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Unformatted text preview: UNIVERSITY OF TORONTO
Faculty of Arts and Science April EXAMINATIONS 2011 PHY 1525
Duration  3 hours Examination Aids: Nonprogrammable/nongraphing calculators and the attached appendix of formulas (last two pages) Write your name, student number; and TA ’s name on ALL examination booklets used. Answer
BOTH Parts A and B. Part A is worth 33.3% and Part B is worth 66.7% of the total. For each
question, the mark breakdown for each subsection is listed in the righthand margin in square
brackets. Show all derivations, and justify each signiﬁcant step in your answer. This exam has 7 pages. PART A (33.3%) Answer FIVE of the following SIX questions. Each question is worth 4 points. If you answer
more than ﬁve questions, then only the ﬁrst ﬁve to be encountered will be graded. If you work on more than ﬁve questions, clearly state which ones you would like graded. 1. A hose with a diameter of 1.0 cm is connected to a pipe with a diameter of 2.0 cm. Water
is ﬂowing in the pipe with a uniform speed of 1.0 m/s. How much water ﬂows through the [4]
hose each hour, and what is the speed of the water in the hose? Assume that the water can be treated as an ideal ﬂuid. (Question 2, over . . .) PHY152S  Foundations of Physics 11 2011 Final Exam, page 1 0f 7 2. Three charges are placed on the yaxis as shown below. There is a charge +q at y = —a, a
charge —2q at y = O, and a charge +q at y = a. Show that the electric ﬁeld on the maxis, for :8 >> a, may be approximated as V [4] 32.
qai, E2 47T€o$4 where i is the unit vector in the arcdirection FIG. 1: Question 2. 3. Select the correct ending, (a) or (b), in the following statement, and explain why it is correct. “The electric potential inside a metal in equilibrium is always: (a) zero, or (b) constant.” [4] 4. A string with tension T = 20 N, mass density n = 2.0 g/m and length L = 1 m is held
ﬁxed at both ends. For the third harmonic (i.e. the harmonic with frequency 3 f0 where f0 is
the fundmental frequency), what is the frequency of Vibration, and at what positions on the [4] string will the antinodes be found? 5. If an astronaut travels to a distant star at a speed of 0.999c and the astronaut ages 4.0 years [4]
during the journey, what is the distance from Earth to the star as measured in the Earth’s rest frame? 6. In an inertial frame S two events occur at ($1 = 0 m,t1 = 0.0 s) and (x2 = 40 km,t2 =
1.0 x 10“4 s). What is the velocity relative to S of the inertial frame 5’ in which these events [4] are simultaneous? (Give the answer as a fraction of 0.) END OF PART A PHYI52S  Foundations of Physics II 2011 Final Exam, page 2 0f 7 PART B (66.7%) Answer FOUR of the following FIVE questions. If you answer more than four questions,
then only the ﬁrst four to be encountered will be graded. If you work on more than four questions, clearly state which ones you would like graded. 7. (a) What is the form of the force vs. displacementfromequilibrium equation for a Simple [1] Harmonic Oscillator? A mass m is attached between two springs with spring constants kl and kg and placed on
a frictionless surface, as shown. Spring 1 is in equilibrium when x = l1, and spring 2 is in equilibrium when L — a: = l2. Considering only horizontal motion in the plane of the page: L
191 k2
mm
(1} FIG. 2: Question 7. (b) In terms of the variables given, when the mass is at :3, what is the net force exterted on [3] the mass by the springs? What is the equilibrium value of :12? (c) Show that the frequency of oscillation of the mass is given by f = t/ ff + f3, where f1 [3]
is the frequency of oscillation when the mass is connected to spring 1 only, and f2 is the frequency when the mass is connected to spring 2 only. At time t = 0 the mass, initially at rest at its equilibrium position, is struck sharply so that
it receives an impulse in the —:c direction. It’s initial velocity is 1.0 m/s. If m = 50 g, k1: 50 N/m and k2 = 80 N/m: (d) Write the equation for :1: as a function oft describing the subsequent motion of the mass. [3] (Question 8, over . . .) PHYI5ZS  Foundations of Physics 11 2011 Final Exam, page 3 of 7 8. (a) Why does the electric ﬁeld have to have the same symmetries as the charge distribution [2] that produces it? (b) If the electric ﬁeld is parallel (or antiparallel) to the z—direction everywhere in space, and
its amplitude is independent of a: and 3/, what symmetries must the charge distribution have? [4] What would be a suitable Gaussian surface for calculating such an electric ﬁeld? (c) The electric ﬁeld from a charge distribution is found to have the form E = 200 2h N / C, where is is the unit vector along the zdirection. What charge density p(:v, y, 2) could pro [4]
duce this ﬁeld? 9. (a) Explain with the aid of sketches why two sound waves with nearly the same frequency [3] roduce “beats”.
P Two strings, ﬁxed at both ends, have identical mass density p, = 4.00 X 10“1 m and tension
T = 50.0 N. When they are both vibrating they produce a sound wave at 176.8 Hz, but it
gets louder and softer with a beat frequency of] Hz. (b) What are the lengths of the two strings? [5] (c) What tension should be applied to the longer string in order for the two strings to produce [2] the same frequency? (Question 10, over . . .) PHY152S  Foundations of Physics II 2011 Final Exam, page 4 of 7 10. In a ﬁctional high energy physics experiment, a proton with an initial velocity of 0.9606 in
the +cc direction in the laboratory frame (i.e. 11' = (0.9600, 0, 0) ) collides with a stationary
proton. Three particles emerge: particle 1 has energy E = 5.557 X 10‘10 J and momentum
1')“: (1.512 X 10‘”, 3.822 X 10‘19,0) kg m/s, but the other two particles are not detected, so they have unknown energy and momentum.
(a) What is the mass of particle 1? (b) Assuming that energy and momentum are conserved, ﬁnd the total momentum and en ergy of the missing particles. (0) Find the velocity of particle 1 in the “zero momentum frame” (i.e. the inertial frame in which the total momentum of the particles is zero). 11. Write short notes on two of the following topics: (i) interference of waves in two dimensions;
(ii) electric ﬁelds and potentials of capacitors; (iii) oscillations of a pendulum. THIS IS THE END OF THE EXAM. PHY152S  Foundations of Physics 11 2011 Final Exam, page 5 of 7 [4]
[3] [3] [5]
[5]
[5] Possibly Useful Facts 4
‘éphere : 3777.37 Asphere = 47TT2
U = if = 27(22) = 6(2) + 13 mm, m) = m) + fti17(t)dt,
Fnet=md=%€a ﬁ=mﬁ’
F=7j+17t, ﬂ=17’+17, 5:61", 5=R6, v=rw=r$—f,
K = $771112, U9 = mgh’
p=Po+pgd FB=prfg QzﬁZf
" K " q’ “ " enc ose
F12: Q;Q2n, E24, 1 d
7" q 60
‘ (I A ' ‘ ~ 1 2Z7 ‘ —1 .5 _. _.
E = = 1', 2 — 2 —, :
47reor2r’ E E 47reo r3 E 471'sO r3 p qs
/\ 77 n
E = E = _ E = _
27m)?" 260’ 50
f _. —. ...
AU” = — F«dl, Um) = qlqz AV = —/E dl,
i 47T€oT12
_ _ _ q
M — n(s wee, vm — 4W
d% 2
mz—w 11;, w=vk/m, w=\/g/La
1
F = —kz, U = 51mg
93(t) = a cos(wt + 45°), v(t) = wa sin(wt + $0), a(t) = ——w2a cos(wt + gbo),
93(t) = Aft/2T cos(wt + $0), = Eoe't/T,
Zn
1): T/a, w=27rf, [9:7, v=fA=w/k,
D(t) = Asin(km — a + (p0), gm, t) 2 k2: — wt + (be,
2L 1) 4L 1)
= — = — = 2 1 .—
)‘m my fm m2L> A2m+1 f2m+1 asin(q§1)+ asin(¢2) = 2a cos(A¢>/2) sin((¢1 + ¢2)/2), A¢ = ZWE A + [3% = 21717, A
D(ac,t) = 2a cos — wmod y(t) = 2a sinUmc) cos(wt),
_ f0 = AL _ ——————~—
ym m /d, 1 q: Usource/Uwave f: t) sin(kav:c — want) , fbeat = f1 — f2! (Continued . . .) PHY152S  Foundations of Physics [I 201] Final Exam, page 6 0f 7 Alf/=77, AL’=‘/1—v2/c2€,
/ __ __ __ I I : uz _ 'U
:0 — 7(33 vt) 3: ~ 7(x +1175) 11 1_umv/02
y’ = y y = y' u = Uy
7(1 — uzv/cz)
z’ = z z = 2’ Hz
u =
_ 2
t’ = 7(t ~ 1293/62) t = 7(t' +vm'/C2) 7(1 uﬂ/c )
(ASV = (61302 — (A502,
1 A5 a 2 2
p = m—AT = 71077111, E = 7pmc , K = (7,, —1)mc , Physical constants: pwater = 1000 kg/m3 g = 9.8 m/s2 K = 1 = 9.0 x 109 Nm2/C2 47160
Usound = III/S ,7 = (1_ v2/62)—1/2
11; +1)
ux =
1 1—119511/c2
“L
uy = I 2
7(1 +UEU/C)
I
uz — uz (m62)2 = E2 _ p202 7p = (1  “2/62)’1/2 pm = 1.013 x 105 Pa
e = 1.60 x 1019 C 60 = 8.85 x 10—12 (32/an2 melectmn = 9.11 x 10'31 kg = 511 keV/c2 = 0.511 MeV/c2 mproton = 1.67 x 10'27 kg = 936.7 MeV/c2
1 Ly = 0(1 yr)
1 eV = 1.6 x 10—19 J c = 3.00 x 108 m/s Mathematics:
1 1 /sin(ax) dz = —E cos(a:c) /cos(aa:) d0: = ; sin(aa:) 1 d y — d =1 / ————— d = ——
/ x x W) 02 + 003/2 y NW 62 gingsinw) 2 6 (191116 cos(6)' 2 1 — —2— (1+ x)" 2 1+ 713: for 1: << 1
A. B‘ = AB 0086 = .4sz + AyBy + ,4sz
am2+bw+c=0 é a:=’bi——— W
PHYISZS  Foundations of Physics 11 2011 Final Exam, page 7 0f 7 ...
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