Physics: Special Relativity: Time Dilation and Length Contraction Worksheet

Question 1

In special relativity, a spaceship passes Earth at a constant speed. From the Earth frame, the spaceship clock appears to tick more slowly than an Earth clock. Explain what this means without saying that the spaceship clock is broken.

Accepted Answer

The spaceship clock is working normally in its own frame. From the Earth frame, time intervals between ticks on the moving spaceship clock are measured to be longer, so the moving clock appears to run slow.

Question 2

A spaceship travels at 0.80c relative to Earth. Calculate the Lorentz factor γ.

Accepted Answer

The Lorentz factor is γ = 1/sqrt(1 - 0.80^2) = 1/sqrt(0.36) = 1/0.60 = 1.67. The Lorentz factor is about 1.67.

Question 3

An astronaut on a spaceship moving at 0.80c measures a trip to take 6.0 years on the spaceship clock. How much time does an observer on Earth measure for the same trip?

Accepted Answer

The spaceship time is the proper time because the astronaut is present at both the start and end of the trip in the spaceship frame. Using γ = 1.67, the Earth time is Δt = γΔt0 = 1.67(6.0 years) = 10.0 years. Earth measures about 10 years.

Question 4

A spacecraft has a proper length of 120 m. It moves past Earth at 0.60c. What length do observers on Earth measure for the spacecraft?

Accepted Answer

At 0.60c, γ = 1/sqrt(1 - 0.60^2) = 1.25. The contracted length is L = L0/γ = 120 m/1.25 = 96 m. Observers on Earth measure the spacecraft to be 96 m long.

Question 5

A muon has a proper lifetime of 2.2 microseconds. If it moves through the laboratory at 0.98c, what lifetime is measured in the laboratory frame?

Accepted Answer

At 0.98c, γ = 1/sqrt(1 - 0.98^2) = 1/sqrt(0.0396) ≈ 5.03. The laboratory lifetime is Δt = γΔt0 = 5.03(2.2 microseconds) ≈ 11.1 microseconds. The muon lifetime in the lab frame is about 11.1 microseconds.

Question 6

Using the laboratory lifetime from the previous problem, estimate how far the muon travels in the laboratory frame before decaying. Use c = 3.00 x 10^8 m/s.

Accepted Answer

The distance is d = vt = (0.98)(3.00 x 10^8 m/s)(11.1 x 10^-6 s). This gives d ≈ 3.26 x 10^3 m. The muon travels about 3260 m in the laboratory frame.

Question 7

Two events happen at the same place in a train passenger's frame: the passenger starts a stopwatch and later stops it. Which frame measures the proper time between these two events, the train frame or the ground frame? Explain.

Accepted Answer

The train frame measures the proper time because both events happen at the same location in the train frame. The ground frame measures a longer dilated time for the moving stopwatch.

Question 8

Calculate γ for speeds of 0.50c and 0.90c. Explain what the results show about how relativistic effects change as speed increases.

Accepted Answer

For 0.50c, γ = 1/sqrt(1 - 0.50^2) ≈ 1.15. For 0.90c, γ = 1/sqrt(1 - 0.90^2) ≈ 2.29. The results show that relativistic effects grow much stronger as speed gets closer to the speed of light.

Question 9

A cube has a proper side length of 10 m. It moves at 0.80c in the x direction relative to an observer. What dimensions does the observer measure for the cube?

Accepted Answer

At 0.80c, γ = 1.67. Only the dimension parallel to the motion contracts, so the x length is 10 m/1.67 = 6.0 m. The y and z dimensions remain 10 m each. The observer measures 6.0 m by 10 m by 10 m.

Question 10

A student says, "A moving object is shorter in every direction because all lengths contract." Identify the error and correct the statement.

Accepted Answer

The error is saying that all directions contract. In special relativity, length contraction occurs only along the direction of relative motion. Dimensions perpendicular to the motion are unchanged.

Question 11

At what speed must an object move for its Lorentz factor to be γ = 2? Give your answer as a fraction of c.

Accepted Answer

Use γ = 1/sqrt(1 - v^2/c^2). If γ = 2, then sqrt(1 - v^2/c^2) = 1/2. Squaring gives 1 - v^2/c^2 = 1/4, so v^2/c^2 = 3/4. Therefore v = sqrt(3/4)c ≈ 0.866c.

Question 12

A star is 4.0 light-years from Earth in the Earth frame. A spaceship travels from Earth to the star at 0.80c. Find the travel time measured on Earth and the travel time measured on the spaceship.

Accepted Answer

In the Earth frame, the time is t = distance/speed = 4.0 light-years/0.80c = 5.0 years. At 0.80c, γ = 1.67, so the spaceship time is t0 = t/γ = 5.0 years/1.67 = 3.0 years. Earth measures 5.0 years, and the spaceship measures 3.0 years.

Physics: Special Relativity: Time Dilation and Length Contraction

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