Chapter 39, Problem 39.91CP

Owen and Dina are at rest in frame S.’ which is moving at 0.600c with respect to frame S. They play a game of catch while Ed. at rest in frame S, watches the action (Fig. P39.91). Owen throws the ball to Dina at 0.800c (according to Owen), and their separation (measured in S') is equal to 1.80 × 10¹² m. (a) According to Dina, how fast is the ball moving? (b) According to Dina, what time interval is required for the ball to reach her? According to Ed, (c) how far apart are Owen and Dina, (d) how fast is the ball moving, and (e) what time inter­val is required for the ball to reach Dina?

To determine

The speed of ball according to Dina.

Answer to Problem 39.91CP

The speed of ball according to Dina is $0.800c$.

Explanation of Solution

Given info: The frame $S^{\prime }$ which is moving at $0.600c$ with respect to the frame $S$. The speed of ball is $0.800c$ with respect to Owen. The distance between Owen and Dina is $1.80\times {10}^{12}\text{m}$.

If both persons are in same frame then the velocity of ball will be same for both persons.

Owen and Dina are in same frame $S$ and Owen throws the ball to the Dina with speed of $0.800c$, then the speed of the ball with respect to the Dina is $0.800c$.

Conclusion:

Therefore, the speed of ball according to Dina is $0.800c$.

To determine

The time interval is required for the ball to reach Dina.

Answer to Problem 39.91CP

The time interval is required for the ball to reach Dina is $7.51\times {10}^3\text{s}$.

Explanation of Solution

Given info: The frame $S^{\prime }$ which is moving at $0.600c$ with respect to the frame $S$. The speed of ball is $0.800c$ with respect to Owen. The distance between Owen and Dina is $1.80\times {10}^{12}\text{m}$.

Write the expression for time taken by the ball to reach Dina.

$t=\frac{L_{\text{P}}}{u}$

Here,

$t$ is the time taken by the ball.

$L_{\text{P}}$ is the distance between Owen and Dina.

$u$ is the speed of the ball.

Substitute $0.800c$ for $u$ and $1.80\times {10}^{12}\text{m}$ for $L_{\text{P}}$ in above equation.

$t=\frac{1.80\times {10}^{12}\text{m}}{0.800c}$

The speed of the light is,

$c=3.0\times {10}^8\text{m}/\text{s}$

Substitute $3.0\times {10}^8\text{m}/\text{s}$ for $c$ in above equation to find the time interval.

$\begin{array}{c}t=\frac{1.80\times {10}^{12}\text{m}}{\left(0.800\right)\left(3.0\times {10}^8\text{m}/\text{s}\right)}\\ =7.51\times {10}^3\text{s}\end{array}$

Conclusion:

Therefore, the time interval is required for the ball to reach Dina is $7.51\times {10}^3\text{s}$.

To determine

The distance between Owen and Dina with respect to Ed.

Answer to Problem 39.91CP

The distance between Owen and Dina with respect to Ed is $1.44\times {10}^{12}\text{m}$.

Explanation of Solution

Given info: The frame $S^{\prime }$ which is moving at $0.600c$ with respect to the frame $S$. The speed of ball is $0.800c$ with respect to Owen. The distance between Owen and Dina is $1.80\times {10}^{12}\text{m}$.

Write the expression of distance between Owen and Dina with respect to the other frame $S^{\prime }$.

$L=L_{\text{P}}\sqrt{1-\frac{v^2}{c^2}}$

Here,

$L$ is the distance between Owen and Dina with respect to Ed.

$v$ is the speed of frame $S^{\prime }$.

Substitute $0.600c$ for $v$ and $1.80\times {10}^{12}\text{m}$ for $L_{\text{P}}$ in above equation.

$\begin{array}{c}L=\left(1.80\times {10}^{12}\text{m}\right)\sqrt{1-\frac{{\left(0.600c\right)}^2}{c^2}}\\ =\left(1.80\times {10}^{12}\text{m}\right)\left(0.8\right)\\ =1.44\times {10}^{12}\text{m}\end{array}$

Conclusion:

Therefore, the distance between Owen and Dina with respect to Ed is $1.44\times {10}^{12}\text{m}$.

To determine

The speed of the ball with respect to Ed.

Answer to Problem 39.91CP

The speed of the ball with respect to Ed is $0.385c$.

Explanation of Solution

Given info: The frame $S^{\prime }$ which is moving at $0.600c$ with respect to the frame $S$. The speed of ball is $0.800c$ with respect to Owen. The distance between Owen and Dina is $1.80\times {10}^{12}\text{m}$.

Write the expression of speed of the ball with respect to the other frame $S^{\prime }$.

$u^{\prime }=\frac{u-v}{1-\frac{uv}{c^2}}$

Here,

$u^{\prime }$ is the speed of ball with respect to Ed.

Substitute $0.600c$ for $v$ and $0.800c$ for $u$ in above equation.

$\begin{array}{c}{u}^{\prime }=\frac{0.800c-0.600c}{1-\frac{\left(0.800c\right)\left(0.600c\right)}{c^2}}\\ =\frac{0.200c}{1-0.48}\\ =0.3846c\\ \approx 0.385c\end{array}$

Conclusion:

Therefore, the speed of the ball with respect to Ed is $0.385c$.

To determine

The time interval required for the ball to reach Dina with respect to Ed.

Answer to Problem 39.91CP

The time interval required for the ball to reach Dina with respect to Ed.

is $4.88\times {10}^3\text{s}$.

Explanation of Solution

Given info: The frame $S^{\prime }$ which is moving at $0.600c$ with respect to the frame $S$. The speed of ball is $0.800c$ with respect to Owen. The distance between Owen and Dina is $1.80\times {10}^{12}\text{m}$.

Write the expression for time taken by the ball to reach Dina with respect to Ed.

$t^{\prime }=\gamma \left(t-\frac{v{L}_{\text{P}}}{c^2}\right)$ (1)

Here,

$t^{\prime }$ is time taken by the ball to reach Dina with respect to Ed.

$\gamma$ is the Lorentz factor.

Write the expression for the Lorentz factor.

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

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$ in equation (1).

$t^{\prime }=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left(t-\frac{v{L}_{\text{P}}}{c^2}\right)$

Substitute $0.600c$ for $v$, $3.0\times {10}^8\text{m}/\text{s}$ for $c$, $7.51\times {10}^3\text{s}$ for $t$ and $1.80\times {10}^{12}\text{m}$ for $L_{\text{P}}$ in above equation.

$\begin{array}{c}{t}^{\prime }=\frac{1}{\sqrt{1-\frac{{\left(0.600c\right)}^2}{c^2}}}\left(7.51\times {10}^3\text{s}-\frac{\left(0.600c\right)\left(1.80\times {10}^{12}\text{m}\right)}{c^2}\right)\\ =1.25\left(7.51\times {10}^3\text{s}-\frac{1.08\text{m}}{3.0\times {10}^8\text{m}/\text{s}}\right)\\ =4.88\times {10}^3\text{s}\end{array}$

Conclusion:

Therefore, the time interval required for the ball to reach Dina with respect to Ed.

is $4.88\times {10}^3\text{s}$.