smc-3699-30-Momentum Dilation and calcs

PHYSICS, M7 2024 HSC 27

The simplified model below shows the reactants and products of a proton-antiproton reaction which produces three particles called pions, each having a different charge.

\(\text{p}+\overline{\text{p}} \rightarrow \pi^{+}+\pi^0+\pi^{-}\)

There are no other products in this process, which involves only the rearrangement of quarks. No electromagnetic radiation is produced. Assume that the initial kinetic energy of the proton and antiproton is negligible.

Protons consist of two up quarks \(\text{(u)}\) and a down quark \(\text{(d)}\) . Antiprotons consist of two up antiquarks \((\overline{\text{u}})\) and a down antiquark \((\overline{\text{d}})\). Each of the pions consists of two quarks.

The following tables provide information about hadrons and quarks.

Table 1: Hadron Information

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \ \ \textit{Particle} & \ \ \textit{Rest mass} \ \ & \quad \textit{Charge} \quad \\
& \left(\text{MeV/c}^2\right)&\\
\hline
\rule{0pt}{2.5ex} \text {proton (p)} \rule[-1ex]{0pt}{0pt} & 940 & +1 \\
\hline
\rule{0pt}{2.5ex} \text {antiproton}(\overline{\text{p}}) \rule[-1ex]{0pt}{0pt} & 940 & -1 \\
\hline
\rule{0pt}{2.5ex} \text {neutral pion }\left(\pi^0\right) \rule[-1ex]{0pt}{0pt} & 140 & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{positive pion }\left(\pi^{+}\right) \rule[-1ex]{0pt}{0pt} & 140 & +1 \\
\hline
\rule{0pt}{2.5ex}\text {negative pion }\left(\pi^{-}\right) \rule[-1ex]{0pt}{0pt} & 140 & -1\\
\hline
\end{array}

Table 2: Quark charges

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \ \ \textit{Particle} \rule[-1ex]{0pt}{0pt} & \quad \textit{Charge} \quad \\
\hline
\rule{0pt}{2.5ex} \text {down quark (d)} \rule[-1ex]{0pt}{0pt} & -\dfrac{1}{3} \\
\hline
\rule{0pt}{2.5ex} \text {up quark (u)} \rule[-1ex]{0pt}{0pt} & +\dfrac{2}{3}\\
\hline
\rule{0pt}{2.5ex} \text {down antiquark}(\overline{\text{d}}) \rule[-1ex]{0pt}{0pt} & +\dfrac{1}{3}\\
\hline
\rule{0pt}{2.5ex} \text{up antiquark }(\overline{\text{u}}) \rule[-1ex]{0pt}{0pt} & -\dfrac{2}{3} \\
\hline
\end{array}

Identify the quarks present in the \(\pi^{-}, \pi^{+}\)and the \(\pi^0\) particles. (2 marks)

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The energy released in the reaction is shared equally between the pions.
Calculate the energy released per pion in this reaction. (2 marks)

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Calculation of the pions' velocities using classical physics predicts that each pion has a velocity, relative to the point at which the proton-antiproton reaction occurred, which exceeds 3 × 10\(^8\) m s\(^{-1}\).
Explain the problem with this prediction and how it can be resolved. (3 marks)

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a. \(\pi^{-}:\ \overline{\text{u}}\text{d}\)

\(\pi^{+}:\ \text{u}\overline{\text{d}}\)

\(\pi^{0}:\ \text{u}\overline{\text{d}}\)

b. \(\text{Initial mass}\ = 940 + 940 = 1880\ \text{MeV/c}^{2} \)

\(\text{Final mass}\ = 3 \times 140 = 420\ \text{MeV/c}^{2} \)

\(\Delta \text{Mass (per pion)}\ = \dfrac{1460}{3} = 487\ \text{MeV/c}^{2} \)

\(\text{Using}\ \ E=mc^2:\)

\(\text{Energy released (per pion)}\ = 487\ \text{MeV}\)

c. Problem with prediction:

→ The calculation shows the pions moving faster than light speed (3 × 10\(^{8}\) m/s), which can’t happen in reality.

Resolving the problem:

→ Since these pions are moving at extremely high speeds, we need to account for relativity.

→ Relativity means that pions’ mass actually increases as they get faster, which prevents them from ever reaching light speed.

→ Part of the energy given to the pions goes into increasing their mass rather than just increasing their velocity.

Show Worked Solution

a. \(\pi^{-}:\ \overline{\text{u}}\text{d}\)

\(\pi^{+}:\ \text{u}\overline{\text{d}}\)

\(\pi^{0}:\ \text{u}\overline{\text{d}}\)

b. \(\text{Initial mass}\ = 940 + 940 = 1880\ \text{MeV/c}^{2} \)

\(\text{Final mass}\ = 3 \times 140 = 420\ \text{MeV/c}^{2} \)

\(\Delta \text{Mass (per pion)}\ = \dfrac{1460}{3} = 487\ \text{MeV/c}^{2} \)

\(\text{Using}\ \ E=mc^2:\)

\(\text{Energy released (per pion)}\ = 487\ \text{MeV}\)

♦ Mean mark (b) 41%.

c. Problem with prediction:

→ The calculation shows the pions moving faster than light speed (3 × 10\(^{8}\) m/s), which can’t happen in reality.

Resolving the problem:

→ Since these pions are moving at extremely high speeds, we need to account for relativity.

→ Relativity means that pions’ mass actually increases as they get faster, which prevents them from ever reaching light speed.

→ Part of the energy given to the pions goes into increasing their mass rather than just increasing their velocity.

♦ Mean mark (c) 43%.

PHYSICS, M7 EQ-Bank 29

A proton travels along a particle accelerator at 10 m s ¯¹ less than the speed of light.

Compare its speed and momentum with a proton travelling at 99% the speed of light. Support your answer with calculations. (4 marks)

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→ First proton speed = `3 xx10^(8)-10 text{m s}^(-1)`, second proton speed = `0.99 xx 3 xx10^(8) text{m s}^(-1)`

→ Comparing the two: `(v_(p1))/(v_(p2))=(3 xx10^(8)-10)/(0.99 xx 3 xx10^(8))=1.0101`.

→ The first proton is travelling 1% faster than the second proton.

→ Calculate the relativistic momentum of the first proton (`p_(1)`):

`p_(1)`	`=(m_(0)v)/(sqrt(1-(v^(2))/(c^(2))))`
	`=(1.673 xx10^(-27)xx(3.00 xx10^(8)-10))/(sqrt(1-((3.00 xx10^(8)-10)/(3.00 xx10^(8)))^(2)))`
	`=1.94 xx10^(-15)\ text{kg m s}^(-1)`

→ Calculate the relativistic momentum of the second proton (`p_(2)`):

`p_(2)`	`=(1.673 xx10^(-27)xx0.99xx3.00 xx10^(8))/(sqrt(1-0.99^(2)))`
	`=3.52 xx10^(-18) text{kg m s}^(-1)`

→ Compare the two: `(p_(1))/(p_(2))=(1.94 xx10^(-15))/(3.52 xx10^(-18))=551`.

→ The momentum of the first proton is 551 times greater than the momentum of the second proton, while only going 1% faster.

Show Worked Solution

→ First proton speed = `3 xx10^(8)-10 text{m s}^(-1)`, second proton speed = `0.99 xx 3 xx10^(8) text{m s}^(-1)`

→ Comparing the two: `(v_(p1))/(v_(p2))=(3 xx10^(8)-10)/(0.99 xx 3 xx10^(8))=1.0101`.

→ The first proton is travelling 1% faster than the second proton.

→ Calculate the relativistic momentum of the first proton (`p_(1)`):

`p_(1)`	`=(m_(0)v)/(sqrt(1-(v^(2))/(c^(2))))`
	`=(1.673 xx10^(-27)xx(3.00 xx10^(8)-10))/(sqrt(1-((3.00 xx10^(8)-10)/(3.00 xx10^(8)))^(2)))`
	`=1.94 xx10^(-15)\ text{kg m s}^(-1)`

→ Calculate the relativistic momentum of the second proton (`p_(2)`):

`p_(2)`	`=(1.673 xx10^(-27)xx0.99xx3.00 xx10^(8))/(sqrt(1-0.99^(2)))`
	`=3.52 xx10^(-18) text{kg m s}^(-1)`

→ Compare the two: `(p_(1))/(p_(2))=(1.94 xx10^(-15))/(3.52 xx10^(-18))=551`.

→ The momentum of the first proton is 551 times greater than the momentum of the second proton, while only going 1% faster.

PHYSICS, M7 EQ-Bank 23

What is the magnitude of the momentum `(\text{in kg m s}^(-1))` of an electron travelling at 0.75`c`? (2 marks)

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`3.10 xx10^(-22) text{kg m s}^(-1)`

Show Worked Solution

`p_(v)`	`=(m_(0)v)/(sqrt((1-(v^(2))/(c^(2)))))`
	`=((9.109 xx10^(-31))(0.75 xx3 xx10^8))/(sqrt((1-((0.75c)^(2))/(c^(2)))))`
	`=((9.109 xx10^(-31))(0.75 xx3 xx10^8))/(sqrt((1-0.75^(2)))`
	`= 3.10 xx10^(-22) text{kg m s}^(-1)`

COMMENT: Note the simplification of the denominator.

PHYSICS, M7 2015 HSC 29

In the Large Hadron Collider (LHC), protons travel in a circular path at a speed greater than 0.9999 `c`.

What are the advantages of using superconductors to produce the magnetic fields used to guide protons around the LHC? (2 marks)

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Discuss the application of special relativity to the protons in the LHC. (3 marks)

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a. Advantages:

→ Strong magnetic fields are necessary to deflect protons travelling around the LHC due to their high velocities and the effect of mass dilation.

→ Superconductors can achieve very high current flow. This produces the very strong magnetic fields that are required.

b. Applications of special relativity:

→ Since the protons are travelling so close to the speed of light, special relativity states that mass dilation will be significant.

→ This effect means that an increasing amount of energy is needed to accelerate the proton.

PHYSICS, M7 2020 HSC 30b

Calculate the wavelength of a proton travelling at 0.1\(c\). (2 marks)

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Explain the relativistic effect on the wavelength of a proton travelling at 0.95\(c\). (2 marks)

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i. `lambda=1xx10^(-14)` m

ii. See Worked Solutions

Show Worked Solution

i.	`lambda`	`=(h)/(mv)`
		`=(6.626 xx10^(-34))/(1.673 xx10^(-27)xx0.1 xx3xx10^(8))`
		`=1xx10^(-14)\ text{m}`

ii. Due to momentum dilation:

→ The momentum of the proton is greater than that predicted by classical mechanics.

→ Since `lambda=(h)/(mv)\ \ =>\ \ lambda prop 1/(mv)`

→ The wavelength of the proton is shorter than would be predicted by classical mechanics.

♦ Mean mark (ii) 45%.

PHYSICS, M7 2021 HSC 28

A spaceship travels to a distant star at a constant speed, `v`. When it arrives, 15 years have passed on Earth but 9.4 years have passed for an astronaut on the spaceship.

What is the distance to the star as measured by an observer on Earth? (3 marks)

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Outline how special relativity imposes a limitation on the maximum velocity of the spaceship. (2 marks)

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12 ly
According to special relativity, as ` v `→ `c`:
→ the momentum of the spaceship approaches infinity
→ the force required to accelerate the spaceship approaches infinity
→ maximum velocity is limited to the speed of light.

Show Worked Solution

a. `t_v =t_0/sqrt((1-(v^2)/(c^2)))`

`15=9.4/sqrt((1-(v^2)/(c^2)))`

`1-(v^2)/(c^2)`	`=((9.4)/(15))^2`
`(v^2)/(c^2)`	`=1-((9.4)/(15))^2`
`v^2`	`=0.60729c^2`
`v`	`=0.779c`

Distance to star from Earth observer:

`s`	`=ut`
	`=0.779 xx15`
	`=12\ text{ly}`

♦ Mean mark part (a) 51%.

b. According to special relativity, as ` v `→ `c`:

→ the momentum of the spaceship approaches infinity

→ the force required to accelerate the spaceship approaches infinity

→ maximum velocity is limited to the speed of light.