Physics 204A FINAL EXAM Chapters 1-14 Fall 2005

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1 Name: Solve the following problems in the space provided Use the back of the page if needed Each problem is worth 10 points You must show our work in a logical fashion starting with the correctl applied phsical principles The equations ou need are on the equation sheet Your score will be maimized if our work is eas to follow because partial credit will be awarded 1 Pumpkins dropped from the edge of Butte Hall fall 60m to the ground below The crowd is 100m back from the edge of the building Find the maimum horizontal velocit the pumpkins can have when released so that the can t reach the crowd v o Using the kinematic equation without the final speed along, = o + v o t + 1 t Putting in the zero values and solving for t, o o = 60m = 100m v o =? v o v = v o v =? a = -980m/s t =? 0 = o + 1 a t " t = # o Using the kinematic equation without the final speed along, = o + v o t + 1 a t Substituting and solving for the initial speed, Putting in the numbers, = v o v o = (100) " o # v o = " o "980 "(60) # v o = 81m /s A racecar goes completel around the racetrack shown at the right moving at a constant speed of 00km/h the entire wa Indicate in the diagram the direction of the acceleration at each of the labeled points and rank the accelerations from largest to smallest Eplain our thinking Since the speed of the car is constant there will be no tangential acceleration The centripetal acceleration is given b, a c = r Since v is constant, the acceleration will be the largest when the radius of the turn is the smallest Where the track is straight, the radius is infinite and the centripetal acceleration is zero So the ranking will be, D>B>C>A=0 a 1

2 3 The car at the right has a mass of 750kg The wheels of the car are broken and cannot roll The will onl slide The coefficient of static friction between the tires and the road is 0800 Find the tension in the cable needed to move the car Note that the normal force in the sketch is the sum of all the normal forces on all the tires The same is true for the friction Appling the Second Law to each direction separatel, "F = ma $ # = "F = m + $ # = $ Using the definition of the COSF and substituting from above, µ " F = t = mg Doing the mountain of algebra to solve for the tension, µmg = +µ = (0800)(750)(980) +(0800) " = 4640N 60 4 The pogo stick at the right uses a spring (k = 50kN/m) At position A the child is at rest while the spring is compressed 100cm At position B, the spring is relaed and the child is moving upward At position C the child is at the top of the jump Find (a)the speed at position B and (b)the height if the jump The mass of the child and pogo stick is 500kg U A = 1 k A " mg A K A U C = mg C K c U B K B = 1 m (a)appling the Law of Conservation of Energ between positions A and B, "U + "K # ( U / B $U A ) + (K B $ K / A ) # K B = U A Putting in the energies and solving, 1 m = 1 k A " mg A # v = k m A " g A v = (0100) " (980)(0100) # v =174m /s (b)appling the Law of Conservation of Energ between positions A and C, "U + "K # (U C $U A ) + ( K / C $ K / A ) # U C = U A Putting in the energies and solving, mg C = 1 k A " mg A # C = k A mg " = (5000)(0100) A (500)(980) " 0100 # C 155m

3 5 Two ice skaters are moving along together at 300m/s The first skater has a weight of 700N The first skater pushes the second skater forward so she speeds up to 400m/s As a result, the first skater slows to 5m/s Find the weight of the second skater The linear momentum before the push is, p o = (m 1 )v o m 1 m v 0 The momentum after the collision is, p = m 1 + Using the Law of Conservation of Linear Momentum, (m 1 )v o = m 1 Solving for the mass of the second skater then multipling both sides b the acceleration due to gravit, m = m 1 v o " " v o # m g = m 1 g v o " " v o # m g = (700N) 300 " " 300 # m g = 55N 6 A frictionless 500g air puck (object A) slides down an incline In addition, the four objects described in the table below roll down the same incline Rank from shortest to longest the time for the object to go from the top to the bottom Eplain our reasoning Object Mass Inner radius Outer radius Length B Solid Rod 450g None 300cm 10cm C Hoop 600g 300cm 500cm 600cm D Disk 00g None 500cm 100cm E Ring 300g 100cm 100cm 300cm Using the Law of Conservation of Energ, the change in potential energ must equal the change in kinetic energ for each object Since the change in potential energ is the same for each object, the change in kinetic energ will also be the same The kinetic energ is the sum of the translational kinetic energ and the rotational kinetic energ Since the rotational kinetic energ depends upon the rotational inertia, the object with the smallest rotational inertia will have the smallest rotational KE and the largest translational KE It is important to note that the radius and mass cancel in the problem and make no difference The winner of the race is the air puck since it spends no KE on rotation and the loser is the ring which spends the largest fraction on rotational KE The ranking is A<B=D<C<E 3

4 7 A 900m kg tree is cut down b a logging compan A machine uses a short cable to lift it b one end until it is in the position shown at the right Find the magnitude and show the direction of each force that acts on the log Appling the Second Law, "F = ma cos(40 +0 ) $ # = "F = m sin(40 +0 ) + $ # = $ "# o = I$ % l sin0 & l cos40 Using the mass weight rule, = mg = (1500)(98) " =147kN Solving the torque equation for the tension, = cos40 sin0 = (147)cos40 " =165kN sin0 Substitute into the equation and solve for the normal force, =147 "165# 443kN = 443N Finall, solve the equation for the friction, = =165" = 85kN 8 The machine in the previous problem continues to lift the log until it is off the ground It gentl swings back and forth pivoted at the end of the cable Find the period of oscillation The log acts like a phsical pendulum The angular frequenc should be given b, " = The rotational inertia of a stick about one end is, I = 1 3 ml " # = 3mgr = 3gr ml l The r is from the pivot point to the cm so it is half the length, r = l " # = The period is related to the angular frequenc b, " = # $ T = # l = # T " 3g Plugging in the numbers, T = " (900) 3(980) # T = 49s 3gl = l 3g l mgr I 4

5 9 The following is a quote from a periodical I read over the summer, It is a bit bigger than Pluto, an astounding 145 billion kilometers from the sun, and the most distant object ever seen in the solar sstem Last week's discover of a "10th planet" Find the time for this new planet to orbit the sun m M r Appling Newton s Second Law to the planet, "F = ma # = ma Using the Universal Law of Gravitation, = G mm " G mm = ma " G M = a r r r Since the motion is circular use the centripetal acceleration, a = r " G M = r r " G M r = The speed will just be the circumference of the orbit divided b the period, v = "r # G M = ( "r T r T ) Solving for the period and plugging in the numbers, T = 4" r 3 GM = 4" ( ) 3 (66710 #11 )( ) $ T = s = 955ears 10 Legend has it that Archimedes was asked b the king to determine whether his crown was made of solid gold without damaging the crown in an wa Archimedes knew that the densit of gold was kg/m 3 So he weighted the crown in air and got 784N He then weighted the crown in water and got 684N Figure out what Archimedes told the king + F s F B When the crown is in the water the three forces on it are the scale pulling upward, gravit pulling downward, and the upward buoant force Appling the Second Law, "F = ma # F s + F B $ # F B = $ F s The buoant force is given b Archimedes Principle, F B = " w Vg The volume of the crown that displaces the water can be written in terms of the densit and mass of the crown, " # m V $ V = m " Substituting into the equation for the buoant force, F B = " w m " g = " w " Substituting into the equation from the Second Law, " w " F = F # g s Solving for the densit of the crown, " w " = F # s F $ " = " g 784 w = (1000) # F s 784 # 684 $ " = kg /m 3 Who wants to tell the king that he has been ripped off? 5