MT2Sol - SOLUTION TO PHYSICS 7C MIDTERM 2 PROBLEM 2 FALL...

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Unformatted text preview: SOLUTION TO PHYSICS 7C MIDTERM 2 PROBLEM 2 FALL 2007 (a). In the final state in the center-of-mass frame, there is a particle of mass M at rest, so E 2- p 2 c 2 = M 2 c 4 . By conservation of energy and momentum, and since E 2- p 2 c 2 is the same in every frame, the initial state in the lab frame must have M 2 c 4 = E 2- p 2 c 2 = ( mc 2 + mc 2 γ ) 2- ( mcvγ ) 2 = m 2 c 4 (1 + 2 γ + γ 2 )- m 2 c 2 v 2 γ 2 . Note that m 2 c 4 = m 2 c 4 γ 2- m 2 c 2 v 2 γ 2 , so substituting this and dividing by c 4 , M 2 = 2 m 2 (1 + γ ) , which given γ = M 2 2 m 2- 1 = 9 2- 1 = 7 2 . The kinetic energy is then Δ A = mc 2 γ- mc 2 = 2 . 5 GeV. (b). In this case, the total momentum is zero, so the final particle is at rest. Thus by conservation of energy, Mc 2 = mc 2 + mc 2 + Δ B , giving Δ B = ( M- 2 m ) c 2 = 1 GeV. (c). Since experiment B requires less kinetic energy, it is more cost effective. [This is why modern particle accelerators smash two beams of particles together rather than sending one beam into a fixed target.]than sending one beam into a fixed target....
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This note was uploaded on 10/05/2009 for the course PHYSICS 7C taught by Professor Lin during the Spring '08 term at Berkeley.

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MT2Sol - SOLUTION TO PHYSICS 7C MIDTERM 2 PROBLEM 2 FALL...

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