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Unformatted text preview: Physics 105B Problem Set'l
May 13, 2008
Jolt Schonert: schoner‘t {at} physics.ucsbcdu Taylor 13.28 a) Since the one—dirru‘msiona1 force is Fm = km, the potential energy is
1 2
U : — Fd:1:= —§k;r The Hamiltonian in this case is the sum of the kinetic and potential energies, which is To examine the possible trajectories, see the attached graph of U If E > 0, then there are
no restrictions on the particle. It can come in from the left or right and then keep traveling in
the same direction unaffected (it will slow down as it approaches :16 = 0, though). However, if
E < 0, then the particle cannot access the region inside the parabola, i.e. the area between
the turning points of the potential. In this case, the particle doesn’t have enough energy to
pass over the parabola, so it just “bounces” off. The solutions to the equation of motion nut = 1m; is a linear combination of sinh and cosh. j ,_
I ‘ a t . f
r e\ L, \ ‘. ,
5 n .3 E?“ . _J I q _
El: ' / '4
‘ ‘ «r
I ‘_.... .n. r 9  r r J
NC I . I _ r . x oer? 2,1119 . . i f _
J Jr ' 4' i i _ .
.‘ a ‘\x \» :— 7 O b) Because the Hamiltonian does not explicitly depend on time, it is a conserved quantity
that is interpreted as the energy. To ﬁnd the phase space orbits in the p — :1: plane, we need
to consider the following equation 2
P 1 2
E 2 —— — 43;: ,
2m 2
which means that p2 = 2mE + mks? First, suppose that E > 0, so that p > 0 for all Then the orbits are hyperbolas centered about the y—axis. Next, suppose that E < 0, so that p is equal to zero at :1; = :I: 2E/kr. Then these orbits are hyperbolas centered about
the :r—axis, as indicated on the graph. This is what we expect from part a), Where it was determined that
k A:
1:05] = A sinh (W —t) + Bcosh ( ——t)
m m Taylor 15.2 First, consider the reaction occurring in some frames. Conservation of momentum in this frame is
TILAVA ‘l‘ mBVB = 771',ch ‘l— TRDVD If this conservation is to be invariant under a Galilean transformation, then it must hold
in any frame 8’ that is moving with velocity V relative to 5. Using equation (15.3): the
conservation of momentum becomes mA(vA — + mB(VB — I maﬁa — + 71%;)(VD “ Now subtract (2) from (1) to get
V(mA + My) = V(m(;' + mg) (3) Since this must hold for all V, we see that mA + m3 = mg + my. Taylor 15 .3 The factor 7 is given by
l 7=—Wt where {3 = v/c. Here, '0 = 8000111/s, so [9’ z 2.66 x 10—5. This gives a 7 of roughly
1.0000000004. The two measured times are related by At = vAtU, where At is the time
measured by the observer on the ground, and Ato is the time measured on the satellite. Then
A150 — At : Atrifl — 7) % 1hr >< 70000000004 = 36008 x 70000000004 = —1.278,us The percent difference between the two times is then (—1.278ns/360[')s) >< 100 = 3.55 x
10T8%. Taylor 15.8 a) The gamma factor for {i = 4/5 is So if Ato = 1.8 X 10' 8s in the rest frame, then the half—life in the moving frame is At = "yAtU.
Plugging in the numbers, this is At = (5/3)(1.8 X 10’8s) = 3.00 X 10‘8s. b) If all the pious are moving at speed .80, then the time it takes them to travel 01 2 36111 is
.8e 2 15 x 10—8s. Since the halflife in the moving frame is 3.00 X 10’8s, the number of
half—lives that occurs in the time to travel the tube length is 15/ 3.00 = 5, which means that
5 complete halflives occur. So if there are 32,000 pions to begin with, the fraction that will remain is
_ 32, 000 N _ :
25 1000 c) If instead we had used that the half—life is 1.8 X 10—8s in the rest frame, then the number
of halflives that occurred in 15 X 10"8s is 8.33. This means that the fraction of pions that remains is 32 000
N = ’ m 100 28.33 Taylor 15.12 In the rest frame of the pious, the tube is moving towards them with a speed .80, so the
factor 7 is the same in 15.8. For the ﬁrst part of the question, We are told in 15.8 that
the half—life in this rest frame is 1.8 x 1058s. Since the tube is moving towards the pions at
a relativistic speed, its length will appear contracted according to L
L’ : — : 21.6111
'1 The time it will take for the tube to pass the pious is then t = 21.6m/8c = 9 X 10’83, which
means that the number of half—lives that occurs in this time is (9 X 10’83) / (1.8 x 10—85) =
5 Therefore, the fraction of pions that remains is 2‘5, which means that the number of remaining pions is
_ 32,000 5 n = 1000, which agrees with 15.8 Taylor 15.14 :1) Consider the picture at the end of this problem that shows the moving rod in which the
observer is ﬁxed at point 0. When the observer makes a measurement, the light from both
ends of the rod must strike his eye at the same time. But consider the two paths that these
light rays must travel. The path from A to O is clearly longer than the path from B to 0,
meaning that the light at point A must have been emitted before the light at B, ie t}, > M. b) There are two distances to consider here: AB and A’ B. After time 163 has elapsed,
the back end of the rod has moved to the point A’, which means that the length of the rod
measured at the instant if; is AB = 1. But since A’ B = l and the observer sees the length AB , this means that AB > I, sinee AB > AUG. (3) Now suppose the observer is standing at a point 0’ just next to the axis long which the
rod is moving. Here we are trying to calculated the length that this observer actually sees,
which is AB. Since we already determined that A’ B : I, it follows that AB = l +'u(t3 — tA).
But the length AB must also equal C((tg — tA), because the light emitted from A must
reaeh point 0 at the same time that the light emitted from B does. These light bursts are
separated by a time (153 — 15A) and travel at speed c, meaning that AB = C(tg 715A). Now
we can solve for At = t1; — L; by setting these two expressions for AB equal to each other. The result of doing this is
l e — v
New plug this result for At back into AB = l + eAt. This length that the observer
actually sees is At: C — I?
l0 1
= _ — 1'
7 (1—5) (‘3)
_ 1 + [3 a
where We used that 1 F—tm Since [3 : e/e is between 0 and l, we see that AB > £0 for all speeds. ...
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 Spring '08
 CLELAND
 mechanics

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