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1 Relativity (Option A) A.4 Relativistic momentum and energy.

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1 1 Relativity (Option A) A.4 Relativistic momentum and energy

2 2  Two basic physical quantities need to be modified  Length (contraction)  Time (dilation)  Other physical quantities will need to be modified as well

3 3 Relativistic momentum  To properly describe the motion of particles within special relativity, Newton’s laws of motion and the definitions of momentum and energy need to be generalized  These generalized definitions reduce to the classical ones when the speed is much less than c i.e. v << c

4 4 Relativistic momentum  To account for conservation of momentum in all inertial frames, the definition must be modified  v is the speed of the particle, m is its mass as measured by an observer at rest with respect to the mass  When v << c, the denominator approaches 1 and so  p approaches mv

5 5 Problem: particle decay  An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is 2.50 × 10– 28 kg, and that of the heavier fragment is 1.67 × 10–27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?

6 6 An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is 2.50 × 10 –28 kg, and that of the heavier fragment is 1.67 × 10 –27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment? Given: v 1 = 0.8 c m 1 =2.50×10 –28 kg m 2 =1.67×10 –27 kg Find: Find: v 2 = ? v 2 = ? Momentum must be conserved, so the momenta of the two fragments must add to zero. Thus, their magnitudes must be equal, or For the heavier fragment, which reduces to and yields

7 7 Relativistic energy  The definition of kinetic energy requires modification in relativistic mechanics E k = KE =  m o c 2 – m o c 2 = (γ-1) m o c 2  The term m o c 2 is called the rest energy of the object and is independent of its speed  The term  m o c 2 is the total energy, E, of the object and depends on its speed and its rest energy

8 8  A particle has energy by virtue of its mass alone  A stationary particle with zero kinetic energy has an energy proportional to its inertial mass E = m o c 2 E = m o c 2  The mass of a particle may be completely convertible to energy and pure energy may be converted to particles

9 9 Energy and Relativistic Momentum  It is useful to have an expression relating total energy, E, to the relativistic momentum, p  E 2 = p 2 c 2 + (mc 2 ) 2  When the particle is at rest, p = 0 and E = mc 2  Massless particles (m = 0) have E = pc  See derivation in class  This is also used to express masses in energy units  mass of an electron = 9.110 x 10 -31 kg = 0.511 MeVc -2  Mass of a proton = 1.673 x 10 -27 kg = 938 MeVc -2  Conversion: 1 u = 931.5 MeVc -2

10 Practice Problems HL Physics 2 nd ed Hamper   Calculate the momentum of an electron accelerated to a total energy of 2.0 MeV [1.9 MeVc -1 ]   Calculate the speed of an electron accelerated theough a potential difference (voltage) of 1.0 MV [0.94c]   Calculate the potential difference (i.e. V) needed to accelerate an electron to a velocity of 0.80c [0.34 MV]   Calculate the speed of an electron with momentum of 2.0 MeVc -1   [v= 0.97c]   Find the momentum of a particle of rest mass 100 MeVc -2 travelling at 0.80c [133MeVc -1 ]   A particle of rest mass 200 MeV is accelerated to a E k of 1.0 GeV. Calculate its momentum and velocity [1183MeVc -1 ; 0.99c]   A proton has momentum 150MeVc -1. Calculate its total energy and E k [950 MeV; 12MeV]

11 Practice Problems Tsokos 5 th ed p. 674   Find the momentum of a pion (rest mass 1.35 MeVc -2 ) whose speed is 0.80c [180 MeVc -1 ]   Find the speed of a muon (rest mass 105 MeVc -2 ) whose momentum is 228 MeVc -1 [0.91c]   Find the kinetic energy of an electron whose momentum is 1.5 MeVc -1 [1.07 MeV]   A particle at rest breaks apart into two pieces of masses 250 MeVc -2 and 125 MeVc -2. The lighter fragment moves away at 0.85c. Using conservation of momentum and total energy find a) the speed of the other fragment and b) the rest mass of the original particle. [0.628c; 558.5MeV]


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