Exploring the Conservation of Momentum and Energy in a Tarzan Swing Collision

  • Thread starter bieon
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In summary, we have a problem involving a pendulum with two masses, where the initial angles and masses are given. The length of the vine remains constant in both the initial and final positions. Conservation of energy and momentum can be applied to the swing down from the starting point to the bottom, the collision as Tarzan grabs Jane (which is a perfect inelastic collision), and the swing up to the final height. The velocity can be calculated using the formula V2=[m1/(m1+m2)]v1.
  • #36
bieon said:
Ok, so I put them in terms of velocity now right?
M##g##(##h_1##)=½ M##v_1##2
2##g####h_1##= ##v_1##2

With Jane,
(M+m)##g##(##h_2##)=½ (M+m)##v_2##2
2##g####h_2##= ##v_2##2
That's good. Now use the relationship between ##v_1## and ##v_2## that you got many posts ago.
 
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  • #37
PeroK said:
That's good. Now use the relationship between ##v_1## and ##v_2## that you got many posts ago.
(##M##+##m##)##√2####gh_2##=##M√2####gh_1##
This?
 
  • #38
bieon said:
so... V2=[m1/(m1+m2)]v1

(Sorry, I don't know how to use fraction here...)

This!
 
  • #39
PeroK said:
This!
Like this?
##√2####gh_2## =[M/(M+m)] ##√2####gh_1##
 
  • #40
bieon said:
(##M##+##m##)##√2####gh_2##=##M√2####gh_1##
This?
It looks like you've done it. Can you finish it off by expressing height in terms of angle.
 
  • #41
PeroK said:
It looks like you've done it. Can you finish it off by expressing height in terms of angle.
I believe I can! Thank you so much! I am sorry for the trouble!
 
  • #42
PeroK said:
It looks like you've done it. Can you finish it off by expressing height in terms of angle.
Uhm I think I had trouble, I know the general concept of l-lcos but I am not sure where it came from...
 
  • #43
bieon said:
Uhm I think I had trouble, I know the general concept of l-lcos but I am not sure where it came from...

It's just trigonometry.
 
  • #44
PeroK said:
It's just trigonometry.
but why is it L - L cos θ ?
Sorry, I just want to understand more
 
  • #47
In summary: $${\left(\frac{m}{m+50}\right)}^2=\frac{h_2}{h_1}$$. We do however perhaps need to question a little the physics of this situation since momentum is conserved in the absence of an external force. If the transfer of momentum in the collision is not instantaneous but rather over a finite time instant Δt , g could act if for a very brief instant. Collision energy in a perfectly inelastic collision is considered lost - here it's possible that some collision energy is not dissipated but rather absorbed by being converted to PE. Jane benefits because Tarzan does not make an instant impact!
 
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