How to Solve a Problem Involving a Collision with Right Angles

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The discussion centers on a physics problem involving a collision between a bullet and a stone on a frictionless surface. Participants clarify the meaning of "right angles" in the context of the bullet's rebound direction, emphasizing that it indicates a perpendicular trajectory rather than a direct reversal. Confusion arises regarding the bullet's final direction and its relationship to the stone's movement, with some asserting that the bullet cannot simply go in the opposite direction to affect the stone. The importance of understanding vector quantities like velocity and momentum is highlighted, as different directions significantly impact the outcome. Ultimately, the conversation seeks to clarify the interpretation of the problem's language and its implications for the collision dynamics.
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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 :).
 
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all this means is that the bullet rebounds parallel to its original trajectory.

(so it is in the opposite direction)
 
srmeier said:
all this means is that the bullet rebounds parallel to its original trajectory.

(so it is in the opposite direction)

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 :).
 
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?
 
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Shifty said:
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...

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.)
 
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Shifty said:
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?

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.
 
I believe cepheid has a most valid point and you should follow his logic. Apologies for any confusion.
 
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