Special relativity — Question 237

Original QCE Vault practice · 6 marks

Q237 · Practice questionComplex unfamiliar6 marks

QUESTION 237 (6 marks)

The graph shows momentum p against

\gamma v

for a particle, where

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

. The fitted line passes exactly through

(1.00\times10^8\,\mathrm{m\,s^{-1}},3.00\times10^{-19}\,\mathrm{kg\,m\,s^{-1}})

. Use

c=3.00\times10^8\,\mathrm{m\,s^{-1}}

Momentum in units of 10 to the minus 19 kg m/s versus gamma v in units of 10 to the 8 m/s. A straight line through the origin passes through displayed points (1,3), (2,6) and (3,9).

Determine the rest mass from the gradient.

[2 marks]

Predict the momentum at

v=0.800c

[2 marks]

Explain why the straight line in this transformed graph does not mean momentum remains finite as the actual speed approaches c.

[2 marks]

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WORKED SOLUTION

Practice marking scheme

6 marks

ANSWER

(a)

3.00\times10^{-27}\,\mathrm{kg}

. (b)

1.20\times10^{-18}\,\mathrm{kg\,m\,s^{-1}}

. (c) As v approaches c,

\gamma v

diverges, so momentum also diverges.

Worked solution

(a)

p=m_0(\gamma v)

, so the SI gradient equals

m_0

. Therefore

m_0=(3.00\times10^{-19})/(1.00\times10^8)=3.00\times10^{-27}\,\mathrm{kg}

. The plotted horizontal variable is

\gamma v

, not v.

(b)

\gamma=5/3

, so

\gamma v=(5/3)(.800)(3.00\times10^8)=4.00\times10^8\,\mathrm{m\,s^{-1}}

. Then

p=m_0\gamma v=1.20\times10^{-18}\,\mathrm{kg\,m\,s^{-1}}

. A value of

\gamma v

greater than c is not a speed greater than c.

\gamma v

. As the actual v approaches c from below,

\sqrt{1-v^2/c^2}

tends to zero and

\gamma v

grows without bound. Consequently

p=m_0\gamma v

also grows without bound.

Equivalent physically justified methods and consistent equivalent units accepted.

Displayed decimals are model answers; accept appropriate significant figures and consistent rounding from stated constants.

Part a: Identify the transformed graph gradient as rest mass.

[1 mark]

Part a: Apply both axis scales and calculate mass.

[1 mark]

Part b: Calculate the correct transformed variable at the given physical speed.

[1 mark]

Part b: Use the rest mass to predict momentum.

[1 mark]

Part c: Distinguish actual v from the transformed horizontal variable.

[1 mark]

Part c: Connect its divergence near c to unbounded momentum.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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Special relativity — Question 237

QUESTION 237 (6 marks)

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