Calculating mass and speed from kinetic energy and linear momentum

In summary: That doesn't solve his problem since this last expression is still a function of m0. He needs m0(p,K) and v(p,K). You gave him m0(\beta,K) and v(\beta,K).That doesn't solve his problem since this last expression is still a function of m0. He needs m0(p,K) and v(p,K). You gave him m0(\beta,K) and v(\beta,K).In summary, Niko has attempted to solve an exercise involving kinetic energy and linear momentum, using the given specifications of Wk = 5713.13 MeV and P = 6584.92 Mev/c0 and the formulas Wk = mc^2*((gamma)-1) and
  • #1
niko2000
51
0
Hi,
I have tied to solve an exercise where I have kinetic energy and linear momentum and I have to calculate the speed and mass (not relativistic mass)
I have these specifications:
Wk = 5713.13 MeV
P = 6584.92 Mev/c0

I have used these formulas:
Wk = mc^2*((gamma)-1)
P = (gamma)*m*v
I have tried to calculate v and m, but I haven't got sensible results.
I have probably done some mistake, but I can't find it. If anyone had time to take a look at this exercise it would be really helpful.
Regards,
Niko
 
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  • #2
niko2000 said:
Hi,
I have tied to solve an exercise where I have kinetic energy and linear momentum and I have to calculate the speed and mass (not relativistic mass)
I have these specifications:
Wk = 5713.13 MeV
P = 6584.92 Mev/c0

I have used these formulas:
Wk = mc^2*((gamma)-1)
P = (gamma)*m*v
I have tried to calculate v and m, but I haven't got sensible results.
I have probably done some mistake, but I can't find it. If anyone had time to take a look at this exercise it would be really helpful.
Regards,
Niko
Usually when someone posts what appears to be a homework question we ask to see what work they have done so far so that we don't do all the work for them. Can you please show us what you did and why you don't think your results were sensible? Thanks.

Until then this is what I got. Let K = kinetic energy, m = proper mass, E = inertial energy = K + mc2. Then

[tex]E^2 - p^2c^2 = m^2c^4 = (K + mc^2)^2 - p^2c^2[/tex]

Multiply the term in the parenthese out to obtain

[tex]K^2 + 2kmc^2 + m^2c^4 - p^2c^2 = m^2c^4[/tex]

Cancel like terms and solve for m to obtain

[tex]m = \frac {p^2c^2 - K^2}{2Kc^2}[/tex]

Now that you have m you can use p = gamma*m*v and solve for v. I'll let you work that part out since I did the first part of the work to obtain m. Let us know if you need more help.

Pete
 
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  • #3
I don't know if the first equation is right:
E = mr*c^2 = m0*c^2 + K,
where mr = relativistic mass and m0 = proper mass, so I thing this can't lead us to the right soluton.
When I was doing this exercise I have got the speed greater then speed of light, so it's not right. I have probably made some calculation mistakes. I think the formulas I have written guide us to the right answer, but I simply couldn't get it because I have made some mistakes I can't find.
 
  • #4
niko2000 said:
I don't know if the first equation is right:
E = mr*c^2 = m0*c^2 + K,
where mr = relativistic mass and m0 = proper mass, so I thing this can't lead us to the right soluton.
When I was doing this exercise I have got the speed greater then speed of light, so it's not right. I have probably made some calculation mistakes. I think the formulas I have written guide us to the right answer, but I simply couldn't get it because I have made some mistakes I can't find.
Using a relativistic mass substitution somewhere you shouldn't have is probably what led to your mistake. Do not use relativistic mass and try again.
There is no place for relativistic mass in modern relativity.
 
  • #5
Here's a simpler way: Square the momentum. That gives you [itex]\beta^2\gamma^2m^2c^2[/itex] and use the identity [itex]\beta^2\gamma^2=\gamma^2-1[/itex]. Divide this by K.E. and note [itex]\gamma^2-1/(\gamma-1)=\gamma+1[/itex]. This gives you gamma, and the rest follows easily.
 
  • #6
niko2000 said:
I don't know if the first equation is right:
E = mr*c^2 = m0*c^2 + K,
where mr = relativistic mass and m0 = proper mass, so I thing this can't lead us to the right soluton.
Yes. That equation is correct, but not useful. I posted the formula for m as a function of kinetic energy and momentum above.
When I was doing this exercise I have got the speed greater then speed of light, so it's not right.
Not so fast. His data does not correspond to a particle which moves at speeds less than the speed of light (i.e. tardyons). This can be seen by noticing that for the proper mass to be positive, (pc)2- k2 must be positive. But with his data this comes out negative. His data there therefore does not describe a tardyon. It describes a tachyon.

Pete
 
  • #7
krab said:
Here's a simpler way: Square the momentum. That gives you [itex]\beta^2\gamma^2m^2c^2[/itex] ..
Sorry but I don't see how you got that. Momentum is given by

[tex]p = mv = \gamma m_0 v[/tex]

Therefore

[tex]p^2 = \gamma^2 m_0^2 v^2[/tex]

You claim this is

[tex] p^2 = \beta^2 \gamma^2 m_0^2 c^2[/tex]

Why?

Kinetic energy is given by

[tex]K = (\gamma - 1)m_0c^2[/tex]

so when you divide p2 by K2 to obtain (p/K)2 you will obtain

[tex](p/K)^2 = \frac{\gamma^2 m_0^2 v^2}{ (\gamma - 1)m_0c^2}[/tex]

Canceling a factor of m0 gives

[tex](p/K)^2 = \frac{\gamma^2 m_0 v^2}{ (\gamma - 1)c^2}[/tex]

Simplify to obtain

[tex](p/K)^2 = (\gamma+1)m_0\beta^2}[/tex]

I don't see that this is useful since you haven't eliminated either v or m0.

Pete
 
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  • #8
Try a rapidity calculation.

[tex] K= m_0 c^2(\cosh\theta - 1) [/tex]
[tex] p= m_0 c\sinh\theta [/tex]

So,
[tex] \cosh\theta = \frac{K}{m_0c^2}+1 [/tex]
[tex] \sinh\theta = \frac{p}{m_0c} [/tex]

Now, given values for K and p,
[tex]1 =\cosh^2\theta - \sinh^2\theta = (\frac{K}{m_0c^2}+1 )^2- ( \frac{p}{m_0c} )^2 [/tex]
solve for [itex] m_0[/itex].
Armed additionally with [itex] m_0 [/itex],
[tex] v = c\tanh\theta = c\frac{\sinh\theta }{\cosh\theta } = c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }[/tex]
 
  • #9
robphy said:
Try a rapidity calculation.

[tex] K= m_0 c^2(\cosh\theta - 1) [/tex]
[tex] p= m_0 c\sinh\theta [/tex]

So,
[tex] \cosh\theta = \frac{K}{m_0c^2}+1 [/tex]
[tex] \sinh\theta = \frac{p}{m_0c} [/tex]

Now, given values for K and p,
[tex]1 =\cosh^2\theta - \sinh^2\theta = (\frac{K}{m_0c^2}+1 )^2- ( \frac{p}{m_0c} )^2 [/tex]
solve for [itex] m_0[/itex].
Armed additionally with [itex] m_0 [/itex],
[tex] v = c\tanh\theta = c\frac{\sinh\theta }{\cosh\theta } = c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }[/tex]
That doesn't solve his problem since this last expression is still a function of m0. He needs m0(p,K) and v(p,K). You gave v(m0,p,K).

Pete
 
  • #10
pmb_phy said:
Sorry but I don't see how you got that. Momentum is given by

[tex]p = mv = \gamma m_0 v[/tex]

Therefore

[tex]p^2 = \gamma^2 m_0^2 v^2[/tex]

You claim this is

[tex] p^2 = \beta^2 \gamma^2 m_0^2 c^2[/tex]

Why?
admin edit: personal attack removed

Kinetic energy is given by

[tex]K = (\gamma - 1)m_0c^2[/tex]

so when you divide p2 by K2 to obtain (p/K)2 you will obtain

[tex](p/K)^2 = \frac{\gamma^2 m_0^2 v^2}{ (\gamma - 1)m_0c^2}[/tex]
admin edit: personal attack removed

Canceling a factor of m0 gives

[tex](p/K)^2 = \frac{\gamma^2 m_0 v^2}{ (\gamma - 1)c^2}[/tex]

Simplify to obtain

[tex](p/K)^2 = (\gamma+1)m_0\beta^2}[/tex]

I don't see that this is useful since you haven't eliminated either v or m0.

Pete
Perhaps you should do the algebra right. You could also stop innappropriately subscipting the mass with a zero as he did not do that and don't claim that p = mv as it was not. It was clearly stated for this thread that m was not relativistic mass.
 
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  • #11
pmb_phy said:
robphy said:
Try a rapidity calculation.

[tex] K= m_0 c^2(\cosh\theta - 1) [/tex]
[tex] p= m_0 c\sinh\theta [/tex]

So,
[tex] \cosh\theta = \frac{K}{m_0c^2}+1 [/tex]
[tex] \sinh\theta = \frac{p}{m_0c} [/tex]

Now, given values for K and p,
[tex]1 =\cosh^2\theta - \sinh^2\theta = (\frac{K}{m_0c^2}+1 )^2- ( \frac{p}{m_0c} )^2 [/tex]
solve for [itex] m_0[/itex].
Armed additionally with [itex] m_0 [/itex],
[tex] v = c\tanh\theta = c\frac{\sinh\theta }{\cosh\theta } = c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }[/tex]

That doesn't solve his problem since this last expression is still a function of m0. He needs m0(p,K) and v(p,K). You gave v(m0,p,K).

Pete

Let me fill in some steps in algebra.
robphy said:
Now, given values for K and p,
[tex]1 =\cosh^2\theta - \sinh^2\theta = (\frac{K}{m_0c^2}+1 )^2- ( \frac{p}{m_0c} )^2 [/tex]
solve for [itex] m_0[/itex].
[tex]\textcolor{red}{m_0=\frac{ (pc)^2-K^2 }{ 2Kc^2} }[/tex]

robphy said:
Armed additionally with [itex] m_0 [/itex],
[tex] v = c\tanh\theta = c\frac{\sinh\theta }{\cosh\theta } = c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }[/tex]
[tex] \begin{align*}
v &=c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }\\
&=\frac{p}{ \frac{K}{c^2}+m_0 }\\
&=\frac{p}{ \frac{K}{c^2}+\textcolor{red}{\frac{ (pc)^2-K^2 }{ 2Kc^2} } } = \frac{2K pc^2}{K^2+p^2c^2}
\end{align*}
[/tex]
 
  • #12
Like DW, I truly do not understand some of pete's comments. But then since the dynamics of relativistic particles is my field of daily work, things that are obvious to me might not be obvious to others...

Anyway, I showed the trick of taking p^2/K to get something proportional to (gamma+1)m. (Sorry, I erred in saying this gave you gamma; I lost sight that it was m that is not known.) Since you already have K, which is prop.to (gamma-1)m, just subtract the two to get mass. In detail,

[tex]{1\over 2}\left({p^2c^2\over K}-K\right)=mc^2[/tex]

This agrees with the red formula given by robphy.
 
  • #13
krab said:
Like DW, I truly do not understand some of pete's comments.
Which ones? If you're referring to the "Why?" question then it was simply a question and nothing else. For some dumb reason I didn't recognize beta2c2 as v2. I don't know WHAT I was thinking. I reserve the right to have a bad day and miss simple things like this. :biggrin:

I also wrote down K2 on the left side instead of K. But I did divide by K on the right. I can't seem to edit that post to correct it. Why is that?

That all seems like a lot of work to get gamma. Gamma can easily be found from the equation in my first post. That gives the proper mass m0 from the given values of p and K. Once you have m0 then use K = (gamma - 1)m0c2 and solve for gamma. Once you have gamma solve for v.

Note: I have to read posts and respond in a hurry for the next several months. Due to a back injury I can't sit in front of the computer for very long otherwise there are severe consequences. The pain is bit distracting too. So please understand that I will make more errors, due to necessary haste (and distraction) in reading and posting, until this gets better. Thanks for your understanding.

Pete
 
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1. How do you calculate mass from kinetic energy and linear momentum?

To calculate mass from kinetic energy and linear momentum, you can use the formula m = p/v, where m is the mass, p is the linear momentum, and v is the velocity.

2. What is the relationship between kinetic energy and linear momentum?

The relationship between kinetic energy and linear momentum is that they are both measures of an object's motion. Kinetic energy is the energy an object possesses due to its motion, while linear momentum is the product of an object's mass and its velocity.

3. Can you calculate an object's speed using kinetic energy and linear momentum?

Yes, you can calculate an object's speed using the formula v = p/m, where v is the velocity, p is the linear momentum, and m is the mass.

4. How does the mass of an object affect its kinetic energy and linear momentum?

The mass of an object directly affects its kinetic energy and linear momentum. As the mass increases, the kinetic energy and linear momentum also increase, assuming the velocity remains constant.

5. Are there any other factors that affect the calculation of mass and speed from kinetic energy and linear momentum?

No, the only factors that affect the calculation of mass and speed from kinetic energy and linear momentum are the mass and velocity of the object. However, it is important to note that these calculations are based on classical mechanics and may not accurately represent objects moving at very high speeds or in extreme conditions.

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