Maximum compression of the spring

  • Thread starter aznboi855
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  • #1
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Homework Statement


360c0f2c5509be337cf879c68faf2ef4.jpg

The two blocks I and II shown above have mass of m and 2m respectively. Block II as an ideal massless spring attached to one side. When block I is placed on the spring as shown in figure (a), the spring is compressed a distance D at equilibrium. Express your answer to all parts of this question in terms of the given quantities and the physical constants.

a) Determine the spring constant of the spring.

Latter the two blocks are on frictionless horizontal surface. Block II is stationary and block I approaches with the speed of V, as shown in the figure b,c,d.

b) The spring compression is a maximum when the blocks have the same velocity. Briefly explain why is this so.

c) Determine the maximum compression of the spring during the collision.

d) Determine the velocity of the block II after the collision, when block I has seperated from the spring.

Homework Equations


F = kd
conservation of energy equation

The Attempt at a Solution


I have already solved for a), which came out to be k = mg/d.
For c), i used the conservation of energy equation and it came out like this :
PEspi + KEi = PEspf + KEf
(1/2)mvi^2 = (1/2)kx^2 + (1/2)3mvf^2

This is where i'm stuck, I know the answer comes out to be SQRT(2D/3g)Vo, but i can't get there.
 

Answers and Replies

  • #2
You need another equation. And an obvious one can be obtained by thinking about linear momentum of the blocks.
 
  • #3
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Thanks for the reply, I just checked in my book and it also said to use the conservation of momentum equation. Solving for vf, I ended up with:
vf = mvi/3m
plugging it back into the energy equation still doesn't yield the result :(, unless i'm doing it wrong somewhere.
 
  • #4
unless i'm doing it wrong somewhere.
I think you are. Just double checked it and the equations do yield the answer you mentioned in the first post
 
  • #5
11
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Nvm, I figured it out, just had some bad algebra.
 

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