Tài liệu Physics exercises_solution: Chapter 37 pptx

37.21 with the primed coordinates replacing the unprimed, and a change of sign of u... We want the velocity v of the cruiser knowing the velocity of the primed frame u and the velocity

37.1: If O sees simultaneous flashes then O will see the A( A) flash first since O would believe that the A flash must have traveled longer to reach O , and hence started first.

)9.0(1

1γ

(1

c

u c

u c

u

s

1000.5)(

)sm102(3.00

(3600)hrs)

4()sm250(2

))(

11()(

9 0

2 8

2 2

2 2

2 0

t c

u t c u t

t

The clock on the plane shows the shorter elapsed time

)978.0(1

1γ

2 2

u c

u

t t

2 2

0

42

6.21

m1020.1γ

b) (0.300s)(0.800c)7.20107m

c) t0 0.300s γ0.180s. (This is what the racer measures your clock to read at that instant.) At your origin you read the original 0.5s

)sm10(3(0.800)

m1020.1

37.7: (1yr)

sm103.00

sm104.801

1

2 8

6 2

37.8: a) The frame in which the source (the searchlight) is stationary is the spacecraft’s

frame, so 12.0 ms is the proper time b) To three figures, u  Solving Eq (37.7) for c

c

u in terms of γ,

.γ2

11)γ1(



c u

Using 1 γt0 t 12.0ms190 msgivesu c0.998

)9860.0(1

1)

(1

1γ

a) 9.17km

6.00

km55γ

651.0

%

km

0.651s)

1020.2)(

9860.0

c t

u d

c) In earth’s frame:

%

1.7km55.0

km90.3

%

km3.90s)

10)(1.32(0.9860

s101.32s)(6.00)10

2.2(γ

5

5 6

c t

u d

t t

0.99540

m10

km

4.3110.44

km45γ

10.44(0.9954)

1

1γ

b

γands,101.440.99540

37.11: a) l0 3600m

m

3568m)(0.991)

(3600

)sm1000.3(

)sm1000.4(1)m3600(

l

)

sm104.00

m

7

0 0

sm104.00

37.12: γ1 0.3048,sou c 1(1 γ)2 0.952c2.86108 m s

37.13:

2 2 0

2

2 0

1

1

c u

l l

c

u l

1

m74.0

37.14: Multiplying the last equation of (37.21) by u and adding to the first to eliminate t

11

γand)(

γ

so x xu t t tu x c2

which is indeed the same as Eq (37.21) with the primed coordinates replacing the

unprimed, and a change of sign of u.

c v u

u v

)600.0()400.0(1

600.0400.0

u v

)600.0)(

900.0(1

600.0900.0

990.0(1

600.0990.0

c v u

u v

37.16: γ1.667(γ5 3if u(4 5)c) a) In Mavis’s frame the event “light on” has space-time coordinates x0 and t5.00 s, so from the result of Exercise 37.14 orExample 37.7, xγ(xu t)and      

2

1 u c

ut x x

)(xut u t x

γγ

γ1

1

11

1

11

γ

γ

1

γγ

1γγγγ

2

2 2

2 2 2

2 2 2

2 2

2

c u

c ux t c

xu t t

c u c

u

c u c

u

c u c

u c

u

u

x t u

x t u

x t

(

)1

(

1

2 2 2

2

c v u v

c uv v v u v

u v c uv v

c uv

u v v

37.19: Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship

We want the velocity v of the cruiser knowing the velocity of the primed frame u and the velocity of the cruiser v in the unprimed frame (Tatooine).

c c

c c

uv

u v

)800.0()600.0(1

800.0600.0

37.20: In the frame of one of the particles, u and v are both 0.9520c but with opposite sign.

.9988.0)9520.0)(

9520.0(1

9520.09520.0)

()(1

)(

c v u

u v

)650.0)(

950.0(1

650.0950.0

c c

c c

v u

u v

400.0700.01

700.0,400

u v v c v

c u

m1000

x

c

v uv v c uv

u v

c u

c v

v v u

v v c

v u

837.0))920.0)(

360.0(1(

920.0360.0

)1

 moving opposite the rocket, i.e., away from Arrakis

37.24: Solving Eq (37.25) foru , (see solution to Exercise 37.25) c

,)(1

)(1

2 0

2 0

f f

f f c

u





and so (a) if f f0 0.98, u c 0.0202, the source and observer are moving away from each other b) if f f0 4, u c 0.882, they are moving toward each other

0

2

0 (c u)f (c u)f f

u c

u c

)1λ)λ((

1)(

)1)((

)(

2 0

2 0 2

0

2 0 2

2 0

2 0 2

f f c f

f

f f

((

)1)575675

37.26: Using u0.600c 3 5c in Eq (37.25) gives

 

525

31

531

0 0

1 2 1 2 2 2

)2(1

c va mv

c v

ma c

v

mv dt

d dt

dp F

)1

()

1(

)1

(

2 2 2

2 2 2 2

2 2

2 2 2 2

F a

c v ma c

v

c v c v ma

b) If the force is perpendicular to velocity then denominator is constant   

dt

dp F

.1

1

2 2 2

m

F a c

34

3

14

11

21

2 2

2

2 2

2

c c

v c v

c

v c

v ma

/ 1 3

1

1so)2(γ2γ2

γ

 12 2 / 3 0.608

37.30: a) γ1.01,so v c 0.140andv4.21108m s b) The relativistic

expression is always larger in magnitude than the non-relativistic expression.

37.31: a) 2 2

2 2

4

121

1

2 2 2

c

v c

36

161

1

2 2

c

v c

v mc

1)

1γ

(

c

mv mv

mc K

s

m1089.4163.0150

44

150

2

02.08

32

102.1

7 2

2

2 2

4 2

v c v

v c

v mv

K K

c) The result of part (b) is far larger than that of part (a)

37.35: a) Your total energy E increases because your potential energy increases;

%103.3)sm10998.2()m30(sm80.9()

(

)(so

)(

13 2

8 2

2

2 2

m

c y mg c E m c

m E

y mg

E

This increase is much, much too small to be noticed

b) (2.00 102 N m)(0.060m)2 36.0J

2 1 2 2

/1

c v

m m

0

434

31

4

1

8 2

2 2 2

c v c

v c v

2 2 0 2

2 0

1

c v

m mc

c m

100

99100

11

8 2

2 2

v

c

v c

v

37.37: a) E mc2 K,so E4.00mc2 meansK 3.00mc2 4 010 10 J

b) E2 (mc2)2 (pc)2; E 4.00mc2,so15.0(mc2)2 (pc)2

smkg1094.1

v c

v mc

1(

2

2 m c c

m K

10.011

1J)1050.1(

11

1)

sm10kg)(3.0010

67.1(

13 10

2 c 0

2 8 27

(0.5) 1 1 10

K

J1032

(0.9) 1 1 10

K

J1094

1   10



37.39: (m6.6410 27 kg, p2.1010 18 kgm s)

a) E  (mc2)2 (pc)2

J

1068

8   10

b) K Emc2 8.6810 10 (6.6410 27 kg)c2 2.7010 10 J

kg)10(6.64

J1070.2

2 27

K

37.40:

2 2 2

2 2 2 4

c p c m

E

2 2

2 2 2

2

2 2

2

12

sm108

2 2

.06.1

J1065.5)

1γ(

K

b) v2.85108 m sγ3.203

.88.4J

1031.31)

(γ

J1078.62

1

0 10

2

11 2

K

mv K

37.42: (5.5210 27 kg)(3.00108 m s)2 4.9710 10 J3105MeV

37.43: a) K qV eV

V1006.2

MeV2.06J

10295.3025

.411

1

6

13 2

2 2 2

mc c

v mc

K

b) From part (a), K 3.3010 13 J2.06MeV

37.44: a) According to Eq 37.38 and conservation of mass-energy

.292.1)7.16(2

75.91212

M

m Mc

mc Mc

Note that since 2 2 ,

1

1

c v





 we have that

.6331.0)292.1(

11

b) According to Eq 37.36, the kinetic energy of each proton is

MeV

274

J101.60

MeV00.1)sm10kg)(3.0010

67.1)(

1292.1()

2

13

)sm10kg)(3.0010

75.9(

37.45: a) E 0.420MeV4.20105 eV

b)

eVJ106.1

)sm1000.3(kg)1011.9(eV1020

2 8 31

5 2

105.11eV1020



c)

2 2 2

v c v

mc E

.sm1051.2836.0eV109.32

eV1011.5

2 5

)eVJ10eV)(1.610

20.4(22

2

1

31

9 5

mv K

3.84108 m s

37.46: a) The fraction of the initial mass that becomes energy is

,10382.6u)2(2.0136

u)0015.4

(

1    3 and so the energy released per kilogram is

J

1074.5)sm10kg)(3.0000

.1)(

J100

37.47: a) E mc2,mE c2 (3.81026 J) (2.998108 m s)2 4.2109 kg

1 kg is equivalent to 2.2 lbs, so m4.6106 tons

b) The current mass of the sun is 1.991030 kg, so it would take it

y101.5s104.7)skg10(4.2kg)

10

99

1

(  30  9   20   13 to use up all its mass

37.48: a) Using the classical work-energy theorem we obtain

m

81.1N)

1020.4(2

)]

sm10)(3.00kg)[(0.92010

00.2(2

)(

4

2 8

12 0

x

b) Using the relativistic work-energy theorem for a constant force (Eq 37.35) we obtain

.)1

920 0 1

1020.4(

)sm10kg)(3.0010

00.2)(

155.2(

4

2 8 12

)sm1010.2(1

kg)10(2.00

N)1020.4(

3 2

2 2

16 2

3 2

2 12

4 2

3 2

v c

v m

)462.0(sm1046.1ii)

2 16

2 16

β a

sm103

But

2 0 2

u c

u t

t

.1011.210

00.4

106.2111

1

1)1(

5

2 2 6 8 2

2 0

2 0 2

106.2

1000.4

6 2

E

 E 2.15104 MeV

37.50: One dimension of the cube appears contracted by a factor of ,

b a

b l



.sm1010.2

.700.040.1

111

8

2 2

yr42.2yr

19γ

yr42.2yr

u c

37.53: a)

100

99γ

1γ)

(1

110

2 2

c

v c

v c

v mc

c

v c m E c v m pc

%

101.0)995.0(10(1

11

1)

(

2 2

2 2

2 2

pc E

37.54: a) Note that the initial velocity is parallel to the x-axis Thus, according to Eqn

37.30,

.sm1049.1)900.01(kg)10(1.67

N)1000.3(

3 2 27

12 2

3 2

N)1000.5(

1 2 27

12 2

1 2

F

b) The angle between the force and acceleration is given by

.5.24

cos 12 122 1214 22 1412 2 15 152 2

) s m 10 31 1 ( ) s m 10 49 1 ( N) 10 00 5 ( N) 10 00 3 (

) s m 10 N)(1.31 10 00 5 ( ) s m 10 1.49 N)(

10 00 3 (

12 2

2

2 2.131 101

v c

c

v c c

v c v c c

v c

1

;)10131.2(

1

2 2

4 2

m

21

4 2

sm(9.80

J)1020.7

3 2

37.57: Heat in QmL f (4.00kg)(3.34105 J kg)1.34106 J

kg

1049.1)sm10(3.00

J)1034.1

2 8

6 2

mc

E m

c E m

p

v   where the atom and the photon have the same

magnitude of momentum, E c.b) c,

mc E

v  so E mc2

37.59: Speed in glass 1.97 10 m s

52.1

eV.101.67MeV0.167MeV)

511.0)(

326.0()

1(

326.11

1

5 2

2 2

c v

sm10

m0

m6.13or

s,1018

2 2 2)(xut  c  tux c



)()(

11

)(

2 2 2

2

t c x

ct x c u t c u x c c

u x

c ux t c ut x

sm1000.3(

)sm10(3.00kg)0.90(8

38

32

1

2 8

4 4 2

K

4

3)

21(

)83

2

2 4

c mv

37.63: (1 v2 c2) 2

m

F dt

F c

F c v

v x

x c x

dx c

0 2 2

2

)(11

)1(

2 2 2 2

Ft c

v m

Ft v

.)(

1 Ft mc 2

m Ft v

(

γ2 2 2 2 2 2 2 2

2

orxc t2 t2 4.53108 m

37.65: a) Want tt2 t1

2

2 2

2

1 2 1

2

1 2

1 2 2

2 2 1 1

1

1γ

γ

,Since

.)(

γ

And

γ)(

γ)(

c

x t c

t t t

c

x t c

x u t

t

t x u c

x x u t t t

t

x t

t

x x u ut

x x ut

1 10

c

ux t c

ux t





.2

t c u x c

u t

u x c u x

x

x

2 2 2

2 2 2

1 2

1

1)

()(11

2

2 2

t c x x

x

t c

c t

37.66: a) (100s)(0.600)(3.00108 m s)1.801010 m b) In Sebulbas frame, the relative speed of the tachyons and the ship is 3.40c and so the time , t2 100s

s

1183.4

118

,43.2,600.0and

43.2

m1012







c and so Sebulba receives Watto’s message before

even sending it! Tachyons seem to violate causality

37.67: Longer wavelength (redshift) implies recession (The emitting atoms are moving

away.) Using the result of Ex 37.26:

1λ)λ(

1λ)λ(0

2 0





 c u

sm10071.13570.01)4.9533.656(

1)4.9533.656

37.68: The baseball had better be moving non-relativistically, so the Doppler shift

formula (Eq (37.25)) becomes f  f0(1(u c)) In the baseball’s frame, this is the

frequency with which the radar waves strike the baseball, and the baseball reradiates at f.

But in the coach’s frame, the reflected waves are Doppler shifted again, so the detected frequency is (1 ( )) (1 ( ))2 0(1 2( )),so 2 0( )

f c u

fractional frequency shift is 2( )

0

c u

hkm154sm42.9

m)1000.3(2

)1086.2(2

8 7

37.69: a) Since the two triangles are similar:

mc A

37.70: a) As in the hint, both the sender and the receiver measure the same distance

However, in our frame, the ship has moved between emission of successive wavefronts, and we can use the time T 1 f as the proper time, with the result that f f0  f0

758.01

758.01MHz345

u c f f

f  f0 930MHz345MHz585MHz.Away:

MHz

217

MHz128758

.01

758.01MHz345

0

2 0

u c f

f

c) γf0 1.53f0 528MHz, f  f0 183MHz.The shift is still bigger than f0, but not as large as the approaching frequency

37.71: The crux of this problem is the question of simultaneity To be “in the barn at one

time” for the runner is different than for a stationary observer in the barn

The diagram below, at left, shows the rod fitting into the barn at time t 0, according to the stationary observer

The diagram below, at right, is in the runner’s frame of reference

The front of the rod enters the barn at time t and leaves the back of the barn at 1

time t2

However, the back of the rod does not enter the front of the barn until the later time t3

37.72: In Eq (37.23), uV,v(c n), and so

.)(1

)(1

)(

2

nc V

V n c nc

cV

V n c

c

nc V n V V n c

nc V V

n c v

11

)(

)()

(

))(1)(

)((

dv a





 d t(dtudx c2)

dv c

u c uv

u v c

uv

dv v

)1

()1

)1

(1

1

c

u c

uv

u v c

uv dv

(

1)

1(

)(1

1

c uv

c u dv

c uv

c u u v c uv dv v

d

)1

(γ

1

)1

(

)1

(.γ

)1

(

)1

(

2 2

2

2 2 2

2 2

2 2

c uv c

uv

c u dt

dv c

dx u dt

c uv

c u dv a

2 2

u a

37.74: a) The speed vis measured relative to the rocket, and so for the rocket and its occupant, v0 The acceleration as seen in the rocket is given to be ag, and so the acceleration as measured on the earth is

.1

2 2

du a

b) With v1 0 when t 0,

.1

)1

(1

)1

(1

2 2 1

1 1

2 2 2

c v g

v t

c u

du g

dt

c u

du g

(

1)1

(1

1

2 2 2 2

2 2 2

du g

c u g

du c

u dt

t d

.arctanh 1

c t

For those who wish to avoid inverse hyperbolic functions, the above integral may be done by the method of partial fractions;

du c

u c u

du t

gd

11

2

1)1)(

cn

v g

c

d) Solving the expression from part (c) for v in terms of 1 t1,(v1 c)tanh(g t1 c), so that

),(cosh1)

)cosh(

1

)tanh(

1 1

1

g

c c t g

c t g g

.sinh 1

gt

If hyperbolic functions are not used, v in terms of 1 t is found to be1

c t g c t g

c t g c t g e e

e e c

1 1 1

1

37.75: a) 4.568110 1014 Hz 4.568910 1014 Hz 4.567710 1014 Hz

f

))(())((

))(())((

)(

)(

)(

)(

2 0 2

2 0 2

0

0

v u c f v u c f

v u c f v u c f f

v u c

v u c f

f v u c

v u c f

f

f f v u

1)(

1)()

0

2 0

f f v u

1)(

1)(

)

0 2

2 0 2

vt R



Also the gravitational force between them (a distance of 2R) must equal the

centripetal force from the center of mass:

.m0.279kg

1055.5kg

mN10672

6

s)m10m)(3.9410

2 11

2 4 9

2 2

mv R

Gm

37.76: For any function f  f ( t x, )and x x(x,t),t t(x,t), let F(x ),t 

)),(

,(),(

,),()

,(),(

t

t t

t x f t

x x

t x f t

t x f

x

t t

t x f x

x x

t x f x

t x f

In this solution, the explicit dependence of the functions on the sets of dependent

variables is suppressed, and the above relations are then

t

t t

f t

x x

f t

2 2

2

x

E x

E x

E x

E t

t x

t v t

x x

E v t

2 2 2

t

E t

x

E v t

E t

x t

E x

E v x

E t

2

2 2

2

2 2

x

E v x

E v

b) For the Lorentz transformation, γ, , / 2 and γ

v x

t v t

x x

The first partials are then

t

E x

E v t

E t

E c

v x

E x

,γ

and the second partials are (again equating the mixed partials)

.γ

2γ

γ

γ2γ

γ

2 2 2

2 2 2

2 2 2 2 2

2 2

2 2

2 4

2 2 2

2 2 2 2

t x

E v t

E x

E v t

E

t x

E c

v t

E c

v x

E x

f x

x x

f x

37.77: a) In the center of momentum frame, the two protons approach each other with

equal velocities (since the protons have the same mass) After the collision, the two protons are at rest─but now there are kaons as well In this situation the kinetic energy of the protons must equal the total rest energy of the two kaons   2  2 

γ(

2 m p c m k c

.526.11

1γ

To get the velocity of this proton in the lab frame, we must use the Lorentz

velocity transformations This is the same as “hopping” into the proton that will be our target and asking what the velocity of the projectile proton is Taking the lab frame to be the unprimed frame moving to the left, uvcm andvvcm(the velocity of the projectile proton in the center of momentum frame)

.MeV2494)

1γ(

658.31

1γ

9619.01

21

2 lab

lab

2

2 lab lab

2

2 cm cm 2

c v

c c

v v c

v u

u v v

p

MeV)7.493(2

MeV24942

k m

K

c) The center of momentum case considered in part (a) is the same as this

situation Thus, the kinetic energy required is just twice the rest mass energy of the kaons

MeV

987.4MeV)