PHYSICS FOR ENGINEERS.
A-EXAMS 2006-2007.
A1
A2
A3
A4
1
AGR
FOR
1.
A student raises one end of a board slowly,
where a block lies. The block starts to move
when the angle is 30º. The static friction
coefficient between the block and the board
surface should be:
30º
2. An object falls freely from rest on a planet without atmosphere where the gravity
acceleration is 10 m s-2. Its velocity when it crashes against the ground is 5 m s-1,
therefore its initial height was:
3. The angle formed by the vectors
a)
3 3
b)
2 3
c)
2 2
a) 2.00 m
b) 0.50 m
d) 2.40 m
e) None of above
c) 1.25 m
a) 45º
  



A  i  j and B  2i  2 j is:
b)  6 rad
c)
4. “An object at rest stays at rest unless acted on by
an external force”. This is:
d) 3 2
e) None of above
d)  6 rad
e) None of above
90º
a) Newton’s third law
d) The work-energy theorem
e) None of above
b) Law of inertia
c) Law of motion along a curved path
5. A punctual mass follows a
circular trajectory with constant
speed. As for its acceleration, it is
true that
a) The punctual mass is not accelerated
b) Its tangential acceleration is positive
d) There is no normal acceleration
e) None of above
c) The modulus of its normal acceleration is constant
6. A particle whose mass is (1.00±0.01)10-2 kg is moving along
a straight path at (1.00±0.10) m s-1. Its momentum is
a) 0.1  0.01 kg  m/s
d) 1.00  0.11 10-2 kg  m/s
b) 0.11  0.01 kg  m/s
e) None of above
c) 1.0  0.11 10-4 kg  m/s
2
GRADING:
EACH CORRECT ANSWER: +0.500
QUESTIONS: 4 POINTS
7.
A spring initially at rest, attached to a mass m, is stretched
to distance x. As for the mass m, it is true that:
EACH WRONG ANSWER: -0.125
PROBLEM: 6 POINTS
a)
The potential energy of m only varies in case the mass hangs vertically on the spring
b)
The potential energy of m varies if the mass hang vertically on the spring, but even if the spring and the mass lie on a
horizontal surface.
c)
The potential energy of m never changes, only its kinetic energy undergoes some variation.
d)
Neither the potential energy of m nor its kinetic energy undergo any variation.
e)
None of above
8. A harmonic oscillator obeys the equation y  5 cos9.90 t  where
every quantity is given in S.I. units. The frequency and the period are:
a) 4.5 Hz and 0.22 s
b) 0.5 Hz and 2.0 s
c) 6.3 Hz and 0.16 s
d) 5 Hz and 9.90 s
e) None of above
PROBLEM
Two blocks of masses m1 and m2 lie on a horizontal table. On m1 we apply a horizontal force F0 as shown in the picture. The
kinetic friction coefficients for both masses are 1 y 2, respectively. Using the numerical values given below, answer the
following questions:
a)
Separately draw the free body diagram for the set of two blocks, for m1 and for m2.
b)
Assuming the force F0 is big enough to move the set of two blocks, find its
acceleration.
c)
Find the force exerted by the first block (m1) on the second one (m2) and the force
exerted by the second one on the first one.
F0 = 2,50 kp
1 = 0.075
2 = 0.040
F0
m1
m2
m1 = 8 kg
m2 = 6 kg
3
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A1.
SOLUTION
1.
A student raises one end of a board slowly,
where a block lies. The block starts to move
when the angle is 30º. The static friction
coefficient between the block and the board
surface should be:
30º
2. An object falls freely from rest on a planet without atmosphere where the gravity
acceleration is 10 m s-2. Its velocity when it crashes against the ground is 5 m s-1,
therefore its initial height was:
3. The angle formed by the vectors
a)
3 3
b)
2 3
c)
2 2
a) 2.00 m
b) 0.50 m
d) 2.40 m
e) None of above
c) 1.25 m
a) 45º
  



A  i  j and B  2i  2 j is:
b)  6 rad
c)
4. “An object at rest stays at rest unless acted on by
an external force”. This is:
d) 3 2
e) None of above
d)  6 rad
e) None of above
90º
a) Newton’s third law
d) The work-energy theorem
e) None of above
b) Law of inertia
c) Law of motion along a curved path
5. A punctual mass follows a
circular trajectory with constant
speed. As for its acceleration, it is
true that
a) The punctual mass is not accelerated
b) Its tangential acceleration is positive
d) There is no normal acceleration
e) None of above
c) The modulus of its normal acceleration is constant
6. A particle whose mass is (1.00±0.01)10-2 kg is moving along
a straight path at (1.00±0.10) m s-1. Its momentum is
a) 0.1  0.01 kg  m/s
d) 1.00  0.11 10-2 kg  m/s
b) 0.11  0.01 kg  m/s
e) None of above
c) 1.0  0.11 10-4 kg  m/s
4
SOLUTION(CONTINUED)
GRADING:
EACH CORRECT ANSWER: +0.500
QUESTIONS: 4 POINTS
7.
A spring initially at rest, attached to a mass m, is stretched
to distance x. As for the mass m, it is true that:
EACH WRONG ANSWER: -0.125
PROBLEM: 6 POINTS
a)
The potential energy of m only varies in case the mass hangs vertically on the spring
b)
The potential energy of m varies if the mass hang vertically on the spring, but even if the spring and the mass lie on a
horizontal surface.
c)
The potential energy of m never changes, only its kinetic energy undergoes some variation.
d)
Neither the potential energy of m nor its kinetic energy undergo any variation.
e)
None of above
8. A harmonic oscillator obeys the equation y  5 cos9.90 t  where
every quantity is given in S.I. units. The frequency and the period are:
a) 4.5 Hz and 0.22 s
b) 0.5 Hz and 2.0 s
c) 6.3 Hz and 0.16 s
d) 5 Hz and 9.90 s
e) None of above
PROBLEM
Two blocks of masses m1 and m2 lie on a horizontal table. On m1 we apply a horizontal force F0 as shown in the picture. The
kinetic friction coefficients for both masses are 1 y 2, respectively. Using the numerical values given below, answer the
following questions:
a)
Separately draw the free body diagram for the set of two blocks, for m1 and for m2.
b)
Assuming the force F0 is big enough to move the set of two blocks, find its
acceleration.
c)
Find the force exerted by the first block (m1) on the second one (m2) and the force
exerted by the second one on the first one.
F0 = 2,50 kp
1 = 0.075
2 = 0.040
F0
m1
m2
m1 = 8 kg
m2 = 6 kg
5
SOLUTION(CONTINUED)
1 = 0.075
F0 = 2,50 kp
m1  m2 a
2 = 0.040
m1 = 8 kg
Newton’s 2nd law
F0  FR1  FR 2  m1  m2 a
Friction forces
FR1  1N1  1m1g
N1
FR 2  2 N2  2m2 g
N2
F0
m1
m2
FR1
FR1
F12
Numerical result
F0  1m1g   2 m2 g
m1  m2
a = 1.162 m/s2
Newton’s 2nd law
F21
F0
a
FR2
m1a
N1
m2 = 6 kg
m2 a
N2
F0  F21  FR1  m1a
F21  F0  FR1  m1a  F0  1m1g  m1a  9.324 N
Newton’s 2nd law
F12  FR 2  m2a
F12 and F21 have to be
equal (action and reaction)
F12  FR 2  m2a  2m2 g  m2a  9.324 N
FR2
6
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A2E
QUESTIONS
P1. A homogeneous sphere is thrown on a rough horizontal floor. Its initial center of
mass velocity is v, meanwhile its initial angular velocity is zero. What happens is:
a) The sphere rolls from the beginning on the horizontal surface.
b) The velocity of the center of mass remains constant, the angular velocity increases.
c) The velocity of the center of mass drops, the angular velocity remains null.
d) The sphere slides, but cannot roll because the friction is not zero.
e) None of the above.
AGR
FOR
SURNAMES
FIRST NAME
P2. A skater on ice has her arms extended while she spins at 2 rps. When she cross
her arms on her body, her moment of inertia halves. As a result, we’ll see that:
a) Her angular velocity halves.
b) Her angular velocity duplicates.
c) Her angular velocity reduces to a fourth.
d) Her angular velocity increases four times.
e) Her angular velocity remains constant, however her angular momentum duplicates.
P3. The figure depicts a ring of radious R and mass M, where the
mass is homogenously shared out along its rim. The momenta of
inertia about the different axis depicted in the figure are:
1
1
MR 2
I XX  MR 2
2
2
1
1
1
2
2
I ZZ  MR
I YY  MR
I XX  MR 2
2
4
4
1
1
1
2
2
I ZZ  MR
I YY  MR
I XX  MR 2
4
2
2
3
1
2
2
I ZZ  MR
I YY  MR
I XX  MR 2
2
2
X
a) I ZZ  MR 2
b)
c)
d)
I YY 
P4. The variation of the angular momentum with time
quantifies:
a) The sum of external forces
b) The sum of the momenta of inertia
Z
Y
c) The sum of the torques of external forces
d) The variation of the angular velocity
e) None of the above
e) None of the above
Warning: to answer questions P1 to P5 you have to use the answer form, only the answers including there will be considered.
For the choosen right answer, use the symbol X . To invalidate a previously marked cell, use the symbol X
For questions P1 to P5, each correct answer adds +1; each wrong answer takes off -0.25
7
P5. A torque of 10 Nm is acting on a spinning disc, which
turns around a normal fixed axis passing through its
simmetry center. If the rotational work is 5.24 J, then we can
say that the angle gone over is (within ±1º):
a) 90º
b) 30º
c) 60º
ANSWER FORM
a
P1

P2
P3
P4
d) 45º
P5
e) 180º

b
c
d



PROBLEM
A thin rigid rod of lenght 2L = 80 cm is composed by two sections, each of lenght L, made on
different materials whose linear densities are A = 2 kg/m and B = 5 kg/m.
The rod is hung horizontally from both ends on two threads A and B, as shown in the
figure. Find:
a) The tension supported by each thread (1.5 p)
b) The moment of inertia about end A (1.5 p)
c) Assume the thread B is cut. Find the angular acceleration and the angular velocity of the rod when it
reaches the vertical position (2 p).
A
A
B
L
L
B
8
e

PROBLEM SOLUTION
Both sections are homogeneous and have the same
xA  L / 2
lenght, so the position of both center of masses will be
a)
Y
TA
A
A
B
L
x0
xCM 
xCM
Total mass: m   A   B L


 A  mg xCM  TB 2 L  0
Fy  mg  TA  TB  0
xCM 1   A  3 B 

 
2 L 4   A   B 
TB
B
mA   A L
mB   B L
Center of mass of the whole rod: xCM 
L
mg
Mass of each section:
xB  3L / 2
m A L / 2   mB 3L / 2 
m A  mB
TB  mg
xCM
2L
xCM 
m A x A  mB x B
m A  mB
 A LL / 2   B L3L / 2
 AL  B L
   3 B  L

xCM   A



B  2
 A
 x 
TA  mg  TB  mg 1  CM 
2L 

TB   A   B L g
1   A  3 B 


4   A  B 
 1    3 B  

TA   A   B  L g 1   A
 4   A   B 
TB 
1
 A  3 B L g
4
TA 
1
3 A   B 9L g
4
PROBLEM SOLUTION
b) Moment of inertia about A
A
A
m   A   B L
dm   dx
dx
L
x0
mg
B
B
   3 B  L

xCM   A
  A  B  2
L
L3
I A   A  7  B 
3
x
L
IA 

L
2L
x 2 dmA 
0

x 2 dmB 

2L
x 2  A dx 
0
L

L
2L
 x3 
 x3 
1
1
2
x  B dx   A     B     A L3   B 8 L3  L3
3
 3 0
 3L 3

L
c) Angular acceleration and angular velocity
TA'
A

 A  xCM mg sin 90     I A 


xCM
mg 90  
3
 A  7  B  L3


xCM mg cos
IA
  A  3 B  L

  A   B L g cos 



B  2
 A
  A  3 B  3g cos 

2L
  A  7B 
  
10

E
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E.
PROBLEM
The pipe shown in figure 1 is used to supply water to a reservoir. It has two open tubes, each in
different parts of the pipe having cross sections S1 and S2 (see numerical values below).
1. The difference of
height between the water
level in both open tubes is
h (see numerical value
below). Find the volume
flux and the mass flux in
the pipe.
Figure 1
h
S1
S2
1
2. There is a pump P (figure 2) which takes
out from the reservoir M kg of water per
minute and carries them up to the height H at
c0 m/s (see numerical values below). Find the
power of the pump (neglect friction losses)
2
P
c0
Figure 2
H
S 1 (cm2) =
350
2
S 2 (cm ) =
h (m) =
M (kg/min) =
c 0 (m/s) =
H (m) =
180
1
4500
9,6
5
Gravity acceleration g = 9.8 m/s2; density of water  = 1.00 g/cm3
Warning: to answer questions P1 to P5 you have to use the answer form, only the answers including there will be considered.
For the choosen right answer, use the symbol X . To invalidate a previously marked cell, use the symbol X
For questions P1 to P5, each correct answer adds +1; each wrong answer takes off -0.25
11
PROBLEM SOLUTION

1
1
P1  c12  gy1  P2  c22  gy2
2
2
Bernoulli’s law
1
P1  P2   c22  c12
2
P1  Patm  gz1
z1
z2
c2
c1
S1
1
y1
c2 
S2
y2
2 gh
2
1  S 2 S1 
P1  P2  g z1  z2 
P2  Patm  gz2
h
V2  V  S 2
P1  P2  gh
2
Continuity equation
2 gh
2
1  S 2 S1 
c1  c2
S2
S1
1  2 2 S 22 
P1  P2    c2  c2 2   gh
2 
S1 
2 gh
m 2  m   S 2
2
1  S 2 S1 
12

PROBLEM SOLUTION (CONTINUED)
2. There is a pump P (figure 2) which takes out from the reservoir M kg of water per minute
and carries them up to the height H at c0 m/s (see numerical values below). Find the power of
the pump (neglect friction losses)
0
Bernoulli’s Law
Pressures in S and 0 are the same
0
P
c0
The surface of the water is at rest
PS
P
1 2
1 2

cS  y S  H P  0 
c0  y0
g 2 g
g 2 g
HP 
1 2
c0  y0  yS
2g
1 2
HP 
c0  H
2g
H
Work done by the pump raising a water mass m to a height H
dW dm

W  g H P m
W 

g HP
Power:
dt
dt
The pump raises M kg a minute:
m 
M (kg) 1

M kg/s
60 (s)
60
Figure 2
H
y0
S
yS
W  m g HP
1
W 
M g HP
60
13
PROBLEM SOLUTION (CONTINUED)
MODEL E
 (kg/m3) =
g (m/s2) =
h (m) =
S1 (m2) =
1000
9,8
0,25
3,50E-02
1000
9,8
1
3,50E-02
1000
9,8
0,5
2,80E-02
1000
9,8
0,4
2,80E-02
S2 (m2) =
2,00E-02
1,80E-02
1,50E-02
2,00E-02
P1-P2 (Pa) =
2450
9800
4900
3920
c2 (m/s) =
2,70
5,16
3,71
4,00
c1 (m/s) =
1,54
2,65
1,99
2,86
V (m3/s) =
m (kg/s) =
5,39E-02
53,95
9,29E-02
92,92
5,56E-02
55,61
8,00E-02
80,02
g (m/s2) =
M (kg/min) =
c0 (m/s) =
H (m) =
9,8
3000
8
4
9,8
4500
9,6
5
9,8
5000
10
2
9,8
800
5
8
HP (m) =
W (watt) =
7,27
3560,00
9,70
7131,00
7,10
5800,00
9,28
1212,00
14
E
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E.
QUESTIONS
P1. A student blows up the tyres of his car in a place where the atmospheric pressure is 980 mb. When the gauge
indicates 2.2 kp/cm2, the absolute pressure inside the tyres is:
a) 284200 Pa
b) 3136 mb
c) 3.05 kp/cm2
d) 3.45 kp/cm2
e) None of above
P2. Three identical glasses are filled up with water. The glass 1 contains just water; a cube of ice floats in the
glass 2 (ice density 0.89 g/cm3), and a piece of wood floats in the glass 3 (wood density 0,75 g/cm3). We call
W1, W2 y W3 the weights of the three glasses including their contents, then it is true that:
a) We cannot say which glass is more weighted without knowing the weights of the ice and the piece of wood.
b) Ordered from more to less weight , the order is W1 > W2 > W3
c) Ordered from more to less weight , the order is W3 > W2 > W1
3
2
1
d) The weight of the three glasses is the same.
e) The weight of the glass containing just water is smaller than the weights of the other two glasses.
P3. The initial level of the water in the system shown in the figure is
indicated by A. The cylinder C is held in a fixed position (it cannot move).
If we fill out the vertical thin tube up to the level B, what happens with the
pressure PC exerted by the liquid against the cylinder C is:
a) Pressure PC multiplies by 3 its initial value.
b) Pressure PC keeps nearly exactly its initial value because the vertical
tube is thin.
c) Pressure PC multiplies by 5 the its inital value.
d) Pressure PC duplicates.
e) None of the above.
B
1m
A
C
1m
PC
5 m2
1 m2
15
0.5 m
E
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E.
QUESTIONS
P4. The arrows depicted in the schemes F1, F2 and F3 indicate the velocities of the
particles of a fluid moving to the right in a pipe. We can say about that:
F1
a) F1 represents laminar flow and F3 turbulent flow
b) The three velocity schemes are characteristic of turbulent flow
c) The three velocity schemes are characteristic of laminar flow
d) The schemes F2 and F3 represent laminar flow; F1 represents turbulent flow
e) None of above
F2
P5. The figure shows a siphon. We try to transvase liquid from the left–hand to the
right-hand vessel. The neccesary condition for the siphon to work is:
a) It is necesary that h << d.
b) It is necesary that d.= 0
c) That siphon only works if the liquid is less dense than the water.
F3
d) The pressure in A must be smaller than the pressure in B.
e) The P5 heading is wrong, because it is impossible that a siphon
setup such as the depicted one could work.
ANSWER FORM
a
P1
A
P2
P3
P4
P5
B
C


b

c
d


16
e
QUESTION 2 SOLUTION
QUESTION 1 SOLUTION
Pa
 92000 Pa
Patm  920 mb  0.920 bar  10
bar
5
Pgauge  2.0 kp/cm 2 2.0  9.8 N/10-4 m 2  196000 Pa
Pabs  Patm  Pgauge
A
Patm
Pgauge
Pabs
B
Patm
Pgauge
Pabs
C
Patm
Pgauge
Pabs
D
Patm
Pgauge
Pabs
QUESTION 3 SOLUTION
kp/cm2 mb
bar
Pa
0,9
882
0,882
88200
2
1960
1,96 196000
2,9
kp/cm2
2842
mb
2,842
bar
284200
Pa
1
2,2
980
2156
0,98
2,156
98000
215600
3,2
3136
3,136
313600
kp/cm2 mb
bar
Pa
0,95
931
0,931
93100
2,1
2058
2,058 205800
3,05
2989
2,989
298900
kp/cm2 mb
bar
Pa
1,05
1029
1,029 102900
2,4
2352
2,352 235200
3,45
3381
Every floating body drives out of the glas a volume of
liquid equal to its weight. Therefore, the weights of the
three glasses are the same.
3,381
338100
The lower part of the cylinder C lies initially 0.5 m
below the free surface of the water (level A). When
the vertical thin tube is fillep up to level B (1 m
besides level A), the new deep of the lower part of
cylinder C is three times the initial one, so the
pressure in C multiplies 3 times.
QUESTION 4 SOLUTION
See theory.
QUESTION 5 SOLUTION
See the siphon problem.
17
A
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A4E.
PROBLEM
10 mol of an ideal gas (initial temperatue 20 ºC) expand reversibly following the law p = aV,
where p, V are pressure and volume, whereas a is a constant. The final volume after the
expansion is twice the initial one. Gasses constant R = 8.314 J/(Kmol).
1. Plot the expansion process in a p-V diagram (1 p).
2. Find the final temperature (1.5 p).
3. Find the work in the expansion (1.5 p).
ANSWER FORM
a
b
c
d
e
P1
P2
P3
P4
P5
P6
QUESTIONS
P
P
P1. About the phase diagram depicted aside, it can be said that:
2
a) 1 is the critical point; 2 is the triple point; AB is a sublimation process.
b) 1 is the triple point; 2 is the boiling point; AB is a melting process.
1
c) 1 is the melting point; 2 is the boiling point; AB is an evaporation process.
A
d) 1 is the triple point; 2 is the critical point; AB is a sublimation process.
B
T
T
e) None of the above.
P2. An ideal gas expands reversibly without changing its temperature. The gas absorbs 200 Kcal. The internal energy variation in
this process is:
a) -200 kJ
b) +200 Kcal
c) -200 Kcal
d) +200 KJ
e) Zero
Warning: to answer questions P1 to P6 you have to use the answer form, only the answers including there will be considered.
For the choosen right answer, use the symbol X . To invalidate a previously marked cell, use the symbol X
For questions P1 to P6, each correct answer adds +1; each wrong answer takes off -0.25
18
A
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A4E.
QUESTIONS
P3. An ideal gas complete a thermodynamics cycle composed by two adiabatics, one isochoric and one isobar. Indicate the correct
option.
a) The internal energy variation of the gas is positive if the cycle is described is clockwise direction.
b) The entropy variation of the gas once the cycle is completed is positive if any of the cycle steps is irreversible.
c) The heat absorbed by the gas is always the same than the work done by the gas.
d) The internal energy variation of the gas once the cycle is completed is not well-defined, it depends on the way.
e) None of the above.
P4. A 20-ton iceberg fall of from a coastal glacier. The temperature of the ice is -5º C. The specific heat of the ice is 0.50
Kcal/kgK, ans its latent heat is 80 Kcal/kg. The energy needed to melt a half of the mass of the iceberg is:
a) 8105 Kcal
b) 1.6105 Kcal
c) 82.5104 Kcal
d) 25105 Kcal
e) None of the above
P5. About the picture on the right, W is work, and about the temperatures, Ta > Tb. The
energies transferred are Qa = 60 KJ and Qb= 50 kJ. Using those data, we can say that
the picture is a representation of:
Ta
Qa
a) A heat engine, whose yield is 20%
W
b) A refrigerator, whose COP is 5.
Qb
c) A heat engine, whose yield is 6/5.
d) A refrigerator, whose COP is 5/6.
Tb
e) None of the above
P6. A system absorbs 300 Kcal from a source at 300 K during a reversible process. The variation of entropy of the universe,
once the process is completed, is:
a) Zero
b) +1 Kcal/K
c) -1 Kcal/K
d) +1 KJ/K
e) None of the above
19
A
PHYSICS FOR ENGINEERS. 2006-2007. EXAM A4.
PROBLEM
10 mol of an ideal gas (initial temperatue 20 ºC) expand reversibly following the law p = aV,
where p, V are pressure and volume, whereas a is a constant. The final volume after the
expansion is twice the initial one. Gasses constant R = 8.314 J/(Kmol).
1. Plot the expansion process in a p-V diagram (1 p).
2. Find the final temperature (1.5 p).
3. Find the work in the expansion (1.5 p).
We know T1 = 20º C = 293 K
We must find a relationship to calculate T2
from T1 and the volume rate V2/V1.
p
2
p  aV
p1  aV1
T1 
p2  aV2
T2 
1
p1 V1

p2 V2
V
V1
V2=2V1
W

V1
V2
pdV 

 
a
a V dV  V 2
2
V1
W
W
V2
V1

(in this case m = 2)
V2  mV1
p2V2
nR
2
2
T1
p1V1  V1   1 
     
T2  m 2 T1

T2 p2V2  V2   m 
T2  m2 T1  22  293  1172 K  899º C
Expansion work:
V2
p1V1
nR







nR 2
a 2
a
m  1 T1
V2  V12  m 2  1 V12 
2
2
2
10  8.314 2
2  1  293  36540 J
2
nRT1
V 
a
2
1
aV12
T1 
nR
20