Proving Acceleration of A ≤ Acceleration of B

In summary, the two masses, MA = 1 kg and MB = 2 kg are connected by a massless spring. A force of 4 N acts on the 2 kg mass. Assuming that the spring is stretchable, we can solve for the net force on the system by taking the difference between the two equations. The net force on the system is 2-kx/3, which is negative. This tells us that it is not necessary that the acceleration of A always be less or equal to the acceleration of B.f
  • #1
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Homework Statement


Question :
two masses MA = 1 kg and MB = 2 kg are connected by a massless spring as shown in figure. A force of 4 N acts on the 2 kg mass.
http://www.vidyamandir.com//displayImageServlet?appletImgId=8a7b221f1fcc7894011ff9cccfe412f0 [Broken]


Is it necessary that the acceleration of A always be less or equal to the acceleration of B?
If yes,then prove it...If no then also prove it,please.



2. The attempt at a solution
I thought this way..Working in centre of mass's frame,both blocks seem to oscillate about it.So,at some instant both will have same accelerations w.r.t. Centre of mass then, using pseudo force concept the net acc. of the block be will get greater than that of block A..
Correct me if i am wrong...
 
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  • #2


What forces act on A? On B?
 
  • #3


Only one force acts and that on B only.
 
  • #4


Only one force acts and that on B only.
There's only one external force acting on the two-block system, but to find the acceleration of each block consider each block separately.

What forces act on each? Compare the expressions for the net force on each, then see if you can draw any conclusions.
 
  • #5


But, the forces on each keep varying??

We can have the equations,
4-kx=2aB
kx=aA

k being the unknown spring constant,and x is extension in it..

If we can prove,
a(B)-a(A)=2-kx/3 can be negative then its done...
 
  • #6


If you are considering block B alone, why do you have twice the mass of B?
 
  • #7


But, the forces on each keep varying??
Right, the spring force can be almost anything. That's the key.

We can have the equations,
4-kx=2aB
kx=aA
OK. So then:
aB = 2 -kx/2
aA = kx

Can you think of a value of kx that makes aA > aB ?
 
  • #8


Ya, kx=1N?

So, what does this imply?
 
  • #9


Ya, kx=1N?
No, that value makes aB = 3/2 & aA = 1. Choose another. (What value would make aB = 0?)
 
  • #10


Earlier i accidentally took its opposite

Taking,kx=4N,will make it..
 
  • #11


Taking,kx=4N,will make it..
OK. So what does that tell you?
 
  • #12


It tells that it is not necessary that the acceleration of A always be less or equal to the acceleration of B...But my fellows says the opposite?He says if 1 force is pulling such a system then acc. of the block A can't be greater than that of the block on which the force is applied i.e.,block B??
 
  • #13


It tells that it is not necessary that the acceleration of A always be less or equal to the acceleration of B...
OK.
But my fellows says the opposite?He says if 1 force is pulling such a system then acc. of the block A can't be greater than that of the block on which the force is applied i.e.,block B??
Ask him why he thinks that.

No information is given about how much the spring is stretched, so nothing stops us from imagining a situation where the stretch is as much as we like.
 
  • #14


Thank you very very much Sir.I'm very satisfied with your valuable views...Thanks
 

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