40 mph v'= 25 mph If the observed horizontal speed of the B  A volley on board is v'=25mph then stationary observers on the train platform, see it travel.

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Presentation transcript:

40 mph v'= 25 mph If the observed horizontal speed of the B  A volley on board is v'=25mph then stationary observers on the train platform, see it travel with a speed (1) -25 mph(3) 15 mph(5) 40 mph (2) -15 mph(4) 25 mph(6) 65 mph

40 mph v'=  25 mph If the horizontal speed of the return (A  B) volley is v'=  25mph then stationary observers on the train platform see the ball’s speed as (1) -25 mph(3) 15 mph(5) 40 mph (2) -15 mph(4) 25 mph(6) 65 mph

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horizontal speed of the B  A volley (6) 65 mph horizontal speed of the A  B volley (3) 15 mph The common sense rule you apply is Though according to spectators on board the train volleys are simply left-right-left-right-left-right-left… the ground-based observers actually see the ball always traveling left, but: fast-slow-fast-slow-fast-slow-fast… (6) 1 and 2 only Sure mass doesn’t seem to be something affected by being in a moving compartment. But if velocity is, why wouldn’t acceleration be? Recall the definition: a=  v/  t =(v final -v initial )/  t. Since v A final = v B final + V rel and v A initial = v B initial + V rel then  v A =  v B ! 5) none of these We call such absolute quantities invariant.