Chapter 25, Problem 49E

(a)

To determine

Distance between $A$ and $B$ relative to rocket.

Answer to Problem 49E

The distance between $A$ and $B$ is $4\text{m}$ relative to rocket.

Explanation of Solution

Rocket travels past observers $A$ and $B$ on platform at speed $0.6c$. Guns that are fired simultaneously strike rocket at points $A^{\prime }$ and $B^{\prime }$.

Write the expression of distance $d^{\prime }$ between $A$ and $B$ with respect to rocket

  $d^{\prime }=d{\left(1-\frac{u^2}{c^2}\right)}^{\frac{1}{2}}$        (I)

Here, $d$ is distance between $A$ and $B$ relative to platform and $u$ is speed of rocket relative to platform.

Conclusion:

Substitute $0.6c$ for $u$ and $5\text{m}$ for $d$ in expression (I).

  $\begin{array}{c}{d}^{\prime }=\left(5\text{m}\right){\left(1-\frac{{\left(0.6c\right)}^2}{c^2}\right)}^{\frac{1}{2}}\\ =4\text{m}\end{array}$

Thus, observers on rocket measure the distance between observers $A$ and $B$ to be $4\text{m}$.

(b)

To determine

Time when observer $A$ fires with respect to rocket.

Answer to Problem 49E

The observer $A$ fires at time $1.25\times {10}^{-8}\text{s}$ with respect to rocket.

Explanation of Solution

Observer $B$ fires at time $t=0$ relative to both the platform and the rocket.

Write the expression of time $t_{\text{A}}$ when $A$ fires with respect to rocket

  $t_{\text{A}}=\frac{du}{c^2}{\left(1-\frac{u^2}{c^2}\right)}^{-\frac{1}{2}}$        (II)

Conclusion:

Substitute $0.6c$ for $u$, $3\times {10}^8\text{m}\cdot {\text{s}}^{-1}$ for $c$ and $5\text{m}$ for $d$ in expression (II).

  $\begin{array}{c}{t}_{\text{A}}=\frac{\left(5\text{m}\right)\left(0.6\left(3\times {10}^8\text{m}/\text{s}\right)\right)}{{\left(3\times {10}^8\text{m}/\text{s}\right)}^2}{\left(1-\frac{{\left(0.6\left(3\times {10}^8\text{m}/\text{s}\right)\right)}^2}{{\left(3\times {10}^8\text{m}/\text{s}\right)}^2}\right)}^{-\frac{1}{2}}\\ =1.25\times {10}^{-8}\text{s}\end{array}$

Thus, observer $A$ fires at time $1.25\times {10}^{-8}\text{s}$ with respect to rocket.

(c)

To determine

Distance between points $A^{\prime }$ and $B^{\prime }$ relative to rocket.

Answer to Problem 49E

The distance between points $A^{\prime }$ and $B^{\prime }$ relative to rocket is $6.25\text{m}$.

Explanation of Solution

Write the expression of distance between points $A^{\prime }$ and $B^{\prime }$ relative to rocket

  $D^{\prime }=\left(d^{\prime }+u{t}_{\text{A}}\right)$        (III)

Here, $D^{\prime }$ is separation between points $A^{\prime }$ and $B^{\prime }$ according to observers on rocket.

Conclusion:

Substitute $4\text{m}$ for $d^{\prime }$, $1.25\times {10}^{-8}\text{s}$ for $t_{\text{A}}$, $0.6c$ for $u$ and $3\times {10}^8\text{m}\cdot {\text{s}}^{-1}$ for $c$ in expression (III)

  $\begin{array}{c}{D}^{\prime }=\left(\left(4\text{m}\right)+\left(0.6\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)\right)\left(1.25\times {10}^{-8}\text{s}\right)\right)\\ =6.25\text{m}\end{array}$

Thus, the distance between points $A^{\prime }$ and $B^{\prime }$ relative to rocket is $6.25\text{m}$.

(d)

To determine

Verification that distance between points $A^{\prime }$ and $B^{\prime }$ relative to platform and the rocket are consistent with formula of length contraction.

Answer to Problem 49E

It is verified that distance between points $A^{\prime }$ and $B^{\prime }$ relative to platform and the rocket are consistent with formula of length contraction.

Explanation of Solution

Write the expression of distance between points $A^{\prime }$ and $B^{\prime }$ relative to platform from length contraction formula

  $D=D^{\prime }{\left(1-\frac{u^2}{c^2}\right)}^{\frac{1}{2}}$        (IV)

Here, $D$ is separation between points $A^{\prime }$ and $B^{\prime }$ with respect to observers on the platform.

Since the distance between points $A^{\prime }$ and $B^{\prime }$ is same as the distance between $A$ and $B$ relative to platform therefore

  $D=5\text{m}$

Conclusion:

Substitute $0.6c$ for $u$, $3\times {10}^8\text{m}\cdot {\text{s}}^{-1}$ for $c$ and $6.25\text{m}$ for $D^{\prime }$ in expression (IV)

  $\begin{array}{c}D=\left(6.25\text{m}\right){\left(1-\frac{{\left(0.6\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)\right)}^2}{{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}^2}\right)}^{\frac{1}{2}}\\ =5\text{m}\end{array}$

Thus, it is verified that distance between points $A^{\prime }$ and $B^{\prime }$ relative to platform and the rocket are consistent with formula of length contraction.