this reduces to:

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


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.)


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


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?


Solution:We use Lorentz transformation equation in the “interval” form:

2'

c

xutt

032.1247.01

1

1

122

c

uFirst, evaluate γ:


We also have:

brbr

br

ttsottt

kmxxx

,0

4.3004.30


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

• 2. The Lorentz Transformation:

2cxutγt'

zz'yy'

utxγx'

SUMMARY

6. The Velocity Transformation:

2cxvu

1

uxv'xv

2c

'xuv

1

'xv

xv

u

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

SUMMARY

10. Useful Formulas for Speed, Energy, and Momentum:

22222 mccpE

E

pc

c

u

2mcEforpcE

• Specific Heat:

• Latent Heat:

• Work done by engine:

• Thermal Efficiency:

TmcQTm

Qc

LmQm

QL

chnetengine QQQW

eng 1h c c

h h h

W Q Q Qe

Q Q Q

this reduces to:

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