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Momentum conservation law: Head on elastic collision

  1. May 31, 2013 #1
    Hi guys,

    I've been thinking on a problem for a while which really bothers me. I've been trying to mathematically solve the following problem:

    A train approaches the station at a velocity of V=50 m/s. Then a tennis ball is thrown with a velocity U=30 m/s, against the approaching train. Assume that the collision is absolutely elastic.
    If the train's velocity after the collision is V=50 m/s, then what is the speed of the ball with respect to the train station?

    So, here is my approach to the problem:

    From the momentum conservation law and the fact that the train will have the same momentum after the collision, we can write:

    MtrainV - mballU = MtrainV + mballUfinal

    so it turns out that the velocity of the ball will be:

    Ufinal = - U = -30 m/s

    Which is incorrect. The correct answer should be:

    Ufinal = 2V+U = 130 m/s

    Can someone provide a rigorous derivation of the later equation?

    Your help will be highly appreciated!
     
  2. jcsd
  3. May 31, 2013 #2
    The easiest way to solve this problem is to do it in the frame of reference of the train. Initially, relative to the train, the ball is traveling -80 m/sec. After the collision, relative to the train, the ball is traveling +80 m/sec. But, relative to the ground, after the collision, the ball is traveling + 130 m/sec.

    If you are going to solve the problem from the ground frame of reference, then you need to assume some huge mass for the train (relative to the tennis ball), and satisfy both conservation of momentum and conservation of energy for the combination of ball and train. When you do this, you will find that, in the limit of infinite ratio of train mass to ball mass, you get the same result for Ufinal. Try doing it this way and see what you get.
     
  4. May 31, 2013 #3
    Thanks Chestermiller!! I appreciate your help!

    I solved the problem using your second suggestion, via Moment. Conser. Law and Energy Conserv. Law.

    However, the first method is kind of obscured to me. I would imagine as the train is approaching at +50 m/s and the ball is flying at a -30 m/s towards it, then the net speed of the ball to be +80 m/s. This is (in my opinion) the same case as throwing the ball towards a stationary train with a velocity of -80 m/s. The it would bounce back with velocity + 80 m/s.
    I guess I am missing something. Is there any physics book where this kind of stuff are explained? For instance, why we first consider the speed of the ball in a reference frame at the train and then we consider the reference frame of the ground?
     
  5. May 31, 2013 #4
    If the train is traveling 50 m/s relative to the ground, and the ball (after collision) is traveling 80 m/s relative to the train, then the ball is traveling 130 m/s relative to the ground. This is the same thing as when a person riding on the train (i.e., stationary on the train) throws a ball forward at 80 m/s. An observer on the ground measures that the ball is traveling 130 m/s. This is the so-called Galillean transformation, relating velocities measured in different inertial frames of reference. Welcome to the world of inertial frames of reference. You will learn a lot more about it when you study relativity.

    It does not matter whether you write down the conservation equations as reckoned by an observer on the ground, or as reckoned by an observer who is moving at a constant velocity relative to the ground. You should get the same answer both ways. So, that being said, I recommend that you redo this problem as reckoned from the frame of reference of an observer who is watching the collision from a car that is traveling parallel to the train with a velocity relative to the ground of 50 m/s. Before the collision, to this observer the train is standing still, and the ball is approaching at 80 m/s. After the collision, the train velocity will be slightly negative relative to this observer, and the tennis ball will be traveling away from the observer at close to 80 m/s. Write down the full momentum and kinetic energy conservation equations as reckoned from the frame of reference of this observer and solve them. I think you will find this exercise very enlightening.

    Chet
     
  6. May 31, 2013 #5
    Thanks Chet,

    Those very nice examples. I will think on them a bit more carefully!
     
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