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A few questions in relation to energy

  1. May 14, 2005 #1
    we were assigned a few to find answers to, the following ones stumped me, hopefully you can help :)

    When two objects of different masses collide and the collision is elastic, it is said that the two objects will always move off at right angles to each other.
    Is this likely to be true? Explain

    Can an object, perhaps in motion, recieve an impulse without having work done on it?
    Conversely, can an object have work done on it without recieving an impulse?


    3) If a satellite in circular orbit around the earth encounters the outer fringes of the atomosphere, it begins to lose energy through friction. Yet the result is that it increases speed, Why is this?

    idea for 3) Is this because it Moves closer to the earth and has a greater force of gravity acting upon it
     
  2. jcsd
  3. May 15, 2005 #2
    1) It is not necessary that the balls always travel prependicularly to each other after collision.In elastic collision both momentum and kinetic energy is conserved and there is no energy loss through heat or sound.Both the cases are possible.


    2)Impulse is basically Force/second. So if the force is applied for a short time in a such a way that no displacement occurs,no work is done by the impulse.Example, drop a ball over a block kept on a smooth surface.


    3) Orbital velocity of a satellite in arbit of radius "r" from the centre of the earth is given by:

    [itex]
    \sqrt{\frac{GM}{r}}
    [/itex]

    As r decreases, its orbital velocity decreases and it loses energy.By losing energy, it does not mean it loses its kinetic energy.Net energy of a satellite in its orbit is given by:

    [itex]
    \frac{-GMm}{r}
    [/itex]

    Therefore as r decreases ,its energy decreases due to the minus sign.As the above energy is the sum of [Kinetic Energy+ Potential Energy] Therefore ,as its orbital velocity increases when it loses some potential Energy.

    Therefore,
    THE SATELLITE GAINS KINETIC ENERGY ON ACCOUNT OF LOSING POTENTIAL ENERGY WHEN IT COMES NEAR TO EARTH.
     
  4. May 15, 2005 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Yes. If you play pool, it is very useful because it allows you to play 'shape'. Otherwise you will not be able to predict where the cue ball goes.

    Since [itex]\vec v_1 + \vec v_2 = \vec v_0[/itex] (momentum conserved) and [itex]v_0^2 = v_1^2 + v_2^2[/itex] (energy conserved) the vector diagram of the collision defines a triangle in which v0 is the hypotenuse whose length is related to the sides by the pythagorean formula. This means the angle between v1 and v2 is a right angle).

    If [itex]F_{net}dt = (dp/dt)dt \ne 0[/itex] then dp is nonzero. [edit:] This means that its velocity must change. But, of course, this does not necessarily its speed changes - the change could be in direction only (speed is unchanged). If the direction of the force and momentum are perpendicular to each other, there is a change of momentum but no change in speed - hence no change in energy.

    It increases vertical speed as it loses its tangential speed. This is because as it slows its orbital speed it falls toward the earth and thus loses gravitational potential energy. This loss of potential results in an increase in kinetic energy (ie energy is conserved).

    AM
     
    Last edited: May 15, 2005
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