Special relativity problem (momentum and velocity)

[Quazswed]

Homework Statement

A particle traveling at 8 E7 m/s is accelerated so that its momentum doubles. What is the final velocity of the particle?

Homework Equations

p=ym0v
y= 1 / sqr(1-(v/c)^2)
p2= 2(p1)
Where p2 is the final momentum and p1 the initial

The Attempt at a Solution

For the initial velocity gamma=1,0376
p1 is therefore, calculated with m0=1 (I put it this way thinking it was not relevant to the problem, and it was meant to be eliminated when finding p2) 83008000 kg m/s
I tried to compare the two equations for the relativistic momentum but the best i came up with was that (v2)x(y2)=166016000 m/s
Given that the result written on the book is 1,5 E8 m/s, y2 must be 1,107
because y2= 2y1v1 / v2
But for a velocity of 1,5 E8 m/s (0,5c), the dilatation factor is 1,154
I really don't know, I keep trying the same method, I can't see a different way to resolve it

Sorry for the bad english, not my lenguage
Thanks to everybody in advance

 

[Doc Al]

A particle traveling at 8 E8 m/s

That can't be right. What is the speed?

 

[Quazswed]

uops, I'm sorry, i meant E7
anyway I used 8 E7 in the equations
Also, edited in the first post, thanks for pointing that out

 

[Doc Al]

I tried to compare the two equations for the relativistic momentum but the best i came up with was that (v2)x(y2)=166016000 m/s

Assuming your arithmetic is correct (I didn't check) write the left hand side entirely in terms of v2. Then solve for v2.

 

[Quazswed]

2y1v1 is = 166011471 m/s, I think it's correct now

so

2y1v1 = 166011471 m/s = k

y2= 1 / sqr(1-(v2/c)2)

v2 / sqr(1-(v2/c)2) = k

v2^2 = k^2 - k^2(v2/c)^2

(k^2/c^2 +1)v2 = k^2

v2 = sqr( k^2 / (k^2/c^2 +1))

which is 145254607 m/s, it's so close to 150000000 m/s

Is this an error caused by bad arithmetic or something else, more important?

 

[Doc Al]

which is 145254607 m/s, it's so close to 150000000 m/s

Is this an error caused by bad arithmetic or something else, more important?

It's not an error. The book just rounded off to 2 digits.

 

[Quazswed]

Seems like I was too pessimistic then :)
Thank you very, very much for your help