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Kinetic Energy and Displacement of a 2 piece system

  1. Nov 16, 2014 #1
    1. The problem statement, all variables and given/known data
    You stand at rest on one end of a 3.00m long platform (also at rest). The platform is on a smooth friction-less surface. If you walk from one end of the platform to the other at the constant rate of 2.00m/s, what is the kinetic energy of the system (you and the platform) right before you reach the other end if your mass is 50kg and the mass of the platform is 100kg. What is the displacement of the platform when you reach the other end?


    2. Relevant equations
    MuVu + MpVp = 0

    3. The attempt at a solution
    Since the system is initially at rest then the momentum is 0, so by conservation of momentum when you move at 2.00m/s the platform must move in a way so that the net momentum is still 0

    Mu = your mass Vu = your velocity
    Mp = mass platform Vp = Velocity Platform

    MuVu+ MpVp = 0
    Vp = -(MuVu)/Mp
    Vp = -(50kg*2.00m/s)/100kg
    Vp = -1.00m/s

    So your velocity is 2.00m/s and the platform velocity is -1.00m/s

    How can we use this information to find the kinetic energy?

    Next the displacement of the platform i just calculated the time it takes for you to get to the other side and then apply that time to velocity of the platform
    3.00m/2.00m/s = 1.5s
    delta x = (-1.0m/s)(1.5s)= -1.5m
    Not sure if that is right but please let me know
    Thanks in advance
     
  2. jcsd
  3. Nov 16, 2014 #2

    gneill

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    Staff: Mentor

    What's the formula for kinetic energy?
    No, you've got a problem with your method of finding the displacement. If you are assuming that the speeds are measured with respect to the frictionless surface then the relative speed between you and the platform is not 2.0 m/s and the time to walk it from one end to the other is not 3/2 s. This is because the platform is moving too. What's the relative speed of you and the platform?
     
  4. Nov 16, 2014 #3
    Kinetic Energy Equation
    Is 1/2MV^2

    M will be just the total mass of the system which is 150kg. Not sure about velocity though.


    And I see that I did make a mistake solving for displacement i can't think of an equation off the top of my head for it but thanks to how the problem is structured I know that at t = 1 you will move 2m positive while the platform moves -1m negativ
     
  5. Nov 16, 2014 #4
    posted before I finished .. but if at t = 1s
    you will be at the other end of it and it will have moved -1m in that time
     
  6. Nov 16, 2014 #5

    gneill

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    Kinetic energy is a scalar quantity. You can sum the individual KE's of all the moving parts.
    You can either wrestle with the relative velocities or consider the behavior of the center of mass of the system ;)
     
  7. Nov 16, 2014 #6
    By sum do you just mean addition?

    So that would be

    1/2(50kg)(2.00m/s)^2 + 1/2(100kg)(-1.00m/s)^2
     
  8. Nov 16, 2014 #7

    gneill

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    Right. The total KE of a system is the sum of the KE's of all its parts.
     
  9. Nov 16, 2014 #8
    Thanks a lot! If you don't mind there is another (final) part of this problem I didn't put up because I thaught that i would have more trouble with this than I actually did. May I ask about the final part of the problem or should I put up a new thread
     
  10. Nov 16, 2014 #9

    gneill

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    Might as well keep it here in this thread if it's part of the same problem.
     
  11. Nov 16, 2014 #10
    Ok here it goes..

    You are back at rest at the left end of the 3.00m board (also at rest) on the frictionless surface but you now have a large 4kg watermelon in hand. If you throw the watermelon to the right, how far will the watermelon land relative to the right end of the platform if you throw it. You are able to throw the watermelon 5.00m on stable ground

    I don't know how to start with this one. It's not like the previous one where you are just walking. I guess what happens is the watermelon is thrown and then the platform and you move backwards again. But I can't think of a good equation for this. Should the watermelon be considered part of the system?
     
  12. Nov 16, 2014 #11

    gneill

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    The watermelon is part of this system since all of the forces acting (horizontally) are internal, acting only on components of the system.

    I think you should consider that you are able to impart the same relative velocity to the watermelon in both cases.
     
  13. Nov 16, 2014 #12
    What effect will throwing the watermelon cause. I know that since the watermelon will move forward the {you + platform} system will move backwards, and that the net momentum must still be 0, but nothing about any velocity is known so I don't know how to start this off.
     
  14. Nov 16, 2014 #13

    gneill

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    When you throw the watermelon on stable ground it achieves a separation of 5 meters. Assume that the relative horizontal speed imparted is some value, say Vo, and that it takes some time t to drop from throwing height to the ground. It doesn't really matter what the particular values of Vo or t are.

    If you can manage to impart the same relative speed (between yourself and the melon) while on the platform, what should be the outcome of the total separation between you and the melon?
     
  15. Nov 16, 2014 #14
    I've been trying to figure out what you said for a while now and I think I get a little of what your saying but unfortunately I am unable to fully understand what you mean. The watermelon is thrown with some velocity Vo and takes a time t to be displaced 5m. Like you said Vo and t don't really matter but I dont see the relationship that would allow for that. Meanwhile the velocity of you and the platform is going in the negative direction with some velocity that makes the net momentum 0.
     
  16. Nov 16, 2014 #15
    What do you mean by the same relative speed? what are you referring to as being the same?
     
  17. Nov 16, 2014 #16

    gneill

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    The RELATIVE velocity of the melon will be the same. That is, the rate of separation between you and the melon will be the same in both cases. And from basic projectile motion, given the same type of throw the time for the projectile to reach the ground will be the same (independence of vertical and horizontal components). So, same speed of separation, same time of flight,....
     
  18. Nov 16, 2014 #17
    Oh ok so what you are saying is that the speed of the watermelon as it moves away from you on stable ground will be the same as the speed of the watermelon as it moves away from you when u are on a platform on the frictionless surface.

    So in both cases the watermelon ends up 5m away from "you" and what your location is at the time of landing
     
  19. Nov 16, 2014 #18

    gneill

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    Yes.
     
  20. Nov 16, 2014 #19
    Thank you
     
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