When and how do I use sin and cos regarding 2-D collisions?

Click For Summary

Homework Help Overview

The discussion revolves around the application of sine and cosine in the context of two-dimensional collisions, specifically regarding the components of velocity in different directions. Participants are exploring when to use these trigonometric functions based on the orientation of angles and the labeling of compass directions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the relationship between angle measurement and the use of sine and cosine in determining vector components. Questions are raised about why sine is used for certain components in specific scenarios and how to appropriately label directions like East and North.

Discussion Status

Some participants are providing insights into the trigonometric definitions related to angles, while others are expressing confusion about the application of these concepts in the context of the problem. There is an ongoing exploration of how different phrasing of angles affects the use of sine and cosine.

Contextual Notes

Participants are discussing the relevance of specific angle measurements and the implications of using different compass directions in their calculations. There is a mention of diagrams that may aid in visualizing the problem, but no consensus has been reached on the best approach to take.

JohnMC
Messages
4
Reaction score
0
I think what I'm doing is overkill because all I want to know is when to use sin and cos in the components regarding (isolated) two-dimensional collisions. I'm just showing more content because it might help for context. Also, there are more steps in order to find the final velocity which is why you can see some steps cut off, but I didn't bother posting more pictures of it because I thought it was irrelevant from then on as I am only looking for why sin and cos are used.

Homework Statement


On QUESTION #1 III., why is sin used in the East component and cos used in the North component, when in QUESTION #2 II., a different but ultimately similar question, the sin and cos are switched? Please feel free to click through the images in order for context. Also, how do I know if I should label the components East and North? Why not other compass directions? I hope I'm making some sense :P

QUESTION #1
I. http://bayimg.com/image/oajpoaaca.jpg
II. http://bayimg.com/image/aajajaacb.jpg
III. http://bayimg.com/image/bajadaacb.jpg
IV. http://bayimg.com/image/cajaaaacb.jpg
V. http://bayimg.com/image/cajaoaacb.jpg


QUESTION #2
I. http://bayimg.com/image/cajcpaacb.jpg
II. http://bayimg.com/image/dajcmaacb.jpg
III. http://bayimg.com/image/eajcjaacb.jpg

Homework Equations


N/A


The Attempt at a Solution


Looked at sine laws, properties of triangles, and so far I have found nothing I can apply it to. Please forgive me if I'm asking such a simple question.
 
Last edited by a moderator:
Physics news on Phys.org
it is because of how the angle is measured.

sin =opp/hyp

cos=adj/hyp

in 3, the east is opposite to the angle, so the east component is hyp*sin, similarly the north is adjacent to the angle. Understand?
 
I would like to say that I understand what you mean, but sadly I don't. Although I think I found out a way how and when to use sin and cos. Wordings such as [12.0 E of N] use East for sin and North for cos; while [12.0 N or E] use East for cos and North for sin. Was this what you meant by "because of the how angle is measured"? My apologies for failing to understand
 
JohnMC said:
I would like to say that I understand what you mean, but sadly I don't. Although I think I found out a way how and when to use sin and cos. Wordings such as [12.0 E of N] use East for sin and North for cos; while [12.0 N or E] use East for cos and North for sin. Was this what you meant by "because of the how angle is measured"? My apologies for failing to understand

See this diagram:

http://img16.imageshack.us/img16/3596/diagrm.jpg

Lets call DAC=α (alpha) and CAB=β

sin=opp/hyp and cos=adj/hyp.

Consider triangle ADC,

cos \alpha =\frac{adjacent}{hypotenuse}= \frac{AD}{AC}

sin\alpha = \frac{opposite}{hypotenuse}= \frac{DC}{AC}

Consider triangle ABC,

sin \beta = \frac{opposite}{hypotenuse}=\frac{BC}{AC}

cos \beta=\frac{adjacent}{hypotenuse}= \frac{AB}{AC}

Do you understand what I meant by how the angle is measured now?
 
Last edited by a moderator:
Thanks so much! I already knew the properties of sin and cos as shown in your second post, but it was just the thought of using East and North as a property of a triangle in 2-D collisions and matching it with sin and cos that confused me. I read your first post again and I finally figured it out. Again, thank you :)
 

Similar threads

Replies
8
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 2 ·
Replies
2
Views
36K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
28
Views
4K