PHY116 From Newton to Einstein
Worked Answers to Problem Sheet 8: Length Contraction
Marked out of 20
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A1). A metre stick moves past you at great speed. Its motion relative to you is parallel
to its long axis. If you measure the length of the moving stick to be 0.31225 m, at
what speed is the stick moving relative to you?
Reasoning: L0 = 1m is the proper length, you observe the contracted length 1m /  =
0.31225.
Answer: So v = c/ Sqrt[2 – 1] = 0.95 c = 2.85  108 m s–1 [1]
Check: This is less than c as expected. Also, compare examples we have seen – 0.6c
gives  = 5/4 (see A2 below); 0.99c or thereabouts gives  ~ 10, so this result fits OK.
Finally, 1/ = 0.3 is  ~3, and 0.952 ~ 0.9, 1–0.9 = 0.1, Sqrt[0.1] ~ 0.3, 1/0.3 ~3 which
checks OK. [1]
A2). A spacecraft flies past a planet at a speed of 0.6c. A scientist on the planet
measures the length of the spacecraft to be 70 m. The spacecraft land on the planet
and the scientist measures the length of the now stationary spacecraft. What value
does he get?
Reasoning: L = 70m is not the proper length, the scientist observes the contracted
length L0 /  = 70m. So L0 is what he measures on the ground. [1]
Answer: So L0 = 70 = 70  1.25 = 87.5m [1]
Check: This is longer than the contracted length as expected.
B3). As measured by an observer on the Earth, a spacecraft runway on Earth has a
length of 3600m. a). What is the length of the runway as measured by a pilot of a
spacecraft flying past at a speed of 4.00  107 m/s relative to the Earth?
Reasoning: The observer on Earth is stationary relative to the runway so at leisure
can measure the proper length.
Answer: For v = 4  107 = 4/30 c,  = 1.0091. Proper length is 3600m, so length for
pilot is the contracted length = 3600/ = 3568 m. [1]
Check: This is shorter than the proper length as expected. [1]
b). An observer on Earth measures the time interval from when the spacecraft is
directly over one end of the runway until it is directly over the other end. What result
does she get?
Reasoning: Proper Length, over speed gives (dilated) time. [1]
Answer: Distance 3600m, speed 4 × 108 m/s so time = 90s [1]
c). The pilot of the spacecraft measures the time it takes him to travel from one end of
the runway to the other. What value does he get?
Reasoning: Contracted Length, over speed gives (proper) time.
Answer: Distance 3568 m, speed 4 × 108 m/s so time = 89.2s. [1]
Check: Shorter that the dilated time of (b), as expected. [1]
B4). A muon is created 55.0 km above the surface of the Earth (as measured in the
Earth’s frame). The average lifetime of a muon, measured in its own rest frame, is
2.20  10–6 s. In the frame of the muon the Earth is moving towards the muon with a
speed of 0.9860c.
Answers: Start by calculating  
1
 6.0
1  2
Check: Looks consistent with -values above.
a) In the muon’s frame, what is its initial height above the surface of the Earth?
Reasoning: The altitude is the proper length L0 in the Earth’s frame, since an
“altitude” is stationary in the Earth’s frame. Then that length is contracted in the
muon’s frame. [1]
Answer: L0/ = 55/6 = 9.17 km. [1]
Check: This is shorter, as expected for length contraction.
b) In the muon’s frame, how much closer does the Earth get during the lifetime of the
muon? What fraction is this of the muon’s original height as measured in the muon’s
frame?
Reasoning: In the muon’s frame, the Earth is travelling towards it at 0.986c for 2.2
s. [1]
Answer: So travels 650 m. This is a fraction f = 0.650/9.17 = 0.071 of the original
height. [1]
c) In the Earth’s frame, what is the lifetime of the muon? In the Earth’s frame, how
far does the muon travel during its lifetime? What fraction is this of the muon’s
original height in the Earth’s frame?
Reasoning: The 2.2 s is the time between creation and decay in the frame in which
both events occur at the same place (the muon’s frame) and so it is the proper time T0
there.
Answer: In the Earth’s frame, this time is dilated, to T0 = 6 × 2.2 = 13.2 s [1]. In the
Earth’s frame, the muon is travelling towards it at 0.986c for 13.2 s, so travels 3.9
km. This is a fraction f = 3.9/55 = 0.071 of the original height. [1]
Check: Same fraction as in part (b).
B5). An unstable particle is created in the upper atmosphere from a cosmic ray and
travels straight down to the surface of the earth with a speed of 0.9950c relative to the
Earth. A scientist at rest on the Earth’s surface measures that the particle is created at
an altitude of 45.0 km.
Answers: Start by calculating  
1
1  2
 10
Check: Looks consistent with -values above.
a) As measured by the scientist, how much time does it take the particle to travel the
45.0 km to the surface of the Earth?
Reasoning: Proper length over speed gives dilated time.
Answer: In the scientist’s frame, the distance is 45 km, the speed is 0.9954c so the
time is 45000 / (.9954 × 3 × 108) = 150s. [1]
Check: Almost c, which is 1 ft per ns, 1000 ft per s, 1 mile per 5s, 30 miles ~ 45
km per 150 s.
b) Use the length contraction formula to calculate the distance from where the particle
is created to the surface of the Earth as measured in the particle’s frame.
Answer: Proper length is 45 km, so length in particle’s frame is 45/ = 4.5 km. [1]
Check: Shorter, consistent with length contraction.
c) In the particle’s frame, how much time does it take the particle to travel from where
it is created to the surface of the Earth? Calculate this time both by the time dilation
formula and also from the distance calculated in part (b). Do the two results agree?
Reasoning: By time dilation, in the particle’s frame, the time taken is the proper time
T0 since then the creation and the arrival of the Earth’s surface are in the same place.
By distance, the particle sees the Earth arrive at 0.9954c from a distance of 4.33km;
both events happen at the particle so this will give the proper time. . [1].
Answer: By time dilation, the scientist’s 150s is the dilated time,  T0, so T0 =
150/10.4 = 14.4s.
By distance, Time is 4.33/(.9954c) = 14.5 s, in agreement apart from rounding
errors [1].
Check: Expected to agree, and they do.