test 1 2007 solutions

Problem 1: Only You Can Save Galactic Eagle!

The pilot of a small spaceship, Galactic Eagle, sees a mirror approaching with speed v =

2c/3. She wants to know how far away the mirror is so that she'll be able to warn the

crew to brace themselves for the inevitable collision. At time 0 on the spaceship's clock,

the mirror is an unknown distance D from the spaceship (as seen in the spaceship's

reference frame), and the captain releases a pulse of light toward the mirror. The light

strikes the mirror with normal incidence. At time  = 1.2 s (on the spaceship's clock), the

pulse of light arrives back at the spaceship. You must calculate how much time the crew

has to prepare for impact. Only you can save Galactic Eagle!



All requested time intervals and distances are those observed in the spaceship's

reference frame.



A. In terms of D, v, and c, determine how long it takes the light pulse to travel from the

spaceship to the mirror.



The distance between the spaceship and the mirror decreases at a rate of v + c, so the

light reaches the mirror after a time interval of D/(v + c).









B. In terms of D, v, and c, determine how long it takes the light pulse to travel from the

mirror back to the spaceship.



The same distance must be covered in the same time: D/(v + c).









C. Determine a numerical value for D.



The roundtrip time is 2D/(v+c) = , so D = (v+c)/2 = (1.2 s)(2c/3+c)/2 = 3108 m.

D. In terms of D, v, and c, determine the distance between the spaceship and the mirror

at the moment the light strikes the mirror.



The distance is D minus the distance the mirror traveled in the time it took the light to get

the mirror. The light strikes the mirror after a time interval of D/(v+c), as shown in part

A, and the mirror travels a distance of vD/(v+c) in this time. So the distance between

the spaceship and the mirror is D – Dv/(v+c), or D(1 – v/(v+c))= Dc/(v + c).









E. Determine a numerical value for how much time the crew has to prepare for impact

after the light arrives back at the ship.



The mirror will strike the ship a time interval of D/v after the light pulse leaves the ship.

The light's roundtrip travel time is 2D/(v+c). So the remaining time is D/v – 2D/(v+c) =

D(1/v – 2/(v+c)) = D(3/(2c) – 6/(5c)) = 0.3D/c = 0.3 s.

Problem 2: an Amazing Archer

An amazing archer shoots an arrow through a hollow tube that is fast approaching. In the

archer's reference frame, the arrow and the tube each have a speed of 0.5c, but they move

in opposite directions. Also in the archer's reference frame, the arrow and the tube are

each 0.866 m in length. Define two events as follows:



Event 1: The back of the arrow enters the tube.

Event 2: The front of the arrow exits the tube.



A. In the archer's reference frame, what is the time interval between the two events?

Which event occurs first (if they are not simultaneous)?









In the archer's frame, the arrow and the tube are exactly the same length, so the two

events are simultaneous: the time interval is zero.









B. What is the speed of the arrow in the tube's reference frame?



We simply use the relativistic velocity addition formula: u' = (u+u)/(1+uu/c2) = c/1.25

= 0.8c.









C. What is the length of the arrow in the tube's reference frame?



First we must find the proper length of the arrow. Its contracted length in the archer's

frame is 0.866 m, so its proper length greater by a factor of : 0.866 m  (1 – u2/c2)-1/2 =

0.866 m  [1 – (0.5c)2/c2]-1/2 = 1 m. Its contracted length in the tube's frame is its proper

length (1 m) divided by ': 1 m/(1 – u'2/c2)-1/2 = 1 m/[1 – (0.8c)2/c2]-1/2 = 0.6 m.

D. In the tube's reference frame, what is the time interval between the two events?

Which event occurs first (if they are not simultaneous)?



The tube's length in the tube's frame is its proper length. Since the arrow and the tube

each have the same speed and same length in the archer's frame, they must each have the

same proper length: 1 m. So in the tube's frame, the tube's length is 1 m, and the arrow's

length is 0.6 m.



0.6 m







1m





In the tube's frame, the arrow is moving to the right with speed u' = 0.8c. From the

figure, we see that when the back of the arrow enters the tube, the arrow has 0.4 m to go

before its front exits the tube. It travels this distance at speed 0.8c, so the time interval is

0.4 m/0.8c = 1.67 ns. Event 1 occurs before Event 2.









E. Is there a reference frame in which the arrow's length is equal to the proper length of

the tube? If so, describe it. If not, why not?



As we saw above, the arrow and the tube each have the same proper length. So the

arrow's length is the proper length of the tube in the arrow's frame.

Problem 3: Those Cuckoo Clocks

Two clocks sit in the S reference frame. The clock C0 sits at 0, and the clock C1 sits at x1

> 0. C0 and C1 are synchronized is S. Three events occur: E0, E1, and E2. E0 occurs at

C0, and E1 and E2 both occur at C1. In the S frame, E0 and E1 and simultaneous, while E2

occurs at a later time.



The S' reference frame moves with speed v in the positive x direction relative to S. As

usual, the x' axis of S' is collinear with the x axis of S, and the coordinates along both x

and x' axes increase in the same direction.



Multiple choice: circle the correct answer among four choices.



A. According to an observer in S', which clock in S is ahead?

C0

C1 C1 is the chasing clock, as viewed from S'.

C0 and C1 are synchronized in S'

Insufficient information is provided to answer this question



B. According to an observer in S', does E1 or E2 occur first?

E1 A timelike interval (same position in S) separates E1 and E2, so they

occur in the same order in all frames

E2

E1 and E2 are simultaneous in S'

Insufficient information is provided to answer this question



C. According to an observer in S', does E0 or E1 occur first?

E0

E1 The chasing clock shows the time of the events first.

E0 and E1 are simultaneous in S'

Insufficient information is provided to answer this question



D. According to an observer in S', does E0 or E2 occur first?

E0

E2

E0 and E2 are simultaneous in S'

Insufficient information is provided to answer this question

In S', the time interval between these two events is

t2' – t0' = [(t2 – t0) – v(x2 – x0)/c2); since t2 – t0 and x2 – x0 are both positive,

t2' – t0' could be either positive or negative



E. In which reference frame is the spacetime interval between E0 and E2 greater?

S

S'

It is the same in both frames It's the same in all frames.

Insufficient information is provided to answer this question

Problem 4: What Do You Serve a Pion?

A pion of rest mass m disintegrates into a muon of rest mass m and an antineutrino of

approximately zero mass. Assume that the magnitude of the antineutrino's momentum is

p and that its total relativistic energy is pc (in the rest frame of the pion). All requested

values are those observed in the rest frame of the pion. Leave answers to parts A, B, C,

and D in terms of as many of the following parameters as necessary: m, m, p, c, and

u, the muon's speed in the rest frame of the pion. (These values are all positive.)



A. What is the pion's momentum before it disintegrates?



We're working in the pion's rest frame, so its momentum is 0.









B. What is the pion's total energy before it disintegrates?



Since it has no kinetic energy, its total energy is its rest mass energy: mc2.









C. Use conservation of momentum to obtain an equation for p in terms of other

parameters. (In other words, solve for p after writing the equation for the conservation

of momentum.)



0 = mu/(1 - u2/c2)-1/2 - p



The minus sign is required because the muon and the antineutrino must go in opposite

directions to conserve momentum.



p = mu/(1 - u2/c2)1/2





D. Use conservation of energy to obtain a different equation for p in terms of other

parameters.



mc2 = mc2/(1 - u2/c2)-1/2 + pc

p = mc - mc/(1 - u2/c2)1/2

E. Find u in terms only of m, m, and c. This is algebraically intensive; do not attempt

this until you have completed the rest of the test.



Evidently we must eliminate p. Setting the answers to D and C equal to each other

yields



mc - mc/(1 - u2/c2)1/2 = mu/(1 - u2/c2)1/2

mc = m(c + u)/(1 - u2/c2)-1/2

(1 - u2/c2)1/2mc = m(c + u)

(1 - u2/c2)m2c2 = m2(c + u)2

(c2 - u2)m2 = m2(c + u)2

(c - u)(c + u)m2 = m2(c + u)2

(c - u)m2 = m2(c + u)

cm2 - um2 = m2c + m2u

u = c(m2 - m2)/(m2 + m2)