Geneal Physics I Exam 3 - Chs. 7,8,9 - Momentum, Rotation, Equilibium Nov. 3, 2010 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults with coect units and significant figues. Patial cedit is available if you wok is clea. Points shown in paenthesis. Fo TF and MC, choose the best answe. 1. (10) In an expeiment, physics students push a 1.80 10 3 -kg ca fo 12.0 s on a level oad. Thee is no fiction. The ca stats at est and eaches a final speed of 6.00 m/s afte the 12.0 s. a) (5) How lage is the magnitude of the impulse impated to the ca by the students? b) (5) What aveage foce magnitude was applied by the students woking togethe? 2. (12) Mass m A = 2.50 kg tavelling at 25.0 m/s cashes head-on into mass m B = 7.50 kg which is oiginally at est. The masses can slide without fiction on the level suface. Afte the collision, m A eveses diection, ecoiling with a speed of 5.00 m/s in the opposite diection. a) (6) Find the velocity of m B afte the collision. b) (6) Detemine whethe the collision is elastic o inelastic. 1
3. (2) T F When a basketball bounces off the floo, its momentum is conseved. 4. (2) T F In a collision of two ocks in oute space, the momentum lost by one is gained by the othe. 5. (2) T F When a 12-kg ock collides with a 6.0-kg ock, the 6.0-kg expeiences the lage magnitude impulse. 6. (2) T F If the time duation of a collision can be educed, the aveage foce involved is also educed. 7. (2) T F A 60-kg peson unning 10 m/s east has the same momentum as a 1200-kg ca going 0.5 m/s noth. 8. (2) A ca taveling at a constant speed along a cuve changes the diection of its velocity fom noth to south. In what diection was the momentum change of the ca? a. noth b. south c. east d. west. 9. (2) A ca taveling at a constant speed changes the diection of its velocity fom noth to east. In what diection was the aveage net foce on the ca? a. notheast b. southeast c. southwest d. nothwest. 10. (16) A 64.0-kg mass initially at est beaks into pieces of 16.0 kg and 48.0 kg, due to an explosion that poduces 2.40 kj of total kinetic enegy in the pieces. a) (2) Which piece acquies the lage magnitude momentum? a. 16.0 kg b. 48.0 kg c. it s a tie. b) (2) Which piece acquies the lage kinetic enegy? a. 16.0 kg b. 48.0 kg c. it s a tie. c) (6) Calculate the KE of the 48.0-kg piece afte the explosion. d) (6) Afte the 48.0-kg piece has taveled 1.00 mete away fom its oiginal position, how fa has the 16.0-kg piece taveled? 2
11. (12) A centifuge acceleates unifomly fom est to its top speed in 6.00 s, while otating though 2440 evolutions. Its oute adius is 2.50 cm. a) (6) How lage was its angula acceleation, in ad/s 2? b) (6) Once unning at its top speed, what is the centipetal acceleation of a point at its oute adius, measued in g s? 12. (16) A 975-kg satellite is a unifom solid cylinde of adius R = 2.50 m. It has thee tangential ockets mounted on the edge, whose thust can change its otation ate. Each ocket povides a tangential thust foce of 175 N when activated. a) (6) When all thee ockets ae tuned on, what total amount of toque do they apply to the satellite aound its cental axis? b) (6) What angula acceleation does the satellite have (in ad/s 2 ) with all thee ockets tuned on? c) (4) If the ockets ae tuned on fo 5.00 minutes, what is the magnitude of the change in angula momentum of the satellite? 3
13. (9) A solid sphee (I = 2 5 mr2 ), a solid cylinde (I = 1 2 mr2 ), and a hoop (I = mr 2 ) have identical masses and adii. They ae eleased togethe at the top of an incline, olling without slipping to the bottom. a) (3) Which one aives at the bottom with the geatest total kinetic enegy? a. sphee b. cylinde c. hoop d. all total KE s ae equal. b) (3) Which one has the geatest otational kinetic enegy at the bottom? a. sphee b. cylinde c. hoop d. all otational KE s ae equal. c) (3) Which one aives at the bottom fist? a. sphee b. cylinde c. hoop d. it s a 3-way tie. 14. (2) T F All points on a spinning wheel have the same angula acceleation. 15. (2) T F All points on a spinning wheel have the same centipetal acceleation. 16. (2) T F The speed of a point on a spinning wheel is popotional to its adius. 17. (2) T F The lage the wheels on a ca, the slowe they otate fo a given ca speed. 18. (2) T F If the net toque on an object is zeo about any axis, the object is in static equilibium. 19. (2) T F The toque due to some paticula foce depends on the choice of axis. 20. (2) T F When a foce passes though the axis of otation, it poduces no toque about that axis. 21. (10) A 960.-newton mass is hanging fom two cods, one connected hoizontally to the wall, and one making a 36.0 -angle to the point whee it connects to the ceiling. a) (5) Find the tension T A in the cod connected to the ceiling. b) (5) Find the tension T B in the cod connected to the wall. 4
22. (12) A woke of mass 80.0 kg stands on the end of a beam weighing 360. N, suppoted on a pivot. The beam s cente of mass is at its cente. W is the weight of a counteweight needed to hold the beam in equilibium. a) (8) By finding the net toque on the beam, with the pivot point as the axis, detemine the counteweight needed, in newtons. b) (4) What vetical foce must the pivot point make on the beam? 23. (8) A ladde weighing 2.00 10 2 N is set against the wall and does not slip. Thee is fiction between the ladde and the floo, but no fiction between the ladde and the wall. What hoizontal foce F W due to the wall is needed to hold the ladde stable? Scoe = /133. 5
Pefixes a=10 18, f=10 15, p=10 12, n=10 9, µ = 10 6, m=10 3, c=10 2, k=10 3, M=10 6, G=10 9, T=10 12, P=10 15 Physical Constants g = 9.80 m/s 2 (gavitational acceleation) M E = 5.98 10 24 kg (mass of Eath) m e = 9.11 10 31 kg (electon mass) c = 299792458 m/s (speed of light) G = 6.67 10 11 N m 2 /kg 2 (Gavitational constant) R E = 6380 km (mean adius of Eath) m p = 1.67 10 27 kg (poton mass) Units and Convesions 1 inch = 1 in = 2.54 cm (exactly) 1 foot = 1 ft = 12 in = 30.48 cm (exactly) 1 mile = 5280 ft 1 mile = 1609.344 m = 1.609344 km 1 m/s = 3.6 km/hou 1 ft/s = 0.6818 mile/hou 1 ace = 43560 ft 2 = (1 mile) 2 /640 1 hectae = 10 4 m 2 Tig summay sin θ = (opp) (hyp), (adj) cos θ = (hyp), (opp) tan θ = (adj), (opp)2 + (adj) 2 = (hyp) 2. sin θ = sin(180 θ), cos θ = cos( θ), tan θ = tan(180 + θ), sin 2 θ + cos 2 θ = 1. Acceleation Equations v = x t, x = x x 0, slope of x(t) cuve = v(t). ā = v t, v = v v 0, slope of v(t) cuve = a(t). Fo constant acceleation in one-dimension: v = 1 2 (v 0 + v), v = v 0 + at, x = x 0 + v 0 t + 1 2 at2, v 2 = v 2 0 + 2a(x x 0 ). Vectos Witten V o V, descibed by magnitude=v, diection=θ o by components (V x, V y ). V x = V cos θ, V y = V sin θ, V = Vx 2 + V 2 tan θ = Vy θ is the angle fom V to x-axis. y, V x. Addition: A + B, head to tail. Subtaction: A B is A + ( B), B is B evesed. Chapte 4 Equations Newton s Second Law: F net = m a, means ΣF x = ma x and ΣF y = ma y. Fnet = Fi, sum ove all foces on a mass. Fiction (magnitude): f s µ s N o F f µ s F N (static fiction). f k = µ k N o F f = µ k F N. (kinetic o sliding fiction) Gavitational foce nea Eath: F G = mg, downwad. Chapte 5 Equations Centipetal Acceleation: a R = v2, towads the cente of the cicle. Cicula motion: speed v = 2π T = 2πf, fequency f = 1 T, whee T is the peiod of one evolution. Gavitation: F = G m1m2 ; 2 g = GM, 2 whee G = 6.67 10 11 Nm 2 /kg 2 ; Obits: v 2 = g = GM 2 ; v = GM. centipetal acceleation = fee fall acceleation. 6
Chapte 6 Equations Wok & Kinetic & Potential Enegies: W = F d cos θ, KE = 1 2 mv2, PE gavity = mgy, PE sping = 1 2 kx2. θ = angle btwn F and d. Consevation o Tansfomation of Enegy: Wok-KE theoem: KE = W net = wok of all foces. Powe: P ave = W t, o use P ave = enegy time. Geneal enegy-consevation law: KE + PE = W NC = wok of non-consevative foces. Chapte 7 Equations Momentum & Impulse: momentum p = m v, impulse p = F ave t. Consevation of Momentum: (2-body collision): m A v A + m B v B = m A v A + m B v B. 1D elastic collision consevation of enegy: 1 2 m AvA 2 + 1 2 m BvB 2 = 1 2 m Av 2 A + 1 2 m Bv 2 B, o v A v B = (v A v B ). Cente of Mass: x cm = m1x1+m2x2+... m 1+m 2+..., v cm = m1v1+m2v2+... m 1+m 2+.... Chapte 8 Equations Rotational coodinates: 1 ev = 2π adians = 360, ω = 2πf, f = 1 T, θ ω = t, ᾱ = ω t, Linea coodinates vs. otation coodinates and adius: l = θ, v = ω, a tan = α, a R = ω 2, (must use adians in these). Constant angula acceleation: ω = ω 0 + αt, θ = θ 0 + ω 0 t + 1 2 αt2, ω + 1 2 (ω 0 + ω), ω 2 = ω 2 0 + 2α θ. θ = ω t. Toque & Dynamics: τ = F sin θ, I = Σm 2, τ net = Iα, L = Iω, L = τ net t, KE otation = 1 2 Iω2. Rotational Inetias about centes: I = MR 2, I = 1 2 MR2, I = 2 5 MR2, I = 1 12 ML2. hoop solid cylinde sphee thin od Chapte 9 Equations Static Equilibium: ΣF x = ΣF y = ΣF z = 0, Στ = 0, τ = F sinθ. 7