What is the funniest decision you can make using impulse and velocity?

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SUMMARY

This discussion centers on the relationship between impulse and velocity, specifically defining impulse as a function of velocity, represented as p=f(V). The author employs Taylor's series expansion to derive the function near a specific velocity (V_0), concluding with the equation p=p_0+m_0V+0.5gV^2, where m represents mass and g is the second derivative of p with respect to V. Two options for decision-making based on impulse and position equations are presented, highlighting the humorous nature of the calculus involved.

PREREQUISITES
  • Understanding of Taylor's series expansion
  • Knowledge of basic physics concepts such as impulse and velocity
  • Familiarity with calculus, particularly differentiation
  • Concept of kinematics, including equations of motion
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  • Explore the relationship between impulse and momentum in classical mechanics
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This discussion is beneficial for physics students, educators, and anyone interested in the mathematical foundations of mechanics and the humorous aspects of theoretical physics.

What is it going to be?

  • Option 1 (look at the original post)

    Votes: 0 0.0%
  • Option2 (look at the original post)

    Votes: 0 0.0%

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Archimedes
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This is the funniest decision you'll ever make.
I say impulse is function of velocity alone i.e. [tex]p=f(V)[/tex]
I use Tayler's differential order to develope that function near some V_0. I assume f(V) is n times differentiable and here is how the developemnet looks:
[tex]p=p_0+\frac{dp}{dV}(V-V_0)+\frac{d^2p}{dV^2}\frac{(V-V_0)^2}{2!}+...+\frac{d^np}{dV^n}\frac{(V-V_0)^n}{n!}[/tex].
Now I assume n=2,
[tex]V_0=0[/tex],
m such that [tex]m=\frac{dp}{dV}[/tex] is mass in kg,
while g such that [tex]g=\frac{d^2p}{dV^2}[/tex] is I don't know what in kg^2.
This is the final developement:
[tex]p=p_0+m_0V+0,5gV^2[/tex]

The decision you are about to make is what is going to be?

Option 1: If for impulse we have
[tex]p=mV[/tex]
then for position we will have
[tex]x=Vt[/tex]

or

Option 2: If for position we have
[tex]x=x_0+V_0t+0,5at^2[/tex]
then for impulse we will have
[tex]p=p_0+m_0V+0,5gV^2[/tex]

I don't think my account should be disabled because of this funny calculus.
Thanks for voting!
 
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if [tex]m=\frac{dp}{dV}[/tex] is a constant then [tex]g=\frac{d^2p}{dV^2}=0[/tex] and both answers are correct !
 
There is no way I can vote ! You short-cuted my brain :bugeye:


:biggrin:
 

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