HW8 - PHYS 4D Solution to HW 8 February 25, 2011 Problem...

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Unformatted text preview: PHYS 4D Solution to HW 8 February 25, 2011 Problem Giancoli 36-5 (II) What is the speed of a pion if its average lifetime is measured to be 4 . 40 × 10 − 8 s ? At rest, its average lifetime is 2 . 60 × 10 − 8 s . Solution: The speed is determined from the time dilation relation, Eq. 36-1a. ∆ t = ∆ t √ 1- v 2 /c 2 ⇒ v = c √ 1- ( ∆ t ∆ t ) 2 = c √ 1- ( 2 . 60 × 10 − 8 s 4 . 40 × 10 − 8 s ) 2 = 0 . 807 c = 2 . 42 × 10 8 m/s. Problem Giancoli 36-7 (II) Suppose you decide to travel to a star 65 light-years away at a speed that tells you the distance is only 25 light-years. How many years would it take you to make the trip? Solution: The speed is determined from the length contraction relation, Eq. 36-3a. l = l √ 1- v 2 /c 2 ⇒ v = c √ 1- ( l l ) 2 ⇒ t = l v = l c √ 1- ( l l ) 2 = 25 ly c √ 1- ( 25 ly 65 ly ) 2 = (25 y ) c c (0 . 923) = 27 y. Problem Giancoli 36-12 (II) A certain star is 18 . 6 light-years away. How long would it take a spacecraft traveling 0 . 950 c to reach that star from Earth, as measured by observers: (a) on Earth, (b) on the spacecraft? (c) What is the distance traveled according to observers on the spacecraft? (d) What will the spacecraft occupants compute their speed to be from the results of (b) and (c)? Solution: (a) l = 18 . 6 ly . t Earth = l v = 18 . 6 ly . 950 c = 19 . 58 yr. (b) The time as observed on the spacecraft is shorter. Use Eq. 36-1a. ∆ t = ∆ t √ 1- v 2 /c 2 = 19 . 58 yr √ 1- (0 . 95) 2 = 6 . 11 yr. (c) To the spacecraft observer, the distance to the star is contracted. Use Eq. 36-3a. l = l √ 1- v 2 /c 2 = 18 . 6 ly √ 1- (0 . 95) 2 = 5 . 81 ly. (d) To the spacecraft observer, the speed of the spacecraft is their observed distance divided by their observed time. v = l ∆ t = 5 . 81 ly 6 . 114 yr = 0 . 95 c. Problem Giancoli 36-17 (II) When at rest, a spaceship has the form of an isosceles triangle whose two equal sides have length 2 l and whose base has length l . If this ship flies past an observer with a relative velocity of v = 0 . 95 c directed along its base, what are the lengths of the ship’s three sides according to the observer? Solution: The vertical dimensions of the ship will not change, but the horizontal dimensions will be contracted according to Eq. 36-3a. The base will be contracted as follows. l base = l √ 1- v 2 /c 2 = 0 . 31 l. 1 When at rest, the angle of the sides with respect to the base is given by θ = cos − 1 0 . 50 l 2 . l = 75 . 52 ◦ . The vertical component of l vert = 2 l sin θ = 2 l sin 75 . 52 ◦ = 1 . 936 l is unchanged. The horizontal component, which is 2 l cos θ = 2 l ( 1 4 ) = 0 . 50 l at rest, will be contracted in the same way as the base.at rest, will be contracted in the same way as the base....
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This note was uploaded on 09/19/2011 for the course PHYS 4d taught by Professor Staff during the Winter '08 term at UCSD.

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HW8 - PHYS 4D Solution to HW 8 February 25, 2011 Problem...

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