Relativity and Equivalence of Mass and Energy

[Alena Selone]

Homework Statement

An electron is accelerated to a speed that is 99 percent the speed of light, and is moving through a 2-km-long tunnel. The rest mass of the electron is 9.11*10^-31 kg. What is the mass of the electron at this speed?
c= speed of light

Homework Equations

t= (tsubscript(o))/ root(1-(v^2/c^2)
L= Lsubscript(o)* root(1-0)= Lsubscript(o)
p= (mv)/ root(1-0) = mv
KE= ((mc^2)/ root(1-(v^2/c^2))-mc^2

The Attempt at a Solution

I tried plugging the rest mass into the last equation on the top which gives me 9.11*10^-31/ root(1-((.99c)^2/c^2
which calculates out to
9.11*10^-31/ root(1-.9801)
9.11*10^-31/root(0.0199)
9.11*10^-31/0.14106736
=6.458*10^-30

I'm unsure if this is correct. It's listed as one of the answers but I don't know if I used the correct equation, so the fact that it's listed as an answer could be a trick. I also don't know how the 2km long tunnel plays into the equation. Please help!

 

[Doc Al]

Looks fine to me. The length of the tunnel is irrelevant. (Note that "relativistic mass" is a rather antiquated concept nowadays.)

 

[Alena Selone]

Looks fine to me. The length of the tunnel is irrelevant. (Note that "relativistic mass" is a rather antiquated concept nowadays.)

So in a different equation using some of the same values, say,
An electron is accelerated to a speed that is 99 percent the speed of light, and is moving through a 2-km-long tunnel.
Could I calculate the length of the tunnel in the frame of reference of the electron or is that too irrelevant?

 

[Doc Al]

Could I calculate the length of the tunnel in the frame of reference of the electron or is that too irrelevant?

Sure you can. For that problem, the "rest" length of the tunnel is very relevant.

 

[Alena Selone]

Sure you can. For that problem, the "rest" length of the tunnel is very relevant.

So how would I do that?

 

[Doc Al]

So how would I do that?

Look up the formula for "length contraction" (one of the key relativistic effects).