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```					                           Interactive Tutorial, Chapter 29 Problem 43
Fundamentals of College Physics, 3rd. Ed.
Dr. Peter J. Nolan
"Fundamentals of College Physics" Third Edition
Dr. Peter J. Nolan, SUNY Farmingdale

Chapter 29 Special Relativity
Computer Assisted Instruction
Interactive Tutorial

43. Length Contraction. The length of a rod at rest is found to be Lo = 2.55 m.
Find the length L of the rod when observed by an observer in motion at a speed
v = 0.25 c.

Initial Conditions
Lo =     2.55 m                                           c = 3.00E+08 m/s
v = 2.50E-01 c      =      7.50E+07 m/s

For speeds that are not given in terms of the speed of light c use the following
converter to find the equivalent speed in terms of the speed of light c. Then place the
equivalent speed into the yellow cell for v above.
v=      1610 km/hr =          447.58 m/s = 1.49E-06 c
v=     2E+06 m/s =          5.37E-03 c

The length contraction is given by equation 29.60 as
L = Lo sqrt[1 - (v2)/(c2)]
L=(        2.55 ) x sqrt[1 -( 7.50E+07 m/s)2 / ( 3.00E+08 m/s)2]
L = 2.47E+00 m

Page 1
Problem 44

"Fundamentals of College Physics" Third Edition
Dr. Peter J. Nolan, SUNY Farmingdale

Chapter 29 Special Relativity
Computer Assisted Instruction
Interactive Tutorial

44. Time Dilation. A clock in a moving rocket ship reads a time duration Dt0 =
15.5 hr. What time elapses, Dt, on earth if the rocket ship is moving at a speed
v = 0.355 c?

Initial Conditions
Dto =     15.5 hr                                        c = 3.00E+08 m/s
v = 3.55E-01 c      =         1.07E+08 m/s

For speeds that are not given in terms of the speed of light c use the following
converter to find the equivalent speed in terms of the speed of light c. Then place the
equivalent speed into the yellow cell for v above.
v=         1610 km/hr =         447.58 m/s = 1.49E-06 c
v = 1.61E+06 m/s         =    5.37E-03 c

The time dilation is given by equation 29.64 as
Dt = Dto / sqrt[1 - (v2)/(c2)]
Dt = (                                          2                 2
15.5 hr) /sqrt[1- ( 1.07E+08 m/s) / ( 3.00E+08 m/s) ]
Dt = 16.57991 hr

Page 2
Problem 45

"Fundamentals of College Physics" Third Edition
Dr. Peter J. Nolan, SUNY Farmingdale

Chapter 29 Special Relativity
Computer Assisted Instruction
Interactive Tutorial

45. Relative velocities. Two spaceships are approaching a space station as in
figure 29.15. Spaceship one has a velocity of 0.55 c to the left and spaceship two
has a velocity of 0.75 c to the right. Find the velocity of rocket ship one as observed
by rocket ship two.

Assume the space station to be at rest. Spaceship one then has the velocity
Vx = - 0.55c. Spaceship two is considered to be a moving coordinate system
approaching with the speed v = 0.75 c.

Initial Conditions
Vx = -4.85E-01 c          =    -1.45E+08 m/s            c = 3.00E+08 m/s
v = 6.55E-01 c           =     1.96E+08 m/s

The relative velocity of approach of spaceship one as observed by the S' observer on
rocket ship two is found from equation 29.75 as
Vx' = [Vx - v] / [1 - (v/c2) Vx]
Vx' = [( -4.85E-01 c)    -       ( 6.55E-01 c)]
/ [(1 - {( 1.96E+08 m/s) / ( 3.00E+08 m/s)2} x ( -1.45E+08 m/s)]
Vx' = -0.86516 c      =    -2.59E+08 m/s

Page 3
Problem 46

"Fundamentals of College Physics" Third Edition
Dr. Peter J. Nolan, SUNY Farmingdale

Chapter 29 Special Relativity
Computer Assisted Instruction
Interactive Tutorial

46. Relativistic mass. A mass at rest has a value mo = 2.55 kg. Find the relativistic
mass m when the object is moving at a speed v = 0.355 c.

Initial Conditions
mo =      2.55 kg                                          c=   3.00E+08 m/s
v = 3.55E-01 c =                1.07E+08 m/s

For speeds that are not given in terms of the speed of light c use the following
converter to find the equivalent speed in terms of the speed of light c. Then place
the equivalent speed into the yellow cell for v above.
v=         1610 km/hr =             447.58 m/s =      1.49E-06 c
v = 1.61E+06 m/s =               5.37E-03 c

The relativistic mass is given by equation 29.86 as
m = mo / sqrt[1 - (v2)/(c2)]
m=(          2.55 kg)/sqrt[1 - ( 1.07E+08 m/s)2 / ( 3.00E+08 m/s)2]
m = 2.72766277 kg

Page 4
Problem 47

"Fundamentals of College Physics" Third Edition
Dr. Peter J. Nolan, SUNY Farmingdale

Chapter 29 Special Relativity
Computer Assisted Instruction
Interactive Tutorial

47. Plot of length contraction and mass change versus speed
The length of a rod at rest is Lo = 1.00 m and its mass is mo = 1.00 kg. Find the length
L and mass m of the rod as its speed v in the axial direction increases from 0.00 c to
0.90 c, where c is the speed of light (c = 3.00 x 108 m/s). Plot the results.

Initial Conditions
Lo =       1m                                          c=     3.00E+08 m/s
v=         0.1 c     =      3.00E+07 m/s                mo =         1 kg

The length contraction is given by equation 29.60 as
L = Lo sqrt[1 - (v2)/(c2)]
L=(           1 ) x sqrt[1- ( 3.00E+07 m/s)2 / ( 3.00E+08 m/s)2]
L = 0.994987 m

The relativistic mass is given by equation 29.86 as
m = mo / sqrt[1 - (v2)/(c2)]
m=(            1 kg)/sqrt[1-( 3.00E+07 m/s)2 / ( 3.00E+08 m/s)2]
m = 1.005038 kg

The contraction in length and increase in mass as the speed increases is calculated
by the above equations and is shown in the table below.

v (%c)       v (m)     L (m)     m (kg)
0   0.00E+00     1.000       1.000
0.1   3.00E+07     0.995       1.005
0.2   6.00E+07     0.980       1.021
0.3   9.00E+07     0.954       1.048
0.4   1.20E+08     0.917       1.091
0.5   1.50E+08     0.866       1.155
0.6   1.80E+08     0.800       1.250
0.7   2.10E+08     0.714       1.400
0.8   2.40E+08     0.600       1.667
0.9   2.70E+08     0.436       2.294

The contraction of the length of the rod L versus the speed v is shown in the graph

Page 5
Problem 47

below.
Length Contraction

1.200

1.000

0.800
Length L

0.600

0.400

0.200

0.000
0   0.2             0.4              0.6   0.8    1
Speed v

The increase in the mass m of the rod versus the speed v is shown in the graph below.

Increase in mass with speed

2.500

2.000
Mass m

1.500

1.000

0.500

0.000
0   0.2             0.4              0.6   0.8    1
Speed v

Page 6
Problem 48

"Fundamentals of College Physics" Third Edition
Dr. Peter J. Nolan, SUNY Farmingdale

Chapter 29 Special Relativity
Computer Assisted Instruction
Interactive Tutorial

48. An accelerated charged particle. An electron is accelerated from rest through
a potential difference V = 4.55 x 105 V. Find (a) the kinetic energy of the electron,
(b) the rest mass energy of the electron, (c) the total relativistic energy of the electron,
(d) the speed of the electron, (e) the relativistic mass of the electron, and (f) the
momentum of the electron.

Initial Conditions
mo = 9.11E-31 kg                                c = 3.00E+08 m/s
q = 1.60E-19 C                                 V = 4.55E+05 V

a. The kinetic energy of the electron is found from equation 19.36 as
KE = work done = q V
KE = ( 1.60E-19 C)        x     ( 4.55E+05 V)
KE =      7.29E-14 J      =     4.56E-01 Mev

b. The rest mass energy of the electron is found from equation 29.101 as
Eo = mo c2
2
Eo = ( 9.11E-31 kg) x        ( 3.00E+08 m/s)
Eo =     8.19E-14 J       =        5.12E-01 Mev

c. The total relativistic energy is found from equation 29.100 as
E = KE + Eo
E = ( 4.56E-01 Mev) + ( 5.12E-01 Mev)
E=     9.67E-01 Mev

d. The speed of the electron is found by solving equation 29.96 for the speed v.
That is,
KE = {(mo c2) / sqrt[1 - (v2)/(c2)]} - mo c2
KE = {(Eo) / sqrt[1 - (v2)/(c2)]} - Eo

and the speed v is found as
v = c sqrt[1 - [1 / {(KE/Eo) + 1}]2]
v = c sqrt[ 1 - [1 /{ (     4.56E-01 Mev) /         ( 5.12E-01 ) + 1} ]2 ]

Page 7
Problem 48

2
v = c sqrt[ 1 - [1 /{ ( 8.90E-01 ) + 1} ] ]
v=      0.8486 c =         2.54E+08 m/s

e. The relativistic mass is given by equation 29.86 as
m = mo / sqrt[1 - (v2)/(c2)]
m = ( 9.11E-31 kg)/sqrt[1- ( 2.54E+08 m/s)2 / ( 3.00E+08 m/s)2]
m = 1.722E-30 kg

f. The momentum of the electron is found from equation 29.90 as
p=mv
p = ( 1.72E-30 kg)       x ( 2.54E+08 m/s)
p = 4.381E-22 kg m/s

Page 8

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