Relativistic momentum is 1% greater than classical, at what speed?

AI Thread Summary
To determine the speed at which relativistic momentum is 1% greater than classical momentum, the equation P_r = mv/sqrt(1-(v^2/c^2)) was set equal to 1.01P_nr. The algebraic manipulation revealed an error in the square root calculation, leading to an incorrect speed of 0.0003884c. The correct answer, as noted in the textbook, is 0.14c. The discussion highlights the importance of careful algebraic handling in relativistic physics problems.
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Homework Statement


How fast must a body be traveling if its forrect relativistic momentum is 1% greater than the classical momentum


Homework Equations


P_r = mv/sqrt(1-(v^2/c^2))
P_nr = mv
p_r = 1.01P_nr

The Attempt at a Solution



mv/sqrt(1-(v^2/c^2)) = 1.01mv
mv = 1.01(mv)[sqrt(1-(v^2/c^2))]
1 = 1.01[sqrt(1-(v^2/c^2))]
1/1.01 = sqrt(1-(v^2/c^2))
(1/1.01)^2 = 1-(v^2/c^2)
0.980296 = 1-(v^2/c^2)
0.980296 - 1 = -(v^2/c^2)
-0.01970 = -(v^2/c^2)
(-0.01970)(c^2) = -v^2
0.01970c^2 = v^2
0.0003884c = v

However, this is not the correct answer. The actual answer in the back of the textbook is 0.14c. If anyone can correct my algebra or point me into the right direction it would be greatly appreciated.

Thanks
 
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chris_0101 said:
0.01970c^2 = v^2
0.0003884c = v

The square root of 0.01970 is not 0.0003884 :wink:
 
I cannot believe I overlooked that, it turned out I was squaring the 0.01970. Thanks a lot :)
 
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