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*From*: "Stanley J. McCaslin" <mccaslin@pscosf.peru.edu>*Date*: Tue, 02 Apr 1996 11:38:08 -0500

At 10:51 AM 4/2/96 EST, you wrote:

-6

A spaceship has a length of 200m in its own reference

frame. It is traveling at 0.95c relative to Earth.

Suppose that the tail of the spaceship emits a flash

of light. (a)In the reference frame of the spaceship,

how long does the light take to reach the nose?

(b)In the reference frame of the Earth, how long does

this take? Calculate the time directly from the motions of

the spaceship and the flash of light, and explain

why you cannot obtain the answer by applying the

time-dilation factor to the result from Part (a).

(from Ohanian's Principles of Physics)

I would rather have the

understanding of the problem and work from there.

I also get 4.16 X 10 sec. Could someone please tell me what I'm missing

in this analysis:

From the earth's viewpoint, the nose is 200sqrt(1 - .95^2) or 62.45 metersahead of the tail when the flash is emitted. Also from earth's view, the

nose travels 0.95ct during the light's flight time. So the distance is

62.45 + 0.95ct

t = distance / c

t = (62.45 + .95ct)/c

-6

t = 62.45/0.05c = 4.16 X 10 sec.

The distance the light travels, from earth viewpoint is 1248 meters.

An observer on the ship would say the light traveled 200 + .95cT

or 200 + .95c(6.67e-7)c = 390 m

1248sqrt(1 - .95*.95) = 390 (it seems consistant to me, making Ohanian's

answer wrong).

Stanley J. McCaslin Inet: mccaslin@pscosf.peru.edu

Assistant Professor, Computer Science Phone:402/872-2208

Peru State College, Peru, NE 68421 Home: 402/872-7595

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