This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1. A 5 kg block slides down an inclined plane 6 m high. What is its speed at the bottom of the plane? (Assume friction is negligible) (cgaruc'hm 6" ska3
\(E‘l 1 9 5‘ =' KE.L*9‘UEL ‘4 ‘ yykgL ~D
6m Odwo‘s “1:1351“ V: FShQE‘Mzd—SO u 1 \\ MKS A)6m/s b)10m/s D)12.5m/s E)7.7m/s 2. If friction is negligible the block will slide up a second inclined plane
to the same height (6 m). With friction present it slides up only to a
"h M we E“ . . .  r), ASA“. WSQUQ 9‘“
height of 4 m. How much energy 1s lost to £1‘1Ct10n.>u“l‘l> ‘ 5.0 0) (£40 a maﬁa "‘5 t ‘ "ﬁlm"? Wkunm UJC ‘ “8 C
A) 50 J B) 37.5 J D) 60 J D) 100 J E) 75 J
3. Suppose that in Problem 1 it is a thin hoop of the same mass rolling down the inclined plane. The speed of the hoop at the bottom of the
O 1 ' The s sbém‘MkV‘“M‘—“M m
p ane ls ngﬁ “30 Vbt‘hmﬂ, V:— «w («ﬁeJ k; A) Lar er than that of the block in Problem 1.
: Smaller than that of the block in Problem 1.
C) The same as that of t e block in Problem 1. D) Cannot be determined because we don’t know the radius of the hoop
E) Cannot be determined because we don’t know the angle of the plane. 4. A 10 kg wagon is rolling along with a constant speed of 4 m/s. A 5 kg
bag of 8 nd is dropped onto the wagon. Its speed after the bag lands is 4%
v:
I Q .M .4 manualuh M. U. : ML UL. moi) =$V
A) 2.67 m/s ‘ B) 3.33 m/s C) 6 m/s D) 2 m/s E) 7 m/s tQQ‘M/
J
5. Is the kinetic energy after the collision E
A) The same as the kinetic energy before the collision. “a ‘C’ “0‘5 1
B) Larger than the kinetic energy before the collision. < XMOLrium'gz‘d
C) Smaller than the kinetic energy before the collision. 2m
m “Mirham (a M Sam C casescell.) x9. iﬂ} Mm . \ soKECosulie..P A pool cue ball (1, on the left) traveling along the X direction at 2 m/s
collides elastically with a stationary object ball (2, in the center of the figure). The cue ball is deﬂected by 30 degrees and has a final
velocity of v1=1.73 m/s. Therefore its velocity components are v1X=+1.50 m/s and v1y=+0.87 m/s. Comséaw‘km immed—km ‘ ‘09. Re am— 90
Y m " ‘waekau coax, £an
L V1 \jQAOCK \%
X V1, : O WWW+MUH :0
I
 .V
.——* 91 “ ’M’ “cr‘ If»?
. V‘w‘ \Itﬁeﬁ‘ “90“
= 087
62 V2,“ = “N/‘g V2 6. The y component of the velocity of the ball (2;) after the collision is A) V2y=0.25 m/s. B) v2y=0.43 m/s. C) v2y=O.50 m/s. D) v2y=—0.66 m/s. (E) v2y=—0.87 m/s. ) 7. The Xcomponent of the velocity of the object ball, after the collision is ' s V yk  .. V\C¢ 30 + U
9M5 % l} + U3} 9 LY V t 30: 23"
A) m =2m/s B) m :15 m/s @D)vzx =0.75 m/s E) Can’t be determined because we don’t know the angle at which the
object ball goes off. 8. A 30 kg block is being pulled along a surface as by a force of 100 N
as shown in the diagram. There is a force of friction acting with a
coefficient of friction u = 0.1. The block is pulled for a distance of 20 m.
The kinetic energy of the block after 20 m is (This problem is harder
than it might seem at first glance. You will need to calculate the net force on the block.) Cm“ 2" \Oou LasCO — Ear A)2000] B)SOOJ C)400J D)600'J E)1000J pErchN 100N \ 1009‘“) '1 M3» 7 = 2am) ~100¥~6Z>
¥\‘ .7 1200 20‘“ gr c. AF = max?
—' «A3
Work = {214 ‘ 29.7 no Fm=£0 33:22.7 EVLOOJ’ A bicycle wheel starts from rest and reached a speed of 6 m/s in 12 s. It
then travels at constant speed. The radius of the bicycle wheel is 0.35 m 9. What is the angular acceleration of the bicycle wheel? A) 0.5 rad/s2 B) 2 rad/s2 D) 6 rad/s2 E) 1 rad/s2
C m :U __’ _ : _ ~ To
9 gogmué u kebab org; small/Si 10. Through what angle will the wheel rotate in the first 12 s? A) 6 rad B) 12 rad C) 72 rad D) 103 rad . E) 204 rad
9: ea+mat+ nutL
, 1
= o + o + mm) 023 9‘ ms «Jim; 11. A black hole absorbs mass from its surroundings. Suppose a black
hole of mass M rotating with an angular speed of 1 revolution every four
days absorbs enough mass to double its mass while maintaining its
shape. Its mass is now 2M. What is its angular speed? A) 1 rev/day B) 1 rev/8 days C) 1 reV/ 16 days D) 2 rev/4days .UL Mme—u:th C 94qu 1M. ‘31”‘5‘k .. . I? “ﬂaw
3" wk =JI 03). LOZFM/i 3' : Goa 12:21:! The mass of the Earth is 5.98 x 1024 kg and its radius is 6380 km. A
story in the Bible tells us that God made the sun stand still or, as we
would say, made the Earth stop rotating on its axis. 12. What is the angular speed of the Earth on its axis?
A) 271 rad/s (B) 7.3 x 10'5 rad/s ) C) 2n rad/year D) 1.5 X 10'4 rad/s _2 «5&an «an. a 1145;: ATVV&&EW A
E) 2.66 x 10 rad/s %, £100 5 w ._. {gm : «BK/0% 13. What force would God have to apply at the Equator to stop the
Earth’s rotation in 10 s? A)1.1x1026N B)7.2x1032N C) 38x1027N D)9.4X1017N
Memﬂln‘kk I «LQAL 31 TzId
E) 3.8x 10 N .1 t 2.4;, H21. _._: 0~q CCZ¢3&*11}H)@§7‘/(f>>—
=37 x1037 My} bf: "w°
1: WT. p r 7.1 N017 ’7? ‘6 Two children of mass 60 kg and 45 kg, respectively are balanced on a
seesaw (teeter totter) as shown. The 60 kg child is 3m from the fulcrum. (Neglect the mass of the seesaw.) f T
. = O WV?an  Mbﬁé)7_
60003633 = QS'CwloQL 45 kg 14. How far is the 45 kg child from the fulcrum? Neglect the weight of
the seesaw. w B)2.25m C)12m D) 1.5m lE)6m
O 15. What is the upward force exerted by the fulcrum? $2,: M\S.+m‘% = “Kim
A) 105N (B)1050N> C) 15N D)150N ]E)1330N :OélxgggN ) 16. The earth rotates west to east on its axis once per day. In July the
average speed of the west to east trade winds across the entire earth is a
few meters per second smaller than it is in January. This would imply that \¢UL. \Mchﬂ eiltum IN.) (a
ghtly larger 1n
f. than in January. 64,12... w J kink C) This has no effect on the rotation of the earth enough to measure. out 17. An object slides down an inclined plane of height H and then travels
inside a circular track of radius R. How high must H be so that the object does not fall off the circular track? Cease?“ .M Cg; E a;
ﬁg“ = #3 cm + n Wu)— BQ‘V 9Q cloJQ—zk. wgk' “TD egg CKC‘oAM L = M5 v L = S (L A) 3R B)1.5R C)2R E)4R 18. Two planets, Armstrong and Pantani, have masses 5M and M.
respectively and are separated by a distance d. The ratio of the gravitational force exerted by Armstrong on Pantani to that exeﬁed by Pantani on Armstrong is Net—3‘93; fairQ LVQUO v 124 maﬁa...“ W
(A Q62“ Acme—TM) ‘AVK {TN/kg Q—LQC‘W an A)1:5 B)1:25 C)5:1 D)25:1<]E)1:1) o dull
Chef».
Eﬁkfaupco 19. A mass of 7 kg is attached to a horizontal spring With spring constant
2000 N/m. The spring is compressed 0.2 m and then released. What is the maximum s eed of the mass? CME 4 5‘0“ l.
P Lb’ “2.: gm v=JZW£ _——— k 0L: YPAKI m
A) 6.8 m/s C) 4. m/s D) 8.9 m/s E) 17.8 m/s ,. 3%]; 20. How much power is needed to carry a 10 kg oEect up a 6 m ﬂight of t' ' 5? 
sa1rs 1n s ©Dmr Wm: \Mj ;_ t0L03glzka¥i
A)6OW B)90W C)180W E)30 ...
View
Full Document
 '06
 DUBSON
 Physics

Click to edit the document details