Physics 218: Final. Class of 2:20pm. May 7th, You have the full class period to complete the exam.

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1 Physics 218: Final Class of 2:20pm May 7th, ules of the exam: 1. You have the full class period to complete the exam. 2. Formulae will be displayed in the big screen. You may NOT use any other formula sheet. 3. When calculating numerical values, be sure to keep track of units. 4. You may use this exam or come up front for scratch paper. 5. Be sure to put a box around your final answers and clearly indicate your work to your grader. 6. Clearly erase any unwanted marks. No credit will be given if we cant figure out which answer you are choosing, or which answer you want us to consider. 7. Partial credit can be given only if your work is clearly explained and labeled. 8. All work must be shown to get credit for the answer marked. If the answer marked does not obviously follow from the shown work, even if the answer is correct, you will not get credit for the answer. Sign below to indicate your understanding of the above rules. SOLUTIONS Name : Student ID : Signature :

2 Part 1: Invasion Question Points Score Invasion 34 Merry-go-round 32 Dragging a suitcase 32 Beam of Photons 32 Trying to escape gravitation 38 Disks 32 Total: 200 (34 points) An alien spacecraft hovering at a height H over a military facility releases a bomb with zero initial velocity. Exactly at the same time the military facility launches a projectile straight up with acceleration a p towards the spacecraft. (a) (12 points) Find the time at the which the missile impacts the bomb. Using a coordinate system with ŷ going up and starting at the ground we get: y b (t) = H g 2 t2 y p (t) = a p 2 t2 At the moment of the collision the heights are the same and so we get y b (t c ) = y p (t c ) H g 2 t2 c = a p 2 t2 c t2 c 2 (a p + g) t 2 c = 2H a p + g

3 (b) (12 points) Find the height at the which the missile impacts the spacecraft. y p (t c ) = a p 2 t2 c y p (t c ) = H a p a p + g (c) (10 points) Find the velocities of the spacecraft and missile at the moment of the collision. v b (t) = gt v p (t) = a p t At the moment of the collision we get v b (t c ) = g 2Ha p + g v p (t c ) = a p 2Hap + g Page 2

4 Part 2: Merry-go-round (32 points) A merry-go-round is spinning with an angular velocity ω 0 = π/4 rad when s it starts malfunctioning and develops a constant angular acceleration of α = π/2 rad in the direction of the angular velocity. s 2 (a) (10 points) Find the linear velocity of an object at radius = 2m from the center at t = 2s. v (t = 2s) = ω(t = 2s) ω(t) = ω 0 + αt = π 4 + π 2 t ω(t = 2s) = π 4 + π = 5 4 πs 1 v(t = 2s) = 5 4 π2m s = 7.85m s (b) (10 points) For an object at radius = 2m find a (magnitude of linear acceleration in the direction of linear velocity) at t = 2s. a = α = π 2 2ms 1 = a = π m s 2 (c) (12 points) At t=2s find the ratio a a where a is the magnitude of the linear acceleration in the direction perpendicular to the linear velocity. a = v2 = ω2 = 25π2 16 2ms 1 = a = 25 8 π2 m s 2 a a = 8 25 Page 3

5 Part 3: Dragging a suitcase (32 points) A tourist drags his suitcase, which weights 130N, a distance of 250m at a constant speed to catch his flight. He exerts a 60N force at an angle 40 degrees above the horizontal. The coefficient of kinetic friction µ k between the suitcase and the floor is unknown. (a) (10 points) What works does the tourist do? W = F S = 60N cos(40)250m W = 11490J (b) (8 points) Find the work done by friction. Explain your reasoning. Because the suitcase is moving with constant velocity in the ˆx direction we know that F x f x = 0, where f x represents the friction force. Therefore the work is just minus the work of the tourist. W = 11490J (c) (14 points) Find the coefficient of kinetic friction µ k. ŷ : N + F sin(40) W = 0 N = W F sin(40) = N = 91.4N ˆx : µ k N + F cos(40) = 0 µ k = F cos(40) N = µ k = 0.5 Page 4

6 Part 4: Beam of Photons (32 points) A uniform spherical ball of mass M and radius is at rest in deep space when a pulse of light from a nearby laser hits the sphere at a distance h from the center. The pulse of light has momentum P γ and is completely absorbed by the sphere. P γ } h M (a) (8 points) Is the linear momentum of the pulse+sphere conserved? Explain why or why not. Yes, because the sum of the external forces in the system is zero. (b) (8 points) Is the angular momentum of the pulse+sphere conserved? Explain why or why not. Yes, because the sum of the external torques in the system is zero. (c) (8 points) Compute the linear velocity of the center of mass of the sphere after it has absorbed the pulse of light. P i = P γ P f = Mv cm v cm = P M (d) (8 points) Compute the angular velocity of the sphere around its center of mass after it has absorbed the pulse of light. L i = P γ h L f = I sphere ω ω = P γh I sphere Page 5

7 Part 5: Trying to escape gravitation (38 points) A scientific satellite of mass m is moving radially away from a spherical planet of radius and mass M. When the satellite is at a distance of d = 5 the propulsion system stops and the satellite is found to have a velocity square of v0 2 = GM. 5 M m v 0 { d = 5 (a) (14 points) Will the satellite break free of the planet? or will it eventually fall back to it? Explain your reasoning. The satellite will break free only of it s mechanical energy is greater than zero, or will fall back if the energy is negative. E i = 1 2 mv2 0 GMm d E i = GMm 10 = 1 GMm 2 5 GMm 5 So the satellite will fall back to the planet. (b) (14 points) Using conservation of energy, if the satellite breaks free compute its velocity when it is very very far away from the planet. If it falls back to the planet compute the velocity it will have when reaching its surface. Ignore air resistance. Using conservation of energy we get E i = E f GMm 10 = 1 2 mv2 f GMm vf 2 = GM ( ) v GM f = 5 Page 6

8 (c) (10 points) Find the work done by gravity in moving the satellite from the initial position to the final position. Explain your reasoning. The only force acting on the satellite is that of gravity. Therefore the Work of gravity is just the change in kinetic energy. W grav = K f K i = 1 2 m9 GM 5 W grav = GMm mgm 5 = GMm ( ) Page 7

9 Part 6: Disks (32 points) In the picture below each disk have radius = 0.10 m, mass M = 50 kg and moment of inertia of a solid disk. Assume the acceleration of gravity is simply g = 10 m. All answers must be a number with proper units. s 2 M ŷ M ẑ ˆx F = 500N 50 kg (a) (15 points) Find the angular acceleration of the disk on the left. F = Iα α = F I = 50Nm kg m = α = 200rads 2 s 2 (b) (17 points) Find the angular acceleration of the disk on the right. First we write all the equations that allow us to completely solve the problem according to our coordinate system above. T mg = ma y (1) a y = α (2) T = Iα (3) Then we solve the system: replacing the second into the first one we get T mg = mα T = mg mα and replacing T into the third we get mg m 2 α = Iα α = mg I + m = mg 2g 2 3 = m2 3 α = 20 m s m = α = 66.6rads s 2 2 Page 8

10 Formulae General math: log (x/y) = log (x) log (y) log (xy) = log (x) + log (y) log (x n ) = n log (x) h h a h o A = A x î + A y ĵ + A zˆk ln x log e x x = 10 (log 10 x) (ln x) x = e h a = h cos = h sin h o = h sin = h cos h 2 = h 2 a + h 2 o tan = h o h a ax 2 + bx + c = 0 x = b ± b 2 4ac 2a + = = 90 A B = A x B x + A y B y + A z B z = AB cos A B = (A y B z A z B y )î + (A z B x A x B z )ĵ + (A x B y A y B x )ˆk = AB sin = A B = AB df dt = natn 1 If f(t) = at n, then t 2 t 1 f(t)dt = a n+1 ( t n+1 2 t n+1 ) 1 Translational otational constant (linear/angular) acceleration only: r(t) = r + v t at2 v(t) = v + at v 2 f = v 2 + 2a(r r ) r(t) = r ( v i + v f )t v = r2 r1 t 2 t 1 a = v2 v1 t 2 t 1 v = d r dt a= d v dt = d2 r dt 2 r(t) = r + t v(t) dt 0 v(t) = v + t a(t) dt 0 W = F r = F d r P = dw dt = F v p cm = m 1 v 1 + m 2 v = M v cm J = F dt = p K trans = 1 2 Mv2 cm Fext = M a cm = d pcm dt Fint = 0 Work-Energy: always true: W = K (t) = + ω t αt2 ω(t) = ω + αt ω 2 f = ω 2 + 2α( ) (t) = (ω i + ω f )t ω = 2 1 t 2 t 1 α = ω2 ω1 t 2 t 1 ω = d dt α= dω dt = d2 dt 2 (t) = + t ω(t) dt 0 ω(t) = ω + t α(t) dt 0 τ = r F and τ = F r W = τ = τd P = dw dt = τ ω L = I 1 ω 1 + I 2 ω = I tot ω K rot = 1 2 I totω 2 τext = I tot α = d L dt τint = 0 U grav = Mgy cm U elas = 1 2 k(x x equilib) 2 E tot,f = E tot,i + W other K f + U grav,f + U elas,f = K i + U grav,i + U elas,i + W other [ F x (x) = du(x)/dx F = U = U x î + U y ĵ + U ˆk ] z Circular motion: s = v tan = ω a tan = α elative velocity: a rad = v2 Constants/Conversions: T = 2π v v A/C = v A/B + v B/C g = 9.80 m/s 2 = ft/s 2 (on Earth s surface) G = N m 2 /kg 2 = m M = kg 1 km = mi 1mi = km 1 ft = m 1 m = ft 1 hr = 3600 s 1 s = hr 1 kg m s 2 = 1 N = lb 1 lb = N 1 J = 1 N m 1 W = 1 J/s 1 rev = 360 = 2π radians 1 hp = W 10 9 nano- n 10 6 micro- µ 10 3 milli- m 10 2 centi- c 10 3 kilo- k 10 6 mega- M 10 9 giga- G Newton s Laws: F = m a FB on A = F A on B w = mgĵ Fspring = k r f s µ s n f k = µ k n Centre-of-mass: Gravity: r cm = m 1 r 1 + m 2 r m n r n m 1 + m m n (and similarly for v and a) F grav = G M 1M U grav = G M 1M 2 12 T = 2πa3/2 GM

I = 1 12 ML2 I = 1 3 ML2 I = 1 12 M(a2 + b 2 ) I = 1 3 Ma2 L L b b a a slender rod, axis through centre slender rod, axis through one end rectangular plate, axis through centre thin rectangular

11 I = 1 12 ML2 I = 1 3 ML2 I = 1 12 M(a2 + b 2 ) I = 1 3 Ma2 L L b b a a slender rod, axis through centre slender rod, axis through one end rectangular plate, axis through centre thin rectangular plate, axis along edge I = 1 2 M( ) I = 1 2 M2 I = M 2 I = 2 5 M2 I = 2 3 M2 1 2 hollow cylinder solid cylinder thin-walled hollow cylinder solid sphere thin-walled hollow sphere SHM: Waves: For a point-like particle of mass M a distance from the axis of rotation, I = M 2 Parallel axis theorem: I p = I cm + Md 2 standing wave ω = 2πf = 2π/T T pend = 2π L/g = 2π I/mgd T spring = 2π m/k T torsion = 2π I/κ k = 2π/λ v = ±λf v 2 string = F T /µ µ = M/L y(x, t) = A SW sin (kx) sin (ωt) λn = 2L n f n = n v, n=1, 2, 3,... 2L x(t) = A cos (ωt + ) v(t) = ωa sin (ωt + ) a(t) = ω 2 A cos (ωt + ) Damped/forced: x(t) = A cos (ω t)e (b/2m)t ω = k/m b 2 /4m 2 A = F max / (k mωd 2)2 + b 2 ωd 2 [ ( travelling x wave : y(x, t) = A cos 2π λ t )] T = A cos (kx ωt) P = µf T ω 2 A 2 sin 2 (kx ωt), P = 1 µft ω 2 A 2 ( 2 power I(r) = = P ) for 3D sphere surf area at r 4πr2

PHYSICS 218 Exam 3 Fall, 2013

PHYSICS 218 Exam 3 Fall, 2013 Wednesday, November 20, 2013 Please read the information on the cover page BUT DO NOT OPEN the exam until instructed to do so! Name: Signature: Student ID: E-mail: Section