Physics 110. Exam #1. September 30, 2016

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1 Phyic 110 Exa #1 Septebe 30, 016 Nae Pleae ead and follow thee intuction caefully: Read all poble caefully befoe attepting to olve the. You wok ut be legible, and the oganization clea. You ut how all wok, including coect vecto notation. You will not eceive full cedit fo coect anwe without adequate explanation. You will not eceive full cedit if incoect wok o explanation ae ixed in with coect wok. So eae o co out anything you don t want gaded. Make explanation coplete but bief. Do not wite a lot of poe. Include diaga. Show what goe into a calculation, not jut the final nube. Fo exaple! p v " = 5kg ( ) ( ) =10 kg Give tandad SI unit with you eult unle pecifically aked fo a cetain unit. Unle pecifically aked to deive a eult, you ay tat with the foula given on the foula heet including equation coeponding to the fundaental concept. Go fo patial cedit. If you cannot do oe potion of a poble, invent a ybol and/o value fo the quantity you can t calculate (explain that you ae doing thi), and ue it to do the et of the poble. All ultiple choice quetion ae woth 4 point and each fee-epone pat i woth 8 point Poble #1 /4 Poble # /4 Poble #3 /4 Total /7 I affi that I have caied out y acadeic endeavo with full acadeic honety.

2 1. Suppoe that you and you younge ibling decide to play a gae of catch. To ake the gae oe inteeting you decide to ue you houe a an obtacle a hown below. You tand on the left ide of you houe and you ibling tand on the ight. Each of you tand 6 fo on you epective ide and the houe i 6 wide. You thow the ball fo a height of 1 above the gound and you ibling catche it at the ae height above the gound. a. With what velocity (agnitude and diection) did you thow the ball o that it jut baely ake it ove top of the houe? We can find the initial y-coponent of the velocity by exaining the vetical otion to the top of the houe fo the point of eleae. We have, v fy = 0 = v iy + a y Δy = v iy gδy v iy = gδy = 9.8 ( 6 1) = 9.9 whee the height of the houe i deteined fo tan 45 = O O = 3 tan 45 = 3 3. h = 3 + O = = 6 To figue out the x-coponent of the velocity, we need to know the tie of flight of the pojectile. To do thi we ue the vetical tajectoy equation. y f = y i + v iy t + 1 a y t = v iy t 1 gt 1 = t 9.8 t 0 = ( t )t t = 0.0 Then the hoizontal coponent of the velocity i: x f = x i + v ix t + 1 a x t = v ix t v ix = x f t = 18.0 = 8.9 The velocity i v i = v ix + v ix = ( 8.9 ) + ( 9.9 ) = 13.3 oiented at φ = tan 1 v iy v ix = 9.9 tan = 480.

3 b. Suppoe that you ibling, tying to outdo you, thow the ball at twice the peed a you thew the ball. Fo whee you wee initially tanding (at a location 6 on the left ide of the houe), whee and by what aount would you have to ove in ode to catch the ball? Aue that you ibling thow the ball at the ae angle a you did. If you could not get an anwe to pat a, aue that you thew the ball with an initial velocity of v i = 0 at θ = The tie of flight of the pojectile with thi new peed ( v i = v i,pat a = 6.6 ) i obtained fo the vetical tajectoy equation y f = y i + v iy t + 1 a y t = v iy t 1 gt = ( v i inθ )t 1 gt t = v inθ i = 6.6 in 48 = 4.03 g 9.8 The hoizontal ditance i x f = x i + v ix t + 1 a x t = v ix t = ( v i coθ )t x f = 6.6 co = Thu the ball will land 71.8 to the left of whee it wa thown, o you ll have to ove to the left by an aount = c. In ode to the ball to be launched at any initial velocity, you have to apply a foce to the ball and acceleate it fo et to the initial velocity that you want. The agnitude of the foce needed to acceleate the ball to any initial velocity i given by 1. F = v, whee t i the tie of flight of the ball fo the font to the back of t the houe.. F = v, whee Δx i the ditance ove which the ball i acceleated. Δx 3. F = v, whee Δx i the hoizontal ditance the ball cove fo the font to Δx the back of the houe. 4. F = vt, whee t i the tie the ball i in contact with you hand.

4 y. A 75kg nowboade ha an initial velocity of 5 at the top of a 8 0 incline a hown below. Afte liding down the 110 long incline (on which thee i fiction between the nowboad and the now) the nowboade attain a peed v. a. What i the peed of the nowboade at the botto of the hill? Auing a tilted coodinate yte on the incline, we beak up the foce in to thei coponent pependicula to and paallel to the incline. We have: F x : F wx F f = F w inθ µ k = a x = a F y : y = coθ = a y = 0 = gcoθ a = F w inθ µ k a = 9.8 in8 0.18co8 = g( inθ µ k coθ ) ( ) = 3.04 The peed of the nowboade at the botto of the hill i given by v fx = v fx + a x Δx. v fx = ( 5 ) = 6.4 F f F W x y x F f F W b. At the botto of the hill, the nowboade then lide aco a flat hoizontal uface and coe to et afte a ditance Δx = x. What i the ditance Δx = x that the nowboade cove and how long doe it take the nowboade to coe to et? Fo the nowboade on the hoizontal uface, we take a taditional Cateian coodinate yte. Beaking the foce up in to coponent pependicula to and paallel to the gound we have. F x : F f = µ k = a x = a F y : = = a y = 0 = g a = µ k = µ kg = = 1.47 Thu the tie fo the nowboade to coe to et i given by v fx = v ix + a x t t = v ix a = = 18. The ditance taveled by the nowboade in coing to et i x f = x i + v ix t + 1 a x t = ( ) 1 ( 1.47 )( 18) = Anothe way: v fx = v ix + a x Δx Δx = v ix = a x ( 6.4 ) 1.47 = 37.1.

5 c. The velocity, a a function of tie, fo the nowboade liding aco the hoizontal uface (with µ k = 0.15 ) i plotted on the gaph below in geen? If the hoizontal uface wee intead fictionle, the velocity veu tie gaph in the fictionle cae (plotted in ed) copaed to the cae of fiction (plotted in geen) would ot cloely eeble v (/) 1.. v (/) t () t () 3. v (/) 4. v (/) t () t ()

6 3. A canival oto ide i hown below. The ide tand in a vetical cylindically walled oo which ha a adiu of 5.5. The ide tat up and when it otating at full peed of v, the floo uddenly dop away and the ide ae agically tuck to the wall. The ide i hown below. a. Fo a caefully labeled fee body diaga of the foce that act on the ide, what ae the coefficient of tatic fiction, µ, between a ide and the wall and the eaction foce of the wall on a ide? Aue that typical a ide ha a of 65kg and that the ide otate at a ate of 30 evolution pe inute. Hint: the peed of an object i the atio of the ditance taveled by the object to the tie it take to tavel that ditance. Fo the fee body diaga and applying a taditional Cateian coodinate yte we have beaking the foce up in the hoizontal and vetical diection F x : = a x = v F y : F f = a y = 0 F f = µ = g = g µ g = v µ µ = g v = = 0.18 ( 17.3 ) The velocity of the ide i given by v = Δx Δt = π π 5.5 = = 17.3 and the T tie fo one evolution i given by T = 1in 30ev 60 1in = ev. The eaction foce of the wall on the ide i given by = g = 65kg 9.8 = 3539N. µ 0.18 F f y x F W

7 b. In pat a, we aued that a typical ide ha a a of 65kg. In fact, anyone with any a can ide the ide and eain attached to the wall. The fact that anyone can ide thi ide i becaue 1. the coefficient of tatic fiction i independent of the a of the ide.. the noal foce fo the wall on the ide i independent of the a of the ide. 3. the noal foce fo the wall on the ide and the fictional foce alway oppoe the weight of the ide. 4. the fictional foce oppoe the weight of the ide and the noal foce fo the floo. c. At oe point the ide begin to low down and eventually will coe to a top when the ide i ove. Duing thi peiod of lowing down, the peed of the ide on the ide will be one half of the axiu peed v. At thi point, what will the acceleation of the ide be down the wall? Aue that the coefficient of kinetic fiction between the ide and the wall i µ k = 0.1. Auing the ae coodinate yte a in pat a, we have in fo the foce in the hoizontal and vetical diection: F x : = a x = v F y : F f = a y a y = µ k v g = a y = F f = ( ) 9.8 = µ k v g

8 Ueful foula: Motion in the = x, y o z-diection Unifo Cicula Motion Geoety /Algeba f = 0 + v 0 t + 1 a t a = v Cicle Tiangle Sphee v f = v 0 + a t F = a = v C = π A = 1 bh A = 4π A = π V = 4 3 π 3 v f = v 0 + a Δ v = π Quadatic equation : ax + bx + c = 0, T F G = G whoe olution ae given by : x = b ± 1 b 4ac a Vecto agnitude of avecto = v x + v y diectionof avecto φ = tan 1 v y v x Ueful Contant g = 9.8 G = N kg N A = ato ole k B = J K σ = W v = 343 K 4 ound Linea Moentu/Foce Wok/Enegy Heat p = v K t = 1 v p f = p + i F Δt K = 1 Iω F = a F = k x F f = µ 1 θ f = θi + ωit + αt M ρ = ω = πf = π T ω V f = ωi + αt F T P = S = π ω f = ω i + αδθ k A τ = Iα = F Pd = P0 + ρgd T P = π l g L = Iω FB = ρgv L τ A1 v1 = Av f = Li + Δt v = ± k A 1 x 1 A Δ = Δθ : v = ω : a α ρ1a1 v1 = ρ Av t = x( t) = Ain( πt T ) 1 1 P1 + ρv 1 + ρgh1 = P + ρv + ρgh a = ω v( t) = A k πt co( T ) v = fλ = ( T) β =10log I I 0 ; I o = W f n = nf 1 = n v L ; f n = nf 1 = n v 4L U g = gh U S = 1 kx W T = FdCoθ = ΔE T W R = τθ = ΔE R W net = W R + W T = ΔE R + ΔE T ΔE R + ΔE T + ΔU g + ΔU S = 0 ΔE R + ΔE T + ΔU g + ΔU S = ΔE di a( t) = A k πt in( T ) v = fλ = F T µ f n = nf 1 = n v L I = π f ρva T C = 5 T 9 [ F 3] T F = 9 5 T C + 3 L new = L old 1 + αδt ( ) ( ) ( ) : β = 3α A new = A old 1 + αδt V new = V old 1 + βδt PV = Nk B T 3 k BT = 1 v ΔQ = cδt P C = ΔQ Δt = ka L ΔT P R = ΔQ ΔT = εσaδt 4 ΔU = ΔQ ΔW Rotational Motion Fluid Siple Haonic Motion/Wave Sound

Test 2 phy a) How is the velocity of a particle defined? b) What is an inertial reference frame? c) Describe friction.

Test 2 phy a) How is the velocity of a particle defined? b) What is an inertial reference frame? c) Describe friction. Tet phy 40 1. a) How i the velocity of a paticle defined? b) What i an inetial efeence fae? c) Decibe fiction. phyic dealt otly with falling bodie. d) Copae the acceleation of a paticle in efeence fae