Physics 122 Midterm (Fall 2010)

University of Waterloo Department of Physics & Astronomy Physics 121 – Midterm Fall 2010 Instructors: Dr. Robert Mann (sections 2,3) Dr. Guenter Scholz (section 1) Date: November 4 , 2009 Time: 19:00-21:00 Duration: 2 hours (120 minutes) rh Instructions Important: Write your name and student ID on each page. If necessary you may use the back of the page to continue your answer but not the back of the previous page. The pages may be separated as part of the marking process. 5 questions constitute a complete paper. Each question is of equal value. All questions will be counted.The last page contains some constants and formula that may be useful. You may remove and keep this page as a souvenir. Aids Permitted calculator writing implements Question Points Score 1 2 3 4 5 6 Total: 20 20 20 20 20 20 1. a Up, up, and away [15] A balloonist can’t resist throwing a drink to another balloonist. The ‘thrower’ is moving at v = -15j m/min while the ‘catcherâ€℠¢ is moving at v = 15j m/min. At the instance the former throws the drink the ‘catcher’ is at (3i – 10j) m from the ‘thrower’. If the throw is horizontal, at what speed does the drink need to be thrown to be caught? 1. Logging [10] Estimate, via a reasonable calculation, the number of trees that need to be cut down to supply the pulp for one day’s edition of the major newspaper â€Å"The Record† in Kitchener-Waterloo2. Circus performance A pivoting pulley hanging from the center tent top allows a lady artist (m = 40 kg) to rotate freely while her partner (M = 100 kg) supports her via a rope over the pulley. he rope length, measured from the pulley, holding the rotating artist is 3. 0 m and the helper is not accelerating. [4] (a) Draw a Free Body Diagram of each performer and the pulley; clearly indicate the forces. 5] (b) What is the tension in the rope? [5] (c) At what angle, with respect to the vertical, is the lady’s supportin g rope? [6] (d) What is the period (time for one revolution) of her revolution? 3. Going Fishing Because of inclement weather, a boater needs to travel as quickly as possible across a channel from a fishing spot at ‘A’ to the harbour at ‘B’. The harbour is 10. 0 km East and 15 km North of his fishing spot. A tide is flowing at 3. 0 km/hr 45 to the South of East, and the boat’s speed is 8. 00 km/hr relative to the water.B A [5] (a) What is the heading of the boater for the shortest trip? clearly indicate this angle on an appropriate diagram) [5] (b) Find the boat’s speed relative to the shore. [5] (c) What is the shortest time for the trip? [5] (d) If there were no tide, how much time would the trip require? m F 4. Blocked Up M ! A small block of mass m rests on the incline of a wedge of mass M and angle ! , whose coefficient of static friction is  µ. The wedge is on a frictionless surface. [8] (a) If m = 1 kg and M = 20 kg, what is the minim um force, F, you need to apply to the wedge that will prevent the small block from sliding down the slope if  µ =0 and ! 45 o ? [12] (b) For general values of m, M ,! and  µ, find the minimum force that you need to apply to the wedge that will cause the small block to just begin to move up the slope. 5. Piano Moving Doofus and Diligent are moving a piano of mass M = 300 kg using the pulley system shown in the diagram. The rope around the pulley holding the piano, is tied to the axle of the top pulley which in turn is fastened to the ceiling. Diligent is holding the rope at the left, suspending the piano 10 metres above the ground. [4] (a) Draw free-body diagrams of the piano and of each pulley.Be sure to include all relevant forces. [7] (b) How much force is Diligent applying to keep the piano suspended? [6] (c) Doofus tries to help Diligent by climbing onto the upper platform and taking the rope off of the hook attaching the uppermost pulley to the platform, thinking he can hel p pull from there. How much force must he exert to keep the piano suspended? [3] (d) Doofus finds that he cannot support the weight and lets go of the rope. How long does Diligent have to dodge the piano before it hits the ground? 6. Safety First Doofus and Diligent are going to a party.They each buckle themselves in with seatbelts Diligent is hold a 25 kg keg of beer on his lap while Doofus drives. [4] (a) While travelling 60 km/hr, Doofus has to make an emergency stop over a distance of 45 m. How much force will Diligent's arms have to exert on the keg during this deceleration period so that it stays on his lap? [4] (b) The trip continues and the car turns a corner onto a highway, going at 90 km/hr. Suddenly Doofus sees a car heading toward them. He panics, locking the brakes and veering off to the right onto a very steep and muddy road allowance with a grade of 35%.The car slides up this nearly frictionless hill and comes to a stop at a cliff edge. How much distance does it cover ? [7] (c) From the top of this cliff they can see the location of the party a 2 kilometers east and 1000 meters above where they are. Diligent wants to walk the rest of the way, but Doofus proposes to use the motorized hang-glider in the back of the car to fly over there with the keg. He says he can fly straight there with a speed of 40 km/hr using its 6 horsepower motor, and sets off with the keg.While he is getting ready, Diligent, knowing that a horsepower is 750 Watts, calculates how much mass the glider can carry. He looks at the package and sees that air buoyancy alone can always support the glider as long as it is not carrying anything; but Doofus weighs 75 kg. Can Doofus carry the keg this way? [5] (d) As Doofus takes off, Diligent shouts out how slow he must fly to get to the party. What does he shout?. Useful Formulae Kinematics (a=const) Work, WKE, Power ! ! ! 1! r2 = r1 + v1 (t2 ! t1 ) + a (t2 ! t1 ) 2 2 ! ! ! v2 = v1 + a (t 2 ! t1 ) ! ! W = F † ! = Fx ! x + Fy ! y + Fz ! z (constant force) ! ! ! 2 v2 ! v12 = 2a † (r2 ! r1 ) Circular Motion K= 1 2 mv 2 Kinetic energy Wnet = K f † K i = ! K ar = ac = v2 r at = dv dt (uniform motion) P= P= !W ! t ! ! dr ! ! dW =F! = F ! v dt dt T U = 2! r dist = speed v Newton’s Laws ! F net = ! F = ma ! ! F12 = ! F21 ! ! Fg = mg Math ! ! ? A + B = ( Ax + Bx )i + ( Ay + B y ) ? j ! ! A ! B = Ax Bx + Ay B y Fs (x) = ! kx sin 2 ! + cos 2 ! = 1 f s ! f s ,max =  µ s n f k = µ k n Relative Motion sin A sin B sin C = = a b c a 2 = b 2 + c 2 ! 2bc cos A ! ! ! v AB = v A ! vB