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The consequences of special relativity questions
Grant Einstein his two postulates and time dilation, length contraction and a limiting speed of c for massive objects all follow: clocks in flight run slow, lengths shrink along the direction of travel, and mass grows with speed until no push, however hard, can carry an object up to c. Experimental tests of these predictions have agreed with them.
18 original questions · 53 marks · the the consequences of special relativity notes · Turning points
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Define the proper time between two events.
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The time between the events measured in the inertial frame in which both events occur at the same place, the time on a clock that rides along with them (1). Every other inertial observer measures a longer time (1).Define the proper length of an object.
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The length measured in the object's own rest frame (1). An observer the object passes measures it shorter along the direction of motion, and unchanged across it (1).Using the idea of relativistic mass, explain why no object can be accelerated to the speed of light.
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As v approaches c the mass m0/√(1 − v2/c2) grows without limit (1), so each further gain in speed costs ever more energy: reaching c would require infinite energy (1).State what is meant by the rest energy of a particle.
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The energy equivalent of the particle's rest mass, E0 = m0c2, measured in the frame in which it is at rest (1).Calculate the rest energy of a proton. Mass of a proton = 1.67 × 10−27 kg; c = 3.00 × 108 m s−1.
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E = m0c2 (1) = 1.67 × 10−27 × (3.00 × 108)2 = 1.5 × 10−10 J (1).A muon's lifetime in its rest frame is 2.2 μs. Calculate its lifetime as measured in a laboratory it passes at 0.95c.
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t = t0/√(1 − 0.952) (1)
= 2.2/0.312 (1)
t = 7.0 μs (1)A spacecraft of proper length 120 m passes Earth at 0.60c. Calculate its length as measured from Earth.
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l = l0√(1 − 0.602) = 120 × 0.80 (1)
l = 96 m (1)
Contracted along the direction of motion only (1)Explain how the arrival of cosmic-ray muons at the Earth's surface provides evidence for time dilation.
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Muons are created about 10 km up, moving near c, with a rest lifetime of 2.2 μs: one undilated lifetime covers only about 650 m, and the descent lasts around fifteen lifetimes, so almost none should survive (1). In fact a large fraction reach the ground (1). Their lifetime as measured from Earth is dilated roughly elevenfold (1), making the descent last barely one and a half lifetimes, and predicting the measured survival exactly (1).An electron (rest energy 8.2 × 10−14 J) travels at 0.98c. Calculate its total energy.
A beam of pions has a rest-frame lifetime of 26 ns. In the laboratory their lifetime is measured as 52 ns. Determine the speed of the pions. c = 3.00 × 108 m s−1.
A muon created 10 km above the ground travels straight down at 0.996c. Calculate the thickness of the atmosphere below it as measured in the muon's rest frame, and state which frame measures the proper 10 km.
An accelerator gives an electron a total energy of 4.1 × 10−13 J. The electron's rest energy is 8.2 × 10−14 J. Determine the electron's speed. c = 3.00 × 108 m s−1.
A spacecraft passes Earth at 0.80c. Its on-board clock records 60 s between two signals, and its proper length is 90 m. Calculate the time between the signals and the craft's length as measured from Earth.
Describe Bertozzi's experiment and explain what it demonstrated directly.
Describe the shape of a graph of an object's mass against its speed, from rest towards the speed of light.
The crew of a spacecraft travelling at 0.70c set off for a star 4.2 light years away, as measured from Earth. The mission plan requires the crew to age less than 4.5 years during the journey. Deduce whether the plan is met. (A light year is the distance light travels in one year; its value in metres is not needed.)
Show that classical physics predicts an impossible speed for electrons accelerated from rest through 4.5 MV, and state what Bertozzi's measurements showed such electrons actually do. e/m = 1.76 × 1011 C kg−1; c = 3.00 × 108 m s−1.
A student sketches the kinetic energy of an electron against its speed as the classical curve Ek = ½m0v2. Describe how the correct relativistic curve differs.
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