Physics I Math Assessment with Answers

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1 Physics I Mth Assessment with Answers The prpose of the following 10 qestions is to ssess some mth skills tht yo will need in Physics I These qestions will help yo identify some mth res tht yo my wnt to review s yo tke Physics I As lwys, select the est nswer from the choices given 1 Consider the following eqtion: (d x) = x where > 0, > 0, nd d > 0 We re looking for soltion for x where 0 < x < d A It is impossile to solve this eqtion except with nmericl techniqes ecse it involves the reciprocls of polynomils nd there is no generl forml for tht B There is exctly one soltion nd it cn e fond sing liner eqtion C There re exctly two soltions tht re fond sing the qdrtic forml D It is possile to write n expression for the soltion(s) sing the qdrtic forml, t it is impossile to know which one(s) (if ny) stisfy the condition on x withot knowing more ot,, nd/or d E None of the ove The correct nswer is B Most people tking this test relize tht the eqtion cn e pt into the following form: x = (d x) There is gret tempttion t this point to pt this eqtion into stndrd qdrtic form nd se the qdrtic forml The prolem cn e solved this wy, t it is esier simply to tke the sqre root of oth sides, which gives x = (d x) Since the conditions on the prolem grntee tht oth sides of this eqtion re positive, we don t worry ot the negtive sqre roots The soltion is x = 1+ d Cold we solve this sing the qdrtic forml? Yes, t neither C nor D is correct After some mnipltion, the soltions sing the qdrtic forml re 1

2 1± x = d 1 = d 1 + 1± 1 There re two soltions to this eqtion One is the correct soltion given ove The other is n incorrect soltion: x = 1 d If <, then x > d If =, then x is ndefined If >, then x <0 In ny cse, this soltion does not stisfy the conditions on x Therefore, there re not two soltions (choice C is incorrect) nd it is possile to select the correct soltion withot knowing more ot nd (choice D is incorrect) We will se this eqtion or similr eqtions to find points in spce where two grvittionl or electric forces cncel ech other In qestions -4, the fnction f(x) is continos nd differentile on the intervl [,] Tht mens tht the grph of f(x) is continos crve with no reks nd the slope of f(x) is welldefined When we spek of the slope of f(x), we men the slope of the grph of f(x) verss x If the slope of f(x) is negtive in [,], then A f (x) 0 t ll points in [,] B f (x) = 0 t some point in [,] C f () > f () D All of the ove E None of the ove C is the correct nswer If the slope of fnction is negtive throghot n intervl, it is decresing A negtive slope does not imply nything ot whether the fnction itself is positive or negtive

3 We stdy grphs of displcement nd velocity in Physics 1 nd drw conclsions from them 3 If f(x) < 0 in [,], then the slope of f(x) A is lwys positive t ll points in (,) B is lwys negtive t ll points in (,) C mst e zero t some point in [,] D All of the ove E None of the ove E is the correct nswer The fct tht fnction is negtive does not imply nything ot its slope We stdy grphs of displcement nd velocity in Physics 1 nd drw conclsions from them 4 In the intervl [,], the slope of f(x) is constnt The verge vle of f(x) in tht intervl is 1 A [f () + f ()] B ( [ + ]) f 1 C the re nder the grph of f(x) etween nd, divided y ( ) D All of the ove E None of the ove D is the correct nswer C is the definition of the verge of fnction A nd B re lso tre for the specil cse of fnction tht hs constnt slope, or in other words the grph is stright line We cn solve mny kinemtics prolems if we plot velocity verss time nd tke the re nder the crve s displcement These mth fcts re prticlrly sefl for motion with constnt ccelertion 3

4 5 In the grph elow (not necessrily to scle), f(x) is defined in the intervl [,] y the stright line segments connecting the points s shown p f(x) 0 c d e -q The re nder the grph A is lwys positive or zero B is zero if p = q C cn e clclted sing re formls for tringles nd rectngles D All of the ove E None of the ove C is the correct nswer A is not tre ecse the re nder the crve of negtive fnction is negtive B is not tre ecse we don t hve enogh informtion ot the loctions of the points long the horizontl xis to drw tht conclsion While in generl we need clcls to compte the re nder crve, in the specil cse of crves tht cn e roken into line segments we cn se simple geometric re formls insted We se the concept of re nder crve in Physics 1 to solve prolems involving displcement, energy, nd implse 4

5 6 The vector v r shown elow hs length c Its ngle mesred conter-clockwise from horizontl xis is θ nd its ngle mesred clockwise from the verticl xis is φ cos( )d +θ φ) The verticl component of v r is given y A c sin(θ) B c cos(90 θ) C c cos(φ) D All of the ove E None of the ove D is the correct nswer The eqivlence of A, B, nd C is sic fct of trigonometry All Physics 1 stdents shold e comfortle with sic trigonometry for finding components of vectors 7 In the digrm elow, the vectors r nd v r re t right ngles to ech other The length of r is c nd the length of v r is d The horizontl nd verticl components of r re nd respectively v e d c The verticl component of v r, e, is 5

6 A d / c B d / c C d / D d c E None of the ove A is the correct nswer In order to solve this prolem, it helps to dd two nmed ngles to the digrm: e d φ θ c We re looking for e, so we need sin of φ We note tht θ is the complementry ngle of φ, so sin(φ) = cos(θ) = /c The finl nswer is d sin(φ) = d / c This qestion reqires mth skills yo will need to solve inclined plne prolems nd to clclte vector directions for grvittionl nd electric forces 8 Consider two vectors, r nd v r, nd their sm, w r = r + v r Which orienttion of the vectors shown elow gives the smllest mgnitde (length) of w r? (The mgnitdes of r nd v r don t chnge in the digrms, t don t ssme tht they re drwn to scle) A v B v C v D It doesn t mke ny difference since the mgnitdes of r nd v r don t chnge E There is no wy to tell withot knowing the mgnitdes of r nd v r 6

7 A is the correct nswer We se vectors lot in Physics 1 This qestion is trying to get yo to think ot how vectors re ligned to give the minimm sm: opposite directions The lignment of vectors is criticl when we consider speeding p / slowing down, centripetl ccelertion, torqe, work, nd mny other topics 9 (Chllenge prolem) The vectors r nd v r re given s shown elow, with lengths c nd d respectively The ngle of r with respect to the reference line is θ nd the ngle of v r with respect to the reference line is φ, going conter-clockwise The mgnitde (length) of the sm, w r = r + v r, is A ( c + d)cos( φ θ) B ccos( θ ) + d cos( φ) C c cos ( θ ) + d cos ( φ) D c + d + cd cos( θ φ) E The mgnitde of w r cnnot e determined nless the direction of the reference line is known D is the correct nswer The esiest wy to get this reslt is to relize tht the mgnitde of vector sm is independent of the choice of coordinte systems In tht cse, pick the X xis to lign with the vector The ngle of the v vector with respect to the X xis is (φ θ) We tke the X nd Y components of v sing trigonometry, dd the X component of, nd finlly se the Pythgoren Theorem to get the length of the sm Jst for lghs, we pt cos(θ φ) insted of cos(φ θ) ecse cos( α) = cos(+α) Note tht (θ φ) nd (φ θ) re independent of the ngle of the reference line 7

8 While we re on the sject, let s go over how to convert vectors from (Length,Angle) coordintes to (X,Y) coordintes nd ck First, ssme tht the ngle hs een mesred in the conter-clockwise direction from the +X xis This is the stndrd yo will see most often nd the one we will se in Physics 1 To convert from (Length,Angle) to (X,Y), we se the following formls (Let length = L nd ngle = α): X = L cos( α) Y = L sin( α) Converting from (X,Y) to (Length,Angle) is little trickier We hve to e crefl to get the ngle in the proper qdrnt First, the esy prt The length is given y the Pythgoren Theorem: L = X + Y Clcltion of the ngle depends on the qdrnt we re in First, the specil cses If X = 0 nd Y = 0, then y convention we sy L = 0 nd α = 0 If X = 0 nd Y > 0, then α = π/ in rdins or 90º If X = 0 nd Y < 0, then α = π/ in rdins or 90º If Y = 0 nd X > 0, then α = 0 If Y = 0 nd X < 0, then α = π in rdins or 180º In cses where X > 0, we re in qdrnts I or IV Use this forml: Y α = rctn X In cses where X < 0 nd Y > 0, we re in qdrnt II Use this forml: Y α = rctn + π (rdins) or +180º X In cses where X < 0 nd Y < 0, we re in qdrnt III Use this forml: Y α = rctn π (rdins) or 180º X Following these rles, α will lwys e in the rnge ( π,+π] or ( 180º,+180º] On compters, yo will find these formls progrmmed into one fnction: tn(y,x) This qestion is more complex thn most vector clcltions we will need in Physics 1, t it rings ot some importnt concepts tht yo will se in lter corses t RPI In ny cse, mke sre yo nderstnd how to convert vector from (Length,Angle) form to (X,Y) form nd ck 8

9 10 Dick nd Jne went into siness selling md pies They set p two md pie stnds t opposite ends of the neighorhood to mximize their potentil cstomers On Mondy, they sold totl of 100 pies One Tesdy, Dick took the dy off ecse his sles were so good on Mondy Jne figred tht she cold triple her sles from Mondy if she copied Dick s slick techniqes Unfortntely, Dick s cstomers from Mondy ll retrned their pies to Jne on Tesdy Jne met her Tesdy sles trget, t she only sold 0 new pies fter reselling ll of Dick s retrned pies from Mondy Wht eqtions wold yo se to determine how mny pies Jne sold on Mondy? A J + D = 100 3J = 10 B J + D = 100 3J D = 0 C J + D = 100 3J + D = 10 D There is not enogh informtion to determine niqe nswer E There is contrdictory informtion so there is no nswer B is the correct nswer The second eqtion cold lso hve een written s 3J = D+0 We often get pirs of liner eqtions in Physics 1 when we solve prolems sing Newton s Second Lw Mny prolems involve ojects connected y ropes A vrile representing tension in rope normlly ppers in two eqtions, one for ech end of the rope With the right choice of coordinte systems, the tension term will e + in one eqtion nd in the other The esiest wy to solve system of eqtions like tht is simply to dd the two eqtions (Esy = less likely to mke n error) To solve the two eqtions in B, dd them together to get J + 3J + D D = Solve for J, then sstitte J ck into either originl eqtion to solve for D 9

adjacent side sec 5 hypotenuse Evaluate the six trigonometric functions of the angle.

adjacent side sec 5 hypotenuse Evaluate the six trigonometric functions of the angle. A Trigonometric Fnctions (pp 8 ) Rtios of the sides of right tringle re sed to define the si trigonometric fnctions These trigonometric fnctions, in trn, re sed to help find nknown side lengths nd ngle