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TEST 1 Spring 2008 solutions1

TEST 1 Spring 2008 solutions1 - PHY3101 Some useful...

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PHY3101 - Some useful formulas Note: t 0 is the elapsed time in the frame considered “stationary” and l 0 is the length of an object at rest relative to the observer. Remember, however, that every inertial frame may be considered to be “stationary”. γ = 1 (1 - v 2 /c 2 ) t = t 0 l = l 0 γ ( u ) = 1 (1 - u 2 /c 2 ) E = γ ( u ) mc 2 p = γ ( u ) mu E 2 - ( pc ) 2 = ( mc 2 ) 2 E = pc for a massless particle u = ( v 1 - v ) / (1 - v 1 v/c 2 ) s ) 2 = ( c Δ t ) 2 - x ) 2 - y ) 2 - z ) 2 t = γ ( t - vx/c 2 ) x = γ ( x - vt ) y = y z = z T = E - mc 2 1
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Intermediate Modern Physics - PHY3101 Exam 1 February 11, 2008 PRINT Your Name SIGNATURE IMPORTANT *** You must show ALL your work to receive full credit *** *** and ALL answers must have the correct units For Instructor Only 1. 2. 3. 4. 3
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1 A proton moving at speed v = 0 . 999 c will appear like a moving disk. (a) Compute the thickness of the disk assuming that at rest the proton’s diameter is 10 - 15 m. (b) Compute the proton’s kinetic energy in MeV/c, assuming its mass is 938 MeV/c 2 . Suppose the proton collides and fuses with a stationary proton to create a new particle. (c) What is the mass of the new particle? (a) Let d be the diameter of the proton at rest in some frame of reference. When the proton moves relative to this frame, its diameter will be Lorentz- contracted in the direction of motion. Therefore, the proton’s thickness will be t = d/γ , where γ = 1 / 1 - β 2 and β = v/c . Setting β = 0 . 999 and d = 10 - 15 m gives t = 4 . 5 × 10 - 17 m = 4 . 5 × 10 - 2 fm. (b) The total energy of a particle is given by E = γmc 2 , therefore, its kinetic energy is T = E - mc 2 , that is, T = ( γ - 1) mc 2 . Therefore, since γ = 22 . 4, T = (22 . 4 - 1) × (938MeV/c 2 ) × c 2 = 20 , 041 MeV, that is, 20.04 GeV. (c) In collisions, in a given frame of reference, energy and momentum are conserved, or, equivalently, 4-momentum P = ( E, p ) is conserved. That is, E 1 + E 2 = E F , and p 1 + p 2 = p F . Since E 1 = γmc 2 and E 2 = mc 2 , the total initial energy is γmc 2 + mc 2 = 21 , 917 MeV, which must equal the total final energy. Likewise, the total initial momentum is p 1 + 0 = γmv + 0 = γmc 2 β/c, = 20 , 958 MeV/c , which must be equal to the total final momentum. Now that we have the energy and momentum of the final state, we can compute its mass using the general relation E 2 = ( pc ) 2 + ( mc 2 ) 2 , that is, using m = E 2 - ( pc ) 2 /c 2 = 21 , 917 2 - 20 , 958 2 /c 2 = 6 , 400 MeV/c 2 .
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