Owen and Dina are at rest in frame S.’ which is moving at 0.600 c 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 × 10 12 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?

Question

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Answer 1

Textbook Question

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?

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

(a)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

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 $0.800c$ .

(b)

Expert Solution

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 $7.51×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t=LPu$

Here,

$t$ is the time taken by the ball.

$LP$ is the distance between Owen and Dina.

$u$ is the speed of the ball.

Substitute $0.800c$ for $u$ and $1.80×1012 m$ for $LP$ in above equation.

$t=1.80×1012 m0.800c$

The speed of the light is,

$c=3.0×108 m/s$

Substitute $3.0×108 m/s$ for $c$ in above equation to find the time interval.

$t=1.80×1012 m(0.800)(3.0×108 m/s)=7.51×103 s$

Conclusion:

Therefore, the time interval is required for the ball to reach Dina is $7.51×103 s$ .

(c)

Expert Solution

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 $1.44×1012 m$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

Write the expression of distance between Owen and Dina with respect to the other frame $S′$ .

$L=LP1−v2c2$

Here,

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

$v$ is the speed of frame $S′$ .

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $LP$ in above equation.

$L=(1.80×1012 m)1−(0.600c)2c2=(1.80×1012 m)(0.8)=1.44×1012 m$

Conclusion:

Therefore, the distance between Owen and Dina with respect to Ed is $1.44×1012 m$ .

(d)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

Write the expression of speed of the ball with respect to the other frame $S′$ .

$u′=u−v1−uvc2$

Here,

$u′$ is the speed of ball with respect to Ed.

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

$u′=0.800c−0.600c1−(0.800c)(0.600c)c2=0.200c1−0.48=0.3846c≈0.385c$

Conclusion:

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

(e)

Expert Solution

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 $4.88×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t′=γ(t−vLPc2)$ (1)

Here,

$t′$ is time taken by the ball to reach Dina with respect to Ed.

$γ$ is the Lorentz factor.

Write the expression for the Lorentz factor.

$γ=11−v2c2$

Substitute $11−v2c2$ for $γ$ in equation (1).

$t′=11−v2c2(t−vLPc2)$

Substitute $0.600c$ for $v$ , $3.0×108 m/s$ for $c$ , $7.51×103 s$ for $t$ and $1.80×1012 m$ for $LP$ in above equation.

$t′=11−(0.600c)2c2(7.51×103 s−(0.600c)(1.80×1012 m)c2)=1.25(7.51×103 s−1.08 m3.0×108 m/s)=4.88×103 s$

Conclusion:

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

is $4.88×103 s$ .

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Answer 2

Textbook Question

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?

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

(a)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

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 $0.800c$ .

(b)

Expert Solution

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 $7.51×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t=LPu$

Here,

$t$ is the time taken by the ball.

$LP$ is the distance between Owen and Dina.

$u$ is the speed of the ball.

Substitute $0.800c$ for $u$ and $1.80×1012 m$ for $LP$ in above equation.

$t=1.80×1012 m0.800c$

The speed of the light is,

$c=3.0×108 m/s$

Substitute $3.0×108 m/s$ for $c$ in above equation to find the time interval.

$t=1.80×1012 m(0.800)(3.0×108 m/s)=7.51×103 s$

Conclusion:

Therefore, the time interval is required for the ball to reach Dina is $7.51×103 s$ .

(c)

Expert Solution

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 $1.44×1012 m$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

Write the expression of distance between Owen and Dina with respect to the other frame $S′$ .

$L=LP1−v2c2$

Here,

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

$v$ is the speed of frame $S′$ .

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $LP$ in above equation.

$L=(1.80×1012 m)1−(0.600c)2c2=(1.80×1012 m)(0.8)=1.44×1012 m$

Conclusion:

Therefore, the distance between Owen and Dina with respect to Ed is $1.44×1012 m$ .

(d)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

Write the expression of speed of the ball with respect to the other frame $S′$ .

$u′=u−v1−uvc2$

Here,

$u′$ is the speed of ball with respect to Ed.

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

$u′=0.800c−0.600c1−(0.800c)(0.600c)c2=0.200c1−0.48=0.3846c≈0.385c$

Conclusion:

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

(e)

Expert Solution

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 $4.88×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t′=γ(t−vLPc2)$ (1)

Here,

$t′$ is time taken by the ball to reach Dina with respect to Ed.

$γ$ is the Lorentz factor.

Write the expression for the Lorentz factor.

$γ=11−v2c2$

Substitute $11−v2c2$ for $γ$ in equation (1).

$t′=11−v2c2(t−vLPc2)$

Substitute $0.600c$ for $v$ , $3.0×108 m/s$ for $c$ , $7.51×103 s$ for $t$ and $1.80×1012 m$ for $LP$ in above equation.

$t′=11−(0.600c)2c2(7.51×103 s−(0.600c)(1.80×1012 m)c2)=1.25(7.51×103 s−1.08 m3.0×108 m/s)=4.88×103 s$

Conclusion:

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

is $4.88×103 s$ .

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Answer 3

Textbook Question

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?

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

(a)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

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 $0.800c$ .

(b)

Expert Solution

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 $7.51×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t=LPu$

Here,

$t$ is the time taken by the ball.

$LP$ is the distance between Owen and Dina.

$u$ is the speed of the ball.

Substitute $0.800c$ for $u$ and $1.80×1012 m$ for $LP$ in above equation.

$t=1.80×1012 m0.800c$

The speed of the light is,

$c=3.0×108 m/s$

Substitute $3.0×108 m/s$ for $c$ in above equation to find the time interval.

$t=1.80×1012 m(0.800)(3.0×108 m/s)=7.51×103 s$

Conclusion:

Therefore, the time interval is required for the ball to reach Dina is $7.51×103 s$ .

(c)

Expert Solution

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 $1.44×1012 m$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

Write the expression of distance between Owen and Dina with respect to the other frame $S′$ .

$L=LP1−v2c2$

Here,

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

$v$ is the speed of frame $S′$ .

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $LP$ in above equation.

$L=(1.80×1012 m)1−(0.600c)2c2=(1.80×1012 m)(0.8)=1.44×1012 m$

Conclusion:

Therefore, the distance between Owen and Dina with respect to Ed is $1.44×1012 m$ .

(d)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

Write the expression of speed of the ball with respect to the other frame $S′$ .

$u′=u−v1−uvc2$

Here,

$u′$ is the speed of ball with respect to Ed.

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

$u′=0.800c−0.600c1−(0.800c)(0.600c)c2=0.200c1−0.48=0.3846c≈0.385c$

Conclusion:

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

(e)

Expert Solution

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 $4.88×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t′=γ(t−vLPc2)$ (1)

Here,

$t′$ is time taken by the ball to reach Dina with respect to Ed.

$γ$ is the Lorentz factor.

Write the expression for the Lorentz factor.

$γ=11−v2c2$

Substitute $11−v2c2$ for $γ$ in equation (1).

$t′=11−v2c2(t−vLPc2)$

Substitute $0.600c$ for $v$ , $3.0×108 m/s$ for $c$ , $7.51×103 s$ for $t$ and $1.80×1012 m$ for $LP$ in above equation.

$t′=11−(0.600c)2c2(7.51×103 s−(0.600c)(1.80×1012 m)c2)=1.25(7.51×103 s−1.08 m3.0×108 m/s)=4.88×103 s$

Conclusion:

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

is $4.88×103 s$ .

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Answer 4

Textbook Question

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?

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

(a)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

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 $0.800c$ .

(b)

Expert Solution

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 $7.51×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t=LPu$

Here,

$t$ is the time taken by the ball.

$LP$ is the distance between Owen and Dina.

$u$ is the speed of the ball.

Substitute $0.800c$ for $u$ and $1.80×1012 m$ for $LP$ in above equation.

$t=1.80×1012 m0.800c$

The speed of the light is,

$c=3.0×108 m/s$

Substitute $3.0×108 m/s$ for $c$ in above equation to find the time interval.

$t=1.80×1012 m(0.800)(3.0×108 m/s)=7.51×103 s$

Conclusion:

Therefore, the time interval is required for the ball to reach Dina is $7.51×103 s$ .

(c)

Expert Solution

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 $1.44×1012 m$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

Write the expression of distance between Owen and Dina with respect to the other frame $S′$ .

$L=LP1−v2c2$

Here,

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

$v$ is the speed of frame $S′$ .

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $LP$ in above equation.

$L=(1.80×1012 m)1−(0.600c)2c2=(1.80×1012 m)(0.8)=1.44×1012 m$

Conclusion:

Therefore, the distance between Owen and Dina with respect to Ed is $1.44×1012 m$ .

(d)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

Write the expression of speed of the ball with respect to the other frame $S′$ .

$u′=u−v1−uvc2$

Here,

$u′$ is the speed of ball with respect to Ed.

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

$u′=0.800c−0.600c1−(0.800c)(0.600c)c2=0.200c1−0.48=0.3846c≈0.385c$

Conclusion:

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

(e)

Expert Solution

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 $4.88×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t′=γ(t−vLPc2)$ (1)

Here,

$t′$ is time taken by the ball to reach Dina with respect to Ed.

$γ$ is the Lorentz factor.

Write the expression for the Lorentz factor.

$γ=11−v2c2$

Substitute $11−v2c2$ for $γ$ in equation (1).

$t′=11−v2c2(t−vLPc2)$

Substitute $0.600c$ for $v$ , $3.0×108 m/s$ for $c$ , $7.51×103 s$ for $t$ and $1.80×1012 m$ for $LP$ in above equation.

$t′=11−(0.600c)2c2(7.51×103 s−(0.600c)(1.80×1012 m)c2)=1.25(7.51×103 s−1.08 m3.0×108 m/s)=4.88×103 s$

Conclusion:

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

is $4.88×103 s$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

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 $0.800c$ .

(b)

Expert Solution

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 $7.51×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t=LPu$

Here,

$t$ is the time taken by the ball.

$LP$ is the distance between Owen and Dina.

$u$ is the speed of the ball.

Substitute $0.800c$ for $u$ and $1.80×1012 m$ for $LP$ in above equation.

$t=1.80×1012 m0.800c$

The speed of the light is,

$c=3.0×108 m/s$

Substitute $3.0×108 m/s$ for $c$ in above equation to find the time interval.

$t=1.80×1012 m(0.800)(3.0×108 m/s)=7.51×103 s$

Conclusion:

Therefore, the time interval is required for the ball to reach Dina is $7.51×103 s$ .

(c)

Expert Solution

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 $1.44×1012 m$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

Write the expression of distance between Owen and Dina with respect to the other frame $S′$ .

$L=LP1−v2c2$

Here,

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

$v$ is the speed of frame $S′$ .

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $LP$ in above equation.

$L=(1.80×1012 m)1−(0.600c)2c2=(1.80×1012 m)(0.8)=1.44×1012 m$

Conclusion:

Therefore, the distance between Owen and Dina with respect to Ed is $1.44×1012 m$ .

(d)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

Write the expression of speed of the ball with respect to the other frame $S′$ .

$u′=u−v1−uvc2$

Here,

$u′$ is the speed of ball with respect to Ed.

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

$u′=0.800c−0.600c1−(0.800c)(0.600c)c2=0.200c1−0.48=0.3846c≈0.385c$

Conclusion:

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

(e)

Expert Solution

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 $4.88×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t′=γ(t−vLPc2)$ (1)

Here,

$t′$ is time taken by the ball to reach Dina with respect to Ed.

$γ$ is the Lorentz factor.

Write the expression for the Lorentz factor.

$γ=11−v2c2$

Substitute $11−v2c2$ for $γ$ in equation (1).

$t′=11−v2c2(t−vLPc2)$

Substitute $0.600c$ for $v$ , $3.0×108 m/s$ for $c$ , $7.51×103 s$ for $t$ and $1.80×1012 m$ for $LP$ in above equation.

$t′=11−(0.600c)2c2(7.51×103 s−(0.600c)(1.80×1012 m)c2)=1.25(7.51×103 s−1.08 m3.0×108 m/s)=4.88×103 s$

Conclusion:

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

is $4.88×103 s$ .

Answer 8

(a)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

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 $0.800c$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The speed of ball according to Dina.

Answer 13

Answer to Problem 39.91CP

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

Answer 14

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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 $0.800c$ .

Answer 15

(b)

Expert Solution

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 $7.51×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t=LPu$

Here,

$t$ is the time taken by the ball.

$LP$ is the distance between Owen and Dina.

$u$ is the speed of the ball.

Substitute $0.800c$ for $u$ and $1.80×1012 m$ for $LP$ in above equation.

$t=1.80×1012 m0.800c$

The speed of the light is,

$c=3.0×108 m/s$

Substitute $3.0×108 m/s$ for $c$ in above equation to find the time interval.

$t=1.80×1012 m(0.800)(3.0×108 m/s)=7.51×103 s$

Conclusion:

Therefore, the time interval is required for the ball to reach Dina is $7.51×103 s$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

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

Answer 20

Answer to Problem 39.91CP

The time interval is required for the ball to reach Dina is $7.51×103 s$ .

Answer 21

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t=LPu$

Here,

$t$ is the time taken by the ball.

$LP$ is the distance between Owen and Dina.

$u$ is the speed of the ball.

Substitute $0.800c$ for $u$ and $1.80×1012 m$ for $LP$ in above equation.

$t=1.80×1012 m0.800c$

The speed of the light is,

$c=3.0×108 m/s$

Substitute $3.0×108 m/s$ for $c$ in above equation to find the time interval.

$t=1.80×1012 m(0.800)(3.0×108 m/s)=7.51×103 s$

Conclusion:

Therefore, the time interval is required for the ball to reach Dina is $7.51×103 s$ .

Answer 22

(c)

Expert Solution

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 $1.44×1012 m$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

Write the expression of distance between Owen and Dina with respect to the other frame $S′$ .

$L=LP1−v2c2$

Here,

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

$v$ is the speed of frame $S′$ .

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $LP$ in above equation.

$L=(1.80×1012 m)1−(0.600c)2c2=(1.80×1012 m)(0.8)=1.44×1012 m$

Conclusion:

Therefore, the distance between Owen and Dina with respect to Ed is $1.44×1012 m$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The distance between Owen and Dina with respect to Ed.

Answer 27

Answer to Problem 39.91CP

The distance between Owen and Dina with respect to Ed is $1.44×1012 m$ .

Answer 28

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

Write the expression of distance between Owen and Dina with respect to the other frame $S′$ .

$L=LP1−v2c2$

Here,

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

$v$ is the speed of frame $S′$ .

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $LP$ in above equation.

$L=(1.80×1012 m)1−(0.600c)2c2=(1.80×1012 m)(0.8)=1.44×1012 m$

Conclusion:

Therefore, the distance between Owen and Dina with respect to Ed is $1.44×1012 m$ .

Answer 29

(d)

Expert Solution

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 $S′$ 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 $1.80×1012 m$ .

Write the expression of speed of the ball with respect to the other frame $S′$ .

$u′=u−v1−uvc2$

Here,

$u′$ is the speed of ball with respect to Ed.

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

$u′=0.800c−0.600c1−(0.800c)(0.600c)c2=0.200c1−0.48=0.3846c≈0.385c$

Conclusion:

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

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The speed of the ball with respect to Ed.

Answer 34

Answer to Problem 39.91CP

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

Answer 35

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

Write the expression of speed of the ball with respect to the other frame $S′$ .

$u′=u−v1−uvc2$

Here,

$u′$ is the speed of ball with respect to Ed.

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

$u′=0.800c−0.600c1−(0.800c)(0.600c)c2=0.200c1−0.48=0.3846c≈0.385c$

Conclusion:

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

Answer 36

(e)

Expert Solution

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 $4.88×103 s$ .

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .

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

$t′=γ(t−vLPc2)$ (1)

Here,

$t′$ is time taken by the ball to reach Dina with respect to Ed.

$γ$ is the Lorentz factor.

Write the expression for the Lorentz factor.

$γ=11−v2c2$

Substitute $11−v2c2$ for $γ$ in equation (1).

$t′=11−v2c2(t−vLPc2)$

Substitute $0.600c$ for $v$ , $3.0×108 m/s$ for $c$ , $7.51×103 s$ for $t$ and $1.80×1012 m$ for $LP$ in above equation.

$t′=11−(0.600c)2c2(7.51×103 s−(0.600c)(1.80×1012 m)c2)=1.25(7.51×103 s−1.08 m3.0×108 m/s)=4.88×103 s$

Conclusion:

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

is $4.88×103 s$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

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

Answer 41

Answer to Problem 39.91CP

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

is $4.88×103 s$ .

Answer 42

Explanation of Solution

Given info: The frame $S′$ 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 $1.80×1012 m$ .