College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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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?

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

Figure. P26.45

(a)

Expert Solution
Check Mark
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' frame is 0.800c .

Dina and Owen are at rest in frame S' . 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 .

(b)

Expert Solution
Check Mark
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 7.50×103m .

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is 1.80×1013m .

The speed of ball in frame S' is 0.800c .

The speed of light is 3.00×108m/s .

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

Δtp=Lpv'ball

  • Δtp is the time interval took for the ball to reach Dina according to her
  • Lp is the distance between Dina and Owen measured at their rest frame
  • v'ball is the speed of ball in frame S'
  • c is the speed of light

Substitute 1.80×1013m for Lp and 0.800c for v'ball to find Δtp .

Δtp=1.80×1013m0.800c=1.80×1013m0.800(3.00×108m/s)=7.50×103s

Thus, the time interval took for the ball to reach Dina according to her is 7.50×103m .

Conclusion:

The time interval took for the ball to reach Dina according to her is 7.50×103m .

(c)

Expert Solution
Check Mark
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 1.44×1012m .

Explanation of Solution

Given Info:

The speed of frame S' is 0.600c .

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

L=Lp1(vc)2

  • L is the distance between Owen and Dina measured by Ed
  • v is the speed of frame S'

Substitute 0.600c for v and 1.80×1012m for Lp to find L .

L=(1.80×1012m)1(0.600cc)2=1.44×1012m

Thus, the distance between Owen and Dina measured by Ed is 1.44×1012m .

Conclusion:

The distance between Owen and Dina measured by Ed is 1.44×1012m .

(d)

Expert Solution
Check Mark
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,

vbE=vbD+vDE1+vbDvDEc2

  • vbE is the speed of ball relative to Ed
  • vbD is the speed of ball relative to Dina
  • vDE is the speed of Dina relative to Ed

Substitute 0.800c for vbD and 0.600c for vDE to find vbE .

vbE=(0.800c)+(0.600c)1+(0.800c)(0.600c)c2=0.385c

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

Conclusion:

The speed of ball relative to Ed is 0.385c

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Chapter 26 Solutions

College Physics

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