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Inellastic collisison

  • Thread starter notabigfan
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  • #1
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I have struggled with this question. I don't understand what do especially that the question says inelastic but there is a spring.

A mass of 12 kg is sliding along a frictionless surface at a speed of 5 m/s.
It collides perfectly inelastically with a stationary mass of 24 kg which is attached to
a spring. If the spring constant is 225 N/m, where are the stuck-together masses 3
seconds after the collision?

I know momentum before collision is 12*5= 60

But I don't know what to do from here ... I know momentum has to be conserved but not energy but I am having trouble understanding what can the next step be.

I hope someone can help me.
 
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  • #2
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You're correct that momentum is conserved, but not energy during the collision, but after they collide, the energy of the spring plus the kinetic energy of the stuck-together masses is conserved.
Calculate the velocity of the stuck-together masses from conservation of momentum and go from there.
 
  • #3
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Well, so (m1+m2)v2 = initial momentum but what happens to energy? or shall I just say that v2 will stay the same and then multiply it by 3 seconds to get the distance ? Why does the question give info about the spring then?
 
  • #4
haruspex
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shall I just say that v2 will stay the same ?
No, certainly not. Momentum conservation gives you the speed immediately after collision. From there it's just a mass compressing a spring. Taking it to be perfectly elastic, that's SHM. Write out the SHM equation and plug in the initial conditions.
 
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  • #5
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No, certainly not. Momentum conservation gives you the speed immediately after collision. From there it's just a mass compressing a spring. Taking it to be perfectly elastic, that's SHM. Write out the SHM equation and plug in the initial conditions.
Thank you ! For some reason it didn't click for me that SHM occurs next. I have one more problem though. I find it very challenging to find the phase constant (Φ0) can you please tell me how to find it in this case. It looks like zero to me because the spring was neither streched nor compressed. Is that right?
 
  • #6
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The velocity you calculated was the max velocity for the SHM. From that and the angular frequency of the oscillator, you should be able to find the amplitude and the phase angle.
 
  • #7
haruspex
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Thank you ! For some reason it didn't click for me that SHM occurs next. I have one more problem though. I find it very challenging to find the phase constant (Φ0) can you please tell me how to find it in this case. It looks like zero to me because the spring was neither streched nor compressed. Is that right?
The phase will depend on whether you write the canonical SHM equation using sine or cosine. Also, t=0 is defined here at the lowest point, not at the midpoint.
 
  • #8
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The phase will depend on whether you write the canonical SHM equation using sine or cosine. Also, t=0 is defined here at the lowest point, not at the midpoint.
Alright, I usually use the expression x=Acos(ωt+ø) How do I find the phase constant in this case?
 
  • #9
haruspex
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haruspex said:
Also, t=0 is defined here at the lowest point, not at the midpoint.
Cancel that - I was confusing two threads.
Alright, I usually use the expression x=Acos(ωt+ø) How do I find the phase constant in this case?
Where it hits the spring will be the midpoint of the oscillation, right? So that is x = 0. If you take that as t=0, you have an equation regarding the phase. There is still some ambiguity, which you resolve by thinking about which direction you are taking as positive, and what the sign of the velocity is at t = 0.
 

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