Two Masses Connected by Spring Against Wall

  • Thread starter Gill987
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  • #1
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Homework Statement



A system consists of two blocks each of mass M, connected by a spring of force constant k. The system is initially shoved against a wall so that the spring is compressed a distance D from its original uncompressed length. Floor is frictionless. The system is now released with no initial velocity.

I had to find the velocity of the right hand block when the spring expands back to its equilibrium position which is the instant the left block leaves the wall (but still has 0 velocity), which was D√(k/M)

V center of mass was (D/2)√(k/M)

a. Eventually as the system moves to the right the spring will compress a maximum S. What will S be equal to?

b. Using energy, explain why S should be less than, greater than, or equal to D.

Homework Equations



F=-kx

F= -dU/dx

U=(1/2)kx^2

The Attempt at a Solution



I don't have much, was trying center of mass equations but things would cancel out. Was trying to use conservation of energy, but I figured the left block now has a velocity and the right block has a different velocity, with the left velocity being greater in order for the spring to be compressed, and I have no idea what the velocities are.

Any help would be appreciated, thanks.
 

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Answers and Replies

  • #2
gneill
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Hi Gill987; Welcome to Physics Forums.

It looks like moving to the center-of-mass frame of reference would be a good idea.

You have calculated the speed of the center of mass and the speed of the right-hand block. The initial speed of the left-hand block is also known, so you should be able to write the speeds of both blocks in the center of mass frame of reference.

In the center of mass frame the KE will trade with the spring PE as the system evolves over time. Fortunately the initial conditions you have have the spring at its natural length...
 
  • #3
TSny
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Hello Gill987. Welcome to PF!

How do the velocities of the two blocks compare at the instant the spring is maximally compressed? If the left block is traveling faster than the right block, then would the spring have reached its max compression?
 
  • #4
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Thanks for responding!

At the moment of max compression, the velocities of both blocks are equal. I tried using conservation of energy, and got (1/2)mv^2 = (1/2)mV^2 + (1/2)mV^2 + (1/2)ks^2, where V is the after velocity of the blocks and v is the before velocity of the right hand block, solving for s and plugging in I got √(D^2 -2(m/k)V^2), but I don't know V. Would I get it by changing the reference frame, because I'm trying that but not quite getting good results...
 
  • #5
TSny
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Can you determine V from your knowledge of the center of mass velocity?
 
  • #6
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I found velocity of center of mass when both blocks are moving at the same speed to get V, set that equal to (D/2)√(k/m), solved for V, plugged in, and got D/√2 = s.

Heh, I actually started typing this and then saw your reply. I hope this is right.
 
  • #7
TSny
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Looks good!
 
  • #8
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Awesome!

For the second part, I know that the energy when it was initially compressed was (1/2)kD^2, and when it compressed again energy was mV^2 + (1/2)ks^2, so can I say total energy came from D initially, and then came from both V and s, so s has a smaller value? Or how could I word that better?

(Thanks for all your help)
 
  • #9
TSny
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Yes, that sounds good. Total energy is conserved. Initially it was all PE and later, when spring is compressed by S, there is both KE and PE. So, final PE is less than initial PE.
 
  • #10
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Great, thanks for all your help!
 

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