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B A momentum question following collision

  1. Jul 29, 2016 #1
    I think this is a basic question, at least I would hope so.

    If two identical motorcycles are going at the same speed but one has a lighter individual. If each individually collides with a stationary object and the driver is ejected. Who will go further and why?
     
  2. jcsd
  3. Jul 29, 2016 #2

    Simon Bridge

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    Not enough information. There are too many variables.

    How are you thinking about this - show us your best reasoning(?) In what context does the question arise?

    I'll simplify the situation ... you have two identical carts travelling at the same speed.
    They each have a loose load, and are stopped suddenly in the same way ... so they come to rest in the same very short time. Then, neglecting friction and air resistance, the determining factor for which load travels the furthest before hitting the ground is the height of the load above the ground. ie. all other things equal, both loads travel the same distance before hitting the ground.

    This is not a conservation of momentum problem - this is 1st law stuff.
    But look at how carefully I had to set up the problem.
     
  4. Jul 29, 2016 #3
    Thanks for the reply Simon!

    My initial thought was that the momentum before collision will be (mass person + mass motorcycle) x velocity. Since the motorcycle comes to a stop, the momentum after collision will be equal to the momentum before collision but just the mass of the person x new velocity. The lighter the person the greater the velocity.

    Then I came back and said that doesn't wound right. If the person on the motorcycle has a certain velocity and the motorcycle is stopped, they will continue moving forward at the same velocity. In this case they would initially move forward at the same velocity but since KE = mv^2. The heavier person should have greater kinetic energy and go further.

    I am sure somewhere my logic is not right but for some reason I cannot see it.
     
  5. Jul 30, 2016 #4

    Simon Bridge

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    The total momentum is conserved - but the motorcycle's momentum is transferred to whatever mechanism caused it to stop.
    In your example, that is some sort of barrier.

    Include the barrier in the conservation of momentum calculation and you will see all the laws get obeyed.

    The initial kinetic energy does not, by itself, determine how far someone goes. You need to know what forces are on the object: ie. what are the mechanisms for energy transfer?

    In the absence of air resistance and friction, the only force on the object is gravity. This is a ballistics problem until impact. If you are thinking that the object then slides along the ground, then you have to add friction to the mix (or sliding does not slow the object down). Friction, in the usual model, is proportional to mass - all else remaining equal, the bigger person has more mass.
     
  6. Jul 30, 2016 #5

    Simon Bridge

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    Again: setting up the problem more carefully...
    Object mass m initially has speed u horizontally, falls distance h to the ground, then slides distance d along the ground and comes to rest.
    Assuming h is low enough that air resistance can be neglected ...

    We just want to know whether a heavier object goes further overall, so we only need d as a function of m.
    So all the initial kinetic energy goes to work against friction.
    Taking the coefficient of kinetic friction to be ##\mu## - write the equation relating initial kinetic energy to the work against friction moving a distance d ... then solve for d in terms of m. What do you see?
     
  7. Jul 30, 2016 #6

    PeroK

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    It seems to me that if the driver is ejected cleanly, then effectively he/she is not involved in any collision. He/she simply becomes a projectile.
     
  8. Jul 30, 2016 #7

    Simon Bridge

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    That is one interpretation of the setup ... unfortunately, the initial problem statement was not so precise.
    See posts #2 and #4.
     
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