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An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is 2.50 × 10 –28 kg , and that of the other is 1.67 × 10 –27 kg. If the lighter fragment has a speed of 0.893 c after the breakup, what is the speed of the heavier fragment?.
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An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is 2.50 × 10–28 kg, and that of the other is 1.67 × 10–27 kg. If the lighter fragment has a speed of 0.893cafter the breakup, what is the speed of the heavier fragment?
An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is 2.50 × 10–28 kg, and that of the other is 1.67 × 10–27 kg. If the lighter fragment has a speed of 0.893cafter the breakup, what is the speed of the heavier fragment? Relativistic momentum of the system of fragments must be conserved. For total momentum to be zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and subscript 1 to the lighter,
An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is 2.50 × 10–28 kg, and that of the other is 1.67 × 10–27 kg. If the lighter fragment has a speed of 0.893cafter the breakup, what is the speed of the heavier fragment?
An unstable particle with a mass of 3.34 × 10–27 kg is initially at rest. The particle decays into two fragments that fly off along the x axis with velocity components 0.987cand –0.868c. Find the masses of the fragments. (Suggestion: Conserve both energy and momentum.)
An unstable particle with a mass of 3.34 × 10–27 kg is initially at rest. The particle decays into two fragments that fly off along the x axis with velocity components 0.987cand –0.868c. Find the masses of the fragments. (Suggestion: Conserve both energy and momentum.)
An unstable particle with a mass of 3.34 × 10–27 kg is initially at rest. The particle decays into two fragments that fly off along the x axis with velocity components 0.987cand –0.868c. Find the masses of the fragments. (Suggestion: Conserve both energy and momentum.) This reduces to:
Suppose a heat engine is connected to two energy reservoirs, one a pool of molten aluminum (660°C) and the other a block of solid mercury (–38.9°C). The engine runs by freezing 1.00 g of aluminum and melting 15.0 g of mercury during each cycle. The heat of fusion of aluminum is 3.97 105 J/kg; the heat of fusion of mercury is 1.18 104 J/kg. What is the efficiency of this engine?
Suppose a heat engine is connected to two energy reservoirs, one a pool of molten aluminum (660°C) and the other a block of solid mercury (–38.9°C). The engine runs by freezing 1.00 g of aluminum and melting 15.0 g of mercury during each cycle. The heat of fusion of aluminum is 3.97 105 J/kg; the heat of fusion of mercury is 1.18 104 J/kg. What is the efficiency of this engine? The heat to melt 15.0 g of Hg is: The energy absorbed to freeze 1.00 g of aluminum is: and the work output is: or
2. (a) Find the rest energy of a proton in electron volts; (b) If the total energy of a proton is three times its rest energy, what is the speed of the proton? (c) Determine the kinetic energy of the proton in electron volts.
Rest energy: E0=mpc2 = (1.67 x 10-27kg) x (3 x 108m/s)2 = 1.5 x 10-10J x (1eV / 1.6 x 10-19J) 2. (a) Find the rest energy of a proton in electron volts; (b) If the total energy of a proton is three times its rest energy, what is the speed of the proton? (c) Determine the kinetic energy of the proton in electron volts. = 938 MeV (b) The total energy K = Etot – E0 = 3mpc2 – mpc2 = 2mpc2 = 2 x (938 MeV) = 1876 MeV (c) Kinetic energy:
A proton in high-energy accelerator moves with a speed c/2. Use the work kinetic energy theorem to find the work required to increase the speed to (a) 0.750c and (b) 0.995c.
An experimenter arranges to trigger two flashbulbs simultaneously, a blue flush located at the origin of his reference frame and a red flush at x = 30.4 km. A second observer, moving at speed 0.247c in the direction of increasing x, also views the flushes. (a) What time interval between them does she find? (b) Which flush does she say occurs first?
An experimenter arranges to trigger two flashbulbs simultaneously, a blue flush located at the origin of his reference frame and a red flush at x = 30.4 km. A second observer, moving at speed 0.247c in the direction of increasing x, also views the flushes. (a) What time interval between them does she find? (b) Which flush does she say occurs first? Solution: We use Lorentz transformation equation in the “interval” form: First, evaluate γ:
An experimenter arranges to trigger two flashbulbs simultaneously, a blue flush located at the origin of his reference frame and a red flush at x = 30.4 km. A second observer, moving at speed 0.247c in the direction of increasing x, also views the flushes. (a) What time interval between them does she find? (b) Which flush does she say occurs first? We also have:
An experimenter arranges to trigger two flashbulbs simultaneously, a blue flush located at the origin of his reference frame and a red flush at x = 30.4 km. A second observer, moving at speed 0.247c in the direction of increasing x, also views the flushes. (a) What time interval between them does she find? (b) Which flush does she say occurs first? We got and therefore since Δt’< 0, we have tb > tr. Hence, the red flush is seen first.
SUMMARY 1. Einstein’s Postulates: • Postulate 1: Absolute uniform motion can not be detected. • Postulate 2: The speed of light is independent of the motion of the source.
SUMMARY • 2. The Lorentz Transformation:
SUMMARY 6. The Velocity Transformation:
SUMMARY • 3. Time Dilation: • 4. Length Contraction: • 5. The Relativistic Doppler Effect: approaching receding
SUMMARY 7.Relativistic Momentum: 8. Relativistic Energy: 9. Rest Energy: 10. Kinetic Energy
SUMMARY 10.Useful Formulas for Speed, Energy, and Momentum:
Specific Heat: • Latent Heat: • Work done by engine: • Thermal Efficiency: