Problem Set 26: 3 | 6 | 8 | 12 | 2 | 4 | 11 | 13 | 14 | 17 | 45 | 47 | 18 | 21 | 23 | 26 | 28 | 31 | Go up

- Scott Parsons, 2003

3.      A beam of a certain type of elementary particle travels at a speed of 2.70 x 10^8 m/s.  At this speed, the average lifetime is measured to be 4.76 x 10^-6 s.  What is the particle’s lifetime at rest?

Use the following equation:

Dt = gDt0

Where

g = 1/Ö1 - (v2/c2)

Put these together and you have

Dt = Dt0 /Ö1 - (v2/c2)

Now replace the variables in the equation and solve for Dt

V = 2.70 x 10^8 m/s

Dt0 = 4.76 x 10^6 s

c = 3.00 x 10^8 m/s

Dt = (4.76 x 10^6 s) /Ö1 – ((2.70 x 10^8 m/s)^2/(c2))  = 2.07 x 10^-6 s

6.      What is the speed of a beam of pions if their average lifetime is measured to be 4.10 x 10^-8 s?  At rest, their lifetime is 2.60 x 10^-8 s.

Use the following equation:

Dt = Dt0 /Ö1 - (v2/c2)

And solve for v to get

V = c Ö1 - (Dt0 2/Dt 2)

Now replace the variables in the equation and find v

Dt = 4.10 x 10^-8 s

Dt0 = 2.60 x 10^-8 s

v = (3.00 x 10^8) Ö1 – ((2.60 x 10^-8 s)^2/(4.10 x 10^-8 s)^2) = (2.13 x 10^8 m/s) / (3 x 10^8) = .773c

8.       At what speed do the relativistic formulas for length and time intervals differ from classical values by 1.00 percent?

Use the following equation:

Dt = Dt0 /Ö1 - (v2/c2)

And solve for v to get

V = c Ö1 - (Dt0 2/Dt 2)

Now replace the variables in the equation and find v

Where Dt = 100 and Dt0 = 99

V = (3.00 x 10^8) Ö1 – ((99)^2/(100)^2) = .196 m/s

12.   How fast must a pion be moving, on average, to travel 10.0 m before it decays?  The average lifetime, at rest, is 2.60 x 10^-8 s.

Use the following equations:

V = s/t    and   Dt = Dt0 /Ö1 - (v2/c2)

Put these to equations together to get

S = vt = vDt0 /Ö1 - (v2/c2)

Now solve for v to get

V = s / Ö Dt02 + (s2 / c2 )

Now replace the variables in the equation and find v

S = 10.0 m

Dt0 = 2.60E-8 s

V = s / Ö Dt02 + (s2 / c2 ) = 10.0 / Ö (2.60E-8 )2 + (10.02 / c2 ) = 2.3655E8 m/s

(2.3655E8 m/s) / (3E8) = .789c

2.         A spaceship passes you at a speed of .850c.  You measure its length to be 48.2 m.  How long would it be at rest?

Use the following equation:

L = L0/g

Where

g = 1/Ö1 - (v2/c2)

Put these together and you have

L = L0 Ö1 - (v2/c2)

Solve  for L0 to get

L0 = L / Ö1 - (v2/c2)

Now replace the variables in the equation and solve for L0

L = 48.2 m

V = .850c = 2.55 x 10^8 m/s

L0 = L / Ö1 - (v2/c2)= 48.2  / Ö1 – ((2.55 x 10^8 m/s)^2)/(c2)) = 91.5 m

4.      If you were to travel to a star 100 light-years from Earth at a speed of 2.60 x 10^8 m/s, what would you measure this distance to be?

Use the following equation from #2:

L = L0 Ö1 - (v2/c2)

Now replace the variables in the equation and solve for L0

L = 100 ly

V = 2.60 x 10^8 m/s

L = L0 Ö1 - (v2/c2)= 100  Ö1 – ((2.60 x 10^8 m/s)^2 /c^2)) = 49.9 ly

11.  A friend of yours travels by you in her fast sports vehicle at a speed of .580c.  You measure it to be 5.80 m long and 1.20 m high.  (a) What will be its length and height at rest?  (b) How many seconds would you say elapsed on your friend’s watch when 20.0 s passed on yours?  (c) How fast did you appear to be traveling to your friend?  (d) How many seconds would she say elapsed on your watch when she saw 20.0 s pass on hers?

(a) Use the following equation:

L = L0 Ö1 - (v2/c2)

Solve  for L0 to get

L0 = L / Ö1 - (v2/c2)

Now replace the variables in the equation and solve for L0

L = 5.80 m

V = .58c

L0 = L / Ö1 - (v2/c2)= 5.80 / Ö1 - (.582)= 7.12 m

The height hasn’t changed so it will still be 1.2 m high

(b) Use the following formula:

Dt = Dt0 /Ö1 - (v2/c2)

Solve for Dt0 to get

Dt0 = Dt Ö1 - (v2/c2)

Now replace the variables in the equation and solve for Dt0

Dt = 20 s

V = .58c

Dt0 = Dt Ö1 - (v2/c2)= 20 Ö1 - (.582)= 16.3 s

(c) .580c

(d) 16.3 s

12.  How fast must a pion be moving, on average to travel 10.0 m before it decays?  The average lifetime at rest is 2.60 x 10-8 s

Well, the pion will be going at a velocity such that
vt = 10.0 m, which sounds simple enough, but the time is of course, dilated, which makes it a bit tricky.  It really is
vto/
Ö(1-v2/c2)= 10.0 m

And this gets a bit messy, but not really too bad. First, square both sides, yielding
v2to2/(
1 - v2/c2) = L2, where L = 10.0 m

Then cross multiply and distribute
v2to2 = L2(
1 - v2/c2) = L - Lv2/c2
v2to2 = L2 - L2v2/c2

Get terms of v on the same side:
v2to2 + L2
v2/c2 = L2

Factor out v2:
v2(to2 + L2
/c2) = L2

Divide both sides by the factor in the ():
v2 = L2/(to2 + L2
/c2)

Finally, square root both sides:
v = L/
Ö(to2 + L2/c2), L = 10.0, to = 2.6 x 10-8 s, c = 3.00 x 108 m/s, v = 2.37 x 108 m/s or .789 c

13.   What is the mass of a proton traveling at v = 0.90c?

Use the following equation:

m = gm0

where g = 1/Ö1 - (v2/c2)

Put these together and you have

m = m0 / Ö1 - (v2/c2)

Now replace the variables in the equation and solve for m

m0 = 1.6726 10^-27 (the rest mass of a proton found in data packet)

v = 0.90c = 2.7 x 10^8 m/s

m = m0 / Ö1 - (v2/c2)= (1.6726 / Ö1 - (2.7 x 10^8 m/s)/ (c2)= 3.8 x 10^27 kg

14.   At what speed will an object’s mass be twice its rest mass?

Use the following equation from #13:

m = m0 / Ö1 - (v2/c2)

Solve for v to get

V = c Ö1 – (m0 2/m 2)

And replace the variables in the equation and solve for v

m = 2

m0 = 1

V = c Ö1 – (m0 2/m 2)= c Ö1 – ((1^2)/(2^2)) = .866c

17.  (a) What is the speed of an electron whose mass is 10,000 times its rest mass?  (b) If the electrons travel in the lab through a tube 3.0 km long, how long is this tube in the electron’s reference frame?

(a) Use the following equation from #13:

m = m0 / Ö1 - (v2/c2)

Solve for v to get

V = c Ö1 – (m0 2/m 2)

And replace the variables in the equation and solve for v

m = 10,000

m0 = 1

V = c Ö1 – (m0 2/m 2)= c Ö1 – (1 2/10,000 2)= .99999c

(b) Use the equation:

L = L0 Ö1 - (v2/c2)

And replace the variables in the equation and solve for L

L0 = 3 km = 3,000 m = 30,000 cm

V = .9999 c

L = L0 Ö1 - (v2/c2)= 30,000 Ö1 - (.999992) = 30 cm

45.   A person on a rocket traveling at 0.50c observes a meteor comes from behind and pass her at a speed she measures as 0.50c.  How fast is the meteor moving with respect to the Earth?

Use the equation:

u'x = (ux - v)/(1 - (uxv/c2))

Since the rockets are going in the same direction the – changes to a +

u'x = (ux + v)/(1 + (uxv/c2))

Now replace the variables in the equation and solve for u'x

V = .50c

ux = .50c

u'x = (ux + v)/(1 + (uxv/c2))= (.50 + .50)/(1 + (.50 x .50))= .8c

47.  An observer on Earth sees an alien vessel approach at a speed of 0.60c.  The Enterprises comes to the rescue overtaking the aliens while moving directly toward Earth at a speed of 0.90c relative to Earth.  What is the relative speed of one vessel seen by the other?

Use the equation:

u'x = (ux - v)/(1 - (uxv/c2))

Since the rockets are going in the same direction the – changes to a +

u'x = (ux + v)/(1 + (uxv/c2))

Now replace the variables in the equation and solve for ux

u'x = .90c

v = .60c

.90 = (ux + .60c)/(1 + (ux  x .60))

ux = .65c

18.  What is the kinetic energy of an electron whose mass is 3.0 times its rest mass?

Use the following equation from the book:

KE = mc2 - m0c2 which is really

KE =(m - m0)c2

Now replace the variables in the equation and solve for KE

m0 = 9.11 x 10^-31 (the rest mass of an electron from data packet)

M = 3(9.11 x 10^-31)

KE =(m - m0)c2 = ((3 x 9.11 x 10^-31) - 9.11 x 10^-31))c2 = 1.6 x 10^-13 J

21.  Calculate the rest energy of an electron in joules and in MeV (1 MeV = 1.60 x 10^-13 J).

Use the following equation:

E = mc2

Now replace the variables in the equation and solve for E

M = 9.11 x 10^-31 (the rest mass of an electron from data packet)

E = mc2 = (9.11 x 10^-31)(3E8)^2 = 8.20E-14

But it also wants it in MeV so divide by 1.60 x 10^-13

(8.20E-14)/( 1.60 x 10^-13) = .511 MeV

23.  The total annual energy consumption in the United States is about 8 x 10^19 J.  How much mass would have to be converted to energy to fuel this need?

Use the following equation:

E = mc2

And solve for m to get

M = E / c2

Now replace the variables in the equation and solve for m

E = 8E19

C = 3.00E8

M = E / c2 = (8E19)/(3E8) 2 = 9 x 102 kg

26.  (a) How much work is required to accelerate a proton from rest up to a speed of 0.998c?  (b) What would be the momentum of this proton?

(a) Use the following equation:

m = m0 / Ö1 - (v2/c2)

And replace the variables in the equation and solve for m

m0 = 1.6726E-27 (the rest mass of a proton from data packet)

v = .998c

m = m0 / Ö1 - (v2/c2)= 1.6726E-27 / Ö1 - (.9982)= 2.6459E-26

Now use the following equation

KE =(m - m0)c2

And replace the variables in the equation and solve for m

M = 2.6459E-26

m0 = 1.6726E-27

KE =(m - m0)c2 = (2.6459E-26 - 1.6726E-27)c2 = 2.23E-9 J

(b) Use the following equation:

p = mv

And replace the variables in the equation and solve for p

M = 2.6459E-26 (from (a) above)

V = 0.998c = 2.994 x 10^8 m/s

p = mv = (2.6459E-26)( 2.994 x 10^8) = 7.91 x 10^-18 kgm/s

28.  Calculate the kinetic energy and momentum of proton traveling 2.50 x 10^8 m/s.

Use the following equation:

m = m0 / Ö1 - (v2/c2)

And replace the variables in the equation and solve for m

m0 = 1.6726E-27 (the rest mass of a proton from data packet)

v = 2.5 x 10^8 m/s

m = m0 / Ö1 - (v2/c2)= 1.6726E-27 / Ö1 - (2.5 x 10^8 m/s 2/3E82)= 3.0258E-27 kg

Now use the equation

KE =(m - m0)c2

And replace the variables in the equation and solve for KE

KE =(m - m0)c2 = (3.0258 - 1.6726E-27)c2 = 1.22E-10 J

Now to find the momentum use the equation:

P = mv

And replace the variables in the equation and solve for p

M = 3.0258E-27 kg

V = 2.5 x 10^8 m/s

P = mv = (3.0258E-27)( 2.5 x 10^8) = 7.55E-19 kgm/s

31.   What is the speed of an electron whose KE is 1.00 MeV?

Use the following equation:

KE = mc2 - m0c2

And solve for m to get

M = KE + m0 c2  / c2

Now replace the variables in the equation and solve for m

KE = 1.00 MeV = 1.60 x 10^-13 J

m0 = 9.11E-31 kg (the rest mass of an electron)

M = KE + m0 = (1.60 x 10^-13) + (9.11E-31) c2 / c2 = 2.6887777E-30

Now to find the speed of the electron use the equation:

m = m0 / Ö1 - (v2/c2)

And solve for v to get

V = c Ö1 – (m0 2/m 2)

Now replace the variables in the equation and solve for v

m0 = 9.11E-31 kg (the rest mass of an electron)

m = 2.6887777E-30

V = c Ö1 – (m0 2/m 2)= c Ö1 – ((9.11E-31 )2/(2.6887777E-30 2)) = .941c