IB Physics Quiz: Apply Galilean And Special Relativity
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Apply Galilean And Special RelativityQuestion 1 of 20

A rocket shaped like a cube with proper side length L0L_0 travels at relativistic speed vv parallel to its x-axis. An observer is at rest. What is the volume of the cube measured by this observer?

L03L_0^3
L03/γL_0^3 / \gamma
L03/γ2L_0^3 / \gamma^2
L03/γ3L_0^3 / \gamma^3
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IB Physics Quiz

IB Physics Quiz: Apply Galilean And Special Relativity

Practice Apply Galilean And Special Relativity in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

A rocket shaped like a cube with proper side length L0L_0 travels at relativistic speed vv parallel to its x-axis. An observer is at rest. What is the volume of the cube measured by this observer?

  1. L03L_0^3
  2. L03/γL_0^3 / \gamma (correct answer)
  3. L03/γ2L_0^3 / \gamma^2
  4. L03/γ3L_0^3 / \gamma^3
Explanation: Length contraction only occurs in the dimension parallel to the direction of motion. The side parallel to the x-axis contracts to a length L=L0/γL = L_0/\gamma. The lengths of the sides perpendicular to the motion (y and z axes) remain L0L_0. The volume measured by the stationary observer is the product of the three side lengths: V=(L0/γ)×L0×L0=L03/γV = (L_0/\gamma) \times L_0 \times L_0 = L_0^3 / \gamma. Distractor D assumes all three dimensions contract, a common misconception.

Question 2

A particle has a proper lifetime of 2.0×10−62.0 \times 10^{-6} s. It is accelerated to a speed such that its lifetime as measured in a laboratory frame is 3.0×10−63.0 \times 10^{-6} s. What is the speed of the particle relative to the laboratory?

  1. 0.44c
  2. 0.67c
  3. 0.75c (correct answer)
  4. 0.82c
Explanation: Time dilation is given by Δt=γΔt0\Delta t = \gamma \Delta t_0, where Δt0\Delta t_0 is the proper time and Δt\Delta t is the dilated time. The Lorentz factor γ=Δt/Δt0=(3.0×10−6)/(2.0×10−6)=1.5\gamma = \Delta t / \Delta t_0 = (3.0 \times 10^{-6}) / (2.0 \times 10^{-6}) = 1.5. The Lorentz factor is also γ=1/1−v2/c2\gamma = 1/\sqrt{1-v^2/c^2}. Rearranging for v/cv/c: (v/c)2=1−1/γ2=1−1/(1.5)2=1−1/2.25=1−4/9=5/9(v/c)^2 = 1 - 1/\gamma^2 = 1 - 1/(1.5)^2 = 1 - 1/2.25 = 1 - 4/9 = 5/9. Taking the square root, v/c=5/3≈0.745v/c = \sqrt{5}/3 \approx 0.745. So, the speed is approximately 0.75c0.75c. Distractor B is the simple ratio of the times, 2/32/3.

Question 3

Spaceship A moves away from Earth at a speed of 0.5c0.5c. Spaceship B moves away from Earth in the same direction at a speed of 0.8c0.8c. What is the speed of spaceship B as measured by an observer on spaceship A?

  1. 0.30c
  2. 0.45c
  3. 0.50c (correct answer)
  4. 1.30c
Explanation: This requires the relativistic velocity subtraction formula. Let Earth be frame S, and spaceship A be frame S'. The velocity of S' relative to S is v=0.5cv = 0.5c. The velocity of spaceship B in S is u=0.8cu = 0.8c. The velocity of B in S' is u′=(u−v)/(1−uv/c2)u' = (u - v) / (1 - uv/c^2). Plugging in the values: u′=(0.8c−0.5c)/(1−(0.8c)(0.5c)/c2)=0.3c/(1−0.40)=0.3c/0.60=0.5cu' = (0.8c - 0.5c) / (1 - (0.8c)(0.5c)/c^2) = 0.3c / (1 - 0.40) = 0.3c / 0.60 = 0.5c. Distractor A is the incorrect Galilean subtraction result (0.8c−0.5c0.8c - 0.5c).

Question 4

An observer in an inertial reference frame S observes two explosions, P and Q, to be simultaneous. The explosions occur at different locations along the x-axis. A second observer in a frame S' moves at a constant velocity vv along the x-axis relative to S. Which statement must be true for the observer in S'?

  1. The explosions P and Q are also simultaneous in S'.
  2. The explosion that occurred at a larger x-coordinate in S will be observed first in S'.
  3. The explosion that occurred at a smaller x-coordinate in S will be observed first in S'.
  4. The order in which the explosions occur in S' depends on the direction of the velocity vv. (correct answer)
Explanation: This is a direct consequence of the relativity of simultaneity. Events that are simultaneous but spatially separated in one inertial frame (S) are not simultaneous in another inertial frame (S') moving relative to it. The Lorentz transformation for time difference is Δt′=γ(Δt−vΔx/c2)\Delta t' = \gamma (\Delta t - v \Delta x / c^2). Since Δt=0\Delta t = 0 in S, Δt′=−γvΔx/c2\Delta t' = -\gamma v \Delta x / c^2. The sign of Δt′\Delta t' (which determines the order) depends on the sign of vv, meaning the direction of motion.

Question 5

In which of the following scenarios is Galilean relativity sufficient to describe the motion accurately, with negligible error compared to special relativity?

  1. Determining the decay rate of a high-energy particle in a particle accelerator.
  2. Calculating the precise precession of the orbit of Mercury around the Sun.
  3. Analyzing the collision of two protons at the Large Hadron Collider (LHC).
  4. Calculating the trajectory of a satellite in low Earth orbit. (correct answer)
Explanation: Galilean relativity is an excellent approximation when speeds are much less than the speed of light (v≪cv \ll c). A satellite in low Earth orbit travels at about 8 km/s, which is a tiny fraction of the speed of light (c≈300,000c \approx 300,000 km/s). In contrast, particles in accelerators (A and C) travel at speeds extremely close to cc, requiring special relativity. The precession of Mercury's orbit (B) is a famous case where even Newtonian mechanics (based on Galilean relativity) fails, and General Relativity is needed for an accurate description.

Question 6

A spaceship travels from Earth to a star 10 light-years away at a constant speed of 0.8c0.8c. An astronaut on the spaceship measures the duration of the trip using a clock on board. An observer on Earth measures the duration using a clock on Earth. Which is a correct statement about the measured times?

  1. The astronaut measures the proper time, which is 12.5 years.
  2. The observer on Earth measures the proper time, which is 12.5 years.
  3. The astronaut measures the proper time, which is 7.5 years. (correct answer)
  4. The observer on Earth measures the proper time, which is 7.5 years.
Explanation: Proper time (Δt0\Delta t_0) is measured in the frame where the two events (departure and arrival) occur at the same location. This is the astronaut's frame. The time measured on Earth is Δt=distance/speed=10 ly/0.8c=12.5\Delta t = \text{distance}/\text{speed} = 10 \text{ ly} / 0.8c = 12.5 years. The Lorentz factor is γ=1/1−0.82=1/0.36=1/0.6=5/3\gamma = 1/\sqrt{1-0.8^2} = 1/\sqrt{0.36} = 1/0.6 = 5/3. The astronaut's time is the proper time: Δt0=Δt/γ=12.5/(5/3)=7.5\Delta t_0 = \Delta t / \gamma = 12.5 / (5/3) = 7.5 years. So, the astronaut measures the proper time, and its value is 7.5 years.

Question 7

A spaceship travels at speed vv past a space station. Observers on the station measure the spaceship's length to be LL and the time between two ticks of a clock on the spaceship to be Δt\Delta t. What are the proper length of the spaceship L0L_0 and the proper time interval Δt0\Delta t_0 between the clock ticks?

  1. L0=L/γL_0 = L/\gamma, Δt0=Δt/γ\Delta t_0 = \Delta t/\gamma
  2. L0=γLL_0 = \gamma L, Δt0=γΔt\Delta t_0 = \gamma \Delta t
  3. L0=γLL_0 = \gamma L, Δt0=Δt/γ\Delta t_0 = \Delta t/\gamma (correct answer)
  4. L0=L/γL_0 = L/\gamma, Δt0=γΔt\Delta t_0 = \gamma \Delta t
Explanation: Proper length L0L_0 is the length in the object's rest frame. The measured length LL is contracted: L=L0/γL = L_0/\gamma, which rearranges to L0=γLL_0 = \gamma L. Proper time Δt0\Delta t_0 is the time interval in the clock's rest frame. The measured time interval Δt\Delta t is dilated: Δt=γΔt0\Delta t = \gamma \Delta t_0, which rearranges to Δt0=Δt/γ\Delta t_0 = \Delta t/\gamma. This question requires correctly rearranging both the length contraction and time dilation formulas.

Question 8

A rocket moves at 0.9c0.9c relative to an observer. It fires a projectile in the forward direction. The speed of the projectile relative to the rocket is also 0.9c0.9c. What is the speed of the projectile relative to the observer?

  1. Exactly cc.
  2. Less than cc. (correct answer)
  3. Exactly 1.8c1.8c.
  4. Greater than cc but less than 1.8c1.8c.
Explanation: According to the second postulate of special relativity, no object with mass can travel at or faster than the speed of light cc. Simple Galilean addition would give 0.9c+0.9c=1.8c0.9c + 0.9c = 1.8c, which is non-physical. The relativistic velocity addition formula, u=(u′+v)/(1+u′v/c2)u = (u'+v)/(1+u'v/c^2), must be used. For any two velocities less than cc, the formula will always yield a result that is also less than cc. In this case, the result is 1.8c/(1+0.81)≈0.994c1.8c / (1 + 0.81) \approx 0.994c, which is less than cc.

Question 9

An object is moving at a speed of v=0.95cv = 0.95c. What is the ratio of the time interval measured in the stationary frame (Δt\Delta t) to the proper time interval (Δt0\Delta t_0)?

  1. 0.31
  2. 1.05
  3. 3.2 (correct answer)
  4. 9.5
Explanation: The relationship between dilated time and proper time is Δt=γΔt0\Delta t = \gamma \Delta t_0. The question asks for the ratio Δt/Δt0\Delta t / \Delta t_0, which is equal to the Lorentz factor γ\gamma. We calculate γ=1/1−v2/c2=1/1−0.952=1/1−0.9025=1/0.0975≈1/0.3122≈3.20\gamma = 1/\sqrt{1-v^2/c^2} = 1/\sqrt{1-0.95^2} = 1/\sqrt{1-0.9025} = 1/\sqrt{0.0975} \approx 1/0.3122 \approx 3.20. Distractor A is the value of 1/γ1/\gamma.

Question 10

Twin A remains on Earth while twin B travels on a spaceship to a distant star and back at a relativistic speed. Upon returning to Earth, twin B is younger than twin A. What is the fundamental reason for this age difference?

  1. The spaceship's clock ran slower only during the periods of acceleration and deceleration.
  2. Twin B was in a single inertial frame for the entire journey, while twin A was not.
  3. The situation is symmetric, so there should be no age difference; it is a paradox.
  4. Twin B changed inertial reference frames during the journey, while twin A did not. (correct answer)
Explanation: The resolution to the twin paradox lies in the asymmetry of the twins' journeys. Twin A remains (approximately) in a single inertial reference frame. Twin B, however, must accelerate to start the trip, decelerate to turn around, and accelerate again to return. This change of inertial frames breaks the symmetry. Because twin B is in non-inertial frames for part of the journey, they are the one who experiences less elapsed proper time. The time difference accumulates during all phases of the journey, not just acceleration.

Question 11

A muon is created in the upper atmosphere and travels towards the Earth's surface at a speed of 0.99c0.99c. In the muon's rest frame, its lifetime is 2.2μ2.2 \mus. From the perspective of the muon, which statement best explains why it reaches the surface?

  1. From the muon's perspective, its own lifetime is dilated, giving it more time to travel.
  2. From the muon's perspective, the distance to the Earth's surface is contracted, allowing it to reach the surface within its lifetime. (correct answer)
  3. From the muon's perspective, the Earth is approaching it at 0.99c0.99c, so the speed of light from Earth is greater than cc.
  4. From the muon's perspective, its mass decreases, allowing it to travel faster than predicted.
Explanation: Special relativity must provide a consistent explanation from all inertial frames. From the Earth's frame, the muon's lifetime is dilated. From the muon's frame, its lifetime is its proper lifetime (2.2μ2.2 \mus), which is not enough to cover the atmospheric distance measured in Earth's frame. However, from the muon's frame, the atmosphere is moving towards it at 0.99c0.99c, so the distance to the Earth's surface is length-contracted, making it short enough to be traversed within the muon's proper lifetime.

Question 12

A thin, square plate with proper side length L0L_0 moves at a relativistic speed vv in a direction parallel to one of its sides. What is the area of the plate as measured by a stationary observer?

  1. L02L_0^2
  2. L02/γL_0^2 / \gamma (correct answer)
  3. L02/γ2L_0^2 / \gamma^2
  4. γL02\gamma L_0^2
Explanation: Length contraction occurs only in the direction of motion. Let the plate move parallel to its x-side. This side's length contracts to Lx=L0/γL_x = L_0/\gamma. The side perpendicular to the motion, the y-side, has its length unchanged, so Ly=L0L_y = L_0. The area measured by the stationary observer is A=Lx×Ly=(L0/γ)×L0=L02/γA = L_x \times L_y = (L_0/\gamma) \times L_0 = L_0^2 / \gamma. Distractor C incorrectly assumes both sides contract.

Question 13

A muon has a proper lifetime of τ0\tau_0. It travels at a speed vv through a laboratory. From the laboratory's frame of reference, the muon travels a distance dd before it decays. Which expression gives the distance dd?

  1. vτ0v \tau_0
  2. γvτ0\gamma v \tau_0 (correct answer)
  3. vτ0/γv \tau_0 / \gamma
  4. cτ0c \tau_0
Explanation: In the laboratory's frame of reference, the muon's lifetime is subject to time dilation. The observed lifetime is Δt=γτ0\Delta t = \gamma \tau_0. The distance the muon travels in the lab frame is its speed multiplied by this dilated lifetime. Therefore, d=v×Δt=v×(γτ0)=γvτ0d = v \times \Delta t = v \times (\gamma \tau_0) = \gamma v \tau_0. Distractor A ignores the effect of time dilation.

Question 14

A spacecraft travels from Earth to a distant star at 0.80c0.80c relative to Earth. According to Earth observers, the journey takes 25 years. The spacecraft then immediately returns to Earth at the same speed. What is the total elapsed time on the spacecraft's clock for the round trip?

  1. 15 years
  2. 30 years (correct answer)
  3. 42 years
  4. 50 years
Explanation: The time dilation factor is γ=11−v2/c2=11−0.82=10.6=1.67\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-0.8^2}} = \frac{1}{0.6} = 1.67. The proper time (spacecraft clock) is Δt0=Δtγ=251.67=15\Delta t_0 = \frac{\Delta t}{\gamma} = \frac{25}{1.67} = 15 years for one way. For the round trip: 2×15=302 \times 15 = 30 years. Choice A gives only one-way time. Choice C incorrectly uses γ×\gamma \times coordinate time instead of dividing. Choice D uses the coordinate time without time dilation correction.

Question 15

A muon is created 15 km above Earth's surface and moves downward at 0.95c0.95c. The muon's proper lifetime is 2.2×10−62.2 \times 10^{-6} s. According to Earth observers, what distance does the muon travel before decaying?

  1. 626 m
  2. 2.0 km
  3. 6.4 km
  4. 21 km (correct answer)
Explanation: Earth observers measure dilated lifetime: Δt=γΔt0=2.2×10−61−0.952=2.2×10−60.312=7.05×10−6\Delta t = \gamma \Delta t_0 = \frac{2.2 \times 10^{-6}}{\sqrt{1-0.95^2}} = \frac{2.2 \times 10^{-6}}{0.312} = 7.05 \times 10^{-6} s. Distance = v×t=0.95c×7.05×10−6=0.95×3×108×7.05×10−6=2.01×104v \times t = 0.95c \times 7.05 \times 10^{-6} = 0.95 \times 3 \times 10^8 \times 7.05 \times 10^{-6} = 2.01 \times 10^4 m = 20.1 km ≈ 21 km. Choice A uses proper lifetime without dilation. Choice B contains calculation error in γ\gamma. Choice C uses incorrect velocity in final calculation.

Question 16

A particle has rest mass m0=1.67×10−27m_0 = 1.67 \times 10^{-27} kg and moves at 0.80c0.80c. What is the particle's relativistic momentum?

  1. 6.0×10−196.0 \times 10^{-19} kg⋅m/s
  2. 1.0×10−181.0 \times 10^{-18} kg⋅m/s (correct answer)
  3. 4.0×10−194.0 \times 10^{-19} kg⋅m/s
  4. 2.0×10−182.0 \times 10^{-18} kg⋅m/s
Explanation: Relativistic momentum: p=γm0v=m0v1−v2/c2p = \gamma m_0 v = \frac{m_0 v}{\sqrt{1-v^2/c^2}}. γ=11−0.82=10.6=1.67\gamma = \frac{1}{\sqrt{1-0.8^2}} = \frac{1}{0.6} = 1.67. p=1.67×1.67×10−27×0.8×3×108=1.0×10−18p = 1.67 \times 1.67 \times 10^{-27} \times 0.8 \times 3 \times 10^8 = 1.0 \times 10^{-18} kg⋅m/s. Choice A uses classical momentum formula p=m0vp = m_0 v. Choice C contains error in γ\gamma calculation. Choice D incorrectly doubles the correct answer.

Question 17

Two spaceships approach each other. Ship A moves at +0.70c+0.70c and Ship B moves at −0.50c-0.50c relative to a space station. What is the velocity of Ship A as measured by observers on Ship B?

  1. +0.89c+0.89c (correct answer)
  2. +1.20c+1.20c
  3. +0.61c+0.61c
  4. −0.89c-0.89c
Explanation: Use relativistic velocity addition: vAB=vA−vB1−vAvBc2=0.70c−(−0.50c)1−(0.70c)(−0.50c)c2=1.20c1+0.35=1.20c1.35=0.89cv_{AB} = \frac{v_A - v_B}{1 - \frac{v_A v_B}{c^2}} = \frac{0.70c - (-0.50c)}{1 - \frac{(0.70c)(-0.50c)}{c^2}} = \frac{1.20c}{1 + 0.35} = \frac{1.20c}{1.35} = 0.89c. Choice B uses classical velocity addition. Choice C incorrectly subtracts velocities instead of adding. Choice D has correct magnitude but wrong sign direction.

Question 18

A clock on a spacecraft shows that 8.0 hours have passed during a journey. Ground observers measure this same journey to take 10.0 hours. If the spacecraft continues at the same speed for a total ground-measured time of 25 hours, how much time will have passed on the spacecraft clock?

  1. 15 hours
  2. 31 hours
  3. 20 hours (correct answer)
  4. 42 hours
Explanation: When you encounter a problem involving clocks on spacecraft versus ground observers, you're dealing with time dilation from special relativity. The key insight is that time passes differently for observers in relative motion. First, establish the relationship between proper time (spacecraft clock) and coordinate time (ground observers). From the given data: spacecraft time = 8.0 hours, ground time = 10.0 hours. This gives us a time dilation factor of 8.010.0=0.8\frac{8.0}{10.0} = 0.8. Since the spacecraft continues at the same speed, this ratio remains constant. For a total ground-measured time of 25 hours, the spacecraft clock time will be: 25×0.8=20 hours25 \times 0.8 = 20 \text{ hours}. Looking at the wrong answers: A) 15 hours incorrectly assumes the spacecraft time should be less than the proportional amount, perhaps confusing which reference frame experiences slower time. B) 31 hours suggests adding the original 8 hours to some incorrect calculation of the remaining time. D) 42 hours appears to reverse the time dilation effect, treating the spacecraft as if time passes faster there rather than slower. The correct answer is C) 20 hours because time dilation maintains a constant ratio between reference frames at constant relative velocity. Study tip: In time dilation problems, always identify which clock is moving relative to the observer. The moving clock (spacecraft) always runs slower than the stationary observer's clock (ground). Set up the ratio from the given information and apply it consistently throughout the problem.

Question 19

A light beam travels from the back to the front of a moving train. The train has proper length 200 m and moves at 0.60c0.60c relative to the ground. According to ground observers, the light beam takes 1.0×10−61.0 \times 10^{-6} s to travel from back to front. What is the speed of light as measured by train passengers?

  1. 1.6×1081.6 \times 10^8 m/s
  2. 3.8×1083.8 \times 10^8 m/s
  3. 3.0×1083.0 \times 10^8 m/s (correct answer)
  4. 5.0×1085.0 \times 10^8 m/s
Explanation: This question tests a fundamental principle of special relativity: the speed of light is constant in all inertial reference frames. When you encounter problems involving light speed measurements from different reference frames, remember that Einstein's second postulate guarantees light always travels at c=3.0×108c = 3.0 \times 10^8 m/s in vacuum, regardless of the observer's motion. The key insight is recognizing what each reference frame measures. Ground observers see a contracted train length and measure the time for light to traverse it, but train passengers measure light speed in their own rest frame. Since the train is an inertial reference frame moving at constant velocity, the speed of light must be cc for train passengers. Let's examine why the other answers represent common misconceptions: Answer A (1.6×1081.6 \times 10^8 m/s) likely comes from incorrectly applying classical velocity addition, subtracting the train's speed from light speed: 3.0×108−0.6×3.0×108=1.2×1083.0 \times 10^8 - 0.6 \times 3.0 \times 10^8 = 1.2 \times 10^8 m/s, or some similar calculation. Answer B (3.8×1083.8 \times 10^8 m/s) probably results from incorrectly adding the train's velocity to light speed using classical mechanics: 3.0×108+0.6×3.0×108=4.8×1083.0 \times 10^8 + 0.6 \times 3.0 \times 10^8 = 4.8 \times 10^8 m/s, or a related computational error. Answer D (5.0×1085.0 \times 10^8 m/s) might come from misusing the given time and distance measurements from the ground frame to calculate speed in the train frame. Study tip: Whenever you see light speed questions in relativity, immediately recall that cc is invariant. The numerical details about train length and timing are often distractors—focus on which reference frame is measuring the light speed.

Question 20

A spaceship emits a pulse of light. According to the second postulate of special relativity, which statement is correct for all observers in inertial reference frames?

  1. The light pulse travels at speed cc relative to the spaceship that emitted it.
  2. The light pulse travels at speed cc relative to the observer measuring it. (correct answer)
  3. The speed of the light pulse depends on the relative motion between the source and the observer.
  4. The frequency of the light pulse is the same for the source and the observer.
Explanation: The second postulate of special relativity states that the speed of light in a vacuum is the same for all observers in inertial reference frames, regardless of the motion of the light source. Therefore, any observer in an inertial frame will measure the speed of the light pulse to be cc. Choice A is the classical expectation. Choice C directly contradicts the postulate. Choice D is incorrect due to the relativistic Doppler effect, which causes a change in the observed frequency.