

    Simon Bridge

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    No worries ... yes I was thinking like this:
    Kinetic energy is ##p^2/2m##, since this is produced by a force acting over a distance you can write ##p^2 = 2mFd##.
    When the constant force acts over the same time, can just use: ##p = F\Delta t## from F=dp/dt.
    (using ##p_i = 0## and ##p_f=p##.)

    You can also do it using the suvat equations, if you've met them.
    ie. ##v^2=u^2+2ad## ... if ##p=mv## and ##q=mu##, then ##p^2 = q^2 + 2Fd##
    In terms of energy and work, that is ##K = K_0 + W## (check)

    That old F=ma and F=dp/dt (if F is constant then ##F = \Delta p / \Delta t## ), those things, are your goto equations for motion. Don't be so quick to dismiss.
    Notice that the last one with the ##\Delta##'s in it turns into the law of conservation of momentum when F=0?
     
  12. Dec 19, 2016 #11
    No, I haven't come across those equations just yet. But yeah, you're right, my mind should've gone there immediately.

    Thanks for the info!
     
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