a) For what of velocities of a particle of (rest) mass m is one justified to use

Question

a) For what of velocities of a particle of (rest) mass m is one justified to use the classical expression for kinetic energy mo2, to within 5% accuracy? (b) The antiproton p was originally discovered in the reaction p + p p+p + p + p in which an accelerated proton p was incident upon a proton at rest in the lab frame of reference. Assume that the products in the reaction move together in unison along the same direction and with the sam velocity and that the rest masses of the protorn and antiproton are identical, at 938 MeV/e. and find out the threshold kinetic energy of the impinging proton for the reaction to go through

Explanation / Answer

a. for 5% accuracy

let the speed be v

then

KEc = 0.5mv^2

KEr = m*c^2/sqroot(1 - v^2/c^2) - mc^2

hence

KEc - KEr = 0.05*Ker

0.5mv^2 - mc^2/sqroot(1 - v^2/c^2) + mc^2 = 0.05mc^2/sqroot(1 - v^2/c^2) - 0.05mc^2

0.5v^2/c^2 + 1.005 = (1.005)/sqrot(1 - v^2/c^2)

let v^2/c^2 = x

(0.5x + 1.005)^2 * (1 - x)= 1.005^2

0.25x^2 + 0.755x + 0.005025 = 0

x^2 + 3.02x + 0.0201 = 0

v = 0.08167c

b. threshold kinetic energy = KE

from energy balance

KE + 2*938 = 4*938

KE = 1876 MeV

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