How Do You Calculate Kinetic Energy in Relativistic Dynamics?

In summary: Yes, I see. There used to be rest mass and relativistic mass. Now, for good reason, there's no meaningful quantity called relativistic mass. Rest mass and m are now one and the same.
  • #1
Scientist94
10
0

Homework Statement

A particle of mass m and kinetic energy K strikes and combines with a stationary particle of mass 2m, producing a single composite particle of mass √17 m. Find the value of K



Homework Equations



E= γmc^2 , p = λmv

The Attempt at a Solution

I have tried many times at this and have used the conservation of energy and conservation of momentum but I seem to get nowhere. The equations are definitely solvable and this seems to just be algebraic manipulation. My question is whether there is any way to simplify these problems as I just cannot get an expression for K without the other variables, this problem keeps cropping up in other questions too!
 
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  • #2
Show us what you've done please.

EDIT:

Also, check out this link. You should get used to writing in latex for physics/math.
 
  • #3
Does the composite particle have zero K.E.?
 
  • #4
rude man said:
Does the composite particle have zero K.E.?
Hi Rudy. Not necessarily.

Chet
 
  • #5
Scientist94 said:

Homework Statement

A particle of mass m and kinetic energy K strikes and combines with a stationary particle of mass 2m, producing a single composite particle of mass √17 m. Find the value of K



Homework Equations



E= γmc^2 , p = λmv

The Attempt at a Solution

I have tried many times at this and have used the conservation of energy and conservation of momentum but I seem to get nowhere. The equations are definitely solvable and this seems to just be algebraic manipulation. My question is whether there is any way to simplify these problems as I just cannot get an expression for K without the other variables, this problem keeps cropping up in other questions too!
You should be able to manipulate and combine the momentum balance and the energy balance into a single quadratic equation involving the variable x = K/(mc2). It takes some playing with the math to bring this about. So, as Strange said, show us what you did, and maybe we can give you some hints on manipulating the math.

Chet
 
  • #6
Chestermiller said:
Hi Rudy. Not necessarily.

Chet

Hi Chet, right. Hadn't realized.

I'll leave this one to you, assuming the OP responds. Lately the only responses I get are from other helpers ... bedides, I'd only invoke rest mass again, stirring that controversy up again probably.

rudy
 
  • #7
rude man said:
Hi Chet, right. Hadn't realized.

I'll leave this one to you, assuming the OP responds. Lately the only responses I get are from other helpers ... bedides, I'd only invoke rest mass again, stirring that controversy up again probably.

rudy
What's wrong with invoking rest mass? That seems to be a large part of what this problem is about.

Chet
 
  • #8
Chestermiller said:
What's wrong with invoking rest mass? That seems to be a large part of what this problem is about.

Chet

I learned 'modern physics' in the early '60's when there was a separate symbol for rest mass: m0. Recently it was pointed out to me in these forums that that is no longer in vogue: there is only one mass, m. If at rest it's m, if moving it's γm. So among other casualties it's no longer E = mc2 but γmc2, and p = mv is no longer correct either. I find this change entirely pointless but found out to my distress that I was in the rank minority ... I suppose I made a mountain out of a molehill but still I see no reason to fix what isn't broke.
 
  • #9
rude man said:
I learned 'modern physics' in the early '60's when there was a separate symbol for rest mass: m0. Recently it was pointed out to me in these forums that that is no longer in vogue: there is only one mass, m. If at rest it's m, if moving it's γm. So among other casualties it's no longer E = mc2 but γmc2, and p = mv is no longer correct either. I find this change entirely pointless but found out to my distress that I was in the rank minority ... I suppose I made a mountain out of a molehill but still I see no reason to fix what isn't broke.
Yes, I see. There used to be rest mass and relativistic mass. Now, for good reason, there's no meaningful quantity called relativistic mass. Rest mass and m are now one and the same.

Chet
 

Related to How Do You Calculate Kinetic Energy in Relativistic Dynamics?

1. What is Relativistic Dynamics?

Relativistic Dynamics is a branch of physics that studies the motion of objects at speeds close to the speed of light. It takes into account the principles of special relativity, which describes how space and time are affected by an object's motion.

2. What are the key concepts in Relativistic Dynamics?

The key concepts in Relativistic Dynamics include time dilation, length contraction, and the equivalence of mass and energy. These concepts are used to explain the behavior of objects moving at high speeds and are essential to understanding the theory of special relativity.

3. How does Relativistic Dynamics differ from classical mechanics?

Relativistic Dynamics differs from classical mechanics in that it takes into account the effects of special relativity, while classical mechanics is based on Newton's laws of motion. At low speeds, the two theories give similar results, but at high speeds, Relativistic Dynamics is necessary to accurately describe the behavior of objects.

4. What are some real-world applications of Relativistic Dynamics?

Relativistic Dynamics has many real-world applications, including in particle accelerators, GPS technology, and understanding the behavior of objects in space. It is also used in the development of advanced technologies, such as nuclear power and space travel.

5. What are some common problems encountered in Relativistic Dynamics?

Some common problems encountered in Relativistic Dynamics include understanding the effects of time dilation and length contraction, calculating the relativistic momentum and energy of objects, and solving for the motion of objects in curved spacetime. These problems require a good understanding of the principles of special relativity and advanced mathematical techniques.

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