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Sample Test 3 Test 3 (11/7/07) STUDENT NAME: ______Key STUDENT Id #: ______ GENERAL PHYSICS PH 221-3A (Dr. S. Mirov) Test 3 (11/7/07) Sample Test 3 STUDENT NAME: ________________________key STUDENT id #: ___________________________ ------------------------------------------------------------------------------------------------------------------------------------------- ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS. NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.) Important Formulas: 1. Motion along a straight line with a constant acceleration v = [dist. taken]/[time trav.]=S/t; vaver. speed= Δx/Δt; vaver.vel.=dx /Δt; ains = Δv /Δt; a aver.= dv/Δtaver.; vel. 2 2 2 v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at ; v = vo + 2ax (if xo=0 at to=0) 2. Free fall motion (with positive direction ↑) g = 9.80 m/s2; y = vaver. t vaver.= (v+vo)/2; 2 2 2 v = vo - gt; y = vo t - 1/2 g t ; v = vo – 2gy (if yo=0 at to=0) 3. Motion in a plane v = v cosθ; vx = vo sinθ; x y= v o t+ 1/2 a t2; y = v t + 1/2 a t2; v = v + at; v = v + at; ox x oy y x ox y oy 4. Projectile motion (with positive direction ↑) vx = vox = vo cosθ; x = vox t; x = (2 v 2 sinθ cosθ)/g = (v 2 sin2θ)/g for y = y ; vmax = v - gto = v sinθ - gt; o in fin y y= v oy t - 1/2 gto2; oy 5. Uniform circular Motion a=v2/r, T=2πr/v 6. Relative motion rrr vvvP APBBA=+ rr aaPA= PB 7. Component method of vector addition 22 -1 A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2; AAA=+xy; θ = tan ⏐Ay /Ax⏐; r The scalar product A abr ⋅ =cos ab φ r r ˆˆˆˆ ˆˆ ab⋅=()() aixy + aj + ak z ⋅ bi xyz + bj + bk r r ab⋅++ = abx xyyzz ab ab r r ˆˆˆˆ ˆˆ The vector product ab×=()() aiajakxyz + + × bibjbk 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. Second Newton’s Law ma=Fnet ; 2. Kinetic friction fk =μkN; 3. Stati c f ri cti on fs =μsN; 4. Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2; 1 5. Drag coefficient DCAv= ρ 2 2 2 mg 6. Terminal speed v = t CAρ 2 7. Centripetal force: Fc=mv /r 2 8. Speed of the satellite in a circular orbit: v =GME/r r r 9. The work done by a constant force acting on an object: WFd= cosφ =⋅ Fd 1 10. Kinetic energy: K = mv2 2 11. Total mechanical energy: E=K+U 12. The work-energy theorem: W=Kf-Ko; Wnc=ΔK+ΔU=Ef-Eo 13. The principle of conservation of mechanical energy: when Wnc=0, Ef=Eo 14. Work done by the gravitational force: Wmgdg = cosφ 1. Work done in Lifting and Lowering the object: Δ=KKf − Kiag = W + WifKKW; f = ia; =− W g 2. Spring Force: F x =−kx (()H ook's law ) 11 1 3. Work done by a spring force: Wkxkxx=−22; if =0 and xxWkx =; =−2 si22 oi f s 2 x f 4. Work done by a variable force: WFxdx= ∫ () x i WdW r 5. Power: P ==; PPFvFv; =cosφ =⋅r avg Δ tdt x 6. Potential energy: Δ=−Δ=−UWU; f Fxdx( ) ∫ x i 7. Gravitational Potential Energy: Δ=Umgyy()fi − = mgyify Δ ; i =0 and U i =0; Uymgy() = 1 8. Elastic potential Energy: Ux()= kx2 2 dU() x 9. Potential energy curves: F ()xKxEUx=− ; () = − () dx mec 10. Work done on a system by an external force: 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 energy: 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 13. Center of mass: rr1 rmrcom= ∑ i i M i = 1 r r 14. Newtons’ Second Law for a system of particles: F net= Ma com r rrdP 1. Linear Momentum and Newton’s Second law for a system of particles: PMv==r and F com net dt r t f r 2. Collision and impulse: J = FtdtJ() ; = FΔ t; when a stream of bodies with mass m and ∫ t avg i nnΔ m speed v, collides with a body whose position is fixed Fpmvv= −Δ=−Δ=−Δ avg ΔΔtt Δ t r r r Impulse-Linear Momentum Theorem: ppJfi− = r r 3. Law of Conservation of Linear momentum: PPif= for closed, isolated system r rrr 4. Inelastic collision in one dimension: pp12ii+= p 1 f + p 2 f 5. Motion of the Center of Mass: The center of mass of a closed, isolated system of two colliding bodies is not affected by a collision. mm12− 2 m 1 6. Elastic Collision in One Dimension: vvvv1121f ==if; i mm12++ mm 12 7. Collision in Two Dimensions: pp12ix+=+ ix p 1 fx ppp 212 fx; iy +=+ iy p 1 fy p 2 fy Rvrel = M a (first rocket equation) 8. Variable-mass system: M i vvvfirel−= ln (second rocket equation) M f S 9. Angular Position: θ = (radian m easure) r 10. Angular Displacement: Δ θ =−θθ21(p ositive for cou n terclockw ise rotation ) Δ θ d θ 11. Angular velocity and speed: ωω==; (p o sitive fo r co u n terclo ck w ise rotatio n ) avg Δ tdt Δ ω d ω 12. Angular acceleration: αα==; avg Δ tdt ω = ωαo + t 1 θθ−=() ω + ωt oo2 1 1. angular acceleration: θθ−= ωtt + ω2 oo2 22 ω =+ωαθθoo2( − ) 1 θθ−= ωtt − ω2 o 2 2. Linear and angular variables related: vr2 22π π s ==θ rv; ωα ra; = ra; ==== ω2 r; T trrvω 3. Rotational Kinetic Energy and Rotational Inertia: 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 5. Torque: τ ==rFt r⊥ F = rF sin φ 6. Newton’s second law in angular form: τ net = I α θ f 7. Work and Rotational Kinetic Energy: W = τθd; W=− τθ(fi θ ) for τ = const; ∫ θ i 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 1122 KI=+comω mv com 8. Rolling bodies: 22 aRcom = α g sin θ a com = 2 for rollingro llin g sm ooothly o th ly dow n th the e ram p 1/+ IMRcom rrr 9. Torque as a vector: ττφ=×r F; = rFsin = rF⊥⊥ = r F r lrpmrv=×r rrr =(); × 1. Angular Momentum 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 Rigid Body: 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 speed of 15m/s. They move along nearby lines and pass each other without colliding. +y What is the height above Earth's surface of the center of mass of the two-ball system at the end of 2.0 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 masses 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 skater travel? (Neglect 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 kg skater moves 3.75 m (c) The 60 kg skater moves 10-3.75=6.25m=6.2m 3.
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