NYA final exam - May 2000 (with solutions)

# NYA final exam - May 2000 (with solutions) - MECHANICS(101...

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Unformatted text preview: MECHANICS (101) NYA. FINAL EXAMINATION.(MRV "91 3000) ANSWER ANY TEN OF THE FOLLOWINGTWELVE QUESTIONS. ‘1.) A stone is thrown vertically upward with an initial velocity of 20 mls from a 60 m high building. a) Find the maximum height. above the ground, reached by the stone. b) Find the time the stone spends in the air. (From‘ the top'of the building to the ground). c) Find the velocity of the stone just before impact with the ground. d) Find the total displacement of the stone and the total distance traveled by the stone. 2.) a) Three vectors A, B and C are each 10 units long. A lies along the x-axis. 3 makes an angle of 30° and 0 makes an angle of 130“ with the positive x-axis.' a) Write each vector in unit-vector notation. b) Find a vectorD such thatA+ B + c + D =0 c) Use graphical methods to ﬁnd vector D. Given the two vectors P = 3 1+ 1.7? and Q = s i — 4.23 . Find the angle between P and Q using the scalar (dot) product of the two vectors. 3.) I ,_ _ _ A cannon ﬁres a ball with an initial velocity of 50 mils at 53° elevation. The cannonball hits a hillside target at a height Of 60m as it is on its way down. Find: a) The maximum height reached by the projectile. 60*" b) The distance from the cannon to the hillside. c) The acceleration of the cannonball just before it hits the target. i 4.) A particle moves in the x—y plane with a constant acceleration. initial position of the particle: r = (4.0 i + 3.0 :11 initial velocity of the particle: v = (2.0:; + 9.0 ‘ ) mls AcceIeration of the particle; a._= (4.0 i + 3.0 J ) We2 3) Find the velocity vector at t = 2.0 s ‘ b) Find the position of the particle in vector form att = 4.0 s 50' A 3.0 kg block hangs from a rope (1) that is tied to two other ropes; a horizontal rope (2) attached to the wall and a third rope. attached to the ceiling (3), that makes an angle of 30° with the horizontal,. Find the tension in all three ropes. 0'9“ A 0.2 kg and aﬁra kg box are placed. one behind the other, as shown. . One cord is attaching the 0.2 kg box to the 0.8 kg box. a second cord from the 0.8 kg box passes over a frictionless pulley and a 0.4 kg block hangs from the free end of this cord. The coefﬁcient of friction is pk = 0.10 between all surfaces, a) Draw free body diagrams for each of the three blocks b) Find the acceleration of all three blocks and the tension in the two ropes. T.) ' A car drives around a smooth banked curve of radius = 40m. ° (The picture shows the car driving towards you). 100 a) Draw a free body diagram for the car, showing all forces and identify the force or its component that is responsible for keeping the car moving around the curve. b) . if the road is banked at an angle of 10° what speed should the car go around the curve? 0) What is this speed in kmlh? A 4 kg block, initially at rest, slides down a 5.0 long inclined plane. The plane makes an angle of 37° with the horizontal and the coefﬁcient of kinetic friction between the block and the plane is pk = 0.25. a) On a carefully drawn free body diagram show all forces acting on the block and calculate the work done by each force during the displacement of 5.0 m. --b) - Using the work—energy theorem calculate the speed of the block at the bottom of the incline. ~— 10.) ‘ I ' _ ill? 11.) A 0.3 kg brick is dropped from a height of 8 m. The brick stops as it hits the ground. a) what is the impulse exerted by the ground on the brick? b) If it takes 1.3 ms from the time the brick ﬁrst touches the ground until it stops completely. what is the average force exerted on the brick by the ground? A roller coaster cart of mass 50.0 kg is accelerated from rest by a compressed spring (point ‘A'). The track is frictionless and the spring. a spring constant k = 2000 Nlm is initially compressed by 2.00 m. If the heights of the various sections of the track are as shown ﬁnd the following: What is the speed of the cart at point B while still at ground level? By how much did the potential energy of the cart change, as it moved fron B to 6? Assuming that the cart stays on the track everywhere, what is its speed at points 0 and at point D? l A 1500 kg cartraveling north at 70 kmlh collides at an intersection. with a 2000 kg car traveling west at 55 kmlh. The two cars stick to each other. a) What is the total momentum of the sytem before the collision? b) Express this total momentum in x and y components. c) Find the magnitude and the direction of the velocity of the wreckage just after the collision. 12.) The engine of an automobile accelerates eta constant rate from 200 revlmin to 3000 rev/min in 7.0 seconds, after which it runs at a constant speed. a) Find the angular velocity and the angular acceleration at t = 0 (just after the acceleration begins) and att 1: 10's (just before the acceleration ends). Express in in _ radls and o in radlsz. b) How many revolutions did the engine make in these 7 seconds? ‘ '“ ‘ 5% Physics NYA Final Exam Solutions ' . May 2000 1. (a)v:—v?=2ay(Ay) m; 0—202=2(—.9.s)(yf'~60) —> y]: (b) Ay=vir+gayt1 —>-—60=20t—4.912 '—> 4.9r2~201—60=0 r: (c)vf=vi+at —> vf=20—9.8(6.09):- 60m “('d) ; d:40.8m+60m= 2' (a) §=IOCOSSOuf+105in30°3 ; E=—1000550°f+103in50°}' = 0.366? + 5.00} =—6.43f +7.66} 3:105 - (b) 1") = —(}i'+ E + E): —[(10 + 8.66 m 6.43)? + (0 +5 + 7.66)}j] g —12.23:°—12.66} OR D =, (12.2327 + 12.661)?é = 17.6 units ; 9 = tan"(12.66/12.23)_= 46.0 ‘ 13 = 17.6 units, 46.0” s ofW (c) by using ruler and protractor, the polygon method (tip—to-tail) gives: _. . . D = 17.6 units, 46.0“ S of W ﬁ-Q P'Q Q: J61 +4.22 =7.32 units ; also, P-Q =3-6—1.7-4.2= 10.86 -. 6656 =ﬁ%=0.4300 —> 9 = cos-1 (0.4300) = -> P (d) F-Q=P-Qcosﬂ => c059: 32+1.72=3.45units ; Final Exam Solutions — May Page 1 of 6 3. (a) vxi =50c0853" =30.09 m/s ; v” =508in53" =39.93 m/s 1}), =v +8: —+ 0=39.93-9.8t —> z=4.07s I Y: 3" ‘ Ay ﬂying);2 —> Ay =39.93(4.07)_-%(9.8)(4.O7)2 = 81.35 m —> (b)Ax2vxit+§axt2 a tzsx=30.09r+§(0):2 —+ Ax=30.09t .(1) Ay=60=39.931—§(9.8)t2 ~—> 4.9r2—39.93r+60=0 —> r=6.168 OR t=1.99s; choose the later time since the earlier time is when cannonballﬁrst reaches a height of 60 m (on the way up); ' Sub into (1) —> Ax=30.09-24.5= ‘(c)a=ax+ay=0—9.8= 4. (a) ﬁf=17i+zit (b) ﬁr'=3-;+ﬁft+%&r2 z (25 +9 3)+ (45+33x2) _ = (45+ 33)+ (25+ 9})(4)+%(4;+ 33(4): = (2? + 93)+ (81° + 63) ‘= (4? + 3})+ (8? + 36})+ (321’ + 243') =<2+8r+<9+w * = = 5. Since we have static equilibrium, ax = O; a), = 0 Draw FBD for 3 kg block: Tn 23:17:81} =0 —> 13—29.4=0 —> T1229.4N Mg = (3x98) = 29.4 N Draw FBD‘for “knot”: T3,: Tan-1530" ' 2 I?1 g max 2 0 —> T100530" - T2 = 0 (1) A T3 I D _ _ ‘ a _ 39 T3,=T3cn530" 2F): — may ‘0 —> T3 81n30 — 29,4_0 (2) 5:29-41“ (2) —> T3 = 58.8 N (3) Suh into (3) —> (’1): T2 = 50.9 N Final Exam Solutions — May ‘ ' Page 2 of 6 6. (a) T1 ' ' fk =pknl = (Lin; 0.2 T1 0.3 T1 T1 ‘ ﬂ = Hz": = 0-3“: _ ,1, mg =(D.2)(9.8)=1.96 N "1 mg = (0.3mm = 1.34 N m mg = (0.4)(931 = 3.92 N . (a I (b) FED for system: I l , 0.1?“ ‘ - 3.92N 0.2 kg box: 0.8 kg box: system: sz =ma'120 2F), =nloyé0 IZFx=ma nl — 1.96 = o - _ n2 — 7.34 = o 3.92 — 0.1(1._96)- 0.1(7.84) = (0.2 + 0.4 + 0.8)a H1=1.96N 113:7.84N azzliom/SZ Now let‘s ﬁnd T1 and T2: lookingr at 0.2 kgmass: 2,14:‘ = max T1 — 0.1(1.96)= (0.2)(2.10) looking at 0.8 kg mass: ' 2F), = may (choosing down as —ve) 3.92 — T2 = (o.4)(2.10) T, = 3.08 N 7. The centripetal component of the normal force (nsinlOD) is responsible for keeping the car moving in a circle. ' 10° - (a) f (b) 21*} 2771.511 ZF, = ma, ' a __ ' 2 “0310 m3 “0 11511110211131— ncole" =mg _ ' r g 2 n= mg a 74 sin10°=J/zv— c0310 c0510“ r v =(rgtan10"yé =(40.9.8-tan10°)‘ = 69.1 11115 Final Exam Solutions — May ' Page 3 of 6 i! 1031er 1h 2 8. ' fl: 2 F), = 1amy 'mgsin37 n n 4 mg cos 37 ” ¥ 0 mg 15... mgcosS? n = mg cos 37" \ 37 (a) =ﬁ-§=ndc0590“= W" ng = mg - a = mgd c0553” '= (4)(9.8)(5)cos§3“ = 117.96 I WI * = fk -d = — pkn - a = — ak (mgcos53")i =—0.25(4)(9.8)cos37" (5): (b) 2W=AK=.K;—KF=Kf-0=Kf also —> 2W=0+117.96—39.13=78.831 . gamma => VI: 9. In this problem, we see transformation of spring potential energy into kinetic energy, then into gravitational potential energyand back into kinetic energy. ' (a) At point A, potential energy is stored in the spring: 2 2 =E=W=4000Nm=40001 A a In going from A to B, the block looses 4000 I in potential energy and gains it in kinetic . _ 2 mvf 2 energy: AU =—4OOOJ=—AK=~(KI “KI)=-Kf =~ (where K1: 0 because the block starts at rest). 2 , Therefore mv’ =4aoaJ=>v, = {2(4000” =12.6 mfs vs = 12-6 111/8 2 50.0 kg ‘ ([3) From point B to point C, there is a gain in gravitational potential energy of AU =mgh= (50.0 kg)(9.s 1 NIngﬁﬂD m)=2943 J AU = 2943 J Final Exam Solutions — May Page 4 of 6 (0) Of the 4000 I of kinetic energy at point B, 2943 I where-used to raise the gravitational energy, so 1057 J of kinetic energy are left at point C: K; -K, =—AU =—29431 therefore: KI =—'29431 + Ki =—2943 J +4000J=1057J 'The speed at point C is vc = 2 10571 :65!) mls 1dc "—“‘ 6-50 IBIS 50.0 kg Between points C and D, the gravitational potential energy decreases andthe kinetic energy increases: - - ‘ AU =~mgh=—(500 kg)(9.31N/kg)(5.00 m): ~2453 J AK=+24531=KJr —KJ =K‘f -10571 = KJr =2453J+10571=35101 The speed at point D is vb = 25:53:] =11.8 m/s VD =11;3 m/S ' - \i - s 10. This is a case of totally inelastic collision. (a) car 1: p1 = mp}; = (1500 kg)(70kmfh).North = 1.05 x 105 kg kin/h North at: 7 ‘ = 29.2 x 10“ kg m/s North - car 2: p2: 12121); = (2000 kg)(55 lonlh) West: 1.10 x 105 kg km/h West = 30.6 x 104 kg mls West . _ . - ‘ Total momentum, magnitude = «(29.2 ><10“)2 + (30.6 >004)2 kg 11115 = 42.3 x 105 k-m/s . 4 V I direction: tanB = m: 0.954 a: 417° North of West 30.6x104 (b) If East in the positive x—direction and North in the positive y—direction, then 13W = (—30.6i+29.23)x104 kg-rn/s ' (C) a ' pm, = (— 30.6 i + 29.2 3)><104 kg ; m/s = newtmme = [(1500 +2000)kg]l7mckﬂge I therefore ‘_ - 1" “. 4 tum a =———-—('30'61 +29'23)X10 kgm/S = (-87.4i+83.4]) In/s =(-315i+300 j')71on/tt E 3500kg The magnitude of the velocity is vmﬂge = 1/(w' 315)2 +(300)2k1n/h= 435 kmlh -.' . . . e 300 . and its Chl'BCLlOIl 1s 9 = tall m"— : 136.4° from the x—ms . "315 Note that 136.4" from the x—axis is the same as 43.60 North of West . Final Exam Solutions — May . Page 5 of 6 ’5‘?“ 11. Initial Velocity, V3: 0 after falling'8.0 m, the velocity will be VJ, = 1klgbéy)=1/2(9.8m/52)(8.0m)=11?‘.51r1ru"s i the momentum of the brick will be zbro when it hits the ground, so , (a) I=AP:_1'Jf—-Pf =0.0-—(0.3kg)(—12.5m/_s) = 3.75 k-mls (b) F=Apm3=LWi = 2.89 x103N 1.30x10"33 12. (a) cum=m = (zooﬂxzmﬂxiﬂ) = 20.9 rad/s mm rev 60 S mm = (Boooﬂmnﬂxiﬂ) = 314 rad/s Imn rev 60 s a) ~01. _ . a=( IA: t):314radf.:,020.9mdls : 41.9Iadjs . S 2 (02 —w-2) .(314radls)z—(20.9md/s)2 9:+;=—#___dm—=1171 d=-186 v (b) 205 2041.9 rad ls“) m E THE END Final Exam solutions — May ’ l Page 6 of 6 ...
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NYA final exam - May 2000 (with solutions) - MECHANICS(101...

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