Physics 204A FINAL EXAM Chapters 1-14 Spring 2006

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1 Nae: Solve the following probles in the space provided Use the back of the page if needed Each proble is worth 0 points You ust show your work in a logical fashion starting with the correctly applied physical principles The equations you need are on the equation sheet Your score will be axiized if your work is easy to follow because partial credit will be awarded A soccer ball is kick with a speed of 80/s at a 300 angle above the horizontal Find (a)the tie it is in the air and (b)the distance it travels before it hits the ground y θ v o x o = 0 y o = 0 x =? y = 0 v ox = v o cosθ v oy = v o sinθ v x = v ox v y =? a x = 0 a y = -980/s 2 t =? t =? x (a)using the kineatic equation along the y-axis, y = y o + v oy t + 2 a yt 2 Putting in the things that are zero and solving for t, 0 = 0 + v oy t + 2 a y t 2 " t = # 2v oy a y = # 2v o sin$ a y t = "2(80)sin300 # t =84s "980 (b)using the kineatic equation along the x-axis, x = x o + v ox t + a 2 xt 2 = v ox t = v o t cos" x = (80)(84)cos300 " x = A gardener pushes a 700kg lawn ower at the constant speed of 0800/s across level ground by exerting a force of 600N directed along the handle, which akes a 500 angle with the horizontal Find the size of (a)the noral force and (b)the frictional force on the ower Applying the Second Law for each direction reebering that the constant velocity eans no acceleration, "F x = a x # cos$ % = 0 and "F y = a y # $ sin% $ = 0 Soling the y-equation for the noral force and the x-equation for the frictional force, = cos" and = sin" + = 600cos500 " = 386N and = 600sin500 +(700)(980) " =5N y θ x

2 3 The golf club shown at the right is 00 long The head has a ass of 700g and the shaft has a ass of 200g The center of ass is 850c below the top of the shaft At the botto of the swing, the club head is oving at 900/s in circular otion about the top of the shaft Find the vertical coponent of the force that the shaft exerts on the head at this instant Applying the Second Law to the head of the club, "F = a # $ = a The acceleration is centripetal so, " = v 2 r Solving for the force due to the shaft, = + v 2 r = g + v 2 r = " g + v 2 % $ ' # r & " % = (0700) $ ' ( = 636N # 00 & 4 The 520kg woan pictured at the right falls into the safety net and stretches it 0500 The net springs back and he flies through the third story window 900 above the net at a speed of 30/s Estiate the height of the window she fell fro As she falls out of the first window, U o = gh o and K o = 0 As she flies into the second window, U = gh and K = 2 v 2 Applying the Law of Conservation of Energy, "U + "K = 0 # (gh $ gh o ) + ( 2 v 2 $ 0) = 0 Solving for the initial height, gh o = gh + v 2 2 " h o = h + v 2 2g h o = (30)2 2(980) " h o =76 shaft + c head 850c 00c 2

3 5 Three equally assive and equally strong astronauts are outside their ship in outer space Two of the get the bright idea to play a gae of catch by throwing the third one back and forth Suppose the gae begins with the first astronaut throwing the third astronaut toward the second astronaut at a speed v o Describe the rest of the gae This eans that you ust find the velocity of each astronaut after each catch and after each throw Sketches of the gae at each stage ight be the best way to explain your answer Be sure to state the principle or principles you use The ain principle is the Law of Conservation of Linear oentu The gae begins with astronaut pushing astronaut 3 to the right at v o Due to the Law of Conservation of oentu, astronaut recoils with a speed v o to the left Astronaut 2 is at rest v o v o v = 0 When astronaut 3 is caught by astronaut 2, the Law of Conservation of oentu requires that they head off to the right at one-half of v o v o 2 v o Since the astronauts are equally strong, when astronaut 2 pushes astronaut 3 back towards astronaut, he can only change astronaut 3 s velocity by v o As a result, astronaut 3 heads back to the left at one-half v o According to the Law of Conservation of oentu, astronaut 2 will ove to the right at three-halves v o v o 2 v o 3 2 v o At this point the gae ends because astronaut3 can never catch up to astronaut 3

4 6 A circular saw blade has a ass of 200g and a radius of 920c It starts fro rest and 200s later is had a rotation rate of 20rad/s Find (a)the torque exerted by the otor and (b)the energy the otor supplied to the blade (a)using the Second Law for Rotation, "# = I$ % # = I$ The saw blade is like a solid disk so, I = 2 r2 The definition of angular acceleration is, " # d$ dt % " = $ &$ o = $ t t # Putting it all together, " = 2 r2 t = 20 2 (0200)(00920)2 2 $ " = 00889N # (b)the energy supplied by the otor is converted into the rotational kinetic energy of the blade, K = 2 I" 2 = 4 r 2 " 2 = 4 (0200)(00920)2 (20) 2 # K =87J 7 A 500g picture frae is propped up at a 530 degree angle by a stick (shown in gray) that contacts the frae one-third of the way down The stick akes a 450 angle with the frae Assuing that the stick only exerts force perpendicular to the frae, find the size of the force it exerts Using the Second Law for Rotation, "# = I$ = 0, F s l l 3 45 since the frae is otionless About the pivot indicated there are only two torques, Substituting into the Second law, " g = #g l 2 cos53 and " s = F s 2l 3 pivot 53 "g l 2 cos53 +F s 2l = 0 # F 3 s = 3 gcos53 4 F s = 3 4 (0500)(980)cos53 " F s = 22N 4

5 8 Shown below is a stick of ass,, hung by a thread fro its center At the end of each stick are two equal asses, The threads are identical The arrangeent is set into oscillatory otion about the axis defined by the thread In other words, the stick and asses ove in a horizontal plane Rank fro greatest to least based on the period of oscillation That is, rank the syste with the longest period first and the shortest period last Be sure to explain your reasoning A B C = 30g = 30g = 30g = 20g = 30g = 0g D E F = 60g = 30g = 60g = 20g = 60g = 0g These are torsional pendulus The angular frequency of a torsional pendulu is " = # I The frequency is related to the period T = 2" # = 2" I $ Since the threads are identical, the torsion constant is the sae in all cases so the period just grows as the rotational inertia grows The rotational inertia for a stick about the center is I = 2 l2, while the rotational inertia due to the other asses is I 2 = 2( l 2 )2 = 2 l2 So the total rotational inertia is, I = I + I 2 = 2 l2 + 2 l2 = ( )l 2 For A: I = ( )l 2 =75l 2 For B: I = ( )l 2 =25l 2 For C: I = ( )l 2 = 75l 2 For D: I = ( )l 2 = 20l 2 For E: I = ( )l 2 =5l 2 For F: I = ( )l 2 =0l 2 The ranking is therefore, D>A>E>B>F>C 5

6 9 Coplete the physics proble alluded to in the newspaper article at the right Assue Xena s oon copletes one orbit of radius R in a tie T Find the ass of Xena R Applying the Second Law to the oon, "F = a Using the Law of Gravitation, G R 2 = a " G R 2 = a Using the centripetal acceleration, G R = v 2 2 R " G R = v 2 The speed is the circuference over the period, G R = 2"R 2 # & % ( $ T ' Solving for the ass of the planet Xena, = 4" 2 R 3 GT 2 Scientists Discover 0th Planet's oon By Alicia Chang Associated Press 0 October 2005 LOS ANGELES (AP) -- The astronoers who clai to have discovered the 0th planet in the solar syste have another intriguing announceent: It has a oon While observing the new, so-called planet fro Hawaii last onth, a tea of astronoers led by ichael Brown of the California Institute of Technology spotted a faint object trailing next to it Because it was oving, astronoers ruled it was a oon and not a background star, which is stationary The oon discovery is iportant because it can help scientists deterine the new planet's ass In July, Brown announced the discovery of an icy, rocky object larger than Pluto in the Kuiper Belt, a disc of icy bodies beyond Neptune Brown labeled the object a planet and nicknaed it Xena after the lead character in the forer TV series ``Xena: Warrior Princess'' The oon was nicknaed Gabrielle, after Xena's faithful traveling sidekick By deterining the oon's distance and orbit around Xena, scientists can calculate how heavy Xena is For exaple, the faster a oon goes around a planet, the ore assive a planet is 0 Treasure divers investigating a shipwreck find a gold bar that is 500c by 00c by 300c The density of gold is 93x0 3 kg/ 3 Find the force they ust exert to lift it off the ocean floor Applying the Second Law, "F = a # + F B $ = 0 # = $ F B The buoyant force is equal to the weight of the water displaced, = g " f g + F B The ass of the displaced fluid and the ass of the gold each can be written in ters of the volue of the gold using the definition of density, = "Vg # " f Vg = (" # " f )Vg = (93x0 3 "00x0 3 )(00500)(000)(0300)(980) # = 269N 6