Question

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. P39.75). Owen throws the ball to Dina at 0.800 c (according to Owen), and their separation (measured in S' ) is equal to 1.80 × 10¹²m .(e) what time interval is required for the ball to reach Dina?

Answer 1

The time interval with respect to Ed is 4.88 ×10³ s.

A frame of reference is a set of reference points—geometric points whose positions are known mathematically and physically—that define the origin, orientation, and scale of an abstract coordinate system.

If a body does not continuously adjust its position in relation to its environment throughout the course of time, it is said to be at rest.

In the frame S', Dina and Owen are at rest.

The speed of the ball with respect to Owen, u =0.800c

The speed of the frame S' with respect to frame S,  v = 0.600c

Distance between Dina and Owen, L(p) = 1.8 × 10¹² m

Speed of light, c = 3 × 10⁸ m/s

Now, for time interval using the Lorentz transformation equation:

Δt = 1 / [ √( 1 - (v²/c²)] ( Δt' + (vΔx'/c²))

Now, v = 0.6c as S' is moving with respect to frame S.

Hence, the time take is:

Δt = 1 / [ √( 1 - ((0.6c)²/c²)] ( 7.5 ×10³ + [(-0.6c)(1.8 × 10¹²)/c²))

Δt = 1.25( 75000 - 3600)

Δt = 4.88 ×10³ s

The total time taken by the ball with respect to Ed is 4.88 ×10³ s.

Learn more about the frame of reference here:

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