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Unformatted text preview: PHY3101  Some useful formulas Note: t is the elapsed time in the frame considered “stationary” and l 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 /γ l = l /γ 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 ) (Δ s ) 2 = ( c Δ t ) 2 (Δ x ) 2 (Δ y ) 2 (Δ z ) 2 t = γ ( t vx/c 2 ) x = γ ( x vt ) y = y z = z T = E mc 2 1 . 2 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 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 / p 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, 4momentum 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...
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This note was uploaded on 02/17/2010 for the course PHY modern phy taught by Professor Prosper during the Fall '09 term at FSU.
 Fall '09
 PROSPER

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