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A particle zips by us with a Lorentz factor (γ) of 1.12. Then another particle zips by us moving at twice the speed of the first particle.
a) What is the Lorentz factor (γ) of the second particle?

b) If the particles were moving with a speed much less than c, the magnitude of the
momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?

Sagot :

Jcherry99idn

Answer:

Explanation:

Part A

Lorentz factor = 1/(sqrt(1 - (v/c)^2 ))

1.12 = 1/(sqrt(1 - v^2/c^2))                         square both sides.

1.2544 = 1 / (1 - v^2/c^2)                          Multiply both sides by 1 - v^2/c^2

1.2544 * (1 - v^2/c^2) = 1                         Remove the brackets

1.2544  - 1.2544 v^2 /c^2 = 1                 Multiply c^2 through the entire equation

1.2544*c^2 - 1.2544v^2 = c^2                Subtract 1.2544 c^2 from both sides

-1.2544v^2 = - 0.2544 c^2                     Divide by - 1.2544

v^2 = 0.2544 c^2 /1.2544

v^2 = 0.2028 c

v =  0.4534 * c

v = 1.360 * 10^8

2*v = 2.272 * 10*8

Lorentz Factor = LF = 1/square root( (1 - 2.272* 10^8/3*10^3))

LF = 2.345

Part B

Momentum is mv

I still get two even though the Lorentz factors are different.