Suppose you have a collection of particles with total momentum p in the x-direct

Question

Suppose you have a collection of particles with total momentum p in the x-direction, and total energy E (a) Show that there is a velocity v such that a person moving with this velocity in the x-direction would see no momentum. This frame s called the center of mass frame. Find a formula for v. Use p for the total momentum, E for the total energy, and for the speed of light. v = (b) Two particles both have mass m. One is stationary, and the other is moving to the right with velocity o.8B0c. What is the total energy (in units of mc^2) and momentum (in units of mc) of this system? E = mc^2 p = mc^2 (c) What is the center of mass velocity for the system described in part (b), as a fraction of the speed of light? v = c

Explanation / Answer

a. for given collection of particles

let velocity of ith particle be vi

then total momentum = (sum of) mivi = p

now, let the velocity of the person moving be v

then velocity of ith particle relative to person = vi - v

so momentum of system as seen by the person = (sum of)(mi(vi - v)) = (sum of)(mivi) - (sum of)(miv) = p - v*(sum of)mi

let toal mass of the system be M

then sum of (mi) = M

so, momentum seen by person = p - Mv

for this momentum to be 0

Mv = p

v = p/M [ where M is sum of masses of all particles of the system]

now, for total energy E, total mass M

E = 0.5Mv^2 = 0.5v*p

v = 2E/p

b. two particles of mass m, one moving with speed 0.88c

total energy, E = 2mc^2 + KE

KE = mc^2[1/sqroot(1-v^2/c^2) -1] = mc^2[1/sqroot(1 - 0.88^2) -1] = 1.1053mc^2

so total energy = 3.10153mc^2

momentum = mv/sqroot(1 - v^2/c^2) = 0.88mc/sqroot(1 - 0.88^2) = 1.8527 mc

c. centre of mass velocity = (KE)/p = 1.1053c/1.8527 = 0.5965777 c

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