IB Physics Quiz: Understand Galilean And Special Relativity

20 questions · exam conditions

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Understand Galilean And Special RelativityQuestion 1 of 20

Event A occurs at $t=0$ at the origin of an inertial frame. Event B occurs at time $t=T$ and position $x=X$ . Under what condition is it impossible for event A to have caused event B?

X > cT

X < cT

X = cT

X/T < 0

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IB Physics Quiz

IB Physics Quiz: Understand Galilean And Special Relativity

Practice Understand 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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This quiz focuses on Understand Galilean And Special Relativity, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Question 1

Event A occurs at $t=0$ at the origin of an inertial frame. Event B occurs at time $t=T$ and position $x=X$ . Under what condition is it impossible for event A to have caused event B?

$X > cT$ (correct answer)
$X < cT$
$X = cT$
$X/T < 0$

Explanation: For event A to cause event B, a signal or influence must travel from the location of A to the location of B. The maximum possible speed for any such signal is the speed of light,

c

. The time available for travel is

T

. The maximum distance a signal can cover in this time is

cT

. If the spatial separation

X

between the events is greater than

cT

, no signal could have reached B from A in time, making a causal link impossible. This corresponds to a space-like spacetime interval.

Question 2

An experimenter in a laboratory measures the speed of light from a laser to be $c$ . A spaceship moves towards the laboratory at a speed of $0.5c$ . According to the postulates of special relativity, what speed will an observer on the spaceship measure for the light from the laser?

$0.5c$
$c$ (correct answer)
$1.5c$
$\sqrt{c^2 - (0.5c)^2}$

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 or the observer. Therefore, the observer on the spaceship will also measure the speed of light to be

c

Question 3

A spaceship has a proper length of 100 m. It travels past a stationary observer at a speed such that its Lorentz factor $\gamma$ is 2. What length does the stationary observer measure for the spaceship?

25 m
50 m (correct answer)
100 m
200 m

Explanation: The length measured by a stationary observer,

L

, is related to the proper length,

L_0

, by the length contraction formula

L = L_0 / \gamma

. Given

L_0 = 100

m and

\gamma = 2

, the measured length is

L = 100 \text{ m} / 2 = 50 \text{ m}

Question 4

An observer is in the exact center of a moving train carriage. Two lights, one at the front and one at the back of the carriage, flash simultaneously according to this observer. What does a stationary observer on a platform conclude about when the flashes occurred?

The two flashes occurred simultaneously.
The flash from the front of the carriage occurred first.
The flash from the back of the carriage occurred first. (correct answer)
Which flash occurred first depends on the platform observer's position.

Explanation: This is a classic example of the relativity of simultaneity. For the light from the two flashes to reach the observer in the middle of the train at the same time, the platform observer deduces that the back of the train was moving away from the point of its flash, while the front was moving towards its flash point. Therefore, the light from the back had to travel further to reach the midpoint of the train. For it to arrive simultaneously with the light from the front, it must have been emitted earlier in the platform's frame.

Question 5

Muons are unstable particles created in the upper atmosphere. They have a short proper half-life. An unexpectedly large number of muons are detected at sea level. Which is the best explanation for this observation from the perspective of the muon's reference frame?

The muons' half-life is increased due to time dilation, allowing more to survive the journey.
The distance from the upper atmosphere to sea level is contracted, shortening the journey's length. (correct answer)
The speed of the muons is measured to be greater than the speed of light.
The Earth's atmosphere is moving towards the muon at relativistic speeds, increasing collision rates.

Explanation: In the muon's reference frame, its half-life is the proper half-life (it is not dilated). Instead, the Earth and its atmosphere are moving towards the muon at a relativistic speed. This causes the distance the muon must travel to reach sea level to be length-contracted, appearing much shorter than it does in Earth's frame. This shorter journey allows more muons to survive.

Question 6

In the twin paradox, one twin travels to a distant star and back at relativistic speed while the other stays on Earth. The travelling twin ages less. What is the key element that resolves the apparent paradox by breaking the symmetry between the twins' frames of reference?

The travelling twin is in a different gravitational potential.
The travelling twin's biological clock physically slows down.
The travelling twin must accelerate to reverse direction. (correct answer)
The distance to the star is contracted for the travelling twin.

Explanation: The situation is not symmetrical because the travelling twin's reference frame is not inertial throughout the entire journey. To return to Earth, the twin must decelerate, turn around, and accelerate. This period of acceleration means their frame is non-inertial, breaking the symmetry with the Earth-bound twin's inertial frame and resolving the paradox.

Question 7

Two events, P and Q, occur in an inertial reference frame. The spacetime interval, $(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2$ , between them is found to be positive. What does this imply about the relationship between P and Q?

It is possible for P to have caused Q, and all observers will agree on the temporal sequence of P and Q. (correct answer)
No causal relationship can exist between P and Q, and some observers may see the events as simultaneous.
The spatial separation $\Delta x$ is greater than the distance light could travel in the time interval $\Delta t$ .
All observers will measure the same time separation $\Delta t$ between the events.

Explanation: A positive spacetime interval is called 'time-like'. This means that

c\Delta t > \Delta x

, so a signal traveling at or below the speed of light could travel between the events. This allows for a causal relationship. For all time-like separated events, the order in time (which event happened first) is preserved for all inertial observers.

Question 8

Two events occur at positions $x_1 = 0$ and $x_2 = 6.0 \times 10^8$ m in frame S. They are simultaneous in S, occurring at $t=0$ . An observer in frame S' moves at $v=0.6c$ in the positive x-direction relative to S. What is the time interval between the second event and the first event ( $t'_2 - t'_1$ ) as measured in S'?

0 s
-1.20 s
-1.50 s (correct answer)
-2.00 s

Explanation: Using the Lorentz transformation for time,

\Delta t' = \gamma(\Delta t - v\Delta x/c^2)

. First, calculate

\gamma = 1/\sqrt{1-0.6^2} = 1/\sqrt{0.64} = 1/0.8 = 1.25

. In frame S, the events are simultaneous, so

\Delta t = 0

. The spatial separation is

\Delta x = x_2 - x_1 = 6.0 \times 10^8

m. Plugging these values in:

\Delta t' = 1.25(0 - (0.6c)(6.0 \times 10^8)/c^2) = -1.25 \times (0.6 \times 6.0 \times 10^8) / (3.0 \times 10^8) = -1.25 \times 0.6 \times 2 = -1.50

s. The negative sign indicates that event 2 occurs before event 1 in frame S'.

Question 9

On a standard spacetime diagram with the time axis ( $ct$ ) vertical and the space axis ( $x$ ) horizontal, the path of an object is its worldline. Which statement correctly describes the worldline of a photon traveling in the positive x-direction, starting from the origin?

A horizontal line starting from the origin.
A vertical line starting from the origin.
A straight line with a slope of +1. (correct answer)
A straight line with a slope greater than +1.

Explanation: A photon travels at speed

v=c

. Its position is given by

x = ct

. On a spacetime diagram with

ct

as the vertical axis and

x

as the horizontal axis, the slope is

\Delta(ct)/\Delta x

. From the equation of motion,

ct/x = 1

, so the slope is exactly +1. A vertical line (infinite slope) represents a stationary object. A line with slope greater than 1 represents an object moving at

v < c

(since slope =

c/v

Question 10

A spaceship is moving at speed $v = 0.8c$ away from a stationary observer. The spaceship turns on a light that points forward, in its direction of motion. From the reference frame of the spaceship, how does the observed speed of the light compare to the speed of light measured by the stationary observer?

It is the same for both observers. (correct answer)
It is greater for the observer on the spaceship.
It is greater for the stationary observer.
The comparison depends on the frequency of the light.

Explanation: This is a direct application of the second postulate of special relativity. The speed of light in a vacuum,

c

, is a universal constant for all observers in inertial reference frames. It does not depend on the motion of the source (the spaceship) or the observer. Both the person on the spaceship and the stationary observer will measure the speed of the light to be exactly

c

Question 11

An astronaut on a spaceship moving at $0.9c$ relative to a space station performs an experiment that lasts for 100 s according to her watch. A scientist at the space station also times the experiment. Which statement correctly identifies the proper time and the relationship between the two measurements?

The scientist measures a time shorter than 100 s; the astronaut's watch measured the proper time.
The scientist measures a time longer than 100 s; the astronaut's watch measured the proper time. (correct answer)
The scientist measures a time shorter than 100 s; the scientist's clock measured the proper time.
The scientist measures a time longer than 100 s; the scientist's clock measured the proper time.

Explanation: Proper time (

\Delta t_0

) is the time interval between two events measured by an observer for whom the events occur at the same location. For the astronaut, the start and end of the experiment happen at her location on the spaceship, so her measurement of 100 s is the proper time. An observer in a different inertial frame (the scientist) will measure a dilated time

\Delta t = \gamma \Delta t_0

. Since

\gamma > 1

for

v>0

, the scientist will measure a time longer than 100 s.

Question 12

A square plate with proper side length $L_0$ rests in an inertial frame S. An observer in a frame S' moves at a relativistic velocity $v$ parallel to one pair of the square's sides. What is the area of the plate as measured by the observer in S'?

$L_0^2 / \gamma^2$
$\gamma L_0^2$
$L_0^2$
$L_0^2 / \gamma$ (correct answer)

Explanation: Length contraction only occurs in the direction of motion. The side of the square parallel to the velocity

v

will be measured to have a contracted length of

L = L_0/\gamma

. The side perpendicular to the motion will still have its proper length

L_0

. The area measured in frame S' is the product of these two lengths: Area' =

(L_0/\gamma) \times L_0 = L_0^2/\gamma

Question 13

A long rod is moving at a relativistic speed parallel to its length. An observer at rest measures the rod's length. How does this measured length compare to its proper length, and in which direction(s) does the contraction occur?

The measured length is shorter, and the contraction occurs only along the direction of motion. (correct answer)
The measured length is shorter, and the contraction occurs equally in all directions.
The measured length is longer, and the expansion occurs only along the direction of motion.
The measured length is the same, as length is an invariant quantity in relativity.

Explanation: According to special relativity, length contraction occurs only for the dimension of an object that is parallel to its direction of motion. Dimensions perpendicular to the direction of motion are unaffected. The measured length is therefore shorter than the proper length.

Question 14

A rocket moves away from Earth at a speed of $0.8c$ . The rocket launches a probe in the same direction at a speed of $0.6c$ relative to the rocket. What is the speed of the probe as measured by an observer on Earth?

0.95c (correct answer)
0.98c
1.00c
1.40c

Explanation: The relativistic velocity addition formula is

v' = (u+v)/(1+uv/c^2)

. Here,

u = 0.8c

and

v = 0.6c

. So,

v' = (0.8c+0.6c)/(1+(0.8c)(0.6c)/c^2) = 1.4c / (1+0.48) = 1.4c / 1.48 \approx 0.946c

. This rounds to

0.95c

. Distractor D is the incorrect Galilean sum.

Question 15

A spaceship of proper length $L_0$ travels at speed $v$ past a space station. An observer on the station measures the time taken for the spaceship to pass a fixed point on the station to be $\Delta t$ . What is the proper length $L_0$ of the spaceship in terms of $\Delta t$ and $v$ ?

$v \Delta t$
$\gamma v \Delta t$ (correct answer)
$v \Delta t / \gamma$
$v \Delta t \sqrt{\gamma}$

Explanation: The station observer measures the contracted length of the spaceship,

L

. This length is equal to the speed of the spaceship multiplied by the time it takes to pass a point:

L = v \Delta t

. The relationship between the contracted length

L

and the proper length

L_0

L = L_0/\gamma

. By equating the two expressions for

L

, we get

L_0/\gamma = v \Delta t

. Rearranging for

L_0

gives

L_0 = \gamma v \Delta t

Question 16

A meter stick lies along the x-axis in its rest frame S. Frame S' moves at $0.866c$ parallel to the x-axis. An observer in S' measures the stick using two synchronized clocks in frame S' to record when the front and back of the stick pass a fixed point. What length does the S' observer measure?

$1.0$ m, since the measurement is performed correctly using synchronized clocks in the observer's frame
$0.866$ m, corresponding to the velocity-dependent contraction factor $v/c$
$2.0$ m, because the stick appears longer when measured with the proper simultaneity procedure
$0.5$ m, due to length contraction with $\gamma = 2.0$ for the relative motion (correct answer)

Explanation: When you encounter special relativity problems involving length measurements, you need to carefully consider which frame is doing the measuring and apply the length contraction formula correctly. The meter stick is at rest in frame S, so its proper length is 1.0 m. Frame S' moves at

0.866c

relative to S, which means the S' observer sees the stick moving past at

0.866c

. To find the Lorentz factor:

\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-(0.866)^2}} = \frac{1}{\sqrt{0.25}} = 2.0

The contracted length measured by the S' observer is:

L = \frac{L_0}{\gamma} = \frac{1.0\text{ m}}{2.0} = 0.5\text{ m}

The measurement procedure described (using synchronized clocks in S' to record when both ends pass a fixed point) is indeed the correct way to measure length in relativity, ensuring simultaneity in the measuring frame. Choice A incorrectly assumes that using the proper measurement technique means no contraction occurs—but length contraction is a real physical effect that occurs regardless of measurement method. Choice B uses

v/c = 0.866

as the contraction factor, but this isn't the correct relativistic formula. Choice C suggests the stick appears longer, which contradicts the fundamental principle that objects contract in the direction of relative motion. Remember: length contraction always makes moving objects appear shorter in the direction of motion, with the contraction factor being

1/\gamma

, not

v/c

. Calculate

\gamma

first, then divide the proper length by

\gamma

Question 17

A particle accelerator increases a proton's speed from $0.1c$ to $0.9c$ . Compare the kinetic energy change calculated using Galilean mechanics versus special relativity. The proton's rest mass is $1.67 \times 10^{-27}$ kg.

Galilean predicts $1.2 \times 10^{-11}$ J while relativistic gives $1.4 \times 10^{-11}$ J, with minimal difference
Both methods give approximately the same result since the average speed is only $0.5c$
Galilean predicts $6.0 \times 10^{-12}$ J while relativistic gives $1.4 \times 10^{-11}$ J, showing significant deviation at high speeds (correct answer)
Relativistic energy is exactly twice the Galilean prediction due to the $\gamma^2$ factor in the energy formula

Explanation: When dealing with particles moving at significant fractions of the speed of light, you must compare classical (Galilean) mechanics with special relativity to see how dramatically they diverge at high speeds. Let's calculate both kinetic energies. For Galilean mechanics:

KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \frac{1}{2}(1.67 \times 10^{-27})[(0.9c)^2 - (0.1c)^2]

. With

c = 3 \times 10^8

m/s, this gives

\Delta KE_{Galilean} = 6.0 \times 10^{-12}

J. For relativistic mechanics, we use

KE = (\gamma - 1)mc^2

where

\gamma = \frac{1}{\sqrt{1-v^2/c^2}}

. At

v = 0.1c

\gamma_i = 1.005

, and at

v = 0.9c

\gamma_f = 2.294

. The relativistic kinetic energy change is

\Delta KE_{rel} = (2.294 - 1.005)(1.67 \times 10^{-27})(3 \times 10^8)^2 = 1.4 \times 10^{-11}

J. Answer C correctly identifies these values and highlights the dramatic difference. Answer A uses incorrect calculations that underestimate the relativistic effect. Answer B incorrectly assumes averaging speeds gives meaningful results—relativistic effects depend on the actual speeds reached, not averages. Answer D mentions a non-existent "

\gamma^2

factor" and claims an exact 2:1 ratio that doesn't match the actual calculations. Study tip: At speeds above

0.3c

, always expect significant deviations between classical and relativistic predictions. The

\gamma

factor grows rapidly, making classical mechanics increasingly inadequate for high-energy particle physics problems.

Question 18

In a thought experiment, twins A and B are initially at rest relative to each other. Twin A accelerates to $0.6c$ , travels for $10$ years (Earth time), then instantly reverses direction and returns at $0.6c$ . During the turnaround, twin A observes twin B's clock. What happens to the observed rate of twin B's clock during this instantaneous turnaround?

Twin B's clock appears to run faster by a factor of $\gamma = 1.25$ during the turnaround phase
Twin B's clock appears to stop completely during the instantaneous acceleration phase
Twin B's clock continues to run slow throughout the turnaround, maintaining the same dilation factor
Twin B's clock rate appears to jump discontinuously from slow to fast, accounting for most of the age difference (correct answer)

Explanation: This classic twin paradox scenario tests your understanding of relativistic effects during acceleration phases, which is often the most misunderstood aspect of special relativity problems. During the constant velocity phases, twin A observes twin B's clock running slow due to time dilation by the factor

\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = 1.25

. However, the crucial insight occurs during the instantaneous turnaround. From twin A's accelerated reference frame, twin B's clock rate appears to jump dramatically from running slow (

\frac{1}{1.25} = 0.8

times normal) to running fast (

1.25

times normal). This discontinuous jump accounts for the majority of the age difference that accumulates during the entire journey. Option A incorrectly suggests the clock simply runs faster by

\gamma

during turnaround, missing the dramatic discontinuous change. Option B is wrong because the clock doesn't stop—acceleration doesn't halt time observation, it changes the observed rate dramatically. Option C fails because it ignores the fundamental asymmetry created by acceleration; the dilation factor doesn't remain constant during the turnaround phase. The correct answer is D because this discontinuous jump in the observed clock rate during acceleration is what resolves the apparent paradox and explains why the traveling twin ages less overall. Study tip: In twin paradox problems, always focus on what happens during the acceleration phases—this is where the asymmetry occurs and where most students make errors. The constant velocity portions are straightforward time dilation, but acceleration creates the dramatic observational changes.

Question 19

In Galilean relativity, a ball is thrown horizontally at $20$ m/s from a train moving at $30$ m/s. In special relativity, if the same scenario involves velocities of $0.6c$ (train) and $0.8c$ (ball relative to train), what is the key difference in the velocity addition?

Galilean gives $50$ m/s while relativistic gives $0.946c$ , demonstrating the non-linear nature of relativistic addition
Galilean gives $1.4c$ while relativistic gives $0.946c$ , preventing superluminal speeds in the ground frame (correct answer)
Both methods give approximately the same result since the velocities are much less than $c$ in the first case
The relativistic result depends on the direction of motion, while Galilean transformation is direction-independent

Explanation: Galilean:

v = 0.6c + 0.8c = 1.4c

. Relativistic:

v = (0.6c + 0.8c)/(1 + 0.6 \times 0.8) = 1.4c/1.48 = 0.946c < c

. The key difference is that relativistic velocity addition prevents speeds exceeding

c

, while Galilean allows superluminal results. Option A incorrectly compares different scenarios. Option C is wrong as the relativistic case involves high speeds. Option D is incorrect as both transformations can be direction-dependent.

Question 20

A spaceship travels to a star $12$ light-years away at $0.8c$ . The trip takes $15$ years as measured on Earth. From the perspective of the astronaut, what is the distance to the star and the travel time?

Distance $7.2$ light-years and travel time $15$ years, since only length contraction affects the measurement
Distance $12$ light-years and travel time $9.0$ years, since only time dilation affects the astronaut
Distance $7.2$ light-years and travel time $9.0$ years, due to length contraction and time dilation (correct answer)
Distance $20$ light-years and travel time $25$ years, due to relativistic expansion effects

Explanation: When you encounter relativistic motion problems, you need to consider how measurements differ between reference frames. Both time dilation and length contraction affect what the astronaut observes compared to Earth-based measurements. From the astronaut's perspective, two key transformations occur. First, time dilation means the astronaut's clocks run slower relative to Earth. Using

\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-0.8^2}} = \frac{5}{3}

, the travel time becomes

t' = \frac{t}{\gamma} = \frac{15}{5/3} = 9.0

years. Second, length contraction shortens distances in the direction of motion:

L' = \frac{L}{\gamma} = \frac{12}{5/3} = 7.2

light-years. Option A incorrectly claims only length contraction affects measurements, ignoring that the astronaut also experiences time dilation. While the distance calculation (7.2 light-years) is correct, keeping the travel time at 15 years misses this crucial effect. Option B makes the opposite error, correctly calculating time dilation (9.0 years) but wrongly assuming distance remains unchanged. The astronaut definitely observes length contraction of the space between Earth and the star. Option D introduces non-existent "relativistic expansion effects" and gives values larger than the Earth-frame measurements, which contradicts the fundamental nature of time dilation and length contraction—both always reduce measured quantities in the moving frame. Option C correctly applies both effects: the astronaut measures a contracted distance of 7.2 light-years and experiences a dilated travel time of 9.0 years. Remember: in special relativity problems, always check whether both time dilation and length contraction apply to the moving observer's measurements.