What is the speed of each particle?

Click For Summary

Homework Help Overview

The problem involves two masses, m1 and m2, connected by a spring, with m1 being twice the mass of m2. The setup includes a frictionless environment and a specific initial angle of the spring. The goal is to determine the speed of each mass when they reach a certain position.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of conservation laws and the interpretation of the problem's phrasing regarding the search for velocity instead of time. There are attempts to clarify the relationship between the masses and their velocities.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of the problem and questioning the assumptions made in the original post. Some guidance has been offered regarding the relevance of conservation laws, but no consensus has been reached on a specific approach.

Contextual Notes

There are indications that the problem may be poorly phrased, leading to confusion about the expected outcome. Additionally, the original post references a solution that may not align with the participants' interpretations of the problem.

OierL
Messages
5
Reaction score
0

Homework Statement


Both m1 and m2 (m1=2m2) masses can slide without friction over parallel and rigid bars that are placed at a distance d from each other. A spring with elastic constant k and with zero natural length connects both masses. The system is placed on a table. The system is released when it is in the state (1) shown in the figure (i.e. when the spring is at 45◦ with the bars). Calculate, when the bodies arrive to state (2) in the figure:
upload_2018-12-21_19-28-50.png
The solution is: v1 =√(k/3m1) ⋅ d

Homework Equations


p=m⋅v
F(spring)=-k⋅x
Ek=1/2⋅m⋅v^2

The Attempt at a Solution


I don't have any idea of how to do this problem... I've been trying to solve it for hours. Please, help!

I have computed that X(enter of mass)=2d/3⋅tan(45º) and that v2=2⋅v1
I don't know how to continue...
 

Attachments

  • upload_2018-12-21_19-26-11.png
    upload_2018-12-21_19-26-11.png
    36.8 KB · Views: 425
  • upload_2018-12-21_19-26-46.png
    upload_2018-12-21_19-26-46.png
    36.8 KB · Views: 377
  • upload_2018-12-21_19-28-50.png
    upload_2018-12-21_19-28-50.png
    5 KB · Views: 443
Last edited:
Physics news on Phys.org
You need to show what you've tried, even if it didn't work out.
 
Have you at least looked into applicable conservation laws?
 
  • Like
Likes   Reactions: jbriggs444
OierL said:
I don't have any idea of how to do this problem..
Conservation of work typically doesn't help in finding a time. Think about forces when in some intermediate position.
 
haruspex said:
Conservation of work typically doesn't help in finding a time.
I concur and wholeheartedly agree. However, couched beside the figure in the original post is that text: "The solution is: v1 =√(k/3m1) ⋅ d", so it would appear that they are looking for the velocity of m1 "when" the bodies are positioned as shown in the second figure. The problem is perhaps phrased a tad unfortunately; They are looking for a velocity rather than a time.
 
gneill said:
I concur and wholeheartedly agree. However, couched beside the figure in the original post is that text: "The solution is: v1 =√(k/3m1) ⋅ d", so it would appear that they are looking for the velocity of m1 "when" the bodies are positioned as shown in the second figure. The problem is perhaps phrased a tad unfortunately; They are looking for a velocity rather than a time.
Ah yes, the all-important comma. Eats, shoots and leaves.
Unfortunately the images meant to show the whole question are truncated.

Thanks!
 
haruspex said:
Ah yes, the all-important comma. Eats, shoots and leaves.
:smile: Also agreed. Well said.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
29
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
16
Views
5K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
2K