Relativistic motion and length contraction

[the farmer]

But my lecturer told us to write in three significant figures

 

[the farmer]

FWIW Anything more than two significant digits is wildly overstating accuracy here.

But let ne write my answers in 2 significant it will be 0.10 m that's a length . Is my answers correct in a) ?

 

[the farmer]

I know the question is asking me to find a length, L = Lo/ y. Y is my gamma factor. In the question I am given kinetic energy. So to find my gamma I need to make it the subject of formula. Y= mc^2 + KE / mc^2
Y = 1 + (50×10^6) × ( 1.6×10^-19) J ÷ 9.11×10^-31 kg ( 3× 10^8m/s)^2
Y = 98. 57287474
L = 10m/98.5787474
=0.10 in to 2 decimal place.
I tried to calculate my length again but this time I am using the given speed and used this formula l= Lo/Y ,Y is gamma factor then L = 10m × (1 -(0.999949)^2)^1/2 ,c^2/c^2 canceled each other then my final answer is 0.10 m in 2 decimal place. I concluded that both these two methods are giving me same answer. Am I on the right truck or there is other method that I can use to solve my a).

 

[Orodruin]

As has already been said, the velocity given has quite obviously been computed by the problem author to correspond to the given kinetic energy. This is done by computing the gamma factor to find what velocity it corresonds to. Obviously that velocity is going to give the correct gamma factor - it was chosen that way by the author.

 

[the farmer]

As has already been said, the velocity given has quite obviously been computed by the problem author to correspond to the given kinetic energy. This is done by computing the gamma factor to find what velocity it corresonds to. Obviously that velocity is going to give the correct gamma factor - it was chosen that way by the author.

Still don't understand what you meant

 

[Orodruin]

Still don't understand what you meant

That is not a constructive response. A constroctive response would be pointing out which part you do not understand.

 

[the farmer]

I don't understand the way you responded. My question was:
Since the question is asking me to find a length, L = Lo/ y. Y is my gamma factor. In the question I am given kinetic energy. So to find my gamma I need to make it the subject of formula. Y= mc^2 + KE / mc^2
Y = 1 + (50×10^6) × ( 1.6×10^-19) J ÷ 9.11×10^-31 kg ( 3× 10^8m/s)^2
Y = 98. 57287474
L = 10m/98.5787474
=0.10 in to 2 decimal place.
I tried to calculate my length again but this time I am using the given speed and used this formula l= Lo/Y ,Y is gamma factor then L = 10m × (1 -(0.999949)^2)^1/2 ,c^2/c^2 canceled each other then my final answer is 0.10 m in 2 decimal place. I concluded that both these two methods are giving me same answer. Am I on the right truck or there is other method that I can use to solve my a)?

 

[Ibix]

Am I on the right truck or there is other method that I can use to solve my a)?

Those are the two methods available. That they give the same answer suggests you've done it right.

 

[jbriggs444]

I don't understand the way you responded. My question was:
Since the question is asking me to find a length, L = Lo/ y. Y is my gamma factor. In the question I am given kinetic energy. So to find my gamma I need to make it the subject of formula. Y= mc^2 + KE / mc^2

As it stands, that formula is not correct. Remember PEMDAS. The formula above parses as: $\gamma = mc^2 + \frac{\text{KE}}{mc^2}$. That is obviously dimensionally incorrect. Gamma is a pure number. You've given a formula for an energy plus a pure number.

Presumably you meant to write $\gamma = \frac{mc^2 + \text{KE}}{mc^2}$ That formula would be correct. Total energy is rest energy plus kinetic energy. Gamma is total energy divided by rest energy.

Let us look at your math.

Y = 1 + (50×10^6) × ( 1.6×10^-19) J ÷ 9.11×10^-31 kg ( 3× 10^8m/s)^2

I had trouble making sense of this. Oh, I think I get it now. You've done a simplification behind the scenes.

You set out to evaluate: $\gamma = \frac{mc^2 + \text{KE}}{mc^2}$
You actually evaluated $\gamma = \frac{mc^2}{mc^2} + \frac{\text{KE}}{mc^2} = 1 + \frac{\text{KE}}{mc^2}$

You convert the $50 \text{ MeV}$ to Joules for the numerator of the right hand side.

You multiply the rest mass of the electron, $9.11\times 10^{-31}\text{ kg}$ by $c^2$ to get Joules for the denominator.

That is a valid calculation for gamma.

Y = 98. 57287474
L = 10m/98.5787474
=0.10 in to 2 decimal place.

Seems sound.

So you have computed $\gamma$ based on the energy ratio that you were given (50 MeV from the problem and $9.11 \times 10^{-31} \text{ kg}$ from other references). Personally I would have looked up the rest mass of the electron in Mev (about 0.5 MeV) and skipped both conversions to joules. Then you can practically read off gamma without even using a calculator.

I tried to calculate my length again but this time I am using the given speed and used this formula l= Lo/Y ,Y is gamma factor then L = 10m × (1 -(0.999949)^2)^1/2 ,c^2/c^2 canceled each other then my final answer is 0.10 m in 2 decimal place. I concluded that both these two methods are giving me same answer. Am I on the right truck or there is other method that I can use to solve my a)?

Yes inceed. Since you are given that $\frac{v}{c} = .999949$ then $\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-.999949^2}$.

As you say, this time you used velocity as the starting point instead of kinetic energy.

Yes, both methods are giving you the same answer. Which is what @Orodruin was trying to point out. The person who set the question purposely chose the kinetic energy (50 MeV) and the velocity (0.999949 c) so that the two figures were consistent.

 

[the farmer]

As it stands, that formula is not correct. Remember PEMDAS. The formula above parses as: $\gamma = mc^2 + \frac{\text{KE}}{mc^2}$. That is obviously dimensionally incorrect. Gamma is a pure number. You've given a formula for an energy plus a pure number.

Presumably you meant to write $\gamma = \frac{mc^2 + \text{KE}}{mc^2}$ That formula would be correct. Total energy is rest energy plus kinetic energy. Gamma is total energy divided by rest energy.

Let us look at your math.


I had trouble making sense of this. Oh, I think I get it now. You've done a simplification behind the scenes.

You set out to evaluate: $\gamma = \frac{mc^2 + \text{KE}}{mc^2}$
You actually evaluated $\gamma = \frac{mc^2}{mc^2} + \frac{\text{KE}}{mc^2} = 1 + \frac{\text{KE}}{mc^2}$

You convert the $50 \text{ MeV}$ to Joules for the numerator of the right hand side.

You multiply the rest mass of the electron, $9.11\times 10^{-31}\text{ kg}$ by $c^2$ to get Joules for the denominator.

That is a valid calculation for gamma.

Seems sound.

So you have computed $\gamma$ based on the energy ratio that you were given (50 MeV from the problem and $9.11 \times 10^{-31} \text{ kg}$ from other references). Personally I would have looked up the rest mass of the electron in Mev (about 0.5 MeV) and skipped both conversions to joules. Then you can practically read off gamma without even using a calculator.

Yes inceed. Since you are given that $\frac{v}{c} = .999949$ then $\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-.999949^2}$.

As you say, this time you used velocity as the starting point instead of kinetic energy.

Yes, both methods are giving you the same answer. Which is what @Orodruin was trying to point out. The person who set the question purposely chose the kinetic energy (50 MeV) and the velocity (0.999949 c) so that the two figures were consistent.

Wow thanks,so meaning I answered the question correctly right?

 

[Herman Trivilino]

Wow thanks,so meaning I answered the question correctly right?

One of the skills needed for success in the course is the ability to judge the validity of your answers for yourself. Focus on sense-making rather than answer-making.

 

[the farmer]

One of the skills needed for success in the course is the ability to judge the validity of your answers for yourself. Focus on sense-making rather than answer-making.

But,I need that to discuss with my colleagues, I want to ensure that is correct

 

[Herman Trivilino]

But,I need that to discuss with my colleagues, I want to ensure that is correct

I know. That's my point. You need to learn how to do that. Otherwise you have to take it on someone else's authority and you never learn. You won't achieve success that way.

 

[the farmer]

I know. That's my point. You need to learn how to do that. Otherwise you have to take it on someone else's authority and you never learn. You won't achieve success that way.

You a right? It correct or i need to redo it again?

 

[PeroK]

Still don't understand what you meant

In classical physics, we have four quantities associated with a particle: mass, kinetic energy, magnitude of momemtum and speed. If you know any two of those, you can calculate the others.

E.g. if you have mass ($m$) and speed ($v$), then you have:

$KE = \frac 1 2 mv^2$ (i.e. kinetic energy is determined by mass and speed), and $p = mv$ (likewise, magnitude of momentum is determined by mass and speed).

And, if you know KE and magnitude of momentum, then:

$v = \frac{2KE}{p} = \frac{mv^2}{mv}$ (i.e. speed is determined by KE and magnitude of momentum)

In SR, you often add the rest energy to the kinetic energy to give energy instead and you have the gamma factor. Technically, we now have six possible quantities. But, again, you only need to know two to determine all six.

I suspect your professor assumes you know this or can work this out for yourself. In any case, I strongly suggest you work through the permutations and check that if you know, say, the magnitude of momentum and the mass, then you can calculate the total energy, kinetic energy, gamma factor and speed.