Physics 2120
Final Examination
Fall 2007
1- The moment of inertia I of a body of mass M about an axis passing thorough its
center of mass is given. Show that the moment of inertial of the body about the
same axis may be written as I=k2M where k is the radius of a uniform hoop of the
same mass M.
2
2
2
2
2
I   r dm   rave
dm  rave
 dm  rave M  k M
Where k=rave.

2- At one instant, a 0.80 kg particle is located at the position r (2.0iˆ  3.0 ˆj )m . The
linear momentum of the particle lies in the xy plane and has a magnitude of 2.4
kg.m/s and a direction of 115° measured counterclockwise from the positive
direction of x. What is the angular momentum of the particle about the origin, in
unit-vector notation?
The momentum components are
px = p cos 
py = p sin 
where p = 2.4 (SI units understood) and  = 115°. The mass (0.80 kg) given in the
problem is not used in the solution. Thus, with x = 2.0, y = 3.0 and the momentum
components described above, we obtain
ˆ
 r  p  (7.4kg  m 2 s )k.
3- A uniform spherical body of radius R and mass M is rotating with angular velocity
Ω about an axis passing through its center. What is the weight of an object of
mass m located on the surface of the sphere? Draw a diagram to clarify your
answer.
Ω
r
R
  MG

g    2  r 2 cos  rˆ
 R

w  mg
θ
Physics 2120
Final Examination
Fall 2007
4- A thermometer of mass 55 grams and specific heat 0.837 cal/kg.K reads 15oC. It
is then completely immersed in 0.3 kg of water, and it comes to the same final
temperature as the water of 44.4oC. What was the initial temperature of the water?
cwater=1cal/g.K.
Let the initial water temperature be Twi and the initial thermometer temperature be Tti.
Then, the heat absorbed by the thermometer is equal (in magnitude) to the heat lost by the
water:
ct mt T f  Tti   cw mw Twi  T f  .
We solve for the initial temperature of the water:
ct mt (T f  Tti)
TW i 
 T f  44.4 o C
c w mw
dp
where p is the pressure and V
dV
is the volume of the gas. For adiabatic processes in an ideal gas, show that (a)
B
RT

B  p and (b) that v s 
; where vs is the speed of sound in gasses.

M
cp
.
 
cV
(a) Differentiating Eq. 19-53, we obtain
5- The bulk modulus for gases is given as B  V
dp

dp

 (constant)  1  B  V
 (constant) 
dV
V
dV
V
which produces the desired result upon using Eq. 19-53 again ( p = (constant)/V).
(b) Due to the fact that v  B /  (from Chapter 17) and p = nRT/V = (Msam/M)RT/V
(from this chapter) with  = Msam/V (the definition of density), the speed of sound in an
ideal gas becomes
v
  M sam / M  RT / V
p
 RT


.

M sam / V
M
6- The temperature of one mol of a monatomic ideal gas is raised reversibly from
300 K to 400 k, keeping the volume constant. What is the entropy change of the
gas?
Physics 2120
Final Examination
Fall 2007
Since the volume of the monatomic ideal gas is kept constant it does not do any work in
the heating process. Therefore the heat Q it absorbs is equal to the change in its inertial
3
energy: dQ  dEint  n R dT . Thus
2
 Tf  3
dQ T f  3nR 2  dT 3
J

  400 K 

 nR ln    1.00 mol   8.31
 ln 

T
i
T
T
2
T
2
mol

K

  300 K 
 i 
 3.59 J/K.
7- A Carnot refrigerator extracts 35.0 kJ as heat during each cycle, operating with a
coefficient of performance of 4.60. What are (a) the energy per cycle transferred
as heat to the room and (b) the work done per cycle?
1
(a) Eq. 20-13 can be written as |QH| = |QL|(1 + 1/KC ) = (35)(1 + 4.6 ) = 42.6 kJ.
S  
(b) Similarly, Eq. 20-12 leads to |W| = |QL|/K = 35/4.6 = 7.61 kJ
8- Knowing that for reference frames S and S’, moving at uniform velocity v relative
to each other, x    ( x  vt) , show that t    (t  vx / c 2 ).
9- A certain particle of mass m has momentum of magnitude mc. What are (a) β, (b)
γ and (c) the ration K/E0? Where K is the kinetic energy and E0 is the total energy.
(a) We set Eq. 37-41 equal to mc, as required by the problem, and solve for the speed.
Thus,
mv
 mc
1  v 2 / c2
leads to   1/ 2  0.707.
(b) Substituting   1/ 2 into the definition of , we obtain

1
1  v 2 / c2
1

b g
1 1/ 2
2  141
. .
(c) The kinetic energy is
K     1 mc 2 


2  1 mc 2  0.414mc 2  0.414 E0 .
which implies K / E0  0.414 .
10- For all irreversible processes involving a system and its environment:
Physics 2120
Final Examination
Fall 2007
a- the entropy of the system does not change
b- the entropy of the system increases
c- the total entropy of the system and its environment does not change
d- the total entropy of the system and its environment increases
e- none of the above
11- A Carnot cycle:
a- is bounded by two isotherms and two adiabats on a p-V graph
b- consists of two isothermal and two constant volume processes
c- is any four sided process on a p-V graph
d- only exists for an ideal gas
e- has an efficiency equal to the enclosed area on a p-V diagram
12- A heat engine in each cycle absorbs energy from a reservoir as heat and does an
equivalent amount of work, with no other changes. This engine violates:
a- the zeroth law of thermodynamics
b- the first law of thermodynamics
c- the second law of thermodynamics
d- the third law of thermodynamics
e- none of the above
13- An observer notices that a moving clock runs slow by a factor of exactly 10. The
speed of the clock is:
a- 0.100c
b- 0.0100c
c- 0.990c
d- 0.900c
e- 0.995c
Physics 2120
Final Examination
Fall 2007
14- An electron is moving at 0.6c. If we calculate its kinetic energy using (1/2)mv2, we
get a result which is:
a- just right
b- just half enough
c- twice the correct value
d- about 1% too low
e- about 28% too low
15- A particle with zero mass and energy E carries momentum:
a- Ec
b- Ec2
c-
Ec
d- E/c
e- E/c2