What is the speed of each particle?

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The discussion revolves around a physics problem involving two masses connected by a spring, where the goal is to calculate the speed of each particle when they reach a specific configuration. The provided solution indicates that the speed of mass m1 can be expressed as v1 = √(k/3m1) ⋅ d. Participants express confusion over the problem's phrasing, noting that it asks for velocity rather than time, which complicates the approach. There is also mention of conservation laws and the need to consider forces at intermediate positions. The conversation highlights the importance of clear problem statements and the challenges posed by truncated images.
OierL
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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...
 

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You need to show what you've tried, even if it didn't work out.
 
Have you at least looked into applicable conservation laws?
 
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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.
 

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