Apply Galilean & Special Relativity - IB Physics

1. On a standard space-time diagram with the time axis (ct) vertical and the space axis (x) horizontal, what does the slope of a worldline for a massive particle represent?
   Answer: The speed of light divided by the particle's velocity, $c/v$. The slope of a line on a graph is the 'rise' divided by the 'run'. On a space-time diagram, the rise is $\Delta(ct)$ and the run is $\Delta x$. So, slope = $\Delta(ct) / \Delta x = c (\Delta t / \Delta x)$. Since velocity is $v = \Delta x / \Delta t$, its inverse is $1/v = \Delta t / \Delta x$. Substituting this into the slope equation gives slope = $c \times (1/v) = c/v$.
2. An object is moving at a speed of $v = 0.95c$. What is the ratio of the time interval measured in the stationary frame ($\Delta t$) to the proper time interval ($\Delta t_0$)?
   Answer: 3.2 The relationship between dilated time and proper time is $\Delta t = \gamma \Delta t_0$. The question asks for the ratio $\Delta t / \Delta t_0$, which is equal to the Lorentz factor $\gamma$. We calculate $\gamma = 1/\sqrt{1-v^2/c^2} = 1/\sqrt{1-0.95^2} = 1/\sqrt{1-0.9025} = 1/\sqrt{0.0975} \approx 1/0.3122 \approx 3.20$. Distractor A is the value of $1/\gamma$.
3. A spaceship travels from Earth to a star 10 light-years away at a constant speed of $0.8c$. 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?
   Answer: The astronaut measures the proper time, which is 7.5 years. Proper time ($\Delta t_0$) 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 $\Delta t = \text{distance}/\text{speed} = 10 \text{ ly} / 0.8c = 12.5$ years. The Lorentz factor is $\gamma = 1/\sqrt{1-0.8^2} = 1/\sqrt{0.36} = 1/0.6 = 5/3$. The astronaut's time is the proper time: $\Delta t_0 = \Delta t / \gamma = 12.5 / (5/3) = 7.5$ years. So, the astronaut measures the proper time, and its value is 7.5 years.
4. A muon is created in the upper atmosphere and travels towards the Earth's surface at a speed of $0.99c$. In the muon's rest frame, its lifetime is $2.2 \mu$s. From the perspective of the muon, which statement best explains why it reaches the surface?
   Answer: From the muon's perspective, the distance to the Earth's surface is contracted, allowing it to reach the surface within its lifetime. 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 ($2.2 \mu$s), 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 $0.99c$, so the distance to the Earth's surface is length-contracted, making it short enough to be traversed within the muon's proper lifetime.
5. A rod has a proper length of 10.0 m. It moves at a constant speed such that its length, measured by a stationary observer, is 8.0 m. What is the speed of the rod?
   Answer: 0.60c Length contraction is given by $L = L_0 / \gamma$. The Lorentz factor is $\gamma = L_0 / L = 10.0 / 8.0 = 1.25$. We can find the speed from $\gamma = 1/\sqrt{1-v^2/c^2}$. Rearranging gives $(v/c)^2 = 1 - 1/\gamma^2 = 1 - 1/(1.25)^2 = 1 - 1/1.5625 = 1 - 0.64 = 0.36$. Taking the square root gives $v/c = \sqrt{0.36} = 0.60$. So the speed is $0.60c$. Distractor C is the simple ratio $L/L_0$. Distractor A is $(v/c)^2$.
6. In which of the following scenarios is Galilean relativity sufficient to describe the motion accurately, with negligible error compared to special relativity?
   Answer: Calculating the trajectory of a satellite in low Earth orbit. Galilean relativity is an excellent approximation when speeds are much less than the speed of light ($v \ll c$). A satellite in low Earth orbit travels at about 8 km/s, which is a tiny fraction of the speed of light ($c \approx 300,000$ km/s). In contrast, particles in accelerators (A and C) travel at speeds extremely close to $c$, 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.
7. Two spaceships approach each other. Ship A moves at $+0.70c$ and Ship B moves at $-0.50c$ relative to a space station. What is the velocity of Ship A as measured by observers on Ship B?
   Answer: $+0.89c$ Use relativistic velocity addition: $v_{AB} = \frac{v_A - v_B}{1 - \frac{v_A v_B}{c^2}} = \frac{0.70c - (-0.50c)}{1 - \frac{(0.70c)(-0.50c)}{c^2}} = \frac{1.20c}{1 + 0.35} = \frac{1.20c}{1.35} = 0.89c$. Choice B uses classical velocity addition. Choice C incorrectly subtracts velocities instead of adding. Choice D has correct magnitude but wrong sign direction.
8. An observer on Earth measures the length of a passing spacecraft to be 120 m when it moves at $0.60c$. The spacecraft then accelerates to $0.90c$. What length does the Earth observer now measure?
   Answer: 52 m First find the proper length: $L_0 = \frac{L}{\sqrt{1-v^2/c^2}} = \frac{120}{\sqrt{1-0.6^2}} = \frac{120}{0.8} = 150$ m. At $0.90c$: $L = L_0\sqrt{1-v^2/c^2} = 150\sqrt{1-0.9^2} = 150 \times 0.436 = 52$ m. Choice B incorrectly applies length contraction directly to 120 m without finding proper length first. Choice C gives the proper length. Choice D incorrectly inverts the contraction formula.
9. A particle has rest mass $m_0 = 1.67 \times 10^{-27}$ kg and moves at $0.80c$. What is the particle's relativistic momentum?
   Answer: $1.0 \times 10^{-18}$ kg⋅m/s Relativistic momentum: $p = \gamma m_0 v = \frac{m_0 v}{\sqrt{1-v^2/c^2}}$. $\gamma = \frac{1}{\sqrt{1-0.8^2}} = \frac{1}{0.6} = 1.67$. $p = 1.67 \times 1.67 \times 10^{-27} \times 0.8 \times 3 \times 10^8 = 1.0 \times 10^{-18}$ kg⋅m/s. Choice A uses classical momentum formula $p = m_0 v$. Choice C contains error in $\gamma$ calculation. Choice D incorrectly doubles the correct answer.
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?
    Answer: Twin B changed inertial reference frames during the journey, while twin A did not. 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.
11. A rocket moves at $0.9c$ relative to an observer. It fires a projectile in the forward direction. The speed of the projectile relative to the rocket is also $0.9c$. What is the speed of the projectile relative to the observer?
    Answer: Less than $c$. According to the second postulate of special relativity, no object with mass can travel at or faster than the speed of light $c$. Simple Galilean addition would give $0.9c + 0.9c = 1.8c$, which is non-physical. The relativistic velocity addition formula, $u = (u'+v)/(1+u'v/c^2)$, must be used. For any two velocities less than $c$, the formula will always yield a result that is also less than $c$. In this case, the result is $1.8c / (1 + 0.81) \approx 0.994c$, which is less than $c$.
12. A spaceship travels at speed $v$ past a space station. Observers on the station measure the spaceship's length to be $L$ and the time between two ticks of a clock on the spaceship to be $\Delta t$. What are the proper length of the spaceship $L_0$ and the proper time interval $\Delta t_0$ between the clock ticks?
    Answer: $L_0 = \gamma L$, $\Delta t_0 = \Delta t/\gamma$ Proper length $L_0$ is the length in the object's rest frame. The measured length $L$ is contracted: $L = L_0/\gamma$, which rearranges to $L_0 = \gamma L$. Proper time $\Delta t_0$ is the time interval in the clock's rest frame. The measured time interval $\Delta t$ is dilated: $\Delta t = \gamma \Delta t_0$, which rearranges to $\Delta t_0 = \Delta t/\gamma$. This question requires correctly rearranging both the length contraction and time dilation formulas.
13. A thin, square plate with proper side length $L_0$ moves at a relativistic speed $v$ in a direction parallel to one of its sides. What is the area of the plate as measured by a stationary observer?
    Answer: $L_0^2 / \gamma$ Length contraction occurs only in the direction of motion. Let the plate move parallel to its x-side. This side's length contracts to $L_x = L_0/\gamma$. The side perpendicular to the motion, the y-side, has its length unchanged, so $L_y = L_0$. The area measured by the stationary observer is $A = L_x \times L_y = (L_0/\gamma) \times L_0 = L_0^2 / \gamma$. Distractor C incorrectly assumes both sides contract.
14. The Lorentz factor, $\gamma$, is defined as $1/\sqrt{1-v^2/c^2}$. Which statement correctly describes the relationship between $\gamma$ and the speed $v$?
    Answer: $\gamma$ is approximately 1 for $v \ll c$ and approaches infinity as $v$ approaches $c$. When $v$ is much smaller than $c$ ($v \ll c$), the term $v^2/c^2$ is very close to zero. Thus, $\gamma \approx 1/\sqrt{1-0} = 1$. As $v$ approaches $c$, $v^2/c^2$ approaches 1, the denominator $\sqrt{1-v^2/c^2}$ approaches 0, and $\gamma$ approaches infinity. This describes the behavior of relativistic effects, which are negligible at low speeds but become extreme as speed approaches the speed of light.
15. 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 $v$ along the x-axis relative to S. Which statement must be true for the observer in S'?
    Answer: The order in which the explosions occur in S' depends on the direction of the velocity $v$. 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 $\Delta t' = \gamma (\Delta t - v \Delta x / c^2)$. Since $\Delta t = 0$ in S, $\Delta t' = -\gamma v \Delta x / c^2$. The sign of $\Delta t'$ (which determines the order) depends on the sign of $v$, meaning the direction of motion.
16. Which of the following describes an observer in an inertial frame of reference?
    Answer: A skydiver who has just reached terminal velocity. An inertial frame of reference is one that is not accelerating (i.e., it is moving at a constant velocity). A skydiver at terminal velocity has zero acceleration because the force of air resistance balances the force of gravity. Therefore, they are moving at a constant velocity and are in an inertial frame. All other options involve acceleration: linear acceleration (A and D) or centripetal acceleration (B).
17. Spaceship A moves away from Earth at a speed of $0.5c$. Spaceship B moves away from Earth in the same direction at a speed of $0.8c$. What is the speed of spaceship B as measured by an observer on spaceship A?
    Answer: 0.50c 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 $v = 0.5c$. The velocity of spaceship B in S is $u = 0.8c$. The velocity of B in S' is $u' = (u - v) / (1 - uv/c^2)$. Plugging in the values: $u' = (0.8c - 0.5c) / (1 - (0.8c)(0.5c)/c^2) = 0.3c / (1 - 0.40) = 0.3c / 0.60 = 0.5c$. Distractor A is the incorrect Galilean subtraction result ($0.8c - 0.5c$).
18. A light beam travels from the back to the front of a moving train. The train has proper length 200 m and moves at $0.60c$ relative to the ground. According to ground observers, the light beam takes $1.0 \times 10^{-6}$ s to travel from back to front. What is the speed of light as measured by train passengers?
    Answer: $3.0 \times 10^8$ m/s 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 $c = 3.0 \times 10^8$ 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 $c$ for train passengers. Let's examine why the other answers represent common misconceptions: Answer A ($1.6 \times 10^8$ m/s) likely comes from incorrectly applying classical velocity addition, subtracting the train's speed from light speed: $3.0 \times 10^8 - 0.6 \times 3.0 \times 10^8 = 1.2 \times 10^8$ m/s, or some similar calculation. Answer B ($3.8 \times 10^8$ m/s) probably results from incorrectly adding the train's velocity to light speed using classical mechanics: $3.0 \times 10^8 + 0.6 \times 3.0 \times 10^8 = 4.8 \times 10^8$ m/s, or a related computational error. Answer D ($5.0 \times 10^8$ 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 $c$ is invariant. The numerical details about train length and timing are often distractors—focus on which reference frame is measuring the light speed.
19. 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?
    Answer: 20 hours 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 $\frac{8.0}{10.0} = 0.8$. 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: $25 \times 0.8 = 20 \text{ hours}$. 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.
20. The Lorentz transformation for the time coordinate is $t' = \gamma(t - vx/c^2)$. What does this equation simplify to in the limit of low speeds, where $v \ll c$?
    Answer: $t' = t$ In the limit of low speeds ($v \ll c$), two approximations can be made. First, the Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$ approaches 1. Second, the term $vx/c^2$ approaches zero because it is divided by the very large number $c^2$. Applying these limits to the transformation gives $t' \approx 1(t - 0)$, which simplifies to $t' = t$. This is the time transformation in Galilean relativity, showing that special relativity reduces to classical physics at low speeds.

Start this quiz