exam2_1997s_sol

exam2_1997s_sol - Name- . Section P221/S97/Pr0blem 2A As a...

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Unformatted text preview: Name- . Section P221/S97/Pr0blem 2A As a practical joke a student on the fifth floor of a dormitory fastens a rubber hose to a window frame and uses it to launch (horizontally) a plastic jug filled with water (total mass M) into the air towards a neighboring dormitory. The hose, which acts as an ideal spring with a spring constant of k, is stretched back across the dorm room, a distance D, just before the jug is launched. [Warning: Do not try this at home] (a) [3. points] Determine an algebraic expression for the magnitude of the force exerted on the window frame by the fully—stretched hose. The magnitude ofthe force is F = 1c(:r — 3:0) = kD (b) [3 points] Determine an algebraic expression for the work done on the jug by the hose during the launch. The work l/V done is the negative ofthe change in elastic potential energy ofthe hose, which is ofthe form U = 516503. As the displacement ofthe hose decreases from its initial value OH) to its final value of O, the change in potential energy is , AU=m—.=§m}wb=%W—D w lV=—AU=. 3 3»: —§kDi kD. [uh—a (c) [4 points] Determine an algebraic expression for the speed ofthe jug as it leaves the window. ' The speed of the jug can be found by the work-energy theorem, W 2 AK. Since the initial kinetic energy of the jug was zero, its final kinetic energy is just equal to the work done on it: [Uh—4 Eff—.7711]??— — W so Ufzx/2W/m:\/kDgl/lm:-DV J‘s/m- Name Section P221/S97/Problern 213 A 2.0-kg mass and a 4.0-kg mass collide elastically in the dark. The 4.0—kg mass has a light attached to it, and is observed to be at rest before the collision and to have a velocity of 2.0 m/s in the + x direction after the collision. (a) [2 points] What physical quantities are conserved during the collision? Because this is a collision, we expect linear momentum to be conserved. Because the collision is elastic, we expect kinetic energy to be conserved. (b) [8 points] What are the initial and final velocities of the (invisible) 2.0-kg mass, assuming that its initial velocity is along the + a: direction? This is a one-dimensional problem. since the initial linear momentum was apparently in the + a: direction. Conservation of linear momentum (CLM) gives m1v1i+ mg’Ugi = mlvlf + mgvgf Of Z’Ulz‘ + O = 2U1f + 4 ‘ m/s) so v1i=v1f+4.Om/s Conservation of kinetic energy (CKE) gives 2 2+; 2-1.2 ii A l 277711111); l 10L; f (2.0 kg)u{f + 5(40 kg)(2.0 mg)? |.—— or %(2.0 kgflfii -i— O = . .x so vi- : vgf + 8.0 mil/s2 Substituting the expression from CLM into the expression for CKE gives (w + 4.0 m/s)2 2 vfif + 8.0 nP/s2 so vi], + (8.0 m/s)v1f + 16 mZ/s2 = off + 8.0 rnQ/s‘2 and (8.0 m/smf = — 8.0 mZ/sz so vlf = — 1.0 m/s and 1111' = Ulf + 4.0 m/s = 3.0 m/s. Section Name P221/S97/Problem 2C The nucleus of-a radium-226 atom, initially at rest, decays into an alpha-particle (helium-4 nucleus) and a radon-222 nucleus. The decay can be represented as follows: 226Ra —-> a + 222Rn The masses of the nuclei are: 225m: 375.202 x 10—27 kg at 6.644 x 10~27 kg 222an 368.549 x 10-27 kg (a) [4 points] Estimate how much mass was converted into energy and estimate the amount of energy (in J) this corresponds to. The change in mass, the final mass minus the initial mass, is: Am 2 (6.644 + 368.549 — 375,202) :< 10'27 kg : — 0.009 x 10—27 kg so 9 x 10"0 kg is converted into energy! ,Ihis corresponds to an energy release of éwiu (.5ng = (9 x 10—30 kg)(3 X 108 m/syl 2 31x 10-14 J x s x 10—13 J (b) [3 points] Explain in words what you can tell about the relative final momenta and speeds of the two products (the alpha particle and the radon-222 nucleus). (“Relative” means you should compare the magnitudes and directions of these quantities for the two products.) " k Since the initial nucleus was at rest, the initial linear momentum was zero. so the final linear momentum must aiso bazero. That means the final linear momenta of the alpha particle and the radon-222, nucleus must be equal'in magnitude and opposite in direction. . 3p. , Since the radon—22Li nucleus is a little over 50 times as massive as the alpha particle, the alpha particle must have a little over 50 times the speed ofthe radon-222 nucleus. (c) [3 points] Explain which of the two products carries away the most kinetic energy, and estimate its fraction (or percentage) of the total final kinetic energy. Since the kinetic energy K = émvz = p3 /‘2m, and the two products have the same magnitude of linear momentum, the less massive product has the greater kinetic energy. That means the alpha—particle has the greater kinetic energy. Since its mass is approximately 1/50 the mass of the radon—222 nucleus, it will have approximately 50 times as much kinetic energy. That means it has about 98% of the final kinetic energy. 31 PM! P221/S97/Problem 2D [10 points] A rope wraps around the inner radius (1'1 = 0.50 m) ofa spool which has an outer radius r2 = 1.00 m, mass 10.0 kg, and moment of inertia 4.0 kg - m2. A 4.0-kg mass is hung from the end of the rope. Find the magnitudes of (1) the acceleration of the mass, (2) the angular acceleration of the spool, and (3) the tensile force in the rope. Take g = 10 m/s2. Circle your three final answers. Let the magnitudes of the acceleration ofthe mass, angular acceleration of the spool, and tensile force in the rope be a, a, and T, respectively. Let the positive y axis be upwards, as usual. The mass will accelerate downwards, so the y component of the acceleration will be negative For the mass, Newton’s law gives For the spool, we have ( ~ Now I?! : Trl and oz 2 a/n. Rewriting the first equation and substituting these into the second. N ‘ I “V T : mtg — a) and Tn : ira/rl or T = Ia/r‘f. a: ( Setting the two expressions for T equal to each other, mtg — a) = we 7 or mg—mazla/rfi or mg 2 (I/fi + m)a ‘ _ mg¢23 (4kg)(10m/Se) _ 40m? _7 wz “’5'- MUM 50 a — mf — m _ (4/(Oi5)2 —-Hkg — 20 ’ - m/S Then a = a/r1 = (2 m/sg)/(0.50 m) = 4 rad/52 g l "’ I and sztjg—a)=(4.0kg)(10m/s£—2w53)=32N.¢-l 6"" fix NIULTIPLE—CHOICE QUESTIONS Each question is worth 2 points. Enter the answers very carefully on your machine-graded form. Note that the first question is number 11, not number 1. U(r) (inJ) r (in m) This diagram above shows the potential energy of an object A in the vicinity of another, stationary object B, where 7' is the distance between the two objects. Suppose the total mechanical energy of the system is equal to 10 J. 11. The closest that object A can approach object B is approximately: *(A) 3m (B) 4m (C) 5m (D) 6m (E) 7m 12. The kinetic energy of object A when the distance between the two objects is 4.0 m is approximately (A) 0 J (B) 3 J *(C) 7 J (D) 10 J (E) 14 J 13. The binding energy of the system is approximately (A) 0 J (B) 3 J (C) 7 J *(D) 10 J (E) 14 J 14. Two metal balls are the same size but one is twice as massive as the other. The two balls roll off a horizontal table with the same speed. In this situation *(A) both balls hit the floor at approximately the same time and with approximately the same speed. (B) the heavier ball hits the floor well before the lighter ball, and does so with a greater speed. (C) the heavier ball hits the floor well before the lighter ball, but the lighter ball hits with the greater Speed. (D) the lighter ball hits the floor well before the heavier ball, and does so with a greater speed. (E) the lighter ball hits the floor well before the heavier ball, but the heavier ball hits with the greater speed. ____________________._____._____._______——-— A constant force 13’ = (3.0 N)? acts continuously at the point 5" = (0.40 m)Ton a rigid object which is free to rotate about an axis that coincides with the z axis and passes through the origin. The moment of inertia of the object about that axis is 0.60 kg - m2. 15. The torque about the origin due to this force is (A) (1.8N-m)k (B) —(1.8N-m)k (C) 0 (D) — (1.2 N - m) fi *(E) (1.2 N - m) fl ?=? x F = (0.40 mfix (3.0 N)? = (1.2N-m)E 16. How much work will this force have done on the object if it increases its angular speed from 0 to 20 rad/s? *(A) 120 J (B) 180 J (C) 240 J (D) 300 J (E) 360 J W = AK = Kf — K. = aw — gm)? 2 (0.5)(060 kg - m3)(2o rad/s)2 = 120 J. 17. A particle of mass m is moving with a constant velocity of magnitude 0.600 as viewed from an inertial reference frame S. The energy of the particle in the frame S is (A) mc2 *(B)1.25mc2 (C) 1.5 me? (D) 1.67 mc2 (E)2mc2 “:1/1-(u/c)3: l 2%:—1w:—1—:175 / I \/ ~ 71 — (0.60)2 \/1 — 0.36 V064 0.8 “‘ so E : e/mc'3 : 1.25mc2 A IO-V force acts tangentially on the edge of a solid sphere of mass 5.0 kg and radius 1.0 m for 2.0 seconds. Assume that the sphere is initially at rest and that there are no other forces acting on it. The moment of inertia of a solid sphere of mass M and radius R is %MR2. 18. What is the final speed of the center of mass of the sphere? (A) 0.40 m/s (B) 2.0 m/s *(C) 4.0 m/s (D) 8.0 m/s (E) 16 m/s 1) : at : (F/m)t = (10 N/5.0 kg)(2.0 s) = 4,0 m/s. 19. What is the magnitude of the final angular velocity of the sphere? (A) 0.40 rad/s (B) 1.0 rad/s (C) 4.0 rad/s *(D) 10 rad/s (E) 10 rad/s w : at = 775/1 2 rFt/%M’rg = 2.5Ft/A/Ir = (2.5)(10 N)(2.0 s)/(5.0 kg)(l.0 m) = 10 rad/s M 20. An observer has two identical stopwatches. After starting them both, he throws one at relativistic velocity to a friend, who catches it and throws it back at the same speed. The observer records the elapsed time for each stopwatch. Which one of the following phrases correctly fill in the blanks in the following statement? “To the observer, the thrown stopwatch measured and appeared when in flight, compared to the stopwatch at rest.” *(A) a shorter time . . . smaller in size (B) a shorter time . . . bigger in size (C) a longer time . . . smaller in size (D) a longer time . . . bigger in size (E) the same time . . . smaller in size A sofa of mass m is pushed horizontally across a floor by a force F. The line of action 13-, of the force points to the center of mass of the P—fi—J sofa, which is in the middle of the sofa. The --------------------------------- -— sofa is supported by the left legs with force N71 and by the right legs with force A72, as shown on the diagram. 21. If there is no friction between the floor and the sofa, which one of the following must be true? N1+N2<mg N1+N2>mg N1 < 1V2 N1 = 1V2 N1 > N2 22. If there is some friction between the floor and the sofa, which of the following must be true? (A) N1+N2 <mg (B) N1+N2>mg N1 < N2 V N1 =N2 N1 > N2 7! m 3m 2m __—O_________O O x <~—1m—~> l <—-—1m—> i ‘— 2m ——> i 23. Where is the center of mass of the system of three masses shown above? (A) :L‘=1.6m (B) m=2.0m (C) :c:2.2m *(D) a: = 2.5 m (E) a: = 3.0 m xcm Z m(l m)+f::);(12nn;);n(2m)(4m) : 156m m A conservative force varies 20 o A With a: as shown. - 2 Via A X o t: 2 3 if 5 .40 x0") -zo 24. If U(3.0 m) = 0 J, where U(:L') is the potential energy associated with F(x), what is UWfimW (A)—5J (B)—1J (C)0J (D)1J ‘kflDSJ The work done by the force during a displacement from x = 0.0 m to :I: = 3,0 m is W = O I ~ 10 J+ 15 J = + 5 J. Thus the change in potential energy AU 2 — W = — 5 J. That means U(0.0 m) must be + 5 J relative to U(3l0 m), so U(0.0 m) = + 5 J. 25. Which one of the following graphs best represents U(x)? xOn) 2.0 T: V Id A o 3 3 5+ ‘5 x 3 —w (m) “I a (E) ...
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This note was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Spring '08 term at Iowa State.

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exam2_1997s_sol - Name- . Section P221/S97/Pr0blem 2A As a...

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