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Simple Harmonic Motion and Energy Problem

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Homework Statement


In the figure, block 2 of mass 2.40 kg oscillates on the end of a spring in SHM with a period of 26.00 ms. The position of the block is given by x = (1.80 cm) cos(ωt + π/2). Block 1 of mass 4.80 kg slides toward block 2 with a velocity of magnitude 6.90 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 6.50 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

Homework Equations


E=KE+PE=0.5k*a^2 where k is spring constant and a is amplitude
KE=0.5m*v^2
T=2π*sqrt(m/k)

The Attempt at a Solution


First rearrange the expression for T:
k=(4*m*π^2)/T^2
We also have initial mechanical energy:
Ei=0.5k*a^2, where a can be read off from the given equation x = (1.80 cm) cos(ωt + π/2), that is 0.018m, and we have managed an expression for k.
Thus we get Ei=45.42286025 J
After the collision, block 1 transfers all its energy to block 2, thus the total mechanical energy increases by the the KE of block 1:
KE1=0.5*M*v^2=0.5*4.8*6.9^2=114.264 J
Now the new mechanical energy is
Ef=Ei+KE1=159.6758603 J
Finally we use the equation E=0.5k*a^2 to solve for the new amplitude and get
a=0.0477 m
This is not the correct answer and I did not use all the information given(t=6.5ms and ø=π/2).
Please point out what I have missed thanks.
 

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Answers and Replies

  • #2
BvU
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Hi,

With a a nickname like that you want to try and outsmart the composer of the exercise. Think what an inelastic collision entails and find a minimum path through the glob of info and relationships. One of your relationships contains all you need for the anser - and involves only a few variables.

[edit] Sorry, misread the aim of the exercise -- when it's late it seems that it is possible to read 'frequency' instead of 'amplitude' :oldconfused: -- . Fortunately you haven't been distracted and still are on the right track in #11 thanks to haru and ehild
 
Last edited:
  • #3
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Hi,

Witha a nickname like that you want to try and outsmart the composer of the exercise. Think what an inelastic collision entails and find a minimum path through the glob of info and relationships. One of your relationships contains all you need for the anser - and involves only a few variables.
Hi
One day I shall prove to the world that I am so much smarter than the the composer of this problem, its just the composer might not live to witness this brilliant moment.

Complete Inelastic collision means all the KE of block 1 is transferred to the new two-blocks system is that right? If so I think my equations follow this
 
  • #4
ehild
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Complete Inelastic collision means all the KE of block 1 is transferred to the new two-blocks system is that right? If so I think my equations follow this
No. Part of the KE of the whole system transforms to non-mechanical energy, like heat. What does complete inelastic collision mean?
 
  • #5
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No. Part of the KE of the whole system transforms to non-mechanical energy, like heat. What does complete inelastic collision mean?
It means the greatest loss in KE, m1*Vi=(m1+m2)*Vf
 
  • #6
haruspex
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It means the greatest loss in KE, m1*Vi=(m1+m2)*Vf
Yes, except that in this question both have an initial velocity.
 
  • #7
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Yes, except that in this question both have an initial velocity.
Does it mean that the actual velocity after collision would be Vf+(velocity of block 2)?
 
  • #8
haruspex
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Does it mean that the actual velocity after collision would be Vf+(velocity of block 2)?
No, it means that you add up the initial momenta and equate that sum to the final momentum.
 
  • #9
ehild
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It means the greatest loss in KE, m1*Vi=(m1+m2)*Vf
What are Vi and Vf? In general, both blocks have their velocities before and after the collision. How are the velocities related after the completely inelastic collision? What was the velocity of block 2 just before the collision?
 
  • #10
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No, it means that you add up the initial momenta and equate that sum to the final momentum.
Right, so M1*V1+M2*V2=(M1+M2)*Vf
 
  • #11
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What are Vi and Vf? In general, both blocks have their velocities before and after the collision. How are the velocities related after the completely inelastic collision? What was the velocity of block 2 just before the collision?
M1*V1+M2*V2=(M1+M2)*Vf where V1 is given and V2 can be calculated from the equation of X(t). Now I am able to get Vf, and Vf=a*w right?
 
  • #12
BvU
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Right
 
  • #13
ehild
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M1*V1+M2*V2=(M1+M2)*Vf where V1 is given and V2 can be calculated from the equation of X(t). Now I am able to get Vf, and Vf=a*w right?
What is a? Are the two block stick together after the collision?
You can calculate Vf, and then you can determine the whole energy of the system. For that, you need to know the elastic energy of the spring at the moment of collision.
 
  • #14
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Right
What is a?
You can calculate Vf, and then you can determine the whole energy of the system. For that, you need to know the elastic energy of the spring at the moment of collision.
a is the amplitude.

K=0.5 • (M1+M2) • Vf
U=0.5k • [0.018cos(ωt + π/2)]^2
K+U=E=0.5 • k • a^2
 
  • #15
ehild
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If the blocks move together the frequency of the SHM will change.
Determine both V1 and V2 before the collision, and the common Vf after it. What is the energy of the spring at the moment of collision?
What is the new ω when the blocks move together?
 
  • #16
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If the blocks move together the frequency of the SHM will change.
Determine both V1 and V2 before the collision, and the common Vf after it. What is the energy of the spring at the moment of collision?
What is the new ω when the blocks move together?
The new w is
w=sqrt(k/M1+M2)
is that right?
 
  • #17
ehild
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The new w is
w=sqrt(k/(M1+M2))
is that right?
if you add a pair of parentheses ...
 

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