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A force perpendicular to velocity does not change speed?

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    There's a ball moving at a constant velocity V and I apply a force perpendicular to the direction of the velocity. Why is it that there's no change in speed?

    If I add a perpendicular force, there would be a new vector thus resulting in a new velocity with a different direction and speed right?

    http://img22.imageshack.us/img22/8261/forceh.jpg [Broken]\[/URL]

    Thanks for any answers :D
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 13, 2009 #2

    Doc Al

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    Staff: Mentor

    In order to keep the speed unchanged and only change the direction of motion, the force must be kept perpendicular to the velocity as the velocity changes. Your picture shows an initial upward velocity and a sideways force. As soon as the velocity starts changing direction, so must the force. This is not illustrated in your diagram.
     
  4. Sep 13, 2009 #3
    thanks for your reply!

    What if the force was instantaneous? Would it still not affect the speed?
     
  5. Sep 13, 2009 #4

    Doc Al

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    I'm not sure what you mean by "instantaneous". If you mean something like an impulsive force (a quick push to the side that lasts a short time), then that could certainly change the speed since it would not be continually adjusted to be perpendicular to the velocity.
     
  6. Sep 13, 2009 #5
    I think you are confusing velocity and speed. A perpendicular force will not increase velocity but it will increase speed.
     
  7. Sep 13, 2009 #6

    Doc Al

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    Staff: Mentor

    A perpendicular force will not change the speed but will change the velocity.
     
  8. Sep 14, 2009 #7
    If an object is traveling upwards at velocity v and at point p it is influenced by force f in the horizontal direction, its velocity [in the upwards direction] is unchanged but its speed has increased to √(v^2 + f^2).
     
  9. Sep 14, 2009 #8

    Hurkyl

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    Among your many clues that this cannot possibly be correct is that you're adding two quantities with different units.
     
  10. Sep 14, 2009 #9
    Yes, that was a pretty stupid mistake. I wasn't thinking about what I was writing. What were the others?
     
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