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GENERAL PH 221-3A (Dr. S. Mirov) Sample Test 3 Test 3 (11/7/07) STUDENT NAME: ______key STUDENT id #: ______------

ALL QUESTIONS ARE WORTH 20 POINTS. OUT FIVE PROBLEMS.

NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)

1 . along along a straight lineaImportant straight with aFormulas: line with a constant acceleration v aver. = [dist. taken]/[ trav.]=S/t; aver.vel. = Δ x/ Δ t; ins =dv x/ Δ t; aver. = vΔ v aver. vel. / Δ t; a Δ t; + aat; = x=dv/ 1/2(v +v)t; x = v t + 1/2 at 2 ; v 2 = v 2 + 2ax (if x =0 at t =0) v =o v o o o o o 2. Free fall motion (with positive direction ↑) g = 9.80 m/s 2 ; y = v aver. t aver. = (v+v o )/2; v - gt; y = v t - 1/2 g t 2 ; v 2 = v 2 – 2gy (if y =0 at t =0) vo = v o o o o 3 . Motion in a plane in a plane v x = v o cos θ ; y = v o sin θ ; x = v v t+ 1/2 a t 2 ; y = v t + 1/2 a t 2 ; v = v + at; v = v + at; ox x oy y x ox y oy Projectile motion (with positive direction ↑) v x =4. v ox = v o cos θ ; x = v ox t ; 2 2 x max = (2 v o sin θ cos θ )/g = (v o sin2 θ )/g for y in = y fin ; = v - gt = v sin θ - gt; y y= v oy t - 1/2 gt o2 ; v oy 5. Uniform a=v 2 /r, π r/v T=2 6. Relative motion rrr vvvP APBBA=+ rr aaPA= PB 7. Component method of vector addition 22 -1 A = A 1 + A 2 ; A x = A x1 + A x2 and A y = A y1 + A y2 ; AAA=+xy; θ = tan ⏐A y /A x ⏐; r abr ⋅ =cos ab φ r r ˆˆˆˆ ˆˆThe scalar product A ab⋅=()() aixy + aj + ak z ⋅ bi xyz + bj + bk r r ab⋅++ = abx xyyzz ab ab r abr ×=()() aiajakˆˆ + +ˆˆ × bibjbk ˆˆ + + The vector product xyz xyz ijkˆˆˆ aaaa aa rrrr ˆˆyzxzˆ xy ab×=−×= ba axyz a a = i − j + k = bbyzbbxz bb xy bbbxyy z ˆˆ ˆ =−()()()abyz ba y z i +− ab zx ba z x j +− ab xy ba x y k 1. ’s Law ma=F net ;

k = μ k N; 2. Kinetic f 3 . St tia ti fc i tif r i c ti on f s = μ s N ; 4. Universal Law of Gravitation: F=GMm/r 2 ; G=6.67x10 -11 Nm 2 /kg 2 ;

1 DCAv= ρ 2 2 5. Drag coefficient 2 mg 6. Terminal speed v = t CAρ

2 c =mv /r 2 7. Centripetal : F =GM E /r 8. Speed of the satellite in a circular orbit: v r r 9. The work done by a constant force acting on an object: WFd= cosφ =⋅ Fd

1 K = mv2 2 10. Kinetic : 11. Total mechanical energy: E=K+U 12. The work-energy theorem: W=K f -K o ; W nc = Δ K+ Δ U=E f -E o

nc =0, E f =E o 13. The principle of conservation of mechanical energy: when W Wmgdg = cosφ 14. Work done by the gravitational force: 1. Work done in Lifting and Lowering the object:

Δ=KKf − Kiag = W + WifKKW; f = ia ; =− W g

F x =−kx (()H ook's law )

2. Spring Force: 11 1 3. Work done by a spring force: Wkxkxx=−22; if = 0 and xxWkx = ; =−2 si22 oi f s 2

x f WFxdx= ∫ () x i

WdW r 5. : P ==; PPFvFv; =cos4. Workφ =⋅done by a variabler force: avg Δ tdt

x Δ=−Δ=−UWU; f Fxdx( ) ∫ x i 6. :

Δ=Umgyy()fi − =7. mgyifyGravitational Δ ; Potential i Energy: =0 and U i =0; Uymgy() =

1 Ux()= kx2 2

8. ElasticdU potential() x Energy: F ()xKxEUx=− ; () = − () dx mec 10. Work done on a system by an external9. force:Potential energy curves:

F ric tio n is n o t in v o lv e d WE= Δ=Δ+Δmec KU

W h e n k in e tic fric tio n fo rc e a c ts w ith in th e sy ste m WE=Δmec +Δ E th

Δ=Efdth k

WEE= Δ=Δ +Δ+Δ E E 11. Conservation of ener gy : mec th in t for isolated system (W =0) Δ EEEmec+Δ th +Δin t = 0

Δ E dE 12. Power: PP==; ; avg Δ tdt

n rr1 rmrcom= ∑ i i M i = 1 r r 14. Newtons’ Second Law for a13. system Center of particles: of : F net= Ma com r rrdP 1. Linear and Newton’s Second law for a system of particles: PMv==r and F com net dt

r t f r 2. Collision and : J = FtdtJ() ; = FΔ t; when a stream of bodies with mass m and ∫ t avg i

nnΔ m Fpmvv= −Δ=−Δ=−Δ avg ΔΔtt Δ t r Impulse-Linear Momentum Theorem: ppJr − r = speed v, collides with a body whosefi is fixed r r PPif= for closed, isolated system

3. Law of Conservation of Linearr momentum:rrr pp12ii+= p 1 f + p 2 f 5. Motion of the : The center of mass of a closed,4. Inelastic isolated collisionsystem of in two one colliding dimension: bodies is

not affected by a collision. mm12− 2 m 1 vvvv1121f ==if; i mm12++ mm 12

7.6. CollisionElastic Collision in Two inDimensions: One Dimension: pp12ix+=+ ix p 1 fx ppp 212 fx; iy +=+ iy p 1 fy p 2 fy

Rvrel = M a (first rocket equation)

M i vvvfirel−= ln (second rocket equation) M f 8. Variable-mass system: S 9. Angular Position: θ = ( m easure) r ωω

Δ θ =−θθ21(p ositive for cou n terclockw ise rotation )

10. Angularαα : Δ θ d θ ==; (p o sitive fo r co u n terclo ck w ise rotatio n ) avg Δ tdt 11. Angular and speed: Δ ω d ω 12. Angular acceleration: ==; avg Δ tdt ω = ωαo + t θθ ω1 ω −=() +t oo2 1 1. angular acceleration: −=θθ ωtt +ω 2 oo2 22 ω =+ωαθθoo2( − ) θθ ω ω1 −=tt − 2 o 2 2. Linear and angular variables related:

vr2 22π π s ==θ rv; ωα ra ; = ra ; ω ====2 r ; T trrvω 3. Rotational and Rotational :

1 22 KIImr==ω ; ∑ ii for body as a system of discrete particles; 2 I = ∫ rdm2 for a bodyyy w ith c o n tin u o u sly distributed m ass.

2 4. The parallel axes theorem: II=+com Mh

τ ==rFt r⊥ F = rF sin φ 5. : τ net = I α 6. Newton’s second law in angular form:

θ W = f τθd τθ; W θ=− τ( ) for = const ; 7. Work and Rotational Kinetic Energy: ∫ fi θi

dW 11 PKKKIIW=Δ=−=−; ωω22 = w o rk e n e rg y th eo re m fo r ro ta tin g b o d ie s dt fi22 f i

vRcom = ω 11 KI=+ω 22 mv 22com com 8. Rolling bodies: α aRcom = g sin a com = 2 for ro rolling llin g sm ooothly o th ly dow n th the e ram p 1/+ IMRcom θ rrr 9. Torque as a vector: ττφ=×r F; = rF sin = rF⊥⊥ = r F r lrpmrv=×r rrr =(); × 1. of a particle: l= rmvsinφ == rp⊥⊥⊥⊥ rmv == r p r mv r dl 2. Newton’s Second law in Angular Form:τr = net dt

r n r Ll= ∑ i i=1 3. Angular momentum of a system of particles: r dL τr = net ext dt

4. AlMAngular Momen tum o fRiidBdf a : LI= ω r r 5. Conservation of Angular Momentum: LLif= (isolated system)

r r Fnet ==0; τ net 0 6. Static equilibrium: if all th e f orces li e i n xy pl ane FFnet,,, x= 0 ; net y==0 ; τ net z 0

7. Elastic Moduli: stress=modulus× strain

FLΔ 8. Tension and Compression: = E , E is the Young's modulus AL

FLΔ 9. Shearing: = G , G is the shear modulus AL

ΔV 10. Hydraulic Stress: p = B , B is the bulk modulus V 1. At the same instant that a 0.50-kg ball is dropped from 25m above Earth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth's surface with an initial +y speed of 15m/s. Theyght move above along Earth's nearby surface lines of and the pass center each of massother ofwithout the two- colliding. ball s ystem at Whatthe end is theof 2.0hei s?

m1

v1=0 v2=15 m/s 10.4m y=25m m2 5.4m

t=0 t=2.0s (a ) Find final y coordinate of the ball 1 after 2.0 s. gt229.8× 2 yt=+×−=−25 0 25 = 5.4 m 1 f 22 (b) Find final y coordinate of the ball 2 after 2.0 s gt229.8× 2 yvt=−=×−15 2 = 10.4 m 22fo22 (()c) Find y coordinate of the center of mass of two balls

my11++ my 2 2 (0.5)(5.4) (0.25)(10.4) ymcom == =7.1 mm12++(0.5 0.25)

2. Two skaters with of 100 kg and 60 kg, respectively, stand 10.0 m apart; each holds one end of a piece of rope. If they pull themselves along the rope until they meet, how far does each skaterskater travel?travel? (Neglect(Neglect friction) friction)

100kg 10m 60kg

0 +x (a ) Since two scaters represent an isolated system and initially theu are at rest their center of mass will be at rest and they will meet at the center of mass

mx11+ mx 2 2 (100)(0)+ (60)(10) xcom ==100+ 60 ==3.75mm 3.8 mm12+ (()b) The 100 k g skater moves 3.75 m (c) The 60 kg skater moves 10-3.75=6.25m=6.2m 3. A 2 kg block of wood rests on a long tabletop. A 5 g bullet moving horizontally with a speed of 150 m/s is shot into the block and sticks in it. The block then slides 270 cm along the table and stops. (a) find thethe speedspeed ofof thethe blockblock just just after after impact. impact. (b) find the friction force between block and table. v =0 v1o 20

m1 m2 2.7 m (a) Consider inelastic bullet-block collision. At the instant of collison bullet-block system is isolated, hence, we can use law of conservation of linear momentum to describe the collision

mv110++ mv 2 20 (0.005)(150) (2)(0) −1 mv110+= mv 2 20 ();mmv12+⇒= v = ==×0.374 4 10 ms / ()mm12+ (0.005+ 2 )

(b) Calculate work done by the on a block-bullet system during its motion along the table There are 3 acting on our system: , Normal reaction, and kinetif friction .

Only kinetic friction force will perform non zero work. ⇒=−WfSnet k (c) Motion of a block-bullet system after the collision can be described by a work-energy theorem ()mmv+++22 ()(00052)(0374)()(0 mmv .005 2)(0.374) 2 WKK=−⇒−=− fS051012 ⇒= f 12 = =×−2 N net f i k2 k 2S 2(2.70) 4. The impact of the head of a golf club on a golf ball can be approximately regarded as an elastic collision. The mass of the head of the golf club is 0.15 kg. The speed of the club before the collision is 46 m/s. The ball acquires a speed of 70 m/s after the collision. The golf club and a ball are in contact for about 0.5 ms. a) What must be the mass of the ball? b) What is the average force exerted by the club on the ball?

2mc (a) For elastic golf club-ball collision vvbf =⇒ci mmcb+

2mvcci 2(0.15)(46) −2 mmbc=−= −=0.15 0.047 =× 4.7 10 kg vbf 70

Δ−P PP− (0.047)(70) 0 (bF )==bf bi = = 6600 N =× 6.6 102 N avg ΔΔtt 0.0005 5. 6. Joe is painting the floor of his basement using a paint roller. The roller has a mass o f 2.4 kg and a radius of 3.8 cm. In rolling the roller across the floor Joe applies a force F=16N directed at an of 35% as shown. Ignoring the mass of the roller handle, what is the magggnitude of the angular acceleration of the roller? 35° y N F

x fs F mg (a) 2nd Newton law x axis: Ffma sinθ − scom=

()2dN()bfRI 2nd New ton Law f or rot ttiation ab out com: αατ = s = α

(c) recall the relationship between aaRcom and : com = ; Iα ()dFRdF sub bistitute eq.(b) (b)d()i() and (c) in (a): si n;θ −=mαR R α FFsinθθ sin 2 F sin θ 2(16)sin 35 ⇒= = = = = I 1 mR++ mR mR 33(24)(0038)33(2mR .4)(0.038) R 2 rad = 67 s2 2 2 2 7. A hoop (Ih=MR ), a uniform disk (Id=1/2MR ), and a uniform sphere (Is=2/5MR ), all with the same mass and outer radius, start with the same speed and roll without up identical inclines. Rank thethe objectsobjects accordingaccording toto howhow highhigh theythey go,go, leastleast toto greatest.

()Durina g rolling up the incline motion the mechanical energy of the hoop-earth, disk-earth, and sphere-earth systems is conserved since 1)frictional force does not transfer any energy to thermal energy because the objects do not slide; 2) normal force is perpendicular to the pass; 3) gravitational force is a conservative one. ⇒ 11 IMvMghω 22+++++00= + 22 (b ) second term is the same for all our objects, ω is also the same, ⇒ the larger I the larger maximum height h reached by the object. ⇒ flttttitfhifrom least to greatest in terms of height att ai ned : sphere, disk, hoop.

9. A uniform seesaw of length 6 m has two youngsters of w 1 =700N and w 2 =400 N sitting on the ends. Find the proper location of the pivot for the seesaw to be just in balance, if (b) th th e seesaw we i h ig h (a)s theM300NM g= weight300N of the seesaw can be ignored 6 m

x w 2 w 1 Mg (a ) Use 2 nd condition of equilibrium:

Wx12−−= W(6 x ) 0; 6W 6(400) xm==2 =2.18 WW12++700 400

(()Useb )Use 2 nd condition of equilibrium:

Wx12−−−−= Mg(3 x ) W (6 x ) 0; 6WMg++ 3 6(400) 3(300) xm==2 =2362.36 WW12++ Mg 700 + 400 + 300