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(PHYS111)[2010](f)final~=tkk933^_65416.pdf
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Physics 111 (Fall 2010) Final Examination Solution
1. (a) Distance of the center of mass from the pivot point: (0.500)(0.500 . 0.100) . (0.240)(0.250)
h .. 0.4865 m

0.500 . 0.240
(b) Let M and m be the mass of the disk and the rod respectively. Rotational inertia of the
pendulum: .12 2 . 12
I . Mr . M (L . r) mL
..2 .. . 3
.12 2 . 1 22
. (0.500)(0.100) . (0.500)(0.500 . 0.100) (0.240)(0.500) . 0.2025 kgm
..2 .. . 3
2(M . m)gh

(c) ..
I
The period:
2.
I
0.2025
T .. 2.
. 2.
.1.505 s
. (M . m)gh (0.74)(9.80)(0.4865)
2. (a) Initial angular momentum of the particle about the pivot point:
2 .1
Li . m0v0(L . r) . (0.1)(10)(0.5 . 0.1) . 0.6 kgm s
Final angular momentum of the particle and the pendulum about the pivot point:

2 .1
Lf ..m0v1(L . r) . I...(0.1)(5)(0.5 . 0.1) . 0.2025.. (.0.3 . 0.2025.) kgm s
Using the conservation of angular momentum,
0.6 ..0.3 . 0.2025.
0.9 .
.. . 4.4444 rads 1
0.2025
(b) Using the conservation of energy, 12
I.. (M . m)gh(1. cos.)2
1 (0.2025)(4.4444)2 . (0.74)(9.8)(0.4865)(1. cos.)
2 (0.2025)(4.4444)2
cos..1.. 0.4331
(2)(0.74)(9.8)(0.4865) .. 64.3o or 1.12 rad
3. (a) Escape speed from Moon: Using the conservation of energy,
12 GMMm
mv .. 0
2 esc rM

2GM M
vesc .
r
M

If the initial speed is . times the escape speed vesc from Moon, then using the conservation of energy when it reaches a distance rM from Moons surface, 13v 2 GM m 1 GM m
. esc . M 2M
m... . mv .
2 . 4 . rM2 2rM 9 GM 9GMGM GM
22M MMM
v . v .. ..
16 esc r 8rr 8r
M MMM
vesc
.. 0.250vesc
4

(b) If the initial speed is . times the escape speed vesc from Moon, then using the conservation of energy when it reaches the maximum distance from Moons surface,
1 . 3v .2 GMm GMm
escM M
m. .. ..
2 . 4 . rM r
2
11 9vesc19 7
.. ...
rr 32GM r 16r 16r
M MMM M
16r
r . M . 2.29rM
7 Hence the maximum distance from Moons surface is 1.29 rM.
(c) If the initial speed is 2 times the escape speed vesc from Moon, then using the conservation of energy when it is far away from Moon, 12 GM M m 12
m.2v .. . mv
2 esc rM2
2GM 8GM 2GM 6GM
22 MMMM
v . 4vesc ....
rrrr
MMM M
v .1.732v
esc esc

4. (a) Amplitude = 2 mm



0.2 .1
Wave velocity: v .. 2 ms 0.1 2. 2..1 .1

Wave number: k ... 7.854 radm . 7.85 radm
. 0.8
.1 .1

Angular frequency: .. vk . (2)(7.854) .15.71rads .15.7 rads v 2
Frequency: f .. . 2.50 Hz
. 0.8

(b)
The displacement of the wave is given by
y(x,t) . ym sin(kx ..t) . 0.002sin[(7.854)(5) . (15.71)(0.8)] . 0.00200 m


(c)
Beat frequency: fbeat . 2.60 . 2.50 . 0.10 Hz
The instants of maximum amplitudes are separated by 1/0.1 = 10 s.




10 s 15 s 20 s 25s 30 s 35 s 40 s
y
t
Fig. 4
5. (a)



Fig. 5

(b) Frequency observed by Porsche: . v . v .. 343 . 25 .
Porsche

fPorsche . f .. ...1200. ..1181Hz
v . v . 343 . 20 .
. Police .

Frequency observed by the police car: . v . v . 343 . 20
Police ..
f