Question

Question:As you read this page (on paper or monitor screen), a cosmic ray proton passes along the left–right width of the page with relative speed v and a total energy of 14.24 nJ. According to your measurements, that left–right width is 21.0 cm. (a) What is the width according to the proton’s reference frame? How much time did the passage take according to (b) your frame and (c) the proton’s frame?

Short answer

Answer

(a) The width in proton’s frame is 0.222mm.

(b) The time of passage for proton in rest frame is 701 ps .

(c) The time of passage for proton in proton’s frame is 7.40 ps.

Step-by-step solution

Identification of given data

The total energy of proton is E = 14.24 nj

The left right width in measurement is $L_0=21\text{cm}=0.21\text{m}$

The Lorentz factor is calculated by using the formula for total relativistic energy of proton then length contraction and time dilation formulas are used to find according to different frames.

Determination of width according to proton’s frame

(a)

The Lorentz factor for proton is calculated as:

$E=\gamma m{c}^2$

Here, m is the mass of electron and its value is $1.67\times {10}^{-27}\text{kg}$, c is the speed of light and its value is $3\times {10}^8\text{m}/\text{s}$

$$\begin{gathered} \left(14.24\text{nJ}\right)\left(\frac{{10}^{-9}\text{J}}{1\text{nJ}}\right)=\gamma \left(1.67\times {10}^{-27}\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2 \\ \gamma =94.74 \end{gathered}$$

The width in proton’s frame is given as:

$L=\frac{L_0}{\gamma }$

Substitute all the values in equation.

$$\begin{gathered} L=\frac{21\text{cm}}{94.74} \\ L=0.222\text{cm} \end{gathered}$$

Therefore, the width in proton’s frame is 0.222 m.

Determination of time of passage for proton in rest frame

(b)

The time of passage for proton in rest frame is given as:

$t_r=\frac{L_0}{\left(\sqrt{1-\frac{1}{{\gamma }^2}}\right)c}$

Substitute all the values in above equation.

$$\begin{gathered} \frac{21\text{cm}}{\left(\sqrt{1-\frac{1}{{\left(94.74\right)}^2}}\right)\left(3\times {10}^{10}\text{cm}/\text{s}\right)} \\ {t}_r=7.01\times {10}^{-10}\text{s}l \\ {t}_r=\left(7.01\times {10}^{-10}\text{s}\right)\left(\frac{1\text{ps}}{{10}^{-12}\text{s}}\right) \\ {t}_r=701\text{ps} \end{gathered}$$

Therefore, the time of passage for proton in rest frame is 701 ps .

Determination of time of passage for proton in proton’s frame

(c)

The time of passage for proton in proton’s frame is given as:

$t=\frac{t_r}{\gamma }$

Substitute all the values in above equation.

$$\begin{gathered} t=\frac{701\text{ps}}{94.74} \\ t=7.40\text{ps} \end{gathered}$$

Therefore, the time of passage for proton in proton’s frame is 7.40ps .