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Page 1 of 10 PHYS 100 Final Exam, April 13, 2007 Version A Last Name__________________ First Name__________________ Student ID__________________ Signature____________________________ INSTRUCTIONS ***READ CAREFULLY*** Students must have their photo ID clearly visible on the desk during the test. A simple Scientific Calculator (Model SC-6108) is allowed. Attempt all questions. Use 9.8 m/s 2 for the gravitational acceleration (g) For part 2, show your work in the exam booklet. Marks Maximum Part 1 20 Part 2 27 Total 47 Bonus 3

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PHYS 100 Final Exam, April 13, 2007 Version A Last Name__________________ First Name__________________ Student ID__________________ Signature____________________________ INSTRUCTIONS ***READ CAREFULLY***

• Students must have their photo ID clearly visible on the desk during the test. • A simple Scientific Calculator (Model SC-6108) is allowed. • Attempt all questions. • Use 9.8 m/s2 for the gravitational acceleration (g) • For part 2, show your work in the exam booklet.

Marks Maximum

Part 1 20

Part 2

27

Total 47 Bonus 3

Page 2 of 10

Part 1: Multiple Choice (20 questions, 1 mark each) Please circle one answer only.

Please refer to the following graph when answering questions 1-3:

I

1,

J

r--+-~~---f--l

~+-'-+-----f--~~=-L ~-+- 1--1

_______J.

5

4

3

2 ..... E 1-­><

0

-1

-2

-3

tIs)

1. A particle is moving along the x-axis. The position of the particle, as a function of time, is shown in the figure above. The average velocity of the particle in time interval

betweenl.0s and 7.0s is A) 2.0 m/s -I - "l :: y ~'lhq.B) 1.0 m/s 7.0 - ,.0C) 15 m/s D) 0.5 m/s ::. - o. ~- WI/;

(§) -05 m/s

2 The average speed of the particle in time interval between 1.0s and 7.0s is A) 2.0 m/s (i3)) 1.5 m/s z + (: + \ ~ 1.0 m/s S ::. --.d.-::

a.v L> AD) 05 m/s E) -0.5 m/s

3. The average acceleration of the particle in time interval between 5.0s and 9.0s is (A)) 1.0 m/s2

)

13) -3.0 m/s2 6 V I - (- ~ _ 4- I > C) -1.0m/s2 Oo.v:: -::. O_!- - It = i 1((\/5 D) -0.5 m/s2 6 1; 7

E) 0.5 m/s2

Page 3 of 10

4. The figure shows the position graph of a car traveling on a straight road as a function of time. At which labeled instant is the speed of the car greatest?

A) A

~~ 'bro E) E .r

D E

c B

5. Ball 1 is thrown straight up in the air and, at the same instant, ball 2 is released from rest and allowed to fall. Which velocity graph (vy versus t) in the figure below best represents the motion of the two balls?

~~ C)C 0) 0 E) None of above (draw your own figure).

Vv Vy

A 1

B 1

2

t

2

t

c D 1

t t

22

6. A boy kicks a football from ground level with an initial velocity of 20 m/s at an angle of 30° above the horizontal. How long does it take for the football to reach its maximum height? Ignore the air resistance.

( e, pro

• 20 -I, . o·r- - {o ~ Page 4 of 10 Va S"~ 30 '" V~o = 6 ..

E- A) 0.67 s - ~j .::UB) 1.50 s V'J = 117° C) 10.0 s -I,D) 17.3 s V;l0 I °

{ " ~ /. o~ [email protected] " 'J 'I. f ""15~

7. A tennis racket exerts an average force of 24 N on a 55 g tennis ball to change its velocity from 15 m/s north to 25 m/s south. Over what period of time does this change of momentum tak~lace?

® 0.092 s ~ 64,}B) 0.023 s A f = ;>

C) 92s = I"'" (II" - (-1.\') F'ot ~ of D) 23 s E) 10.9s - "'" '(4-0) ~

:: C. 0 rr- X <j.o " l· 2. fr:J' "'/j . ot ::- ~.:. 'Z;~.v 8. A child and a sled with a combined mass of 45.0 kg slide down a frictionless hill. If fue sled starts from rest and has acquired a speed of 23.0 m/s by the time it reaches the bottom of the hill, what is the height of the hill?

A) 54.0 m I I V .. B) 1215m ..... '" "-~ ... C) 2.35 m

([j} 27.0 m ~ 1.17 m z. • ;. g

9. A person of weight w is in an eievator going up when the cable suddenly breaks. What is the person's apparent weight after the cable breaks?

A) 9.8w B) greater than w

ce1)0 D) less than w E) -9.8w

Page 5 of 10

v. 11. A race car undergoes uniform circular motion around a banked race track with a radius of 92m. The track is banked at an angle of 45 degrees towards the centre of the track is indicated below. At what speed will the force of friction on the car's tires equal zero?

A) 10m/s '\ () v"'::- f2 J .t,,,,, 8 B) 15m/s Ncd1(J " W'd C) 20m/s

V=~(D) 30m/s tJ = ..:..a. 10 40m/s t<fl-(J ::.~.&'

rJ· 5;'" 0 -=- ...Y­12­

Wl 6·to.. .... 19 "'" 1z-....

12. A spring being compressed by x meters has a potential energy Vo. When the compression is doubled to 2x meters, the potential energy stored in the spring is

13. An iron block with mass lIlB slides down a frictionless hill of height H. At the base of the hill,

f;. it collides with and slicks to a magnet with mass lIlM . What is the speed vof the block and

magnet immediately after the collision?

A) lIl B+ lIlM ~2gH lIlB

® lIlB ~2gH VB -= bdl-l

mfl+1nM

':'C) lIl B ~2gH Me V/J (""'/""f"'I) . V m,l,f

WlI3 lIl B +lIlM 2gH 1}-= V~D)

lIl B WI!} oj. 1'1,..,

lIl ""'IJ~E) B2gH lIlM ""'G -\ M; hblt

14. A baseball of mass 0.15 kg moving at 20.0 m/s strikes the glove of a catcher. The

G . glove recoils a distance of 5.0 em. The magnitude of the average force applied by the ball on the glove is

667N w = .t1\lE

o· I

A) avo B)O C) 2Vo

):>.l.. Vo

®4Vo

j. 0'/)'7 •

~B 600 N ) 60 N

0) 3 N E)0.15N

o·ot'

Page 6 of 10

15. A car with a mass of 1000 kg and a speed of 20 m/s heading north approaches an intersection. At the same time, a minivan with a mass of 1500 kg and a speed of 10 m/s heading east is also approaching the intersection. The car and the minivan collide and stick together. The speed of the wrecked vehicles just after the collision is: It ....U

A) 15 m/s. oJ..>'" ll"0 ••0 " l-<J, ~ ~ 1

B) 14 m/s r:: f' H1. 1~o~ ~

D) 20 m/s f " J\,,'1 po" r'r~l f-= l-OVtIo'"E) 25 m/s - • ~

'2 S •o'ifll v~ . f'" ... S - "2. ~ <!TO V -= l..----''-3'." () """" .- . . -..., .f'"'" : 10"/1 . '

16. The speed of light in a specific material is measured to be 2x 1OBm/s. Therefore the index of refraction (n) of the media is:

A) 1.0 C.B) 0.5 C 11\ .,. - =r· s- .(j:=- V(C))1.5 m2.0 '" E) 1.25

17. The average sun-earth distance is 1.5x10"m . How many light years is this? A) 25 x 10'. /} I r- WI

@j)1.6xlO,5 I LJ '" prf} ~ ]brt ..... 7--'!-x ~'".f~ i.lfb XIO

C)6.8x 10·5

,I D)4x lO'15 r.~ 1I10 -j

E) 8x 10'o t/..t' r :: I- r 8 flO '! 7' ~ '0

18. Light traveling from a media of n=1.8 travels to air. What ;s the value of the critical angle A where total internal reflection takes place? (A!J33.7 degrees. 1J)10.0 degrees. C) 42 degrees.

, lJ.7 b

-... = • D) 90 degrees. E) There is no critical ang Ie for this situation.

A 19. A '~. to Glass (n=1.50). The ray will ht ray propagates from Water (n=1.33) A) efract towards the normal.

refract at the same angle as it incidents. C) not refract. D) refract away from the normal. E) none of the above.

20. The magnitude of electric force between two charges is F. If the distance between the two y charges is doubled, what is the magnitude of the electric force?

. A) 2F B)4F -,1._QF/2

®F/4 E) F

Page 7 of 10

Part 2: Written questions (9 questions, 3 marks each). Please show your work in the exam booklet. Please show complete solutions and explain your reasoning, stating any principles that you have used. 21. A swimmer heads directly across a river, swimming at 2.00 m/s relative to still water. He arrives at a point 60.0 m downstream from the point directly across the river, which is 80 m wide. a) What is the speed of the river current? b) What is the swimmer's speed relative to the shore? c) In what direction should the swimmer head so as to arrive at the point directly opposite his starting point? Express your answer in terms of the angle with respect to a line drawn directly across the river. Indicate the angle in a schematic diagram. 22. A car traveling at a constant speed of 108 km/hr passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets in a chase after the car with a constant acceleration of 6.0 m/s2. How far does the trooper travel before he overtakes the speeding car? 23. A ball is thrown from the roof of a building at a 50° angle above the horizontal with a speed of 15 m/s. 4.0 seconds later, the ball hits the ground. Ignore the air resistance.

A) Determine height of the building. B) Find the horizontal distance x.

50°

x

24. A cart of mass 0.12 kg moving to the right with a velocity 0.25 m/s collides elastically with a cart of mass 0.36 kg moving to the left with a velocity of -0.25 m/s. What are the final velocities of the two carts?

m 3m

v -v

Page 8 of 10

25. A block of mass 4.00 kg moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in the figure. a)How much work is done by the force as the block moves from the origin to 8.0 m? b) If the block has a velocity of 1.2 m/s at the origin, what is the velocity of the block at +8.0 m?

-15

-10

-5

0

5

10

15

0 1 2 3 4 5 6 7 8

Position (m)

Forc

e (N

)

26. An object of mass 10.0kg is on a frictionless floor. Two forces are acting on the object. The first force, of 110 N, acts at an angle of 30° upward from the horizontal to the right. The second force, of 75.0 N, acts at an angle of 20° upward from the horizontal to the left.

a) Draw a free body diagram for the object. b) What is the acceleration of the object? c) If the mass is changed to 4.0kg, what will be the acceleration?

20° 30°

110N 75N

Page 9 of 10

27. As shown in the figure below, the two boxes have identical masses of 40 kg. The angle of inclination is θ=35°. Ignore the mass of the tie cord and all the frictional forces. (a) Draw free body diagrams for each of the two masses and write down Newton’s second law for each diagram in component form. (b) Find the acceleration of the boxes and the tension in the tie cord.

θ

28. Three charges are aligned as shown in the figure below.

a) Calculate the MAGNITUDE and DIRECTION of the force exerted on q3 by q1 and q2. b) Assume you can add a NEGATIVE charge (q4) to the above system. Draw the the position of this added charge in the diagram provided that would result in q3 being in static equilibrium from the forces of q1, q2 and q4. Include the NET force on q3 from part a) in your diagram and quickly state why you chose the position you picked. NO CALCULATIONS NEEDED FOR THIS PART, JUST APPROXIMATE THE POSITION.

Page 10 of 10

29. A light ray travels from glass to diamond and then to air. The angle of incident (Ө) is 62.5 degrees. Using the law of reflection and refraction determine the angles of reflection and refraction at both interfaces and draw the rays (if any) on the diagram provided below.

30 (Bonus question, up to 3 marks). A block of mass m=1.8 kg on a table is pushed against a spring that has a force constant of k=250 N/m, compressing it 20 cm (the block is not attached to the spring). The block is then released, and the spring pushes the block to move to the right. The coefficient of kinetic friction between the block and the table is 0.45. (a) Determine the distance the block will travel before it stops. (b) Sketch the F~x curve, i.e., the net force acting on the block as a function of the block’s position. Choose the equilibrium position as x=0. (c) Find the maximum speed of the block.

m

0 x

....i ~

~ V t- VitJG.- 5vJ~I. VSG.

{./J ..,.

()o..) . VIJJ~ be ... Vsw

:: ~

T = -~.. If. VWb "

'.t80'" ...>

LK'lI W6. =- { V~ LV :::1 (2...0) :: (. ~ ..../f ...>

~ VS&'t

~ .• '1~J80 ItII 6° ....O~~ k-:::: :: •

VwGo,1/sw

60 ...VWq . VSW :: /. \ ..../~ -8"''''''

':; ~ - .:= L.S- ....~j 11. 2- 72" +(.J' L

b) VSG + f/ w("Sw "

c.) .

I----------::;r====------- II.s :o} 0 ...fr

;f (f)tl 2- 3 r 7 K~ " ,,( ... ) ~=1/sj~JoJ-

100

o:I~~-:..~:-J-~4-:r:--;' -:,-~g-'7.L--( ~L..-.J.II---'-I"'1_-'4-3---+(5 )

{ D·H '. r"f·f.f'l( s '" "j; r (2..t.~ t ::- ---- :::

Jot.::: J (*-1) 1 II' ~ f. - ~"~wp..-

Co.t ::. e..-U+ I ':\1"'''' :5(f1,~_I)I..,::~>I"'"

t 2..._ r~*-+ \ ~O

( 0 Il ,

--t-'"::l3 _

'"I' Q ,,0 •

~v :: H

Vy.~ :: 1/" t.rJ1.{)

::. (r "'II$ . (vz.J1>" ~o = H

::: 1. 6 tJ -/$ .

lIy" ." 1/0 ~: ... e '-a ~

=: ('(' ..../~ .::; ,.... )

(;<'" =0 .

::. I',.r -j; . () '1;; - 3 ~ -1- f ~;J~

"/.. ~ ...V'(d = 1· (,'1- .(t)

~ =- ~ I> -J. 1/;Q ..t - ~ J ;( 1. c:=- H + (1_\ O) - 4. 1 .:t L .

H"=' 4. ~ 1- /6 - / /. r Xr.,., =F . If WI

b)

A, , ,"V, + J V'Z.

V1- -::. - '\/, .

{

o -::: 6 tI,'lIz,' + {,1I·l1­

o ::: & h' (11,' .{ 1/./ )

o ~ 111.' ('fI,' + v/ )

{; ,,' ::: 0 ,

,I r ,v, + 1/1. :00 . (. e -1!L

1/... / ::: (/'1- =- - o· "r ....1$

VII ::. - V.' ::. 0 .'2-S'" '"vIJvr'=- -D.) ""-Is

( h>/hf?vf - }It> (0 1I"lilA~) .. (- 0 V"J, ­

0.) .

::: (7-1-)) (rN'x 1m)

- r-·r X ~ (J)

fir ~) '2. VII ,1/ .1 i l;t-tv(

~ J2 (~.r)

-.: j I7. 7r 7- f. If.~

....I

N ~ ..... F; F.

~~0.c-..... =+:", ) ~" • . ' .... ~)

F, (<lL jo 0 _ F.. (dl.- zo ° '" r;1 ,;::1.)(

F, c<11 J0 °- F.. {<-'/. 2.0 "

a 'X =. m :::: '2.. Cl-tf ~ •

~ -{o-p : Fi 0 I ( Sf"]O -I- F.. ~, ... ],O~ .. N - d = ,." ~

fAJ -= -j;, (F, S," 10°f fi s,.. u)°-t-tV--J)

= -f;; [ tV - 17· 3 (N) )

(

=0 5 ,' .. (~ 11,(' f lo<rr ('" ... fl'tl V/,(e.) {;\ "'01'.." ... 1 rre of 17·3 N.

"",0' dl to? ?(- t/tl-l'r1t,;.,

c). {f WI :: '1.0 leJ .

•• ~

~d ':: -d; (rror';"}()·+7S- S,·.. 20·-fN- fI.~,,~J

~~( (lfl,rN-ftl)4<0 f<J

tf!·r t.t -:::: (0.1f- ...!s~ 4: 0 ftJ

...:.

tJ,

..... . w1" T, -fx

~~'J -:. .....

Nl.T2. ,

' . ....e ~~

If, WI~-S

VI" ,: ')(- (0"'\' : T::.M,o..

(V, - WI'J = 0

"'1- (~"" f : W/1.J S,'", B - T =' 1t'I.0.

~- {~"'r: IV2 - WllJ (11Z () =- 0

}'YI.,. J 5'r'", 8 = /1'1, (). + W1z Ii\ .

fl\ -= Wh-J s .... ~ -;:: 7· 9 .S, 'M iF () := 2, (1 ".~ 2

W1.+IVI1. 2.

T-= m, 0. = q..o (~J . 2.,f ~.. = 1/2 /1/

J ~OI, .

/ I r..

~ 8. GI}. .... ....

F:> Fn ' -J.

f1.3 "I ..J.

F~

f)i

-:=

~

F"

F; l

s Ivd)

(62 {j

- G.,

-... 11. ~ !J'.!.l-HO . 1- ::

1;"'"//

/

/

,/

; .... .... F,}

== S'l.f-1ro-~. ~_ - ~lIro-~

:= -)-.76 -3

ilo N

Ff}

,~

,••

b). (Vtje-iiVf C~(j\l)~ t lJ.

sh\iVl.(J bt' rlAt d.1~,,'J

fJ -' l'fA .x I t'. t. ,.VI - F­ oUrt ( (I'"

2J­t",tfVf(~ (D.. f,(MS -) 9""lNt o"J,

Of::' ts- 11

&t -:: P\' :: t s- ~ (r.. fI.., ...1( ~~ )

19 + =. ~V' )11' "f rt-fl?- ~ 7L~~

('). s'-", t~ :: 2·r ,5,'., Of

S' 10 ('I.", \l{".o H,)','.. tfO =oSCt-""ji

&t- :: ~"I'I-' o.r'-f.J8 :: H 0 (im..( to~ )

t",teyfo-~ 0: 1)"Ii\W1~..,J ~ C/lr- ,

u. ...)t ~ rt-t!-eJcJr; (/r =f),' ~ 33°

TI,,'f~" ,'s no Y'e{r-,,(r(~.,

(Tof... l ,\,t~~,,-I 'A!f!'t',~~,,),

"30 .

b)

-IJ .2.

1i. 0'2.­ 0.&

F~-~x-f F :: - f =- 7· '!if- N

-= - h -].'11+- .

( F:o-o ")

'J.o '=. -.i.. ::. ],9 4 :=-0.0 p... WI

-f.. - '2,. ro

Wfltf := AH'OI (,<t\d-er F-.,r; ll.LYld. -fen'tt. X(.X.

-= i (rf.~-~)(-{'f.l· -.{)=i (cJ.16g)(zS'O~O .. Z -7·~V-)=3.s3;

IAJ htt '=:: J. ...... Vt ~

..... (JS3)

( .. ~