1 PHYSICS FOR ENGINEERS. A-EXAMS 2006-2007. A1A2A3A4.

1

PHYSICS FOR ENGINEERS.

A-EXAMS 2006-2007.

A1 A2 A3 A4

2

A1A 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

1. a)

b)

c)

d)

e) None of above

33

32

22

23

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:

2. a)

b)

c)

d)

e) None of above

m 00.2

m .500

m .251

m 40.2

3. The angle formed by the vectors jiA

jiB

22 and is:

a)

b)

c)

d)

e) None of above

º45

rad 6

º90

rad 6

“An object at rest stays at rest unless acted on by an external force”. This is:

4.a)

b)

c)

d)

e) None of above

Newton’s third law

Law of inertia

Law of motion along a curved path

The work-energy theorem

A punctual mass follows a circular trajectory with constant speed. As for its acceleration, it is true that

a)

b)

c)

d)

e) None of above

The punctual mass is not accelerated

Its tangential acceleration is positive

The modulus of its normal acceleration is constant

There is no normal acceleration5.

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

6. a)

b)

c)

d)

e) None of above

m/skg 1011.000.1 2- m/skg 0.01.10

m/skg 0.01.110

m/skg 100.110.1 4-

AGR

FOR

3

A1A spring initially at rest, attached to a mass m, is stretched to distance x. As for the mass m, it is true that:

7.

a)

b)

c)

d)

e) None of above

The potential energy of m only varies in case the mass hangs vertically on the spring

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.

The potential energy of m never changes, only its kinetic energy undergoes some variation.

Neither the potential energy of m nor its kinetic energy undergo any variation.

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:

m1

m2

F0

Assuming the force F0 is big enough to move the set of two blocks, find its acceleration.

b)

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.

1 = 0.075 2 = 0.040 m1 = 8 kgF0 = 2,50 kp m2 = 6 kg

Separately draw the free body diagram for the set of two blocks, for m1 and for m2. a)

PROBLEM

GRADING:

PROBLEM: 6 POINTS

QUESTIONS: 4 POINTS

EACH CORRECT ANSWER: +0.500

EACH WRONG ANSWER: -0.125

A harmonic oscillator obeys the equation ty 90.9cos 5 where

every quantity is given in S.I. units. The frequency and the period are:

8. 4.5 Hz and 0.22 s a) 0.5 Hz and 2.0 s b)

6.3 Hz and 0.16 s c) 5 Hz and 9.90 s d)

None of abovee)

4

A1A 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

1. a)

b)

c)

d)

e) None of above

33

32

22

23

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:

2. a)

b)

c)

d)

e) None of above

m 00.2

m .500

m .251

m 40.2

3. The angle formed by the vectors jiA

jiB

22 and is:

a)

b)

c)

d)

e) None of above

º45

rad 6

º90

rad 6

“An object at rest stays at rest unless acted on by an external force”. This is:

4.a)

b)

c)

d)

e) None of above

Newton’s third law

Law of inertia

Law of motion along a curved path

The work-energy theorem

A punctual mass follows a circular trajectory with constant speed. As for its acceleration, it is true that

a)

b)

c)

d)

e) None of above

The punctual mass is not accelerated

Its tangential acceleration is positive

The modulus of its normal acceleration is constant

There is no normal acceleration5.

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

6. a)

b)

c)

d)

e) None of above

m/skg 1011.000.1 2- m/skg 0.01.10

m/skg 0.01.110

m/skg 100.110.1 4-

PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A1.

SOLUTION

5

A1A spring initially at rest, attached to a mass m, is stretched to distance x. As for the mass m, it is true that:

7.

a)

b)

c)

d)

e) None of above

The potential energy of m only varies in case the mass hangs vertically on the spring

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.

The potential energy of m never changes, only its kinetic energy undergoes some variation.

Neither the potential energy of m nor its kinetic energy undergo any variation.

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:

m1

m2

F0

Assuming the force F0 is big enough to move the set of two blocks, find its acceleration.

b)

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.

1 = 0.075 2 = 0.040 m1 = 8 kgF0 = 2,50 kp m2 = 6 kg

Separately draw the free body diagram for the set of two blocks, for m1 and for m2. a)

PROBLEM

GRADING:

PROBLEM: 6 POINTS

QUESTIONS: 4 POINTS

EACH CORRECT ANSWER: +0.500

EACH WRONG ANSWER: -0.125

A harmonic oscillator obeys the equation ty 90.9cos 5 where

every quantity is given in S.I. units. The frequency and the period are:

8. 4.5 Hz and 0.22 s a) 0.5 Hz and 2.0 s b)

6.3 Hz and 0.16 s c) 5 Hz and 9.90 s d)

None of above

SOLUTION(CONTINUED)

e)

6

A1m1 m2

F0

F21

FR1

gmNFR 11111

gmNFR 22222

F0

F12

FR2

N2

FR1 FR2

ammFFF RR 21210

N1 N2

amm 21 Newton’s 2nd law

21

22110

mm

gmgmFa

Friction forces

Newton’s 2nd law

amFFF R 11210

amFFF R 11021

1 = 0.075 2 = 0.040 m1 = 8 kgF0 = 2,50 kp m2 = 6 kg

Numerical result a = 1.162 m/s2

N 324.91110 amgmF

Newton’s 2nd law

amFF R 2212

amFF R 2212 N 324.9222 amgm

am1

am2

F12 and F21 have to be equal (action and reaction)

N1

SOLUTION(CONTINUED)

7

A2a) 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.

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:

a)

b)

c)

d)

e) None of the aboveX

Y

Z222

2

1

2

1 MRIMRIMRI XXYYZZ

222

4

1

4

1

2

1MRIMRIMRI XXYYZZ

222

2

1

2

1

4

1MRIMRIMRI XXYYZZ

222

2

1

2

3 MRIMRIMRI XXYYZZ

P4. The variation of the angular momentum with time quantifies:

a) The sum of external forces

b) The sum of the momenta of inertia

c) The sum of the torques of external forces

d) The variation of the angular velocity

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 . To invalidate a previously marked cell, use the symbol

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:

X X

For questions P1 to P5, each correct answer adds +1; each wrong answer takes off -0.25

PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A2E

QUESTIONS AGR

FOR

SURNAMES

FIRST NAME

8

A2P5. A torque of 10 Nm 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

d) 45º

e) 180º

a b c d e

P1

P2

P3

P4

P5

ANSWER FORM

A B

LL

A B

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)

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).

b) The moment of inertia about end A (1.5 p)

PROBLEM

9

A2A B

0xLL

A B

CMx

mg

AT BT

a) Both sections are homogeneous and have the same lenght, so the position of both center of masses will be

2/LxA 2/3LxB

Center of mass of the whole rod:

Mass of each section:Lm AA

Lm BB

BA

BBAACM mm

xmxmx

BA

BACM mm

LmLmx

2/32/

LL

LLLLx

BA

BACM

2/32/

2

3 Lx

BA

BACM

Total mass: Lm BA

Y

02 LTxmg BCMA L

xmgT CM

B 2

0 BAy TTmgF

L

xmg CM

21 BA TmgT

BA

BACM

L

x

3

4

1

2

BA

BABAB gLT

3

4

1 gLT BAB 3

4

1

BA

BABAA gLT

3

4

11 gLT BAA 3

4

1

PROBLEM SOLUTION

10

A2A

LL

A B

0x

A B

L

L

B

L

A

xx2

3

0

3

3

3

333 8

3

1

3

1LLL BA

73

3

BAA

LI

mg

CMx

90

90sin ACMA Imgx

'AT

cos

A

CM

I

mgx

Lm BA

2

3 Lx

BA

BACM

cos

2

3

7

33

gLL

L BABA

BA

BA

L

g

BA

BA

2

cos3

7

3

b) Moment of inertia about A

mg

dx

x

dxdm

L

L

B

L

AA dmxdmxI

2

2

0

2

L

L

B

L

A dxxdxx

2

2

0

2

c) Angular acceleration and angular velocity

PROBLEM SOLUTION

11

A3The 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.

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)

S1

h

1 2S2

H

0c

P

Figure 1

Figure 2

PROBLEM

S 1 (cm2) = 350

S 2 (cm2) = 180h (m) = 1

M (kg/min) = 4500c 0 (m/s) = 9,6

H (m) = 5

Gravity acceleration g = 9.8 m/s2; density of water = 1.00 g/cm3

PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E.

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 . To invalidate a previously marked cell, use the symbolX X


E

12

A3Bernoulli’s law

S1

y1

h

1 2

1211 2

1gycP

S2

y2

2222 2

1gycP

c1c2

z2

z1

11 gzPP atm

22 gzPP atm

21

2221 2

1ccPP

2121 zzgPP

ghPP 21

1

221 S

Scc

21

222

22221 2

1

S

SccPP

Continuity equation

gh

212

21

2

SS

ghc

212

221

2

SS

ghSVV

212

221

2

SS

ghSmm

PROBLEM SOLUTION

13

A3H

0c

P

S

0

SSS yc

gg

P 2

2

1

020

0

2

1yc

gg

P

PH

Sy

0y0

SP yycg

H 0202

1

PHgmW

PHgdt

dm

dt

dWW mHgW P

kg/s 60

1

(s) 60

(kg) M

Mm PHgMW

60

1

Bernoulli’s LawPressures in S and 0 are the same The surface of the water is at rest

H

Hcg

H P 202

1

Work done by the pump raising a water mass m to a height H

Power:

The pump raises M kg a minute:

Figure 2

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)

14

A3 (kg/m3) = 1000 1000 1000 1000g (m/s2) = 9,8 9,8 9,8 9,8

h (m) = 0,25 1 0,5 0,4S1 (m

2) = 3,50E-02 3,50E-02 2,80E-02 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) = 5,39E-02 9,29E-02 5,56E-02 8,00E-02m (kg/s) = 53,95 92,92 55,61 80,02

g (m/s2) = 9,8 9,8 9,8 9,8M (kg/min) = 3000 4500 5000 800

c0 (m/s) = 8 9,6 10 5

H (m) = 4 5 2 8

HP (m) = 7,27 9,70 7,10 9,28

W (watt) = 3560,00 7131,00 5800,00 1212,00

MODEL E

PROBLEM SOLUTION (CONTINUED)

15

A3PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E. E

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 e) None of aboved) 3.45 kp/cm2

1 2 3

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:

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

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.

a) We cannot say which glass is more weighted without knowing the weights of the ice and the piece of wood.

C

1 m25 m2

1 m0.5 m

1 m

A

B

CP

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.


QUESTIONS

16

A3F1

F2

F3

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:

b) The three velocity schemes are characteristic of turbulent flow

a) F1 represents laminar flow and F3 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

A

B C

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:

e) The P5 heading is wrong, because it is impossible that a siphon setup such as the depicted one could work.

a) It is necesary that h << d.

b) It is necesary that d.= 0

d) The pressure in A must be smaller than the pressure in B.

c) That siphon only works if the liquid is less dense than the water.

a b c d e

P1

P2

P3

P4

P5

ANSWER FORM

PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E. E

QUESTIONS

17

A3gaugeatmabs PPP

mb 920atmP bar 920.0 bar

Pa 105 Pa 92000

2kp/cm 0.2gaugeP 2-4 m N/10 8.90.2 Pa 196000

A kp/cm2 mb bar PaPatm 0,9 882 0,882 88200Pgauge 2 1960 1,96 196000

Pabs 2,9 2842 2,842 284200

B kp/cm2 mb bar PaPatm 1 980 0,98 98000Pgauge 2,2 2156 2,156 215600

Pabs 3,2 3136 3,136 313600

C kp/cm2 mb bar PaPatm 0,95 931 0,931 93100Pgauge 2,1 2058 2,058 205800

Pabs 3,05 2989 2,989 298900

D kp/cm2 mb bar PaPatm 1,05 1029 1,029 102900Pgauge 2,4 2352 2,352 235200

Pabs 3,45 3381 3,381 338100

QUESTION 1 SOLUTIONEvery 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.

QUESTION 2 SOLUTION

QUESTION 3 SOLUTION

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.

18

A410 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/(Kmol).

1. Plot the expansion process in a p-V diagram (1 p).

PROBLEM

A

2. Find the final temperature (1.5 p).

3. Find the work in the expansion (1.5 p).

P1. About the phase diagram depicted aside, it can be said that:

a) 1 is the critical point; 2 is the triple point; AB is a sublimation process.

b) 1 is the triple point; 2 is the boiling point; AB is a melting process.

c) 1 is the melting point; 2 is the boiling point; AB is an evaporation process.

d) 1 is the triple point; 2 is the critical point; AB is a sublimation process.


QUESTIONS

T

P

T

P

1

2

AB

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 e) Zerod) +200 KJ

ANSWER FORM

a b c d eP1P2P3P4P5P6

PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A4E.

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 . To invalidate a previously marked cell, use the symbolX X


19

A4QUESTIONS

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.


P3. An ideal gas complete a thermodynamics cycle composed by two adiabatics, one isochoric and one isobar. Indicate the correct option.

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/kgK, ans its latent heat is 80 Kcal/kg. The energy needed to melt a half of the mass of the iceberg is:

a) 8105 Kcal b) 1.6105 Kcal c) 82.5104 Kcal d) 25105 Kcal e) None of the above

aQ

bQ

W

aT

bT

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:

a) A heat engine, whose yield is 20%

b) A refrigerator, whose COP is 5.

c) A heat engine, whose yield is 6/5.

d) A refrigerator, whose COP is 5/6.

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 e) None of the aboved) +1 KJ/K

APHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A4E.

20

A4PROBLEM

PHYSICS FOR ENGINEERS. 2006-2007. EXAM A4. A

p

V

1

2

V1 V2=2V1

aVp 11 aVp

22 aVp

nR

VpT 11

1

nR

VpT 22

2

We know T1 = 20º C = 293 K

We must find a relationship to calculate T2 from T1 and the volume rate V2/V1.

22

11

2

1

Vp

Vp

T

T

2

1

2

1

V

V

p

p

2

2

1

V

V

12 mVV

21

m1

22 TmT

(in this case m = 2)

C º899K 11729322 21

22 TmT

Expansion work:

2

1

2

1

2

1

2 2

V

V

V

V

V

V

Va

dVVapdVW 21

221

22 1

2

2Vm

aVV

aW

nR

aVT

21

1 a

nRTV 12

1

12 1

2Tm

nR

J 36540932 12 2

314.810 2

W

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/(Kmol).

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).

1 PHYSICS FOR ENGINEERS. A-EXAMS 2006-2007. A1A2A3A4.

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