How to Solve a Problem Involving a Collision with Right Angles

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Homework Help Overview

The problem involves a collision scenario where a bullet strikes a stone resting on a frictionless surface, with specific attention to the directions of motion before and after the collision. The subject area includes concepts of momentum and vector analysis in physics.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the interpretation of the bullet's rebound direction, questioning whether it is parallel or at right angles to the original trajectory. There are attempts to clarify the meaning of "right angles" in the context of the problem and its implications for the directions of the bullet and the stone.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the bullet's direction after the collision. Some guidance has been offered regarding the meaning of "right angles," but there is no explicit consensus on the bullet's final direction relative to its initial trajectory.

Contextual Notes

Participants are grappling with the implications of the problem's wording, particularly regarding the directions involved in the collision and how they relate to vector quantities like velocity and momentum. There is a noted confusion about the relationship between the bullet's direction and the stone's movement.

Shifty
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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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