Question
A $\Sigma^{+}$particle decays from rest into a neutron and another particle $X$ according to the reaction
$
\Sigma^{+} \rightarrow \mathrm{n}+\mathrm{X}
$
The following data are available.
$
\begin{array}{ll}
\text { Rest mass of } \Sigma^{+} & =1190 \mathrm{MeV} \mathrm{c}^{-2} \\
\text { Momentum of neutron } & =185 \mathrm{MeV} \mathrm{c}^{-1}
\end{array}
$
Calculate, for the neutron,
a(i)the total energy.[1]
a(iithe speed.[2]
b. Determine the rest mass of $X$.[3]
▶️Answer/Explanation
Ans:
a(i)neutron energy $=\sqrt{185^2+940^2}=958 « \mathrm{MeV} »$
NOTE: Allow $1.5 \times 10^{-10} « \mathrm{~J} »$
a(ii)ALTERNATIVE 1
«use of $E=\gamma E_0$ »
$
\begin{aligned}
& « 958=940 \gamma \text { so» } \gamma=1.019 \\
& v=0.193 c
\end{aligned}
$
ALTERNATIVE 2
«use of $p=\gamma m v »$
$
\begin{aligned}
& 185=940 \frac{\frac{v}{c}}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \checkmark \\
& v=0.193 c
\end{aligned}
$
ALTERNATIVE 3
«use of $p=\gamma \quad m v$ »
$
\begin{aligned}
& v=\frac{p c}{E} \boldsymbol{V} \\
& v=\frac{185}{958}=0.193 c
\end{aligned}
$
NOTE: Allow $v=5.8 \times 10^7 \ll \mathrm{ms}^{-1} \gg$
b. momentum of $\mathrm{X}=185 \ll \mathrm{MeV} \mathrm{c}^{-1} » \mathscr{V}$ energy of $X=1190-958=232 \ll \mathrm{MeV} \gg \boldsymbol{V}$ $m_0=« \sqrt{232^2-185^2}=» 140 \ll \mathrm{MeV} \mathrm{c}^{-2} »$ NOTE: Allow mass in kg – gives $2.5 \times 10^{-28} \ll \mathrm{kg} »$
Question
The particle omega minus $\left(\Omega^{-}\right)$decays at rest into a neutral pion $\left(\pi^0\right)$ and the xi baryon $\left(\Xi^{-}\right)$according to
$
\Omega^{-} \rightarrow \pi^0+\Xi^{-}
$
The pion momentum is $289.7 \mathrm{MeV} \mathrm{c}^{-1}$.
The rest masses of the particles are:
$
\begin{aligned}
& \Omega^{-}: 1672 \mathrm{MeV} \mathrm{c}^{-2} \\
& \pi^0: 135.0 \mathrm{MeV} \mathrm{c}^{-2} \\
& \Xi^{-}: 1321 \mathrm{MeV} \mathrm{c}^{-2}
\end{aligned}
$
a. Show that energy is conserved in this decay.[3]
b. Calculate the speed of the pion.[2]
▶️Answer/Explanation
Ans:
a. momentum of $x i$ baryon is also $289.7 « \mathrm{MeVc}^{-1} »$
total energy of xi baryon and pion is $\sqrt{289.7^2+1321^2}+\sqrt{289.7^2+135.0^2}=1672$ «eV»
which equals the rest energy of the omega
Allow a backwards argument, assuming the energy is equal.
b. $\gamma \ll=\frac{\sqrt{289.7^2+135.0^2}}{135.0} »=2.367$
$
v \ll=\sqrt{1-\frac{1}{2.367^2}} c »=0.903 c
$
Award [2] for bald correct answer.
Question
Nature of science:
Pure science: Einstein based his theory of relativity on two postulates and deduced the rest by mathematical analysis. The first postulate integrates all of the laws of physics including the laws of electromagnetism, not only Newton’s laws of mechanics. (1.2)
Understandings:
The two postulates of special relativity
Clock synchronization
The Lorentz transformations
Velocity addition
Invariant quantities (spacetime interval, proper time, proper length and rest mass)
Time dilation
Length contraction
The muon decay experiment
Applications and skills:
Using the Lorentz transformations to describe how different measurements of space and time by two observers can be converted into the measurements observed in either frame of reference
Using the Lorentz transformation equations to determine the position and time coordinates of various events
Using the Lorentz transformation equations to show that if two events are simultaneous for one observer but happen at different points in space, then the events are not simultaneous for an observer in a different reference frame
Solving problems involving velocity addition
Deriving the time dilation and length contraction equations using the Lorentz equations
Solving problems involving time dilation and length contraction
Solving problems involving the muon decay experiment
Guidance:
Problems will be limited to one dimension
Derivation of the Lorentz transformation equations will not be examined
Muon decay experiments can be used as evidence for both time dilation and length contraction
Data booklet reference:
- $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{e^2}}}$
- $x^{\prime}=\gamma(x-v t) ; \Delta x^{\prime}=\gamma(\Delta x-v \Delta t)$
- $t^{\prime}=\gamma\left(t-\frac{v x}{c^2}\right) ; \Delta t^{\prime}=\gamma\left(\Delta t-\frac{v \Delta x}{c^2}\right)$
- $\boldsymbol{u}^{\prime}=\frac{u-v}{1-\frac{\mathrm{vw}}{e^2}} \mid$
- $\Delta t=\gamma \Delta t_0$
- $L=\frac{L_0}{\gamma}$
- $\left(c t^{\prime}\right)^2-\left(x^{\prime}\right)^2=(c t)^2-(x)^2$
Utilization:
- Once a very esoteric part of physics, relativity ideas about space and time are needed in order to produce accurate global positioning systems (GPS)
Aims:
- Aim 2: the Lorentz transformation formulae provide a consistent body of knowledge that can be used to compare the description of motion by one observer to the description of another observer in relative motion to the first
- Aim 3: these formulae can be applied to a varied set of conditions and situations
- Aim 9: the introduction of relativity pushed the limits of Galilean thoughts on space and motion
Question
A train is moving across a bridge with a speed $v=0.40 c$. Observer $A$ is at rest in the train. Observer $B$ is at rest with respect to the bridge.
The length of the bridge $L_B$ according to observer $B$ is $2.0 \mathrm{~km}$.
According to observer $\mathrm{B}$, two lamps at opposite ends of the bridge are turned on simultaneously as observer A crosses the bridge. Event $\mathrm{X}$ is the lamp at one end of the bridge turning on. Event $Y$ is the lamp at the other end of the bridge turning on.
Events X and Y are shown on the spacetime diagram. The space and time axes of the reference frame for observer B are x and ct. The line labelled ct’ is the worldline of observer A.
a(i)Calculate, for observer $A$, the length $L_A$ of the bridge[2]
a(ii)Calculate, for observer $A$, the time taken to cross the bridge.[2]
b. Outline why $L_B$ is the proper length of the bridge.[1]
c(i)Draw, on the spacetime diagram, the space axis for the reference frame of observer A. Label this axis $x^{\prime}$.[1]
c(ii)Demonstrate using the diagram which lamp, according to observer $\mathrm{A}$, was turned on first.[2]
c(iii)Demonstrate, using the diagram, which lamp observer A observes to light first.[2]
c(iv)Determine the time, according to observer $\mathrm{A}$, between $\mathrm{X}$ and $\mathrm{Y}$.[2]
▶️Answer/Explanation
Ans:
$
\begin{aligned}
& \mathrm{a}(\mathrm{i}) \gamma=1.09 \\
& L_{\mathrm{A}}=« \frac{2.0}{1.09}=» 1.8 « \mathrm{~km} »
\end{aligned}
$
a(ii)ALTERNATIVE 1
$
\begin{aligned}
& \text { time }=\frac{1.8 \times 10^3}{1.2 \times 10^8} \\
& 1.5 \times 10^{-5} « s »
\end{aligned}
$
ALTERNATIVE 2
$
\begin{aligned}
t_B & =\frac{2 \times 10^3}{1.2 \times 10^8}=1.66 \times 10^{-5} \text { 《S》 } \\
t_{\mathrm{A}} & =\frac{t_{\mathrm{B}}}{\gamma}=1.5 \times 10^{-5} \text { «S》 }
\end{aligned}
$
b. $L_B$ is the length/measurement «by observer $B$ » made in the reference frame in which the bridge is at rest
NOTE: Idea of rest frame or frame in which bridge is not moving is required.
x′ axis drawn with correct gradient of 0.4 ✔
NOTE: Line must be 1 square below Y, allow ±0.5 square.
Allow line drawn without a ruler.
lines parallel to the x′ axis through X and Y intersecting the worldline ct′ at points shown ✔
so Y/lamp at the end of the bridge turned on first ✔
NOTE: Allow lines drawn without a ruler
Do not allow MP2 without supporting argument or correct diagram.
light worldlines at 45° from X AND Y intersecting the worldline ct′ ✔
so light from lamp X is observed first ✔
NOTE: Allow lines drawn without a ruler.
Do not allow MP2 without supporting argument or correct diagram.
ALTERNATIVE 1
$
\begin{aligned}
& \Delta t^{\prime}=1.09 \times\left(0-\frac{0.4 \times 2.0 \times 10^3}{3.0 \times 10^8}\right) \\
& =«-» 2.9 \times 10^{-6} \text { «s» }
\end{aligned}
$
ALTERNATIVE 2
equating spacetime intervals between $X$ and $Y$
relies on realization that $\Delta x^{\prime}=\gamma(\Delta x-0)$ eg:
$\left(c \Delta t^{\prime}\right)^2-(1.09 \times 2000)^2=0^2-2000^2$
$
\Delta t^{\prime}=\left\langle \pm \gg \frac{\sqrt{(1.09 \times 2000)^2-2000^2}}{3.0 \times 10^8}=\left\langle \pm » 2.9 \times 10^{-6} \ll \mathrm{S} \gg\right.\right.
$
ALTERNATIVE 3
use of diagram from answer to $4(\mathrm{c})(\mathrm{ii})(1$ small square $=200 \mathrm{~m}$ ) counts 4.5 to 5 small squares (allow $900-1000 \mathrm{~m}$ ) between events for A seen on B’s $c t$ axis $\frac{950}{\gamma c}=2.9 \times 10^{-6} \pm 0.2 \times 10^{-6} \ll \mathrm{S} \gg$
Rocket A and rocket B are travelling in opposite directions from the Earth along the same straight line.
In the reference frame of the Earth, the speed of rocket A is 0.75c and the speed of rocket B is 0.50c.
a.i.Calculate, for the reference frame of rocket A, the speed of rocket B according to the Galilean transformation.[1]
▶️Answer/Explanation
Markscheme
a.i.
1.25c
[1 mark]
ALTERNATIVE 1
\(u’ = \frac{{(0.50 + 0.75)}}{{1 + 0.5 \times 0.75}}c\)
0.91c
ALTERNATIVE 2
\(u’ = \frac{{ – 0.50 – 0.75}}{{1 – ( – 0.5 \times 0.75)}}c\)
–0.91c
[2 marks]
nothing can travel faster than the speed of light (therefore (a)(ii) is the valid answer)
OWTTE
[1 mark]
Muons are created in the upper atmosphere of the Earth at an altitude of 10 km above the surface. The muons travel vertically down at a speed of 0.995c with respect to the Earth. When measured at rest the average lifetime of the muons is 2.1 μs.
a.i.Calculate, according to Galilean relativity, the time taken for a muon to travel to the ground[1]
▶️Answer/Explanation
Markscheme
a.i.
«\(\frac{{{{10}^4}}}{{0.995 \times 3 \times {{10}^8}}} = \)» 34 «μs»
Do not accept 104/c = 33 μs.
[1 mark]
time is much longer than 10 times the average life time «so only a small proportion would not decay»
[1 mark]
\(\gamma = 10\)
\(\Delta {t_0} = \) «\(\frac{{\Delta t}}{\gamma } = \frac{{34}}{{10}} = \)» 3.4 «μs»
[2 marks]
the value found in (b)(i) is of similar magnitude to average life time
significant number of muons are observed on the ground
«therefore this supports the special theory»[2 marks]
The diagram shows the motion of the electrons in a metal wire carrying an electric current as seen by an observer X at rest with respect to the wire. The distance between adjacent positive charges is d.
Observer Y is at rest with respect to the electrons.
a.State whether the field around the wire according to observer X is electric, magnetic or a combination of both[1]
▶️Answer/Explanation
Markscheme
a.
magnetic field
[1 mark]
«according to Y» the positive charges are moving «to the right»
d decreases
For MP1, movement of positive charges must be mentioned explicitly.
[2 marks]
positive charges are moving, so there is a magnetic field
the density of positive charges is higher than that of negative charges, so there is an electric field
The reason must be given for each point to be awarded.
[2 marks]
Outline the conclusion from Maxwell’s work on electromagnetism that led to one of the postulates of special relativity.
▶️Answer/Explanation
Markscheme
light is an EM wave
speed of light is independent of the source/observer
a.State one prediction of Maxwell’s theory of electromagnetism that is consistent with special relativity.[1]
A proton is at rest relative to the laboratory and the wire.
Observer X is at rest in the laboratory. Observer Y moves to the right with constant speed relative to the laboratory. Compare and contrast how observer X and observer Y account for any non-gravitational forces on the proton.[3]
▶️Answer/Explanation
Markscheme
a.
the speed of light is a universal constant/invariant
OR
c does not depend on velocity of source/observer
electric and magnetic fields/forces unified/frame of reference dependant
[1 mark]
observer X will measure zero «magnetic or electric» force
observer Y must measure both electric and magnetic forces
which must be equal and opposite so that observer Y also measures zero force
Allow [2 max] for a comment that both X and Y measure zero resultant force even if no valid explanation is given.
[3 marks]
Question
A black hole has a Schwarzschild radius $R$. A probe at a distance of $0.5 R$ from the event horizon of the black hole emits radio waves of frequency $f$ that are received by an observer very far from the black hole.
a. Explain why the frequency of the radio waves detected by the observer is lower than $f$. [2]
b. The probe emits 20 short pulses of these radio waves every minute, according to a clock in the probe. Calculate the time between pulses as
measured by the observer.[2]
▶️Answer/Explanation
Ans:
a. ALTERNATIVE 1
as the photons move away from the black hole, they lose energy in the gravitational field
since $E=h f$ «the detected frequency is lower than the emitted frequency»
ALTERNATIVE 2
if the observer was accelerating away from the probe, radio waves would undergo Doppler shift towards lower frequency by the equivalence principle, the gravitational field has the same effect as acceleration
ALTERNATIVE 3
due to gravitational time dilation, time between arrivals of wavefronts is greater for the observer since $f=\frac{1}{T}$, «the detected frequency is lower than the emitted frequency»
NOTE: The question states that received frequency is lower so take care not to award a mark for simply re-stating this, a valid explanation must be given.
b. time between pulses $=3 \mathrm{~s}$ according to the probe
$
\Delta t=\ll \frac{3}{\sqrt{1-\frac{1}{1.5}}}=» 5.2 \text { « } » \checkmark
$
Question
A rocket is accelerating upwards at $9.8 \mathrm{~m} \mathrm{~s}^{-2}$ in deep space. A photon of energy $14.4 \mathrm{keV}$ is emitted upwards from the bottom of the rocket and travels to a detector in the tip of the rocket $52.0 \mathrm{~m}$ above.
a. Explain why a change in frequency is expected for the photon detected at the top of the rocket.[3]
b. Calculate the frequency change.[2]
▶️Answer/Explanation
Ans:
a. ALTERNATIVE 1
detector accelerates/moves away from point of photon emission
so Doppler effect / redshift
so $f$ decreases
ALTERNATIVE 2
equivalent to stationary rocket on earth’s surface
photons lose «gravitational» energy as they move upwards
$\mathrm{h} f$ OR $f$ decreases
b. «Using $E=h f » f=\frac{14.4 \times 10^3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}=3.48 \times 10^{18}$ «z» $\checkmark$
$
\begin{aligned}
& \| \Delta f=f \frac{g \Delta h}{c^2} » \\
& \Delta f=3.48 \times 10^{18} \frac{9.81 \times 52}{\left(3 \times 10^8\right)^2}=1.97 \times 10^4 \text { «z » } \quad
\end{aligned}
$