Chapter 26, Problem 45AP

Owen and Dina are at rest in frame S′, which is moving with a speed of 0.600c with respect to frame S. They play a game of catch while Ed, at rest in frame S, watches the action (Fig. P26.45). Owen throws the ball to Dina with a speed of 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, and (d) how fast is the ball moving?

Figure. P26.45

To determine

The speed of ball according to Dina.

Answer to Problem 45AP

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

Explanation of Solution

Given Info:

The speed of ball in $S\text{'}$ frame is $0.800c$.

Dina and Owen are at rest in frame $S\text{'}$. The speed of the ball relative to the frame will be the speed of the ball. The magnitude of speed of ball is $0.800c$. The speed of the ball relative to Dina is $-0.800c$.

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

Conclusion:

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

To determine

The time interval took for the ball to reach Dina according to her.

Answer to Problem 45AP

The time interval took for the ball to reach Dina according to her is $7.50\times {10}^3\text{m}$.

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is $1.80\times {10}^{13}\text{m}$.

The speed of ball in frame $S\text{'}$ is $0.800c$.

The speed of light is $3.00\times {10}^8\text{m/s}$.

Formula to calculate the time interval took for the ball to reach Dina according to her is,

$\Delta t_p=\frac{L_p}{v{\text{'}}_{ball}}$

• $\Delta t_p$ is the time interval took for the ball to reach Dina according to her
• $L_p$ is the distance between Dina and Owen measured at their rest frame
• $v{\text{'}}_{ball}$ is the speed of ball in frame $S\text{'}$
• $c$ is the speed of light

Substitute $1.80\times {10}^{13}\text{m}$ for $L_p$ and $0.800c$ for $v{\text{'}}_{ball}$ to find $\Delta t_p$.

$\begin{array}{c}\Delta t_p=\frac{1.80\times {10}^{13}\text{m}}{0.800c}\\ =\frac{1.80\times {10}^{13}\text{m}}{0.800\left(3.00\times {10}^8\text{m/s}\right)}\\ =7.50\times {10}^3\text{s}\end{array}$

Thus, the time interval took for the ball to reach Dina according to her is $7.50\times {10}^3\text{m}$.

Conclusion:

The time interval took for the ball to reach Dina according to her is $7.50\times {10}^3\text{m}$.

To determine

The distance between Owen and Dina measured by Ed.

Answer to Problem 45AP

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

Explanation of Solution

Given Info:

The speed of frame $S\text{'}$ is $0.600c$.

Formula to calculate the distance between Owen and Dina measured by Ed is,

$L=L_p\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

• $L$ is the distance between Owen and Dina measured by Ed
• $v$ is the speed of frame $S\text{'}$

Substitute $0.600c$ for $v$ and $1.80\times {10}^{12}\text{m}$ for $L_p$ to find $L$.

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

Thus, the distance between Owen and Dina measured by Ed is $1.44\times {10}^{12}\text{m}$.

Conclusion:

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

To determine

The speed of ball relative to Ed.

Answer to Problem 45AP

The speed of ball relative to Ed is $0.385c$.

Explanation of Solution

Given Info:

The speed of ball relative to Dina is $-0.800c$.

The speed of Dina relative to Ed is $0.600c$.

Formula to calculate distance covered by the cheetah is,

$v_{bE}=\frac{v_{bD}+v_{DE}}{1+\frac{v_{bD}{v}_{DE}}{c^2}}$

• $v_{bE}$ is the speed of ball relative to Ed
• $v_{bD}$ is the speed of ball relative to Dina
• $v_{DE}$ is the speed of Dina relative to Ed

Substitute $-0.800c$ for $v_{bD}$ and $0.600c$ for $v_{DE}$ to find $v_{bE}$.

$\begin{array}{c}{v}_{bE}=\frac{\left(-0.800c\right)+\left(0.600c\right)}{1+\frac{\left(-0.800c\right)\left(0.600c\right)}{c^2}}\\ =0.385c\end{array}$

Thus, the speed of ball relative to Ed is $0.385c$.

Conclusion:

The speed of ball relative to Ed is $0.385c$