Two objects attached with a spring

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In summary, the conversation discusses the problem of finding the maximal acceleration that can be reached by the second object on a frictionless surface. The objects involved in the problem have masses of 0.9 kg and 1.6 kg, respectively, and are held by a person while a spring is shrunk to a distance of 10 cm. The first object is then released, and when it reaches equilibrium, the second object is released. The relevant equations and attempt at a solution are also discussed, with emphasis on finding the velocity of the center of mass and considering the relative velocities of the masses in order to calculate the acceleration of the second object.
  • #1
giokrutoi
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1

Homework Statement


on frictionless surface there are two objects m1 = 0.9 kg m2 =1.6 kg
at first these two are held by person an spring is shrined L = 10 cm . After that first objet is released and when this object is in it's equilibrium moment another object is released . Find maximal acceleration that can be reached by second objet

Homework Equations

The Attempt at a Solution


mv2/2= kx2/2 m1v1/m1+m2 = v2 v2 is the velocity of the center of mass (m1+m2)a=kx1 k1 is found by conservation of energy mv12 = (m1+m2)v22/2 + kx12/2 are my conclusion write?
 
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  • #2
giokrutoi said:

Homework Statement


on frictionless surface there are two objects ##m_1 = 0.9 ## kg, ##m_2 =1.6 ## kg

at first these two are held by a person and the spring is shrinked over a distance ##L = 10 ## cm.

After that, the first object is released and when this object is in its equilibrium position the other object is released .

Find the maximum acceleration that can be reached by the second object.

Homework Equations



There is nothing here ?

The Attempt at a Solution


## mv^2/2= kx^2/2 \qquad ## In general, or at some specific point ?
## m_1v_1/m_+m_ = v_2## ##v_2## is the velocity of the center of mass Can't be: dimensions do not match

##(m_1+m_2)a=kx_1## k1 is found by conservation of energy Don't know. What is a ?

##mv_1^2 = (m_1+m_2)v_2^2/2 + kx_1^2/2## How do you derive that ?

are my conclusion right ?

I don't think so. Perhaps you want to work more systematically:
  • name all variables and make clear how you derive the unknowns from the knowns
  • check dimensions for every equation
  • clearly list all the relevant equations
 
Last edited:
  • #3
BvU said:
I don't think so. Perhaps you want to work more systematically:
  • name all variables and make clear how you derive the unknowns from the knowns
  • check dimensions for every equation
  • clearly list all the relevant equations
relative equations: Kx12/2=m1v12/2
 
  • #4
oh I forgot k is also given
 
  • #5
BvU said:
I don't think so. Perhaps you want to work more systematically:
  • name all variables and make clear how you derive the unknowns from the knowns
  • check dimensions for every equation
  • clearly list all the relevant equations
 

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  • #6
I decipher
  • now ##(m_1+m_2)a = kx_1\quad(2)##
What is ##x_1## and what is ##a## ?

Then: "I guess..."
No need to guess: you already calculated the velocity of the center of mass.
But it seems to me you are still in the dark about where you are going and how you want to go there ?
 
  • #7
BvU said:
I decipher
  • now ##(m_1+m_2)a = kx_1\quad(2)##
What is ##x_1## and what is ##a## ?

Then: "I guess..."
No need to guess: you already calculated the velocity of the center of mass.
But it seems to me you are still in the dark about where you are going and how you want to go there ?
a is acceleration of center of mass and I think that initial energy - energy of the center of mass should be potential energy of spring which has it's new x1
 
  • #8
And how would the center of mass be able to undergo acceleration ?
 
  • #9
BvU said:
And how would the center of mass be able to undergo acceleration ?
it should be because net force on system isn't zero
 
  • #10
it's motion should be linear but It can also gain acceleration
 
  • #11
giokrutoi said:
it should be because net force on system isn't zero
What external forces are there on the system?
 
  • #12
giokrutoi said:
relative equations: Kx12/2=m1v12/2
What, exactly, do x1 and v1 represent there? (There is no value in being able to quote equations if you cannot say what the terms in it represent.)
 
  • #13
haruspex said:
What external forces are there on the system?
ok guess that only external force is kx which cause objects to move with acceleration
but it can also be that I don't understand their motion too well
 
  • #14
haruspex said:
What, exactly, do x1 and v1 represent there? (There is no value in being able to quote equations if you cannot say what the terms in it represent.)
this formula meant to be in general but in this cases all variables are given
 
  • #15
I can't figure out how do they move and I don't have to find system acceleration but maximal of the second objet
 
  • #16
Can you see that ##kx## is NOT an external force ? According to Isaac the force one exercises on the other is opposite and equal to the force the other exercises in the one. So for the motion of the center of mass this has no significance !
 
  • #17
giokrutoi said:
this formula meant to be in general but in this cases all variables are given
You have not understood what I asked. Even though you have not stated it, I'm happy to guess that x1 is a spring compression (or extension) and that v1 is a velocity. But what compression and what velocity? Hint: it does not give you the velocity of an attached mass when the extension is x1.
As BvU notes, in the system consisting of the two masses and the spring, forces between those entities are internal to that system. What forces act on that system from outside it?
 
  • #18
i don't understand what forces are from outside
 
  • #19
giokrutoi said:
i don't understand what forces are from outside
Exactly, there aren't any. So would the common mass centre accelerate?
 
  • #20
you're write I understand what you're saying but. How can I calculate m2 acceleration
 
  • #21
maybe m1a1=m2a2
 
  • #22
so center has no acceleration
 
  • #23
giokrutoi said:
you're write I understand what you're saying but. How can I calculate m2 acceleration
If you find the velocity of the common mass centre then you can consider the motions of the masses relative to that. When m2 is at its maximum acceleration, what can you say about the acceleration of m1 and the relative velocities of the masses?
 
  • #24
maybe relative velocity of m1 due Tue center of mass is v-vcenter and we can calculate v2 which is velocity of m2 from cocervation of energy
 
  • #25
giokrutoi said:
maybe relative velocity of m1 due Tue center of mass is v-vcenter
Sorry, but I have no idea what you mean. Maybe it will make sense if you post some equations.
 
  • #26
what about this
 

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  • #27
Are we going full circle ?
##mv^2/2= kx^2/2 \qquad ## was in post 1 as well. My question was "In general, or at some specific point ?"

Now you post (I deduce) $$m_1 v_1^2/2 = (m_1+m_2) v_{\rm\, c.o.m.}^2 /2 + m_2 v_2^2/2 $$
What relevant equation can possibly lead to this ? Any idea yourself what you mean with these variables ? Or at what time ?

Why don't you start over and present a complete problem statement with a complete list of symbols/variables/given/known data (plus explanation what they mean) and the relevant equations (not that many anyway). Very helpful and instructive. Just coming up with ever more new "what about this" isn't going to solve this exercise for you !
 
  • #28
BvU said:
Are we going full circle ?
##mv^2/2= kx^2/2 \qquad ## was in post 1 as well. My question was "In general, or at some specific point ?"

Now you post (I deduce) $$m_1 v_1^2/2 = (m_1+m_2) v_{\rm\, c.o.m.}^2 /2 + m_2 v_2^2/2 $$
What relevant equation can possibly lead to this ? Any idea yourself what you mean with these variables ? Or at what time ?

Why don't you start over and present a complete problem statement with a complete list of symbols/variables/given/known data (plus explanation what they mean) and the relevant equations (not that many anyway). Very helpful and instructive. Just coming up with ever more new "what about this" isn't going to solve this exercise for you ![/ to you QUOTE]
i have question to you
do you know Russian
 
  • #29
do you know Russian
 
  • #30
?
 
  • #31
hey I guess I found solution I realized that I had to choose reference frame on center of mass
are my conclusion write
 

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  • #32
giokrutoi said:
hey I guess I found solution I realized that I had to choose reference frame on center of mass
are my conclusion write
Those images are simply unreadable. Please make the effort to type your equations in. Use parentheses, superscript (X2) and subscript (X2) as appropriate.
Diagrams would be good, but draw them in black and try to get the lighting even on the page.
 
  • #33
I rewrote it
is it readable?
 

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  • #34
giokrutoi said:
I rewrote it
is it readable?
Yes, this readable.
Check the signs in the first equation on the right hand page.
Some places you write "due to centre of mass" where I think you mean "relative to centre of mass".
Other than that, you seem to be on the right track now.
 
  • #35
I think all signs are write
in relation to center of mass I think i have to subtract velocity of center of mass
 

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