Calculating mass and speed from kinetic energy and linear momentum

[niko2000]

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

 

[pmb_phy]

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

E^2 - p^2c^2 = m^2c^4 = (K + mc^2)^2 - p^2c^2

Multiply the term in the parenthese out to obtain

K^2 + 2kmc^2 + m^2c^4 - p^2c^2 = m^2c^4

Cancel like terms and solve for m to obtain

m = \frac {p^2c^2 - K^2}{2Kc^2}

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

 

[niko2000]

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.

 

[DW]

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.

 

[krab]

Here's a simpler way: Square the momentum. That gives you \beta^2\gamma^2m^2c^2 and use the identity \beta^2\gamma^2=\gamma^2-1. Divide this by K.E. and note \gamma^2-1/(\gamma-1)=\gamma+1. This gives you gamma, and the rest follows easily.

 

[pmb_phy]

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

 

[pmb_phy]

Here's a simpler way: Square the momentum. That gives you \beta^2\gamma^2m^2c^2 ..

Sorry but I don't see how you got that. Momentum is given by

p = mv = \gamma m_0 v

Therefore

p^2 = \gamma^2 m_0^2 v^2

You claim this is

p^2 = \beta^2 \gamma^2 m_0^2 c^2

Why?

Kinetic energy is given by

K = (\gamma - 1)m_0c^2

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

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

Canceling a factor of m₀ gives

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

Simplify to obtain

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

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

Pete

 

[robphy]

Try a rapidity calculation.

K= m_0 c^2(\cosh\theta - 1)
p= m_0 c\sinh\theta

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

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

 

[pmb_phy]

Try a rapidity calculation.

K= m_0 c^2(\cosh\theta - 1)
p= m_0 c\sinh\theta

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

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

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

 

[DW]

Sorry but I don't see how you got that. Momentum is given by

p = mv = \gamma m_0 v

Therefore

p^2 = \gamma^2 m_0^2 v^2

You claim this is

p^2 = \beta^2 \gamma^2 m_0^2 c^2

Why?

admin edit: personal attack removed

Kinetic energy is given by

K = (\gamma - 1)m_0c^2

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

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

admin edit: personal attack removed

Canceling a factor of m₀ gives

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

Simplify to obtain

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

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.

 

[robphy]

Try a rapidity calculation.

K= m_0 c^2(\cosh\theta - 1)
p= m_0 c\sinh\theta

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

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

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.

Now, given values for K and p,
1 =\cosh^2\theta - \sinh^2\theta = (\frac{K}{m_0c^2}+1 )^2- ( \frac{p}{m_0c} )^2
solve for m_0.

\textcolor{red}{m_0=\frac{ (pc)^2-K^2 }{ 2Kc^2} }

Armed additionally with m_0,
v = c\tanh\theta = c\frac{\sinh\theta }{\cosh\theta } = c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }

\begin{align*}<br /> v &amp;=c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }\\<br /> &amp;=\frac{p}{ \frac{K}{c^2}+m_0 }\\<br /> &amp;=\frac{p}{ \frac{K}{c^2}+\textcolor{red}{\frac{ (pc)^2-K^2 }{ 2Kc^2} } } = \frac{2K pc^2}{K^2+p^2c^2}<br /> \end{align*}<br />

 

[krab]

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,

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

This agrees with the red formula given by robphy.

 

[pmb_phy]

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.

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