Question

Anna is on a railroad flatcar moving at 0.6c relative to Bob. (Their clock's read 0 as Anna’s center of mass passes Bob's.) Anna’s arm is outstretched in the direction the flatcar moves, and in her hand in a flashbulb. According to the wristwatch on Anna's hand, the flashbulb goes off at 100 ns. The time of this event according to Bob differs by 27 ns. (a) Is it earlier or later than 100 ns? (b) How long is Anna's arm (i.e., from her hand to her center of mass.)?

Short answer

(a) The event is later than the 100 ns.

(b) The length of Anna’s arm is $0.8\mathrm{m}$.

Step-by-step solution

Identification of the given data

The given data is written as:

• The speed of railroad flat car is $\mathrm{v}=0.6\mathrm{c}$.
• The time for flashbulb is $\mathrm{t}=100\mathrm{ns}$.
• The difference in time is $∆\mathrm{t}=27\mathrm{ns}$.

Significance of the duration for event

The duration of the event is based on time dilation, and the length of Anna’s arm is calculated by length contraction. Anna observes the time dilation in the railroad car.

(a) Determination of whether event is earlier or later

The duration for the event is calculated as:

$\mathrm{T}=\frac{\mathrm{t}}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{T}=\frac{100\mathrm{ns}}{\sqrt{1-{\displaystyle \frac{{\left(0.6\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{T}=125\mathrm{ns} \end{gathered}$$

Therefore, the event is later than the 100 ns.

(b) Determination of the length of Anna’s arm

The length of Anna’s arm is given as:

$\mathrm{L}=\frac{{\mathrm{c}}^2}{\mathrm{v}}\left(\left(\mathrm{t}+∆\mathrm{t}\right)\left(\sqrt{1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)-\mathrm{t}\right)$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{L}=\frac{{\mathrm{c}}^2}{\left(0.6\mathrm{c}\right)}\left(\left(100\mathrm{ns}+27\mathrm{ns}\right)\left(\sqrt{1-\frac{{\left(0.6\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}\right)-100\mathrm{ns}\right) \\ \mathrm{L}=\frac{\mathrm{c}\left({\displaystyle \frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\mathrm{c}}}\right)}{0.6}\left(\left(100\mathrm{ns}+27\mathrm{ns}\right)\left(0.8\right)-100\mathrm{ns}\right) \\ \mathrm{L}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{0.6}\left(101.6\mathrm{ns}-100\mathrm{ns}\right)\left(\frac{{10}^{-9}\mathrm{s}}{1\mathrm{ns}}\right) \\ \mathrm{L}=0.8\mathrm{m} \end{gathered}$$

Therefore, the length of Anna’s arm is data-custom-editor="chemistry" $0.8\mathrm{m}$.