Chapter 28: Special RelativityContent check — College Physics (AP), Chapter 28: Special Relativity

(a) 1.0328

(b) 1.15

(a) $0\text{.}\text{745}c$

(b) $0\text{.}\text{99995}c$ (to five digits to show effect)

(a) 0.996

(b) $\gamma$ cannot be less than 1.

(c) Assumption that time is longer in moving ship is unreasonable.

(a) 1.387 km = 1.39 km

(b) 0.433 km

(c) $$\begin{array}{l}L=\frac{L_0}{\gamma }=\frac{1.387\times {10}^3\text{m}}{3.20}=\mathrm{433.4\; m}=\text{0.433 km}\end{array}$$

Thus, the distances in parts (a) and (b) are related when $\gamma =\text{3.20}$.

(a) 4.303 y (to four digits to show any effect)

(b) 0.1434 y

(c) $\text{Δt}={\text{γΔt}}_0\Rightarrow \gamma =\frac{\text{Δt}}{{\text{Δt}}_0}=\frac{4\text{.}\text{303 y}}{0\text{.}\text{1434 y}}=\text{30}\text{.}0$

Thus, the two times are related when $\text{γ=}\text{30}\text{.}\text{00}$.

(a) 0.250

(b) $\gamma$ must be ≥1

(c) The Earth-bound observer must measure a shorter length, so it is unreasonable to assume a longer length.

(a) $0\text{.}\text{909}c$

(b) $0\text{.}\text{400}c$

a) $\text{658 nm}$

b) red

c) $v/\text{c}=9\text{.}\text{92}\times {\text{10}}^{-5}$ (negligible)

$u\prime =c$, so

$$\begin{array}{lll}u& =& \frac{v+u\prime }{1+(vu\prime /c^2)}=\frac{v+c}{1+(vc/c^2)}=\frac{v+c}{1+(v/c)}\\ & =& \frac{c(v+c)}{c+v}=c\end{array}$$

a) $0\text{.}\text{99947}c$

b) $1\text{.}\text{2064}\times {\text{10}}^{\text{11}}\text{y}$

c) $1\text{.}\text{2058}\times {\text{10}}^{\text{11}}\text{y}$ (all to sufficient digits to show effects)

(a) $3\text{.}\text{000000015}\times {\text{10}}^{\text{13}}\text{kg}\cdot \text{m/s}$.

(b) Ratio of relativistic to classical momenta equals 1.000000005 (extra digits to show small effects)

(a) $1\text{.}\text{121}\times {\text{10}}^{\text{–8}}\text{m/s}$

(b) The small speed tells us that the mass of a proton is substantially smaller than that of even a tiny amount of macroscopic matter!

$8.20\times {\text{10}}^{-\text{14}}\text{J}$

0.512 MeV

(a) $1\text{.}\text{11}\times {\text{10}}^{\text{27}}\text{kg}$

(b) $5\text{.}\text{56}\times {\text{10}}^{-5}$

$7\text{.}1\times {\text{10}}^{-3}\text{kg}$

$7\text{.}1\times {\text{10}}^{-3}$

The ratio is greater for hydrogen.

208

$0.999988c$

$6.92\times {\text{10}}^5\text{J}$

1.54

(a) $0\text{.}\text{914}c$

(b) The rest mass energy of an electron is 0.511 MeV, so the kinetic energy is approximately 150% of the rest mass energy. The electron should be traveling close to the speed of light.

(a) $$\begin{array}{l}{E}^2=p^2{c}^2+m^2{c}^4={\gamma }^2{m}^2{c}^4,\text{so that}\\ p^2{c}^2={\gamma }^2-1m^2{c}^4\text{, and therefore}\\ \frac{{\text{pc}}^2}{{{\text{mc}}^2}^2}={\gamma }^2-1\end{array}$$

(b) yes

$6\text{.}\text{56}\times {\text{10}}^{-8}\text{kg}$

$4.37\times {\text{10}}^{-\text{10}}$

$0.314c$

$0.99995c$

(a) 1.00 kg

(b) This much mass would be measurable, but probably not observable just by looking because it is 0.01% of the total mass.

(a) $6\text{.}3\times {\text{10}}^{\text{11}}\text{kg/s}$

(b) $4\text{.}5\times {\text{10}}^{\text{10}}\text{y}$

(c) $4\text{.}\text{44}\times {\text{10}}^9\text{kg}$

(d) 0.32%

(a) $L=$ $1000.0\sqrt{1-\frac{{(0.6c)}^2}{c^2}}=$ $800.00\text{ m}$

(b) $L=$ $1000.0\sqrt{1-\frac{{(0.6c)}^2}{c^2}}=$ $800.00\text{ m}$

(c) No.

(d) Yes.

Relativistic kinetic energy is given as ${\text{KE}}_{\text{rel}}=$ $(\gamma -1)m{c}^2$

where $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Classical kinetic energy is given as ${\text{KE}}_{\text{class}}=$ $\frac{1}{2}m{v}^2$

At low velocities $v=0$, a binomial expansion and subsequent approximation of $\gamma$ gives:

$\gamma =1+\frac{1v^2}{2c^2}$ or $\gamma -1=\frac{1v^2}{2c^2}$

Substituting $\gamma -1$ in the expression for ${\text{KE}}_{\text{rel}}$ gives

${\text{KE}}_{\text{rel}}=$ $\left[\frac{1v^2}{2c^2}\right]m{c}^2=$ $\frac{1}{2}m{v}^2=$ ${\text{KE}}_{\text{class}}$

Hence, relativistic kinetic energy becomes classical kinetic energy when $v\ll c$.