Physics 4A FINAL EXAM Chapters 1-16 Fall 1998

Transcription

1 Name: Posing Code Solve he following problems in he space provided Use he back of he page if needed Each problem is worh 10 poins You mus show our work in a logical fashion saring wih he correcl applied phsical principles which are on he las page Your score will be maimized if our work is eas o follow because parial credi will be awarded 1 A marble rolls off he edge of an 800cm high able and srikes he ground 50cm horizonall from he edge of he able ind (a)he ime ha he marble is in he air and (b)he speed of he marble as i lef he able Use he kinemaic equaions separael for each direcion: o = 0 o = 0800m = 050m = 0 =? = 0 v = v =? a = 0 a = -980m/s =? 800cm 50cm (a)using he kinemaic equaion wihou he final speed for he -direcion, = o a 0 = o + 1 a = o = (0800) = 0404s a 980 (b)using he kinemaic equaion wihou he final speed for he -direcion, = o a = = = = 0619m/s A 500kg block is pulled along a horizonal floor wih an acceleraion of 00m/s b a cord ha eers a force of 150N a a 300 angle as shown ind he coefficien of kineic fricion beween he block and he able Appling he Second Law o each direcion separael, Σ = ma cos30 fr = ma fr = cos30 ma, Σ = ma sin30 + n = 0 n = sin30 500kg n fr 30 = 150N Subsiuing ino he definiion of coefficien of fricion, µ fr = cos30 ma n sin30 = cos30 ma mg sin30 Plugging in he numbers, µ = (150)cos30 (500)(00) (500)(980) (150)sin30 µ =

2 3 An Awood s Machine is made from a 600g mass, a 400g mass and a 00g pulle of radius 400cm The 600g mass is released from res and falls 00m ind he angular speed of he pulle when he 600g mass reaches he ground The iniial energ of he ssem is, K o = 0 U o = gh The final energ of he ssem is, K = 1 v + 1 m v + 1 Iω U = m gh Using he Law of Conservaion of Energ, K + U = W nc ( 1 v + 1 m v + 1 Iω 0) + (m gh gh) = 0 Doing some algebra, 1 ( )v + 1 Iω = ( m )gh 400g r 600g 00m The linear speed of he masses is equal o he angenial speed of he pulle which is relaed o is angular speed, v = rω 1 ( )(rω) + 1 Iω = ( m )gh Subsiuing he roaional ineria of he pulle which is a disk, I = 1 mr 1 ( )(rω) ( mr )ω = ( m )gh Solving for he angular speed, 4(m ω = 1 m )gh [ ( ) + m]r = 4( )(980)(00) [ ( ) + 000](00400) ω = 667rad/s 4 An 800kg car collides head-on wih a saionar 100kg ruck Epers esimae ha he speed of he combined wreckage jus afer impac is 40±5km/h ind (a)he epers esimae of he speed of he car before collision and (b)he epers esimae of he uncerain in his speed m v Before Afer (a)the momena before and afer he collision are p o = and p = ( )v Using he Law of Conservaion of Linear Momenum, = ( )v = m v = (40) = 100km/h 800 (b)using he muliplicaion rule for uncerain, Using proper significan figures, = 100 ±10km/h = v v v = 100 o v = v 40 (5) = 15km/h

3 5 A golf club designer wans he roaional ineria abou he handle end of he shaf o be 115kg m She assumes he shaf can be reaed as a sick 10m long wih a mass of 300g and he head can be reaed as a poin mass Esimae he required mass of he head of he club Roaional ineria adds so I = I shaf + I head The roaional ineria of he shaf is I shaf = 3 1 ml The roaional ineria of he head is I head = Mr = Ml Subsiuing, I = 1 3 ml + Ml M = I l 1 3 m = 115 (10) 1 3(0300) M = 0699kg = 699g shaf 10m head 6 A 100g golf ball is sruck b he club described in problem 5 The veloci of he club head is 400m/s jus before he collision and 30m/s jus afer ind (a)he angular momenum of he golf club abou he end of he shaf jus before he collision, (b)he angular momenum of he golf club abou he end of he shaf jus afer he collision and (c)he veloci of he ball assuming ha he orques eered on he handle during he shor ime of he collision are small (a)the angular momenum of a rigid bod is, = Iω o The angular speed of he club head is relaed o is linear speed, v = rω = I l 10m Plugging in he numbers, = = 383kg m s (b)similarl, L a = I u 30 = 115 l 10 L a = 307kg m s (c)the angular momenum of he golf ball afer collision can be found from he definiion of angular Before momenum, r L r r p L b = mul Using he Law of Conservaion of Angular Momenum, = L a + L b = L a + mul u = L a ml Plugging in he numbers, u = (0100)(10) u = 633m/s 10m v Afer u 3

4 7 A 00kg sign 300m wide hangs from a horizonal rod ha is suppored b a cable ha makes a 300 angle wih he rod ind (a)he ension in he cable and (b)he horizonal and verical componens of he force ha he wall eers on he rod Ignore he mass of he rod Appling he Second Law, Σ = ma h cos30 = 0 h = cos30, Σ = ma sin30 + v = 0 v = sin30, Στ o = Iα 4 sin30 (5) = 0 = (5) 4sin30 (a)solving he orque equaion for he ension, = (5) 4sin30 = (5)mg 4sin30 = (5)(00)(980) = 45N 4sin30 (b)solving he equaions from he force componens, h = cos30 = (45)cos30 h = 1N, v = mg sin30 = (00)(980) (45)sin30 v = 735N v h 30 Phsics Problems Solved 5 100m 300m 8 The Inernaional Space Saion ha is jus saring o be buil will orbi a an aliude of 350km ind he period of orbi for he space saion Using he Second Law and he Law of Universal Graviaion, Σ = ma G Mm = m v r r G M r = v, where he equaion for cenripeal acceleraion has been used The definiion of speed involves he period, v = πr T G M r = πr T T = 4π r 3 GM The radius of orbi is he radius of Earh plus he aliude, r = R + h = = m Calculaing he period, T = 4π ( ) 3 ( )( ) T = 5480s = 913min 4

5 9 A 500g cork is held underwaer b a sring as shown a he righ Cork has a densi of 560kg/m 3 and waer has a densi of 1000kg/m 3 ind he ension in he sring b Appling he Second Law, Σ = ma b = 0 = b = b mg According o Archimedes Principle, he buoan force is equal o he weigh of he displaced fluid, = m f g mg The mass of he displaced fluid is relaed o he densi of he fluid Using he definiion of densi, ρ m V m f = ρ f V =ρ f Vg mg, where V is he volume of he cork which can also be found from he definiion of densi, ρ m V V = m ρ m =ρ f ρ g mg = ρ f ρ 1 mg Puing in he numbers, = (00500)(980) = 0385N 500g 10 The golf club designer wans o es he club she has designed and buil She holds i a he end of he shaf and les i swing ind (a)he cener of mass of he club and (b)he prediced period of oscillaion (a)assuming he cm of he shaf is in he middle and using he definiion of he cm, r cm = m l + Ml (300)(0600) + (699)(10) = r cm = 10m m + M (b)the angular frequenc of a phsical pendulum is, ω = mgr ( m + M)gr = cm I p I The angular frequenc is relaed o he period, ω = πf = π T T = π ω = π I ( m + M)gr cm Puing in he values, 115 T = π ( )(980)(10) T = 13s 5