Finding the mass of Tarzan?

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  • #26
PeroK
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So.
##E_1##=##m##g##h##+½mv12
##E_2## = ½(m+50) v22

then
##E_1## = ##E_2##
No.

##E_1 = Mgh_1 = \frac 1 2 Mv_1^2##

##E_1 \ne E_2##

##E_2 = (M + m)gh_2 = \frac 1 2 (M+m)v_2^2##

Where ##M## and ##m## are the masses of Tarzan and Jane.

I suspect this problem is a lot harder than the ones you are used to. Solving this problem involves a strategy, not just plugging numbers into an equation!
 
  • #27
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No.

##E_1 = Mgh_1 = \frac 1 2 Mv_1^2##

##E_1 \ne E_2##

##E_2 = (M + m)gh_2 = \frac 1 2 (M+m)v_2^2##

Where ##M## and ##m## are the masses of Tarzan and Jane.

I suspect this problem is a lot harder than the ones you are used to. Solving this problem involves a strategy, not just plugging numbers into an equation!
I am sorry, I am really not used to these questions... I believe I need more practices!
 
  • #28
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No.

##E_1 = Mgh_1 = \frac 1 2 Mv_1^2##

##E_1 \ne E_2##

##E_2 = (M + m)gh_2 = \frac 1 2 (M+m)v_2^2##

Where ##M## and ##m## are the masses of Tarzan and Jane.

I suspect this problem is a lot harder than the ones you are used to. Solving this problem involves a strategy, not just plugging numbers into an equation!
So...
(##M##+##m##)##v_2##=##Mv_1##

Without Jane,
##Mg##(1-##cos##53)=½ ##Mv_1##2

With Jane,
##(M+m)g##(1-##cos##37)=½ ##(M+m)v_2##2
 
  • #29
PeroK
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So...
##(M+m)v_2=Mv_1##

Without Jane,
##Mg##(1-##cos##53)=½ ##Mv_1##2

With Jane,
##(M+m)g##(1-##cos##37)=½ ##(M+m)v_2##2
That's the idea.

Although the length of the vine should be in there somewhere.
 
  • #30
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That's the idea.

Although the length of the vine should be in there somewhere.
Ah dang... :/
 
  • #31
PeroK
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Ah dang... :/
Try leaving it as ##h_1, h_2## for now. That's simpler. We can sort out these in terms of angles later.
 
  • #32
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Try leaving it as ##h_1, h_2## for now. That's simpler. We can sort out these in terms of angles later.
Like this?

Without Jane,
##Mg##(##h_1##)=½ ##Mv_1##2

With Jane,
##(M+m)g##(##h_2##)=½ ##(M+m)v_2##2
 
  • #33
PeroK
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Like this?

Without Jane,
##Mg##(##h_1##)=½ ##Mv_1##2

With Jane,
##(M+m)g##(##h_2##)=½ ##(M+m)v_2##2
Yes. Now you need the equation for ##v_1## and ##v_2## you got from conservation of monentum.
 
  • #34
PeroK
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Yes. Now you need the equation for ##v_1## and ##v_2## you got from conservation of monentum.
PS note that the masses cancel in those equations, so that simplifies things a bit.
 
  • #35
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PS note that the masses cancel in those equations, so that simllifies things a bit.
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
 
  • #36
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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.
 
  • #37
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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
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so... V2=[m1/(m1+m2)]v1

(Sorry, I dont know how to use fraction here...)
This!
 
  • #39
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This!
Like this?
##√2####gh_2## =[M/(M+m)] ##√2####gh_1##
 
  • #40
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(##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
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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
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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
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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
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It's just trigonometry.
but why is it L - L cos θ ?
Sorry, I just want to understand more
 
  • #47
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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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