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

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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.
  • #1
bieon
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Homework Statement
Jane thinks that Tarzan has been gaining too much mass. However, he doesn’t want to tell her what he weighs. One day Tarzan is swinging in the forest. He starts from rest from one side at an angle of 53° from the vertical. He swings down and grabs Jane at the bottom of the swing. They swing together on the other side up to a maximum angle of 37 °. Jane knows that her mass is 50 kg. Calculate Tarzan’s mass. (Ans.120.8kg)
Relevant Equations
Refer Below.
My Attempt So Far:

1) Drawing a diagram based on the question.

Tarzan.png

Diagram 1: Pendulum Diagram Based On Question

2) List down pieces of information found in the question.

θ1=53°
θ2=37°
m1=m1
m2=m1+50kg
v
i=0

3) Since 53°+37°=90°, I redraw another diagram making it into a right angle triangle in order to find the length of the vine.
Since the length of the vine is the same in the initial location and final location, I assumed that the adjacent and opposite sides are equal.

Triangle.png

Diagram 2: Right-Angled Triangle Derived From Question
4) I then tried to find the length of x by applying the Cosine Rule.

c2=a2+b2-2ab(cos C)
Equation 1: Cosine Rule


x2=a2+x2-2ax(cos 45)

Substitute the unknowns with the numbers or unknowns that are shown in the diagram drawn.

-a2=-2ax(cos 45)
Cancel like terms and try to put everything in term of a

a=√2 x
Let this equation be the first equation.

5) After getting the first equation, I tried finding the length of the vine by using sin θ = Opposite/Hypotenuse

sin 45 = x/a

sin 45 = x/√2 x

sin 45 = √2/2

√2=x

2=a

I was stuck from this point on... Any help is appreciated...
 
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  • #2
Why not let the length of the vine be ##l## and see what you get. Perhaps ##l## will cancel out.

Hint: think about conservation of momentum and energy. Where and when do these apply in this problem?
 
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  • #3
In addition to following @PeroK's advice about the length of the vine, compare the situation immediately before and immediately after Tarzan grabs Jane. What's different? What remains the same?
 
  • #4
So I am assuming that the initial position has mgh,
PeroK said:
Why not let the length of the vine be ##l## and see what you get. Perhaps ##l## will cancel out.

Hint: think about conservation of momentum and energy. Where and when do these apply in this problem?

So, at the initial position, there is mgh, but no 1/2mv2? Since it started at rest? then at the middle point when he grabbed Jane, the mgh is 0? After that I am not sure
 
  • #5
Doc Al said:
In addition to following @PeroK's advice about the length of the vine, compare the situation immediately before and immediately after Tarzan grabs Jane. What's different? What remains the same?

The mass and velocity? I am assuming that as the mass increase the velocity decreases?
 
  • #6
bieon said:
So, at the initial position, there is mgh, but no 1/2mv2? Since it started at rest? then at the middle point when he grabbed Jane, the mgh is 0?
Yes, at the initial and final positions, there is potential energy but no kinetic. And it's OK to measure PE from the lowest point.
 
  • #7
bieon said:
So I am assuming that the initial position has mgh,So, at the initial position, there is mgh, but no 1/2mv2? Since it started at rest? then at the middle point when he grabbed Jane, the mgh is 0? After that I am not sure
In physics terms what can you say about the process of Tarzan picking up Jane? Is energy conserved? Is momentum conserved?
 
  • #8
PeroK said:
In physics terms what can you say about the process of Tarzan picking up Jane? Is energy conserved? Is momentum conserved?

So, um Energy cannot be destroyed but it can be transfer? I assume the momentum is conserved as the momentum before and the momentum after is equal when there are no external forces
 
  • #9
Doc Al said:
Yes, at the initial and final positions, there is potential energy but no kinetic. And it's OK to measure PE from the lowest point.

By that, since the energy is conserved, I can assume that the PE of initial and final to be equal right?
 
  • #10
Consider this in three parts: The swing down from the starting point to the bottom; The "collision" as Tarzan grabs Jane (what kind of collision is this?); The swing up to the final height.

In each of these parts, some quantities are conserved while others are not.
 
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  • #11
Doc Al said:
Consider this in three parts: The swing down from the starting point to the bottom; The "collision" as Tarzan grabs Jane (what kind of collision is this?); The swing up to the final height.

In each of these parts, some quantities are conserved while others are not.
When Tarzan grab Jane, it stick together, so it would be... perfect inelastic collision?
 
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  • #12
bieon said:
When Tarzan grab Jane, it stick together, so it would be... perfect inelastic collision?
Exactly! Make use of that insight.
 
  • #13
Doc Al said:
Consider this in three parts: The swing down from the starting point to the bottom; The "collision" as Tarzan grabs Jane (what kind of collision is this?); The swing up to the final height.

In each of these parts, some quantities are conserved while others are not.
Doc Al said:
Exactly! Make use of that insight.
so... V2=[m1/(m1+m2)]v1

(Sorry, I don't know how to use fraction here...)
 
  • #14
bieon said:
so... V2=[m1/(m1+m2)]v1

(Sorry, I don't know how to use fraction here...)
What about the velocity though?
 
  • #15
bieon said:
What about the velocity though?
Can you relate ##v_1## and ##v_2## to the initial and final positions?
 
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  • #16
PeroK said:
Can you relate ##v_1## and ##v_2## to the initial and final positions?
##v_1## the initial velocity is at zero, since it's stated that Tarzan was at rest...
##v_2## the final velocity is it zero too? Since he stopped?

(I am really sorry for taking so much of both of your time...)
 
  • #17
bieon said:
##v_1## the initial velocity is at zero, since it's stated that Tarzan was at rest...
##v_2## the final velocity is it zero too? Since he stopped?
##v_1## and ##v_2## are the velocities immediately before and after the inelastic collision.
 
  • #18
Doc Al said:
##v_1## and ##v_2## are the velocities immediately before and after the inelastic collision.
Hmm... This is slightly confusing...
 
  • #19
Doc Al said:
##v_1## and ##v_2## are the velocities immediately before and after the inelastic collision.
So there will be a ##v_f## for final velocity of both right?
 
  • #20
bieon said:
So there will be a ##v_f## for final velocity of both right?
In your momentum equation, that "final" velocity is ##v_2##.
 
  • #21
Doc Al said:
In your momentum equation, that "final" velocity is ##v_2##.
##v_2##=[##m_1##/(##m_1##+##m_2##)]##v_1##
This equation?

How do I use this to obtain the final velocity when I don't know the ##v_1##?
 
  • #22
bieon said:
##v_2##=[##m_1##/(##m_1##+##m_2##)]##v_1##
This equation?

How do I use this to obtain the final velocity when I don't know the ##v_1##?
As @PeroK suggested, express those velocities in terms of the initial and final positions. (What's conserved during those swings?)
 
  • #23
Doc Al said:
As @PeroK suggested, express those velocities in terms of the initial and final positions. (What's conserved during those swings?)
The momentum is conserved?
Position as in displacement?
 
  • #24
bieon said:
##v_1## the initial velocity is at zero, since it's stated that Tarzan was at rest...
##v_2## the final velocity is it zero too? Since he stopped?

(I am really sorry for taking so much of both of your time...)
To give you a bit more help, let me show you how I would set up this problem. There are three phases.

The first phase is Tarzan swinging. Energy is conserved. Let's say the energy is ##E_1##. We start with GPE and end with KE.

The second phase he collides with Jane and energy is lost. After the collision the energy is ##E_2##.

We can relate ##E_1## and ##E_2## by conservation of momentum.

The third phase is both swinging up. Again energy is conserved and KE becomes GPE.

Note: you have to be careful if you like using "initial" and "final" terminology. These apply to each phase of the motion. Especially for more complicated problems I prefer using ##v_0, v_1, v_2, v_3## etc.

In this problem ##v_0 = 0## and ##v_3 = 0##, so we don't even need those.

But we do need ##v_1## and ##v_2##, as the velocities just before and after the collision.
 
  • #25
PeroK said:
To give you a bit more help, let me show you how I would set up this problem. There are three phases.

The first phase is Tarzan swinging. Energy is conserved. Let's say the energy is ##E_1##. We start with GPE and end with KE.

The second phase he collides with Jane and energy is lost. After the collision the energy is ##E_2##.

We can relate ##E_1## and ##E_2## by conservation of momentum.

The third phase is both swinging up. Again energy is conserved and KE becomes GPE.

Note: you have to be careful if you like using "initial" and "final" terminology. These apply to each phase of the motion. Especially for more complicated problems I prefer using ##v_0, v_1, v_2, v_3## etc.

In this problem ##v_0 = 0## and ##v_3 = 0##, so we don't even need those.

But we do need ##v_1## and ##v_2##, as the velocities just before and after the collision.
So.
##E_1##=##m##g##h##+½mv12
##E_2## = ½(m+50) v22

then
##E_1## = ##E_2##
 
  • #26
bieon said:
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
PeroK said:
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
PeroK said:
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
bieon said:
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
PeroK said:
That's the idea.

Although the length of the vine should be in there somewhere.
Ah dang... :/
 
  • #31
bieon said:
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
PeroK said:
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
bieon said:
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 said:
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
PeroK said:
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
 
<h2>1. What is conservation of momentum?</h2><p>Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant. This means that in any interaction or collision between objects, the total momentum before and after the collision must be the same.</p><h2>2. How does conservation of momentum apply to a Tarzan swing collision?</h2><p>In a Tarzan swing collision, the conservation of momentum applies because the system is considered closed. This means that the only external force acting on the system is gravity, and the total momentum of the system (Tarzan and the swing) must remain constant before and after the collision.</p><h2>3. What is the role of energy conservation in a Tarzan swing collision?</h2><p>Energy conservation is also a fundamental principle in physics that states that energy cannot be created or destroyed, only transferred or converted from one form to another. In a Tarzan swing collision, energy is conserved because the total energy of the system (kinetic and potential) remains constant before and after the collision.</p><h2>4. How are momentum and energy related in a Tarzan swing collision?</h2><p>Momentum and energy are related in a Tarzan swing collision because both are conserved. This means that as momentum is transferred from Tarzan to the swing, the energy of the system is also transferred and converted from kinetic to potential energy.</p><h2>5. What factors affect the conservation of momentum and energy in a Tarzan swing collision?</h2><p>The conservation of momentum and energy in a Tarzan swing collision can be affected by factors such as the mass and velocity of Tarzan and the swing, the angle of the swing, and the elasticity of the rope. These factors can impact the transfer and conversion of momentum and energy in the collision.</p>

1. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant. This means that in any interaction or collision between objects, the total momentum before and after the collision must be the same.

2. How does conservation of momentum apply to a Tarzan swing collision?

In a Tarzan swing collision, the conservation of momentum applies because the system is considered closed. This means that the only external force acting on the system is gravity, and the total momentum of the system (Tarzan and the swing) must remain constant before and after the collision.

3. What is the role of energy conservation in a Tarzan swing collision?

Energy conservation is also a fundamental principle in physics that states that energy cannot be created or destroyed, only transferred or converted from one form to another. In a Tarzan swing collision, energy is conserved because the total energy of the system (kinetic and potential) remains constant before and after the collision.

4. How are momentum and energy related in a Tarzan swing collision?

Momentum and energy are related in a Tarzan swing collision because both are conserved. This means that as momentum is transferred from Tarzan to the swing, the energy of the system is also transferred and converted from kinetic to potential energy.

5. What factors affect the conservation of momentum and energy in a Tarzan swing collision?

The conservation of momentum and energy in a Tarzan swing collision can be affected by factors such as the mass and velocity of Tarzan and the swing, the angle of the swing, and the elasticity of the rope. These factors can impact the transfer and conversion of momentum and energy in the collision.

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