Classical mechanics confusions

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Discussion Overview

The discussion revolves around classical mechanics, specifically addressing the behavior of objects during collisions and the implications of changing mass on kinetic energy calculations. Participants explore the principles of momentum conservation, the definition of objects in mechanics, and the application of kinetic energy formulas in scenarios involving mass change.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question why two identical objects interchange their velocities during a head-on collision instead of simply reversing direction while maintaining their speed, suggesting that momentum and kinetic energy would still be conserved.
  • Others argue that momentum is not conserved in the proposed scenario due to the vector nature of momentum, where direction plays a crucial role.
  • There is a discussion about the applicability of the kinetic energy formula for objects with changing mass, with some participants suggesting that the formula can still be used while others express uncertainty about its validity in such cases.
  • One participant provides an example of a rocket losing mass as it combusts fuel, raising questions about how to calculate its kinetic energy accurately.
  • Concerns are raised about energy losses, including the kinetic energy of the exhaust stream and the implications of changing mass on kinetic energy calculations.
  • Participants discuss whether there is a specific type of energy associated with changing mass, with some asserting that kinetic energy is defined independently of mass constancy.

Areas of Agreement / Disagreement

Participants express differing views on the conservation of momentum in collisions and the use of kinetic energy formulas for changing mass scenarios. The discussion remains unresolved, with multiple competing perspectives presented.

Contextual Notes

Participants highlight the importance of definitions in classical mechanics, particularly regarding what constitutes an "object" and how mass changes affect calculations. There is also mention of the need for specific examples to clarify points raised.

Faiq
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Homework Statement


Q1. What's the reason behind two identical objects interchanging their velocities upon head-on collision?
Why can't just each individual particles just reverse its direction and keep traveling at its original speed? Kinetic energies and momentum would still be conserved.

Q2. Why can't we use the formula of Kinetic energy where kinetic energy is half mass times velocity squared to find kinetic energies of objects whose mass is changing? Is there a way to modify the formula to get the desired output?
 
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Faiq said:
Kinetic energies and momentum would still be conserved.
Are you sure about that?

Faiq said:
Q2. Why can't we use the formula of Kinetic energy where kinetic energy is half mass times velocity squared to find kinetic energies of objects whose mass is changing? Is there a way to modify the formula to get the desired output?
How is the mass changing?
 
Q1: No, momentum is not conserved in your scenario. Momentum is a vector, it has magnitude and direction, and though in your scenario the magnitude of the momentum of each particle remains the same, the direction reverses for each particle, so the direction of the total momentum also reverse, so momentum is not conserved.

Q2: Not sure here, I think you can use the formula for kinetic energy but you can't use the work-energy theorem.
 
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Faiq said:
Q1. What's the reason behind two identical objects interchanging their velocities upon head-on collision?
Why can't just each individual particles just reverse its direction and keep traveling at its original speed? Kinetic energies and momentum would still be conserved.

I don't understand the distinction you are making between "interchanging their velocities" and "each individual ... just reverse direction...". If particle A is assigned the velocity of particle B after the collision, does particle A "just reverse direction" ?

Q2. Why can't we use the formula of Kinetic energy where kinetic energy is half mass times velocity squared to find kinetic energies of objects whose mass is changing?

What is your definition of "object"? The state of an object in classical mechanics includes a description of its mass. If the "object" represents a stick and you break it in half, which of the halves is the original "object"? It would help if you gave an specific example of a problem in classical mechanics where an "object" has a changing mass.
 
^ I think he means something like an accelerating rocket that is losing mass by combusting its fuel...
 
Delta² said:
Q1: No, momentum is not conserved in your scenario. Momentum is a vector, it has magnitude and direction, and though in your scenario the magnitude of the momentum of each particle remains the same, the direction reverses for each particle, so the direction of the total momentum also reverse, so momentum is not conserved.

Q2: Not sure here, I think you can use the formula for kinetic energy but you can't use the work-energy theorem.

thank you for question 1
 
Stephen Tashi said:
I don't understand the distinction you are making between "interchanging their velocities" and "each individual ... just reverse direction...". If particle A is assigned the velocity of particle B after the collision, does particle A "just reverse direction" ?
What is your definition of "object"? The state of an object in classical mechanics includes a description of its mass. If the "object" represents a stick and you break it in half, which of the halves is the original "object"? It would help if you gave an specific example of a problem in classical mechanics where an "object" has a changing mass.
An object which is constantly gaining or losing mass, like spaceship losing mass by combusting its fuel
 
Faiq said:
An object which is constantly gaining or losing mass, like spaceship losing mass by combusting its fuel
So you are asking why we cannot use ##E=\frac{1}{2}mv^2## to determine the current kinetic energy of a rocket ship that is losing mass as it fires its thrusters?

We can use ##E=\frac{1}{2}mv^2## to determine the kinetic energy of such a rocket. That formula will give a perfectly correct answer.

But if you add the starting kinetic energy of the rocket to the chemical energy in the fuel being burned you get a result that is (almost always) larger than the resulting kinetic energy of the rocket. There is some missing energy. Can you imagine where that "missing" energy might have gone?
 
Energy losses to such as friction?
 
  • #10
Like if I have a problem like
A rocket has a mass of 16000kg. It loses mass at a rate of 15kg/s. It maintains a steady velocity of 200m/s. What is the Kinetic energy of the rocket?
What will be the value of m here?
 
  • #11
Faiq said:
Energy losses to such as friction?
That is not it. The rocket can be in space and the only force can be that of its own thrust. What about the kinetic energy of the exhaust stream?

Faiq said:
Like if I have a problem like
A rocket has a mass of 16000kg. It loses mass at a rate of 15kg/s. It maintains a steady velocity of 200m/s. What is the Kinetic energy of the rocket?
What will be the value of m here?
What will be the value of m when? The mass of the rocket is changing over time. The kinetic energy of the rocket at a particular time depends on its mass at that time.
 
  • #12
Yes that's what I was talking about. So is there a specific kind of energy for objects whose mass changes over time. Like kinetic energy is related with changes in motion, is there a type of energy related to changes in mass?
 
  • #13
Faiq said:
Yes that's what I was talking about. So is there a specific kind of energy for objects whose mass changes over time. Like kinetic energy is related with changes in motion, is there a type of energy related to changes in mass?
Kinetic energy is not specifically related to changes in motion. It is defined by ##KE=\frac{1}{2}mv^2##. The velocity there need not change. Nor must the mass be a constant.
 
  • #14
Oh okay thanks
 

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