122_t1_soln

122_t1_soln - - u MS Physics 122 Section 2 (1:25—2:15)...

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Unformatted text preview: - u MS Physics 122 Section 2 (1:25—2:15) Name- 30" T '0 Exam 1 I Thursday, September 21, 2006 Emall: Part 1 (50 points): Multiple Choice: Please circle the best answer. No partial credit. Note: use the acceleration of gravity as 10 ml s2 1. The correct procedure for graphically adding two vectors A and B is to a.) Place vector A’s tail on vector B’s head and draw the resultant vector from vector B to vector A. b.) Place vector A’s tail on vector B’s head and draw the resultant vector from vector A to vector B. 0.) Place vector A’s head on vector B’s tail and draw the resultant vector from vector A to vector B. Both a and c e.) Both a and b 2. A car travels with a constant acceleration of —3.00 rn/s2 for 20s, if the car’s initial velocity is 100 m/s, what is the car’s final velocity? a.) ~20m/s \/x = \/,v +qx‘t b.) 60m/s f L JiggrilnS/s \/,(Jf \oth + ('3,OOM(S'L)(ZOS) e.) none of the above VP tom‘s - moms : 90M]: .& 3. Compute the sum 346.335 + 590.12. The answer to the correct number of significant figures is I $936455 3%, 3333 .) 936.46 ‘ c.) 936.5 50‘ O ' 20 d.) 936.45 0! £3 A HS 5 e.) 936.00 K fl ’\ ww§+ beer 4:: ‘htt'l' (JGCH Mud VMCQ Vow/rd ‘4? 4‘0 WQQVQ§+ ‘QQUA vtvtwtloer ha 5. 4. For a car traveling in one dimension, the car slows down if a.) The direction of the car’s velocity is to the right and the direction of the car’s acceleration is to the right. b.) The direction of the car’s velocity is to the left and the direction of the car’s acceleration is to the left The direction of the car’s velocity is to the right and the direction of the car’s acceleration is to the left d.) Both a and b e.) None of the above 5. Calculate the maximum range (R) of a golf ball if it is hit at a magnitude of initial velocity of 100.0 m/s. @1000m mam («Me 9=i§iflfls 90° L b.) 100,000m Zx / >2 . 0.) 10m R: V? Sm (43°) ~ L ; QoOowh) d.) 100m —L-—-—— — ’——————-z e.) 10,000m 23 f] \0 M18 6. How high does the golf ball go in question 5 (hint: take sin245° = 1/2)? 7. a.) one quarter its range distance 2 x Z x ’l 0 Q the value of the range distance lA : g “A a 91‘ Vi 5‘“ L} S 0.) one third its range distance x : M d.) one half its range distance 2 2 j e.) not enough information 7, k - WOO) ‘ ' L (moo M) 2 (\o\ Ll 7. Given vector A = 69-73— mg, what angle does this vector make relative to the positive x-axis if projected onto the x,y plane (hint: draw this vector in a Cartesian (x—y) coordinate system)? Y a) —tan'1(10/7) b) tan" (7/6) @ -tan‘1 (7/6) d) tan'1(10/7) e) —tan'l (6/7) 7. 6: Jim/{4(3) f Mazda/e (Maths/e +0 (JOSH-Ne {vouch 8. If an object’s initial velocity is 320 m/s to the right and its final velocity is 420 m/s t0 the right and travels under constant acceleration, what is the displacement vector if the car travels for 10s? (pmgmvfi QCC€\ . 1 a.) 740mright X4;— X‘ + Vx‘f + lick-t E— b.) 4,200 m right L t g 3,700 m right i ’- QIAIOOmright AX : X4“ = Vx-J‘ 1“ taxi M . + L \ * e.) 3,200 m right \I ~ ’- 4 t Ct ' Wx ’ pl) 0; 5 Ll 3 fo - th + ( K +____ L O O 620w]: VQZO'M!‘ +1000” {v a {[35 (XX: 5 3 [3i ‘ {520104 )(\03 I I05 to 5 ' /s 2 4 u M = + to “451 + ‘%(+l0‘“(gl)(w$\ q 9. If the mass of a penny is 0.01kg and the volume of a jar of pennies is 10L (remember 1L = 1000cm3), approximately how many 3pennies are in the jar if the mass density of copper is approximately 10,000 kg/m ? h: a» e M 0.0 i 3 ; ‘ 0 M3 _ M t] : ‘_ = 3 @1x10“ - —- W‘s ["4 1x106 Va 3 \I P 53 3 c. 1x109 » _ 3 d3 1x105 VI. (comm—5K “0 .L ( ‘ 4., 1; 1M C.) 1X1010 T (oocm [OOCm \OOQM '31 “2: 3 Lt Apamnies : it: @ L(?:Q’—-W‘3 1‘“) Pei/mi“ I 10. What is the acceleration vector of an object whose position vector is r = 1.0 t44- 2.0 [zf-l- 10 exp (—01? m? (Hint the derivative with respect to time of exp(—t) is —exp(t)). 1.: a: A; a.) 4.0 t3¢- 4.0 :5». 10 exp (4);: m/s . 12.0 tzf- 4.0 tf— 10 exp (4)1; m/s2 @120 an 4.0“ 10 exp (—09 'm/sz d.) 12.0 at- 40,— 10 exp (—00 m/s2 e.) 12.0 :29- 4.0,. Partial Credit Problem (100 points). Please be neat and show your work. A particle moves from point A to B under the influence of a variable force. Starting from rest, it has an acceleration of 5 ft/s2 for 6s, and then its acceleration instantly increases to 7 ft/s2 until it has reached a velocity of 100 ft/s. a.) Find the time it takes for the object to reach 100 ft/s after changing its acceleration. V ‘2 F'W *3 :09 *0 t=§g FVCM/r was“ «K: 7 JEHS an is COL—Le'l'f.vk+ VXC: V; = Vxl. + C(Kt for: ‘60 1'; at . : O E StmtS ROM VK : VXL + ka At {65+ 4 O) z \/x - ,<~ UOO't , ' T ‘E ,1 GS AJC t L : fo" “KT (3 {’3 > «X ynéilsz After sometime its velocity remains constant, then the particle slows down with constant acceleration and within 103 it stops a point B. b.) Find the acceleration during the 103 it takes to stop the object. \/X,=IOO+t[S VK L f= OHIS t=<os V r. - K4, chcl—‘qK—k —U C.) The total time travel from A to B is 46s. Determine the time for which the object has zero acceleration. hm fir Q=S£+isz 2) C95 lime kw a=?¥elsz => IDS m at a: 40 et 52 => lbs {,th L0( acceleration =7 265 lime azo *2 Llés~Zé=s : '205! (1.) Draw the acceleration vs. time curve, indicating the times on the graph where the acceleration changes and the values of the acceleration. 4) Determine the distance from point A to B. L J_ 2 CF K : X- + + GHQ-t xi, 2 90% ((37948; >4 _ X +Vw+ 912—60—fo telos If L L 2< ' (10% *(BO‘O‘Ct/SB 005) ‘F \ , 13((OS)L X’%=; Qoa + “g BOOPV : :56on + Lahocaflfl + 3504* 5%; 74o£+ PE V 53:0 K ; Xx '+ Vflk‘e : $804+ + ( {mm/sts) 6: 20s «C L L Wtf 54 ; ¥£€o¥++ ZOOO¥+ : 2?“) ( f2, v} '» 'ZwoxCH (roomswm “(’2 (we/SWW wax/#1 K'F '1: 2:;qu 41 “900% ’ 7009+ : M 4+ 2 wow} xi + w + w ...
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This note was uploaded on 04/29/2008 for the course PHYS 122 taught by Professor Pope during the Spring '08 term at Clemson.

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122_t1_soln - - u MS Physics 122 Section 2 (1:25—2:15)...

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