PHYSICS 6
Practice Final Exam (2003)
SOLUTIONS
I.
1e ; 2c; 3a; 4b; 5c; 6b; 7c; 8d; 9e; 10d
II.
a)
distance = v(horiz) t
so t = d/v(horiz) = 200 Km/ 1 (Km/s) =
200 sec
b) reach max ht in half the total time or 100 sec
c) max ht = ½ g t
2
= 0.5
×
9.8(m/s
2
)
×
(100 s)
2
= 4.9
×
10
4
m =49 Km
d) Consider the clock on the ground at the time of impact. It reads 200 sec.
According to the rocket, that clock is moving and runs slow. The corresponding rocket
clock will read
t
×
γ
= t
÷
√
(1v
2
/c
2
) =200
÷
√
(1(0.8)
2
) = 200/0.6 =330 sec
e) distance measured by the rocket is contracted
= d/
γ
= 200 Km
×
0.6 = 120 Km
III. a) E
n
= 13.6 eV / n
2
so E
3
=  13.6/9 eV and E
1
= 13.6 eV
Then E(photon) = 13.6
×
(1/9 – 1) = 12.1 eV
b) E(photon) = h f , so f = E(photon)/h = 12.1eV/4.1x10
15
eV·s =
2.95
×
10
15
s
1
c) 1 Watt=1 J/s. To get 1 Joule of photon energy note that
1 Joule=(1/1.6
×
10
19
)eV = 6.25
×
10
18
eV
Then have number of photons is
6.25
×
10
18
eV/(12.1 eV/photon) = 5.17
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This note was uploaded on 03/10/2008 for the course PHY 6 taught by Professor Garyr.goldstein during the Spring '06 term at Tufts.
 Spring '06
 GaryR.Goldstein
 Physics

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