Physics 218: Exam 1. Class of 2:20pm. February 14th, You have the full class period to complete the exam.

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Phyic 218: Exam 1 Cla of 2:20pm February 14th, 2012. Rule of the exam: 1. You have the full cla period to complete the exam. 2. Formulae are provided on the lat page. You may NOT ue any other formula heet. 3. When calculating numerical value, be ure to keep track of unit. 4. You may ue thi exam or come up front for cratch paper. 5. Be ure to put a box around your final anwer and clearly indicate your work to your grader. 6. Clearly erae any unwanted mark. No credit will be given if we cant figure out which anwer you are chooing, or which anwer you want u to conider. 7. Partial credit can be given only if your work i clearly explained and labeled. 8. All work mut be hown to get credit for the anwer marked. If the anwer marked doe not obviouly follow from the hown work, even if the anwer i correct, you will not get credit for the anwer. Sign below to indicate your undertanding of the above rule. Name : Student ID :...................................... Signature :......................................

Part 1: Baic Quetion Point Score Baic 12 Velocity and Acceleration 12 Aircraft Carrier 18 Acceleration in both component 24 Reading DVD 20 Alien 14 Total: 100 Solve the following problem in baic unit converion. (12 point) (a) (2 point) What are the baic unit of time, ma and length in the International Sytem of unit? For time i econd, for ma i kg, and for length i meter. (b) (5 point) Convert 56 gram cm 3 to kg m 3 Knowing that : 10 3 gram = 1kg 1gram = 10 3 kg 10 2 cm = 1m 1cm = 10 2 m We get : 56 gram = 56 10 3 kg cm 3 (10 2 m) 3 = 56 kg kg 10 3 = 56 10 6 103 m3 m 3 (c) (5 point) Compute the magnitude of the vector c = a b, where the vector a ha a magnitude of 3 m traight in the North direction, and the vector b 2 ha a magnitude of 5 m at an angle of 37 North of the Eat direction. 2 a = (0, 3) m 2 b = (5co(37), 5in(37)) m 2 c = a b = ( 5co(37), 3 5in(37)) m 2 c = ( 5co(37)) 2 + (3 5in(37)) 2 4 m 2

Part 2: Velocity and Acceleration (12 point) An object i moving in the (x,y) plane. The two plot below how the x and y poition of the object a a function of time. 4 x [m] 4 y [m] 3 3 2 2 1 1 0 0 1 2 3 4 5 time [] 0 0 1 2 3 4 5 time [] (a) (4 point) Etimate the magnitude of the object velocity at time t = 2. From the lope of the firt plot at t = 2 we get : v x (t = 2) = 1 m From the lope of the econd plot at t = 2 we get : v y (t = 2) = 0.5 m from which we get : v = ( 1) 2 + 0.5 2 = 1.118 m (b) (4 point) If y repreent North and x repreent Eat, what direction i the object moving at t = 4? Explain your reaoning. From the lope of the firt plot at t = 4 we get : v x (t = 4) = 0 m From the lope of the econd plot at t = 4 we get : v y (t = 4) < 0 m from which we get that the object i moving in the South direction: (c) (4 point) At time t = 4 determine the ign of the acceleration in the x and y component. From the concavity of the firt plot at t = 4 we get : a x (t = 4) > 0 m From the concavity of the econd plot at t = 4 we get : a y (t = 4) < 0 m Page 2

Part 3: Aircraft Carrier (18 point) An AV-8B II plane need to land on it aircraft carrier. A it i approaching the landing trip the plane i heading outh at a peed of 50 m with repect to the air. The carrier meaure the peed of the air to be 30 m at a direction of 30 degree wet of north with repect to the carrier. (a) (4 point) Draw a baic diagram indicating the relevant reference frame and draw ALL relevant velocitie. ŷ v A/C 30 ŷ Plane Carrier Air v P/A ˆx ˆx (b) (6 point) Compute both component of the velocity of the plane with repect to the carrier v P/C = v P/A + v A/C = (0, 50) m + ( 30in(30 ), 30co(30 )) m = ( 30in(30 ), 50 + 30co(30 )) m v P/C = ( 15, 24)) m (c) (4 point) If the only condition for a ucceful landing i that the peed of the plane at the moment of the landing i le than 25 m with repect to the carrier hip, will the plane land uccefully? v P/C = ( 30in(30 )) 2 + ( 50 + 30co(30 )) 2 = 28.3 m o the plane will not land ucefully. (d) (4 point) If before landing the direction of the wind change to be now in the north direction while it peed remain at 30 m a before, will the plane be able to land now? o the plane will land uccefully. v P/C = 20 2 = 20 m Page 3

Part 4: Acceleration in both component (24 point) On the urface of a moon with magnitude of gravity a cientific device i launched at an angle α with an initial peed of v 0. The device ha a et of burner on it ide providing the device with a contant horizontal acceleration in the poitive x direction of a x. Aume the ma of the device i contant and the atmophere of the moon can be neglected. All anwer mut be in term of the known variable, α, v 0, and a x. (a) (4 point) Draw the problem and indicate a coordinate ytem α v 0 (b) (5 point) Write the equation of motion of the device for that coordinate ytem. That i the horizontal and vertical component a a function of time, X(t) and Y(t). If you put term that are zero indicate o. x(t) = v 0 co(α)t + a x t 2 2 y(t) = v 0 in(α)t t 2 2 (c) (5 point) Find the time it take the device to fall to the ground. y(t g ) = 0 = v 0 in(α)t g t 2 g 2 t g = 2v 0in(α) Page 4

(d) (5 point) Find the range of the device, thi i the horizontal ditance from where it launched to where it landed. t 2 g x(t g ) = v 0 co(α)t g + a x 2 plugging t g from the previou point we get [ x(t g ) = 2v2 0 co(α)in(α) + a ] x in 2 (α) (e) (5 point) Find the horizontal poition of the device at the point it reache the maximum height. If you make any aumption jutify it. v y (t m ) = 0 = v y in(α) t m t m = v 0in(α) t 2 m x(t m ) = v 0 co(α)t m + a x 2 = v 0 co(α) v 0in(α) v + a 0in 2 2 (α) x 2g 2 m [ x(t m ) = v2 0 co(α)in(α) + a ] x in 2 (α) 2 Page 5

Part 5: Reading DVD (20 point) A DVD dik i read by pinning the dik over a laer ytem whoe radial poition from the center of the dik can be controlled from a radiu of 2.2cm and up to a radiu of 5.7cm. However, in a DVD the frequency of revolution mut be controlled uch that the linear velocity of the dik at the radiu where the laer i poitioned mut alway be a contant 3.5 m. Uniform circular motion can be aumed in thi problem. All anwer mut be a number with proper unit. R L 2.2 cm 5.7 cm R (a) (6 point) Find the frequencie at which the DVD mut pin when the laer ytem i reading at it inner radiu and at it outer radiu. Expre in unit of revolution per econd. v = 2πR L T dvd = 2πR L f dvd = 3.5 m f dvd = 3.5 m 2πR L at inner radiu: at outer radiu: 3.5 m = 25.3 revolution/econd 2π0.022m 3.5 m = 9.8 revolution/econd 2π0.057m Page 6

(b) (4 point) Find the acceleration at the outer radiu of the dik when the dik i pinning at the two frequencie find in the previou problem. Expre in unit of m 2. a = 4π2 5.7cm T 2 dvd = 4π 2 0.057mf 2 dvd at inner radiu a inner = 1440 m 2 at outer radiu a outer = 216 m 2 (c) (10 point) Now a little ant hop into the DVD and it at radiu R ant = 3cm. If the laer ytem i reading at radiu R L = 4cm what i the acceleration experienced by the ant? a ant = v2 ant R ant = 4π2 R ant T 2 dvd = 4π2 R ant 4π 2 R 2 L v 2 = R ant v 2 = a RL 2 ant = 229 m Page 7

Part 6: Alien (14 point) The pacehip you command i about land on a planet. The computer find the landing olution and when the pacehip i at a height h (t = 0) it et the velocity of the pacehip to be v y (t) = v 0y + kt 2, where h, v 0y and k are all poitive and known and the coordinate ytem aume that the ŷ axi goe from the urface of the planet up to the pacehip. All anwer mut be in term of the known variable h, v 0y, and k. (a) (2 point) Find the time it take to land if the landing i uppoed to be perfectly mooth (i.e. the vertical velocity mut be zero at landing) v y (tg) = v 0y + kt 2 g = 0 t g = v0y k (b) (4 point) I the vertical acceleration in thi problem contant or not? I it poitive or negative at the moment of landing? a y (tg) = d( v 0y + kt 2 ) dt = 2kt acceleration poitive and not contant (c) (8 point) Find the poition a a function of time and the height of the pacehip when half the landing time ha elaped. y(t) = y 0 + = h + t 0 t 0 v y (t)dt = 2kt ( v0y + kt 2) dt y(t) = h v 0y t + k t3 3 t g y(t g /2) = h v 0y 2 + k t3 g 2 3 3 y(t g /2) = h v 0y v0y 2 k + k 24 ( v0y ) 3/2 k Page 8