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Date	November 2019	Marks available	2	Reference code	19N.3.SL.TZ0.4
Level	Standard level	Paper	Paper 3	Time zone	0 - no time zone
Command term	Calculate	Question number	4	Adapted from	N/A

Question

A train is moving across a bridge with a speed v = 0.40c. 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 km.

According to observer B, two lamps at opposite ends of the bridge are turned on simultaneously as observer A crosses the bridge. Event 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.

Calculate, for observer A, the length L_A of the bridge

[2]

a(i).

Calculate, for observer A, the time taken to cross the bridge.

[2]

a(ii).

Outline why L_B is the proper length of the bridge.

[1]

Draw, on the spacetime diagram, the space axis for the reference frame of observer A. Label this axis $x$ '.

[1]

c(i).

Demonstrate using the diagram which lamp, according to observer A, was turned on first.

[2]

c(ii).

Demonstrate, using the diagram, which lamp observer A observes to light first.

[2]

c(iii).

Determine the time, according to observer A, between X and Y.

[2]

c(iv).

Markscheme

$γ = 1.09$ ✔

$L_{A} = « \frac{2.0}{1.09} = »$ 1.8 «km» ✔

a(i).

ALTERNATIVE 1

time = $\frac{1.8 \times 10^{3}}{1.2 \times 10^{8}}$ ✔

1.5 × 10^–5«s» ✔

ALTERNATIVE 2

$t_{B} = \frac{2 \times 10^{3}}{1.2 \times 10^{8}} = 1.66 \times 10^{- 5}$ «s» ✔

$t_{A} = \frac{t_{B}}{γ} = 1.5 \times 10^{- 5}$ «s» ✔

a(ii).

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.

c(i).

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.

c(ii).

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.

c(iii).

ALTERNATIVE 1

$∆ t' = 1.09 \times (0 - \frac{0.4 \times 2.0 \times 10^{3}}{3.0 \times 10^{8}})$ ✔

= «–»2.9 × 10^–6«s» ✔

ALTERNATIVE 2

equating spacetime intervals between X and Y

relies on realization that $∆ x' = γ (∆ x - 0)$ eg:

${(c ∆ t')}^{2} - {(1.09 \times 2000)}^{2} = 0^{2} - 2000^{2}$ ✔

$∆ t' = « \pm » \frac{\sqrt{{(1.09 \times 2000)}^{2} - 2000^{2}}}{3.0 \times 10^{8}} = « \pm » 2.9 \times 10^{- 6}$ «s» ✔

ALTERNATIVE 3

use of diagram from answer to 4(c)(ii) (1 small square = 200 m)

counts 4.5 to 5 small squares (allow 900 – 1000 m) between events for A seen on B’s ct axis ✔

$\frac{950}{γ c} = 2.9 \times 10^{- 6} \pm 0.2 \times 10^{- 6}$ «s» ✔

c(iv).

Examiners report

[N/A]

a(i).

[N/A]

a(ii).

[N/A]

c(i).

[N/A]

c(ii).

[N/A]

c(iii).

[N/A]

c(iv).

Syllabus sections

Option A: Relativity » Option A: Relativity (Core topics) » A.2 – Lorentz transformations

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