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(PHYS111)[2008](f)final~=tkk933^_66698.pdf
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Physics 111 (Fall 2008)
Final Examination Solution


1.
(a) Since r .a(1. e) and r . a(1. e) , we have
max min
r . r 102 . 0.5
max min

e . .. 0.990
r . r 102 . 0.5
max min

(b) We will use Keplers law of periods, T 2 . a3.
11

For the comet Ikeya-Zhang, a . (r . r ) . (102 . 0.5) . 51.25 AU , whereas
max min

22
for Earth, T = 1 y and a = 1AU. Hence for the comet,

3 3
. aIZ .2 . 51.25 .2

TIZ . TE .. ... (1). .. 367 y
a . 1 .
. E .

GMm
(c) The orbital energy of the comet is given by E ..
2a GMm GMm 12
Using the conservation of energy, . ... mv
2a rmac 2

22GM GM
v ..
rmac a

.11 30

. 21 .
(6.67 .10 )(1.99 .10 ) . 21 ..1
v .
GM . ... 291ms
.. ... 11 .

. rmac a . 1.5.10 .102 51.25 .
2.
(a) Using the parallel axis theorem, 12 . 2 .217 217 . 22
I . MR . M . R.. MR ... .(0.2)(0.3) . 0.017 kgm
2 . 3 . 18 .18 .

(b) The period of small oscillations of the pendulum:
I
0.017

T . 2.

. 2.
.1.31s pivot
Mgh (0.2)(9.8)(0.2)

centre
(c) Consider the period of the simple pendulum:
l lI

T . 2.

..
g g Mgh

17 2 I 18 MR 17 ..17 .

. l .. . R ..(0.3) . 0.425 m
Mh . 2 . 12 .12 .
M . R .
. 3 .

3.
(a) Angular frequency: .. 2.f . (2. )(300) . 1885 rads.1
. 1885

Wavenumber: k .. . 5.495 radm.1
v 343

(b) Consider the superposition of y1 and y2.
y . y . y . y sin(kx ..t . kx ) . y sin(kx ..t . kx )

120 00 0

. 2 y0 sin kx cos(.t . kx0)
Amplitude . 2y0 | sin kx |. ymax | sin kx |
. y | sin[(5.495)(0.7)]|.| .0.648 | y . 0.648y

max max max
n.

(c) Positions of maximum amplitude: kx . , where n = .1, .3,
2
n.

x .. 0.2859n ..0.2859 m,.0.8576 m,.1.4293 m
2k

4. . v . v .. 343 . 20 .
(a) f '. f .. D .. .1600. ..1744 Hz
v . v 343 .10
. S .. .
. v . vS .. 343 .10 .


(b) f ". f ' .. ...1744. .. 1906 Hz
v . v 343 . 20
. D .. .

5.

v 1

Mach cone angle: sin.. .. 0.5 . .. 30o vS 2
5000 5000 5000

tan.. . t .. . 22.0 s
(v . v )t (v . v ) tan. (2 . 0.85)(343) tan 30o
BC BC

6.
p

pAVA (2)(1.013.105)(0.001)
C
(a)
n .. . 0.04873 mol

RTA (8.315)(500)
57



(b)
For a diatomic gas, CV .R and CP . R


22

BCP 7D
... ..1.4
C 5

V A
In adiabatic processes, TV ..1 . constant . Therefore,
..1

1.4.1

.VA .. 0.001 . VB VA V
TB . TA .. ... 500. ..1256 K
. VB .. 0.0001.
..1
1.4.1
. VC .. 0.0001.

T . T .. ... 3500. ..1393 K
DC V . 0.001 .
. D .

(c) How much heat is absorbed during BC?
5

Q . nC (T .T ) . nR(T .T )
BC V CB CB
2
.. 5 .

..(0.04873)(8.315)(3500 .1256) . 2273 J
. 2 .

(d) Using the first law of thermodynamics,
5

WCD . ..Eint,CD ..nCV (TD .TC) .. nR(TD .TC)
2
. .. 5 .

..(0.04873)(8.315)(1393 . 3500) . 2134 J
. 2 .
5

WAB . ..Eint,AB ..nCV (TB .TA) .. nR(TB .TA)
2

. ... 5 ..(0.04873)(8.315)(1256 . 500) ..766 J
. 2 .

WCD .WAB 2134 . 766

(e) Efficiency: ... . 60.2%
QBC 2273
Alternatively, one may calculate the heat r