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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?. - PowerPoint PPT Presentation
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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 0.893c after 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 0.893c after 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,
2 1p p
28
2 2 2 1 1 1 2
2.50 10 kg0.893
1 0.893mu mu c
27
2 28
22
1.67 10 kg4.960 10 kg
1
uc
u c
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 0.893c after the breakup, what is the speed of the heavier fragment?
2 2272 2
28 2
1.67 101
4.960 10
u uc c
22212.3 1u
c
2 0.285u c
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 0.987c and –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 0.987c and –0.868c. Find the masses of the fragments. (Suggestion: Conserve both energy and momentum.)
1 2
12.01
1 0.868
2 2
16.22
1 0.987
1 2 totalE E E 2 2 2
1 1 2 2 totalmc m c m c
271 22.01 6.22 3.34 10 kgm m
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 0.987c and –0.868c. Find the masses of the fragments. (Suggestion: Conserve both energy and momentum.)
27
1 22.01 6.22 3.34 10 kgm m This reduces to:
271 23.09 1.66 10 kgm m
1 2p p 1 1 1 2 2 2mu mu
1 22.01 0.868 6.22 0.987c m c m
1 23.52m m
281 8.84 10 kgm
282 2.51 10 kgm
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:
3 415 10 kg 1.18 10 J kg 177 Jc fQ mL
The energy absorbed to freeze 1.00 g of aluminum is:
3 510 kg 3.97 10 J/ kg 397 Jh fQ mL
and the work output is: eng 220 Jh cW Q Q
eng 220 J0.554
397 Jh
We
Q or 55.4%
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.
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.
smccuc
u
c
u
cu
cu
cmcmEE pptot
/1083.2943.03
8
9
8
9
11
1
13
1
33
82
2
2
2
2
2
2
2
22
0
(a)Rest energy:E0=mpc
2 = (1.67 x 10-27kg) x (3 x 108m/s)2 = 1.5 x 10-10J x (1eV / 1.6 x 10-19J)
= 938 MeV
(b) The total energy
(c) Kinetic energy:K = Etot – E0 = 3mpc
2 – mpc2 = 2mpc
2 = 2 x (938 MeV) = 1876 MeV
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:
2'
c
xutt
032.1247.01
1
1
122
c
uFirst, 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:
brbr
br
ttsottt
kmxxx
,0
4.3004.30
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?
sssm
m
c
xutt 8.251058.2
/103
)104.30)(247.0(0)032.1( 5
8
3
2'
We got br ttt and therefore '''
br ttt
since Δt’< 0, we have tb > tr. Hence, the red flush is seen first.
MeVeV
eVJJmcE
JeV
Jsmkgcm
kgm
electronFor
reste
e
e
511.010511.0
)/10602.1/()102.8(
10602.11
102.8)/103)(1011.9(
1011.9
:
6
19142,
19
1428312
31
MeV
eVeVJJE
Jsmkgcm
kgm
protonFor
restp
p
p
6.938
106.938.0)/10602.1/()1051.1(
1051.1)/103)(10673.1(
10673.1
:
91910,
1028272
27
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
• 3. Time Dilation:
• 4. Length Contraction: • 5. The Relativistic
Doppler Effect: 0fuc
uc'f
approaching
0fuc
uc'f
receding
02
o
c
u1
TT T
0'L
L
SUMMARY
2
2
1cu
mup
)1(
1
22
2
2
2
mcmc
c
u
mcK
7. Relativistic Momentum:
8. Relativistic Energy:
9. Rest Energy: 20 mcE
10. Kinetic Energy
2
2
222
1c
u
mcmcKmcE