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Understaing A Problem - Easy

  1. Nov 17, 2009 #1
    Hey guys, I'm having some trouble understanding the highlighted text below. Just need some quick clarification on the right angle stuff, don't get how to draw my diagram.

    Problem:
    0.100 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 310 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 210 m/s.

    Shifty :).
     
    Last edited: Nov 17, 2009
  2. jcsd
  3. Nov 17, 2009 #2
    all this means is that the bullet rebounds parallel to its original trajectory.

    (so it is in the opposite direction)
     
  4. Nov 17, 2009 #3
    Thank you for the reply, sounds reasonable :). What confuses me is the bit of "right angles" to the original direction. Is it talking about 2 right angles adding to 180 degrees?

    Either way, horizontal gives it away I guess. I would have understood understood it better if it just said its when in the opposite direction of its original one.

    Thanks anyways :).
     
  5. Nov 17, 2009 #4
    Well I couldn't get the answer, and a different part asked me to input the direction of the rock and it's not the same to the bullet's initial direction.

    This means the bullet can't be going in the opposite direction, because in order to change the direction of the rock, the final direction of the bullet has to not be opposite (or angled to ini. dir.). :frown:

    If it doesn't go back, then where is the bullet going?
     
    Last edited: Nov 17, 2009
  6. Nov 17, 2009 #5
    You say that you entered the bullets direction to be "the same as the bullet's initial direction" and got the question wrong. Then you say that the bullet can't be going in the opposite direction because of this. How can you rule out the opposite direction based on the same direction being the wrong direction?

    Initial direction: + x-axis
    Finial direction: - x-axis

    (two completely different directions. This is very important when dealing with a vector quantity like velocity or momentum.)
     
    Last edited: Nov 17, 2009
  7. Nov 17, 2009 #6

    cepheid

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    No. "At right angles to" is just an English expression meaning "forming an angle of 90 degrees with (something)." So, the sentence should be interpreted in the same way as it would have been if it had read, "rebounds horizontally at a right angle to the original direction..."

    So, I dispute srmeier's claim that the bullet is going in the exact opposite direction (antiparallel) after the collision. I think it's going in an orthogonal (perpendicular) direction to the original.
     
  8. Nov 17, 2009 #7
    I believe cepheid has a most valid point and you should follow his logic. Apologies for any confusion.
     
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