Conservation of Energy in Conjunction With Momentum

In summary, a problem involving an elastic collision between two bodies of equal mass was discussed. The conservation of momentum and energy were used to prove that the final velocities after the collision are orthogonal. The law of cosines was applied to show that the angle between the final velocities is 90 degrees.
  • #1
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Homework Statement


Consider an elastic collision between two bodies of equal mass, one of which is initially at rest. Let their velocities be [itex]\vec{v_1}[/itex] and [itex]\vec{v_2} = \vec{0}[/itex] before the collision, and [itex]\vec{v_1 '}[/itex] and [itex]\vec{v_2 '}[/itex] after the collision. Write down the vector equation representing conservation of momentum and the scalar equation which expresses that the collision is elastic. Use these to prove that [itex]\vec{v_1 '}[/itex] and [itex]\vec{v_2 '}[/itex] are orthogonal.

Homework Equations


The Attempt at a Solution



If the collision is elastic, then KE = KE'.

m1 = m2 = m

[itex]\vec{p} = \vec{p'}[/itex]

If the two velocity vectors are orthogonal, then [itex]\vec{v_1 '} \cdot \vec{v_2 '} = 0[/itex].

Conservation of energy:

[itex]\frac{1}{2} mv_1^2 = \frac{1}{2} m v_1^{'2} + \frac{1}{2} m v_2^{'2}[/itex]

[itex]v_1^2 = v_1^{'2} + v_2^{'2}[/itex], which can be written as [itex]| \vec{v_1} |^2 =| \vec{v_1 '} |^2 + | \vec{v_2 '} |^2[/itex], although I am not sure of how helpful this will be.

Conservation of momentum:

[itex]m \vec{v_1} + \vec{0} = m \vec{v_1 '} + m \vec{v_2 '}[/itex]

[itex]\vec{v_1} = \vec{v_1 '} + \vec{v_2 '}[/itex]

Taking the magnitude, [itex]| \vec{v_1} | = | \vec{v_1 '} |+ | \vec{v_2 '} |[/itex]; again, I am not certain of how helpful this will be.

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Up to this point, I am not sure as to how to proceed. Could someone provide me with a hint?
 
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  • #2
Hi Bashyboy!

Bashyboy said:
Taking the magnitude, [itex]| \vec{v_1} | = | \vec{v_1 '} |+ | \vec{v_2 '} |[/itex]; again, I am not certain of how helpful this will be.

Nooo. :redface:

##|\vec{v_1}|=| \vec{v_1 '} + \vec{v_2 '} | ≠ | \vec{v_1 '} |+ | \vec{v_2 '} |##. :wink:
 
  • #3
Yes, of course. I actually have the correct expression on my paper, but somehow I failed to properly type it up.
 
  • #4
Bashyboy said:
Yes, of course. I actually have the correct expression on my paper, but somehow I failed to properly type it up.

Well...##|\vec{v_1'}+\vec{v_2'}|## means the magnitude of resultant of ##\vec{v_1'}## and ##\vec{v_2'}##. Can you proceed further?
 
  • #5
Well, I now have that [itex]| \vec{v_1} | = | \vec{v_r '} | [/itex], and I also have that [itex] |\vec{v_1}|^2 = | \vec{v_1 '} |^2 + | \vec{v_2 '}|^2[/itex]. I am to make some sort of substitution?
 
  • #6
Bashyboy said:
Well, I now have that [itex]| \vec{v_1} | = | \vec{v_r '} | [/itex], and I also have that [itex] |\vec{v_1}|^2 = | \vec{v_1 '} |^2 + | \vec{v_2 '}|^2[/itex]. I am to make some sort of substitution?

If two vectors, ##\vec{A}## and ##\vec{B}## make an angle ##\theta## with each other, what is the magnitude of resultant of the two vectors?
 
  • #7
I don't quite follow. I have never seen the sum of two vectors in terms of their angle.
 
  • #8
Bashyboy said:
I don't quite follow. I have never seen the sum of two vectors in terms of their angle.

I don't know how it is at your place but that's something taught at the very beginning here.

Okay, if you haven't seen it, you can use cosine rule to derive it. Draw a vector diagram.
 
  • #9
No, I would much rather see your method, please.
 
  • #10
Bashyboy said:
No, I would much rather see your method, please.

Apply cosine rule to the triangle shown in the attachment.
 

Attachments

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  • #11
When you say cosine rule, do you also mean the law of cosines? The law of cosines for vectors is simply the dot product, is it not?
 
  • #12
Bashyboy said:
When you say cosine rule, do you also mean the law of cosines? The law of cosines for vectors is simply the dot product, is it not?

Yes, very sorry, it is the law of cosines. I am used to calling it the "cosine rule". :redface:

To avoid confusion, I am talking about this: http://mathworld.wolfram.com/LawofCosines.html
 
  • #13
So, you are suggesting that by solving the law of cosines for theta, it will equal 90 deg.?
 
  • #14
Bashyboy said:
So, you are suggesting that by solving the law of cosines for theta, it will equal 90 deg.?

You can see it yourself. Apply the law of cosines. Show your work.
 
  • #15
Sorry it took awhile to reply. Well, after applying the Law of Cosines, the problem became rather trivial: it's all a matter of memorizing those tricks and formulas of mathematics. At any rate, Pranav-Arora, I thank you very much for time.
 
  • #16
Bashyboy said:
Sorry it took awhile to reply. Well, after applying the Law of Cosines, the problem became rather trivial: it's all a matter of memorizing those tricks and formulas of mathematics. At any rate, Pranav-Arora, I thank you very much for time.

Glad to help. :)
 
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What is conservation of energy in conjunction with momentum?

Conservation of energy in conjunction with momentum is a fundamental principle in physics that states that the total energy and momentum of a closed system remain constant over time. In other words, energy and momentum cannot be created or destroyed, only transferred or transformed.

Why is conservation of energy in conjunction with momentum important?

Conservation of energy in conjunction with momentum is important because it allows us to predict and understand the behavior of physical systems. It also helps us to develop technologies that are efficient and sustainable, as we can use this principle to optimize energy usage and minimize waste.

How is conservation of energy in conjunction with momentum applied in real life?

Conservation of energy in conjunction with momentum is applied in many areas of daily life. For example, it is used in the design of vehicles and transportation systems to ensure efficient use of energy and minimize collisions. It is also used in renewable energy technologies such as wind turbines and solar panels, where energy is converted from one form to another.

What are some common misconceptions about conservation of energy in conjunction with momentum?

One common misconception is that conservation of energy and conservation of momentum are the same thing. While they are related, they are separate principles that govern different aspects of a system's behavior. Another misconception is that energy and momentum can be created or destroyed, when in fact they can only be transferred or transformed.

How does the conservation of energy in conjunction with momentum relate to the laws of thermodynamics?

The conservation of energy in conjunction with momentum is closely related to the laws of thermodynamics, particularly the first law which states that energy cannot be created or destroyed. This principle also helps to explain the second law of thermodynamics, which states that the total entropy (disorder) of a closed system increases over time. Conservation of energy and momentum play a crucial role in understanding and applying thermodynamic principles.

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