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Relativistic kinetic energy By Lisa Jacobson

Relativistic kinetic energy By Lisa Jacobson . The Basics. Mass increases as the speed of an object increases Therefore energy also increases (mass is a form of energy)

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Relativistic kinetic energy By Lisa Jacobson

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  1. Relativistic kinetic energy By Lisa Jacobson

  2. The Basics • Mass increases as the speed of an object increases • Therefore energy also increases (mass is a form of energy) • Einstein tried to find mathematical mass-energy relationship, but KE=1/2 mv2 doesn’t work at high speeds .......................................................

  3. But this formula does: KE = mc2 – moc2 E = mc2(energy of object with velocity) Eo = moc2 (rest energy) E = moc2 + KE (total energy = rest energy +kinetic energy) DE = (Dm)(c2)(changing energy in ANY system will change the mass in the system)

  4. An electron (mo = 9.11 x 10-31 kg) travels at a speed v = 0.85c = 2.5 x 108 m/s. What is the kinetic energy? Compare to a classical calculation.

  5. mo 9.11 x 10-31 kg An electron (mo = 9.11 x 10-31 kg) travels at a speed v = 0.85c = 2.5 x 108 m/s. What is the kinetic energy? Compare to a classical calculation. m = = = 1.73 x 10-30 kg v2c2 (1 – )1/2 (1 – 0.852 )1/2

  6. mo 9.11 x 10-31 kg m = = = 1.73 x 10-30 kg v2c2 (1 – )1/2 (1 – 0.852 )1/2 An electron (mo = 9.11 x 10-31 kg) travels at a speed v = 0.85c = 2.5 x 108 m/s. What is the kinetic energy? Compare to a classical calculation. KE = (m – mo)c2 = (-9.11 x 10-31 kg + 1.73 x 10-30 kg)c2 KE = 7.37 x 10-14 J

  7. mo 9.11 x 10-31 kg m = = = 1.73 x 10-30 kg v2c2 (1 – )1/2 (1 – 0.852 )1/2 An electron (mo = 9.11 x 10-31 kg) travels at a speed v = 0.85c = 2.5 x 108 m/s. What is the kinetic energy? Compare to a classical calculation. KE = (m – mo)c2 = (-9.11 x 10-31 kg + 1.73 x 10-30 kg)c2 KE = 7.37 x 10-14 J 12 12 KE = mov2 = (9.11 x 10-31 kg)(2.5 x 108)2 KE = 3.0 x 10-14 J [less then half of correct result]

  8. How much energy would be released if a po meson (mo = 2.40 x 10-28 kg) decayed completely into electromagnetic radiation? 2.16 E -11

  9. Eo = moc2 How much energy would be released if a po meson (mo = 2.40 x 10-28 kg) decayed completely into electromagnetic radiation? 2.16 E -11

  10. Eo = moc2 = (2.40 x 10-28 kg)c2 = 2.16 x 10-11 J or kgm2/s2 How much energy would be released if a po meson (mo = 2.40 x 10-28 kg) decayed completely into electromagnetic radiation?

  11. Relativistic momentum

  12. mov v2c2 (1 – )1/2 E = mc2 E2 = m2 c4 = m2 c2 (c2 + v2 – v2 ) = m2c2v2 + m2c2(c2 – v2) (since p = mv and m = mo/(1- v2/c2)1/2 ) p = mv = (Relativistic momentum) = p2c2 + m2oc4(1 – v2/c2)/(1 – v2/c2) E2 = p2c2 + mo2c4(Energy-momentum relation)

  13. An electron’s mass gets ten times as large as its rest mass (9.11 E -31) when it is traveling at 2.00 E 8 m/s. What is its momentum? E2 = p2c2 + m2oc4 (10*9.11E-31 *c2)2 = p2c2 + (9.11E-31)2 c4 = 2.72 E -21 kgm/s 2.72 E -21 kgm/s

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