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Conceptual Check - Momentum

  1. Jan 12, 2014 #1
    Hi,

    Im new to PF and Physics so pardon my likely extremely basic question (teaching myself physics).

    1.) Question: (a) an empty sled is sliding on frictionless ice when susan drops vertically from a tree above onto the sled. When she lands, does the sled speed up, slow down, or keep the same speed? (b) Later, susan falls sideways off the sled. When she drops off, does the sled speed up, slow down, or keep the same speed?

    2.) Relevant Equation(s)
    Conservation of momentum?
    m1v1 + m2v2 = m1v1' + m2v2'

    3.) Attempts:
    (a) this makes sense to me - I think. Initially the sled has some momentum but susan does not. When susan lands on the sled all she has is vertical velocity and not horizontal. Thus, the equation might look something like (assume 1 is the sled and 2 is susan):
    m1v1 + 0 = v(m1+m2)
    -when solving for v using some numbers the speed has decreased - I think thats right.
    (b) honestly do not know where to start with this one - if someone could point me in the right direction in order to solve this I would greatly appreciate it. I know that susan will for a few seconds (or ms) have a horizontal velocity equal to what she previously had on the sled but im stuck after that. If her momentum doesnt change then neither does the sleds?? So confused...
     
  2. jcsd
  3. Jan 12, 2014 #2
    Also credits to Physics by Giancoli (6th edition) for question - forgot to post on there and cannot edit from my phone.
     
  4. Jan 12, 2014 #3
    Well, in part a the sled should slow down. You could view your calculations as
    [tex]\frac{m_1}{m_1 + m_2}\times v_1 = v[/tex]in which case the sled would only maintain its initial velocity if the girl had no mass, seeing that is not possible, the sled will slow down.

    Part b sounds tricky though. I assume she falls off the sled not affecting sled's direction and is not giving it a "push" by falling off. As the law states though, the system's total momentum cannot change. Therefore both will maintain their velocity of v and continue their own separate ways.
     
  5. Jan 12, 2014 #4

    CWatters

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    I agree. Consider two sledges tied together side by side. Push them off a hill with no riders on them. Cut the string holding them together using a high power long range laser. Both sledges will still keep going at the same speed after being separated (conservation of momentum). Then one hits a tree! The other will keep going at the same speed it had before.
     
  6. Jan 13, 2014 #5
    I see, could one also think of it as driving a car in a world without any nonconservative forces and jumping out of the car at 30 mph? Both the person jumping put and the car will
    have the same velocities but different momentums?

    V(M+m)=m1v1 + m2v2
     
  7. Jan 13, 2014 #6
    If the velocity of the girl with respect to the sled just after "falling" is zero then the speed will remain the same. If this is not the case then the sled will speed up.
     
  8. Jan 13, 2014 #7

    CWatters

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    Yes, but I would suggest the equation is..

    Momentum before jumping out = Momentum after jumping out
    or
    V(M+m) = VM + Vm
     
  9. Jan 14, 2014 #8
    Prove your theory. It does not make sense. Sure the surface is frictionless, but momentum is not dependant on such phenomena. What you say would break the conservation of momentum.

    If the girl falls at 0 velocity with respect to the sled's direction then the sled would for sure have to slow down, because the same momentum must then support an extra mass, no doubt, a much larger mass.

    If the girl is moving at the same speed before falling on the sled then you would have to be careful how you think about it. If you are chasing a sled running down the hill and then jump on it your own velocity will be quite a bit faster than the sled's and your own momentum will be humongous compared to the sled. You feel the thrust when you land, because the sled is nothing in comparison to the body's mass. However, for there to Not be any extra things to keep in mind, she would have to kind of float at a velocity of V and drop on the sled from pointblank range, considering the velocities are equal, then:
    [tex]m_1 V + m_2 V = (m_1+m_2)V_x\Rightarrow V = V_x [/tex]
    This does not reflect reality, though, so don't get confused - imagine a fairy tale world where every law of nature is applied in its most simple/perfect form.
     
  10. Jan 14, 2014 #9

    CWatters

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    I think "consciousness" was referring to the girl falling OFF the sledge not onto it.
     
  11. Jan 14, 2014 #10
    I was helping with part (b), because the OP has answered part (a) already. :biggrin:
     
    Last edited: Jan 14, 2014
  12. Jan 14, 2014 #11
    Oh pardon me, I mis-understood - I thought "falling" meant the falling in part a, onto the sled.
     
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