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Physics 2220 – Module 16 Homework
A firecracker explodes in reference frame S at t
1 same position at t
2
= 1.0 seconds. A second firecracker explodes at the
= 3.0 seconds. In reference frame S', which moves in the x -direction at speed v , the first explosion is detected at x
1
' = 4.0 m and the second at x
2
(a)
(b)
' = - 4.0 m.
What is the speed of frame S' relative to frame S?
What is the position of the two explosions in frame S?
Use the Galilean Transformation Equations
For the first firecracker: x
1 x
1
' = x
1
− vt
' = 4.0 m
1 t
1
= 1.0 s
For the second firecracker: x
2 x
2
' = x
2
− vt
' = − 4.0 m
2 t
2
= 3.0 s
(a)
(b)
Both firecrackers explode at the same point in the S frame, so x
1
= x
2 v = x
2 t
1
' − x
1
− t
2
'
=
(− x
1
= x
2 x
1
' + vt
1 v ( t
1
− t
2
= x
) = x
2
2
4.0 m
1.0 s −
' + vt
' − x
) − (
1
2
'
4.0 m
3.0 s
)
= 4.0 m/s
Using one of the previous relations to find the position in the S frame: x
1
= x
1
' + vt
1
= ( 4.0 m ) + ( 4.0 m/s ) ( 1.0 s ) = 8.0 m
Can verify with the other equation: x
2
= x
2
' + vt
2
= (− 4.0 m ) + ( 4.0 m/s ) ( 3.0 s ) = 8.0 m
A baseball pitcher can throw a ball with a speed of 40.0 m/s. He is in the back of a pickup truck that is driving away from you. He throws the ball in your direction, and it floats toward you at a lazy 10.0 m/s.
What is the speed of the truck?
Assume the truck (the S' frame) is traveling to the right, so its velocity is positive. Therefor both the velocity of the ball in the S' frame and the S frame will be to the left and negative.
Use the Galilean Addition Law for Velocity u ' = u − v v = u − u ' = (− 10.0 m/s ) − (− 40 m/s ) = 30.0 m/s
An out-of-control alien spacecraft is diving into a star at a speed of 1.5 × 10 8 relative to the spacecraft, is the starlight approaching?
m/s. At what speed,
Recall the 2 nd postulate of the special theory of relativity
• The speed of light in a vacuum has the same value (2.99272458 × 10 8 m/s) in all inertial frames, regardless of the velocity of the observer or the velocity of the source emitting the light

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A starship blasts past the Earth at 2.5 × 10 8 m/s. Just after passing the Earth, it fires a laser beam out of the back of the starship. With what speed does the laser beam approach the Earth?
Recall the 2 nd postulate of the special theory of relativity
• The speed of light in a vacuum has the same value (2.99272458 × 10 8 m/s) in all inertial frames, regardless of the velocity of the observer or the velocity of the source emitting the light
A positron moving in the positive x -direction at 2.3 × 10 8 m/s collides with an electron at rest. The positron and electron annihilate, producing two gamma-ray photons. Photon 1 travels in the positive x direction and photon 2 travels in the negative x-direction. What is the speed of each photon?
Recall the 2 nd postulate of the special theory of relativity
• The speed of light in a vacuum has the same value (2.99272458 × 10 8 m/s) in all inertial frames, regardless of the velocity of the observer or the velocity of the source emitting the light
You are standing at x = 9.0 km and your assistant is standing at x = 3.0 km. Lightning bolt 1 strikes at x
= 0 and lighting Bolt #2 strikes at x = 12.0 km. You see the flash from Bolt #2 at t = 10 μs and the flash from Bolt #1 at t = 50 μs. According to your assistant, were the lightning strikes simultaneous? If not, which occurred first, and what was the time difference between the two?
07.
The clock starts at t = 0
• At t = 10 μs, you see the flash from Bolt #2
• At t = 50 μs, you see the flash from Bolt #1
Light from both bolts travel at the speed of light
3.00
× 10
8 m/s = 300 m/ μ s
You are 9000 m from Bolt #1, so calculate the time it takes to reach you:
9000 m
300 m/ μ s
= 30 μ s
• Since you see Bolt #1 at t = 50 μs, this means that Bolt #1 struck at 20 μs
You are 3000 m from Bolt #2, so calculate the time it takes to reach you:
3000 m
300 m/ μ s
= 10 μ s
• Since you see Bolt #2 at t = 10 μs, this means that Bolt #1 struck at 0 μs
Therefore the strikes are NOT simultaneous. There is a 20 μs difference between the strikes.
Bolt #2 was first. Your assistant is in the same reference frame as you, therefore the bolts strikes are not simultaneous to them either.
You are flying your personal rocketcraft at 0.9c from Star A toward Star B. The distance between the stars, in the stars' reference frame, is 1.0 light year. Both stars happen to explode simultaneously in your reference frame at the instant you are exactly halfway between them. Do you see the flashes simultaneously? If not, which do you see first, and what is the time difference between the two?
The key phrase in the question is the stars explode at the same time in your reference frame (in the spaceship) instead of in the stars' reference frame. So in your reference frame, you will see the flashes simultaneously

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A cosmic ray travels 60 km through the Earth's atmosphere in 400 μs, as measured by the experimenters on the ground. How long does the journey take according to the cosmic ray?
Note the velocity of the cosmic ray in the frame of the Earth's atmosphere: v = d t
=
60 km
400 μ s
= 1.5
× 10
8 m/s
The cosmic ray, at rest with its internal clock, will experience the proper time. So:
Δ t ' =
Δ t
γ = Δ
√
1 − v c
2
2
Δ t = γ Δ t '
= ( 400 μ s )
1 −
( 1.5
× 10
( 3.0
× 10 8
8 m/s )
2 m/s )
2
= 346 μ s
An astronaut travels to a star system 4.5 light years away at a speed of 0.9c. Assume that the time needed to accelerate and decelerate is negligible.
(a)
(b)
(c)
How long does the journey take according to Mission Control on Earth?
How long does the journey take according to the astronaut?
How much time elapses between the launch and the arrival of the first radio message from the astronaut saying that she has arrived?
(a)
(a)
(b)
(c)
The distance to the star system in the Earth and star system frames is 4.5 light years (ly). This is the proper length. The speed they travel according to Mission Control is 0.90c. Calculate the time it takes:
• 1 light year = 9.46 × 10 15 m t = d v
=
( 4.5 ly ) ( 9.46
× 10 15
( 0.9
) ( 3.00
× 10 8 m/ly m/s )
)
= 1.57667
× 10 8 s = 5.0 years
Since the proper length is experienced by Mission control, the proper time is experienced by the astronaut.
Δ t ' =
Δ t
γ = Δ t
√
1 − v c
2
2
= ( 5.0 years )
√ 1 − ( 0.9
)
2
= 2.2 years
Five years to make the journey according to Mission Control plus the speed of light travel time back to carry the message:
Δ t = 5.0 years + 4.5 years = 9.5 years
A muon travels 60 km through the atmosphere at a speed of 0.9997c. According to the muon, how thick is the atmosphere?
In this case, 60km is the proper length of the atmosphere.
Use the equation for length contraction:
L =
L p
γ =
L p
√
1 − v 2 c 2
= ( 60 km )
√ 1 − ( 0.9997
)
2
= 1.5 km
Our Milky Way galaxy is 100,000 light years in diameter. A spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.0 light year.
(a)
(b)
What is the speed of the spaceship relative to the galaxy?
How long is the crossing time as measured in the galaxy's reference frame?
In this case, the proper length of the galaxy is 100,000 light years.
Use the proper length equation to solve for the fraction of speed of light

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(b)
( L
L p
2 )
= 1 − v 2 c 2
L =
L p
γ =
L p
√
1 − v c
2
2
→ v 2
= c 2
(
1 −
( L
L p
2 ) ) v =
(
1 −
( L
L p
2 ) ) c =
(
1 −
( 1.0 ly
100000 ly
) 2 )
= 0.999999999c
Find the time in the galaxy's reference frame: t = d v
=
( 100000 ly ) ( 9.46
× 10 15
( 0.999999999
) ( 3.00
× 10 8 m/ly ) m/s )
= 3153333336487 s ≈ 100,000 years
A rocket cruising past Earth at 0.8c shoots a bullet out of the back door, opposite the rocket's motion, at
0.9c relative to the rocket. What is the bullet's speed relative to the Earth?
Set the frames up as the following:
• Earth is the S – Frame
• The rocket is the S' – Frame traveling at 0.8 c with respect to the S – Frame
• Firing the bullet out of the rocket, opposite the direction of the rocket's direction, in the S' –
Frame is the “event” The velocity of – 0.9c according to the S' – Frame
Solve for the speed of the bullet relative to the Earth: u x
=
1 u x
+
' + u x v
' v c 2
=
1
(−
+
0.9c
(−
) + (
0.9c
c
) (
2
0.8c
)
0.8c
)
= − 0.36c
A distant quasar is found to be moving away from the Earth at 0.8c. A galaxy closer to the Earth and along the same line of sight is moving away from us at 0.2c. What is the recessional speed of the quasar as measured by astronomers in the other galaxy?
Set the frames up as the following:
• Earth is the S – Frame
• The galaxy is the S' – Frame traveling at 0.2 c with respect to the S – Frame
• The moving quasar as measured in the S – Frame is the “event” The velocity of 0.8c is according to the S – Frame.
Solve for the speed of the quasar relative to the galaxy: u x
' =
1 u x
− v
− u x c 2 v
=
1
( 0.8c
) − ( 0.2c
)
−
( 0.8c
) ( 0.2c
) c 2
= 0.71c
14.
A proton is accelerated to 0.999c.
(a) What is the proton's momentum?
(b) By what factor does the proton's momentum exceed its Newtonian momentum?
(a) Relativistic momentum: p = γ mv =
1 mv
− v 2 c 2
=
( 1.67
× 10 − 27 kg ) ( 0.999
) ( 3.00
× 10 8
√ 1 − ( 0.999
) 2 m/s )
= 1.12
× 10 − 17 kg m / s

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(b) Newtonian momentum: p = mv = ( 1.67
× 10 − 27 kg ) ( 0.999
) ( 3.00
× 10 8 m/s ) = 5.00
× 10 − 19 kg m / s
Ratio: p
R p
N
=
1.12
× 10 − 17
5.00
× 10 − 19 kg m / s kg m / s
= 22.4
At what speed is the particle's momentum twice its Newtonian value?
Set the relativistic momentum equal to twice the Newtonian momentum and solve for its velocity. The equation will look like: p
1
2 v c
R
2
2
=
= p
N
√
1 − v 2
1 mv
→
√
1 −
1
→
4 c 2
= 1 − → v 2
4 v 2 c 2
= 2mv
= 1 −
= 0.75c
v 2 c 2
2 v = 0.866c
What are the kinetic energy, the rest energy, and the total energy of a 1.0 g particle with a speed of
0.8c?
Rest Energy:
E
Rest
= mc 2
= ( 0.001 kg ) ( 3.00
× 10 8 m/s )
2
= 9.0
× 10 13 J
Total Energy:
E
Total
= γ mc
2
= mc 2
√
1 − v c
2
2
=
( 0.001 kg ) ( 3.00
× 10 8
√ 1 − ( 0.8
)
2
Kinetic Energy:
E
KE
= E
Total
− E
Rest
= 1.5
× 10
14
J − 9.0
× 10
13 m/s )
2
= 1.5
× 10
14
J
J = 6.0
× 10
13
J
How fast must an electron move so that its total energy is 10% more than its rest mast energy?
Total Energy = γ mc 2 → Want this 10% more than the rest mass
Rest Mass Energy = mc 2
Total Energy = Rest Mass Energy + Kinetic Energy
1.1
=
√
1 −
1 v 2 c 2
→
1.1
( mc 2
) = γ mc 2
1 − v 2 c 2
=
1
( 1.1
)
2
→ v = c
1 −
1
( 1.1
)
2
= 0.42c
v 2 c 2
= 1 −
1
( 1.1
)
2

18.
(b)
A starship voyages to a distant planet 10 light years away. The explorers stay 1 year, then return at the same speed, and arrive back on Earth 26 years after they left. Assume that the time needed to accelerate and decelerate is negligible.
(a) What is the speed of the starship?
(b) How much time has elapsed on the astronauts' chronometers?
Note the following from the Earth's reference frame:
• t
1
= Time the starship travels to the planet
• On the planet, assume the same reference frame as the Earth
◦ t
2
= 1 year
• t
3
= Time the starship travels back to Earth
• 26 years total time, 25 years of which to go:
◦ 10 light years + 10 light years = 20 light years
(a) The speed of the starship is: v = d t
=
20 light years
25 years
=
( 20 years ) c
25 years
= 0.80c
The astronauts measure proper time:
Δ t = γ Δ t '
Δ t ' =
Δ t
γ = Δ t
√
1 − v c
2
2
= ( 25 years ) √ 1 − ( 0.80
)
2
= 15 years
So 15 years of travel + 1 year on the planet = 16 years