Physics 6A. Practice Final (Fall 2009) solutions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Transcription

1 Phyic 6A Practice inal (all 009) olution or Capu Learning Aitance Service at UCSB

2 . A locootive engine of a M i attached to 5 train car, each of a M. The engine produce a contant force that ove the train forward at acceleration a. If 3 of the car are reoved, what will be the acceleration of the horter train? We jut ue Newton nd law here: Net force (total a) x (acceleration) Initially, the total a i 6M (5 car, plu the engine) After reoving 3 car the total a i 3M ( car, plu the engine) In both cae the Net force i the ae (the engine didn t change). Here i the forula in both cae: Initial (6M)(a) inal (3M)(a final ) Setting thee equal give a final a Another way to think about thi one i that ince the a i cut in half, the acceleration ut double. or Capu Learning Aitance Service at UCSB

3 . Two boxe are placed next to each other on a ooth flat urface. Box A ha a kg and Box B ha a 3 kg. A contant horizontal force of 8 N i applied to Box A. ind the force exerted on Box B. We can think of thi a a ingle box with total a 4kg. Then uing a we get the acceleration of the whole yte. 8 a N 4 yte kg a yte Now we do the ae thing, but jut for box B: A B kg B 3 B 6 N Reeber Box B ha a a of only 3kg or Capu Learning Aitance Service at UCSB

4 3. Two boxe are placed next to each other on a flat urface. Box A ha a kg and Box B ha a 3kg. The coefficient of friction are 0.3 and 0.4 for kinetic and tatic friction, repectively. A contant horizontal force of 8N i applied. ind the acceleration of Box B. We can think of thi a a ingle box with total a 4kg. Then uing a we get the acceleration of the whole yte. We need to account for friction, o firt find the axiu force of tatic friction: tatic,ax µ N ( N ( ) ( ) kg N tatic,ax 0 8N A B tatic Thi friction force i ore than the 8N force trying to ove the boxe, o they are held in place by friction. Thu the acceleration of both boxe i 0. *Note that the actual force of friction holding the boxe in place i only 8N jut enough to keep the fro oving. or Capu Learning Aitance Service at UCSB

5 4. A 0. kg piece of wood i being held in place againt a vertical wall by a horizontal force of 5 N. ind the agnitude of the friction force acting on the wood. ro the force diagra, we can ee that the friction force ut equal the weight of the piece of wood to keep it fro falling. friction friction friction friction g ( )( 0.kg 9.8 ).96N 5N N g or Capu Learning Aitance Service at UCSB

6 5. A 4 kg block on a horizontal urface i attached to a pring with a force contant of 50 N/. A the pring and block are pulled forward at contant peed, the pring tretche by 5 c. ind the coefficient of kinetic friction between the block and the table. The key phrae here i contant peed. Since the block i oving at contant peed (and direction) we know that it acceleration i 0. Thu the net force i 0 a well. Here i the force diagra. Since the net force i 0, we know that the Noral force ut equal the weight, and the riction force ut equal the Spring force. friction N pring The pring force obey Hooke Law: The weight i: ( 50 N)( 0.5).5N pring k x ( 4kg)( 9.8 ) 39.N g inally, we can put thi together to find the friction:.5nµ µ pring k 0.3 friction k µ k ( 39.N) N g or Capu Learning Aitance Service at UCSB

7 6. A 90 kg an drive hi car at a contant peed of 5 / over a all hill that ha a circular cro ection of radiu 40. ind hi apparent weight a he cret the top of the hill. (Hint: the apparent weight i the ae a the noral force on the an.) 5 / When the car reache the top of the hill, it will have only force: it weight, and the noral force upplied by the road. Since the an i itting in the car, he feel a noral force a well. The car (and an) ut be accelerating toward the center of the circle, o the net force on the an will be equal to the centripetal force required to keep hi oving along the circle. v cent g Noral r v Noral g r Noral Noral 376N ( 90kg ( )( ) )( 5 ) 90kg N g or Capu Learning Aitance Service at UCSB

8 7. Planet X ha a radiu that i 3 tie a large a Earth and a a that i 6 tie that of Earth. A NASA atronaut who weigh 550 N here on Earth i planning to ebark on a anned iion to Planet X. What will be the atronaut weight when he land there? The weight i the ae a the gravitational force. On Earth the weight i 550N: G M Earth grav 550N ( rearth) In thi forula, G i contant, and will not change when the atronaut goe to a new planet. So the only part that will change are the a and radiu of the planet. The eay way to do thi proble i to iply ubtitute the new planet X value into the forula, and rearrange it: MX 6 MEarth rx 3 rearth G( 6 M ) Earth grav,x ( 3 rearth) 6 G M Earth grav,x 9 ( rearth) grav,x 3 G MEarth ( rearth) 3 { 550N} 367N Earth Planet X or Capu Learning Aitance Service at UCSB

9 8. A car i traveling at a peed of 40 /. The brake are applied, and a contant force bring the car to a coplete top in a tie of 6. econd. The tire on the car have a diaeter of 70 c. How any revolution doe each tire ake while the car i braking? Given: v040 /; vf0; t6.; dia70c r v0 rad ω0 4.3 r rad ω α 8.4 t 6. θω t+ αt 355rad 56 revolution π rad rev 0 rad ( )( ) rad ( 8.4 )( 6.) rad 355 radian or Capu Learning Aitance Service at UCSB

10 9. Two block of equal a M are attached by a ale rope, with one block on a frictionle table, and the other block hanging down below, a hown. When the block on the table i oving in a circular path at revolution per econd, the hanging block i tationary. ind the radiu of the circle. a)0 c b) 5 c c) 50 c d) 50 c Tenion in rope weight of hanging block Mg Tenion in rope i alo the centripetal force on the oving block Set thee equal and olve for R: Mg MRω g R we ut convert the given peed fro revolution per econd to radian per econd ω rev rad π rev ω π cent rad Mv R MR ω ue vrω Plug in to get R: g R ω 9.8 ( rad π ) 0.5 5c or Capu Learning Aitance Service at UCSB

11 0) A erry-go-round i initially rotating at a rate of revolution every 8 econd. It can be treated a a unifor dik of radiu eter and a 400 kg. A 50 kg child run toward the erry-go-round at a peed of 5.0 /, juping on to the ri (tangentially, a hown). ind the child linear peed after juping onto the erry-go-round. a). / b).3 / c) 5.0 / d) 7. / We can ue conervation of angular oentu for thi one. rev πrad ω rev rad Initial angular peed of dik I dik L dik MR Iω 68 ( 400kg)( ) 800kg kg I child L child R vr ( 50kg)( ) 00kg ( )( 50kg 5 )( ) 500 kg L I total L total I dik L dik + I child + L 000kg child 8 kg ( )( ).8 rad. kg rad final Itotal ωf 8 ω f.8 v f,child 3 or Capu Learning Aitance Service at UCSB

12 ) A ballitic pendulu conit of a olid block of titaniu with a 5 kg, upended fro a light wire. A bullet of a 5 g i launched toward the block at an unknown peed. The bullet bounce back at half it original peed, and the block rie to a height of.8 c above it tarting point. What wa the initial peed of the bullet? a) 00 / b) 00 / c) 300 / d) 400 / Before Colliion After Colliion Highet Point v 0 ½v 0 v block 5kg 5kg 5kg.8c Ue conervation of oentu for the colliion, then conervation of energy for the winging to the highet point. ( 0.005kg) ( v ) ( ) ( kg v0) + ( 5kg) v block ( 0.005kg) ( v ) ( 0.005kg) ( v ) + ( 5kg) ( 0. ) v ( )( ) ( )( )( ) 5kg v block 5kg vblock (Round up to get 400/) 0 Now we can put thi value into our oentu forula to get the initial peed of the bullet or Capu Learning Aitance Service at UCSB

13 ) Two car are oving toward an interection. Car A i traveling Eat at 0 /, and Car B i traveling North at /. The a of Car A i 000 kg and the a of Car B i 000 kg. Driver A i applying acara to her eyelahe, and driver B i reading a text eage, o neither of the low down a they approach the interection. When the car crah into each other, they tick together. ind the coon velocity of the car jut after the colliion. a) 3.0 / at an angle of 45 North of Eat b).0 / at an angle of 30 North of Eat c) 3.3 / at an angle of 3 North of Eat d) 0.4 / at an angle of 50 North of Eat Moentu i conerved in each direction, o we get two forula: ( ) a + b vf,x vf,x 6.67 ( ) a + b vf,y vf,y 8 x : v y : a b v a b Cobine the coponent with the Pythagorean theore to find the final peed, and ue tangent to find the angle: v final tanθ ( 6.67 ) ( + 8 ) θ o / A y B A/B / v final x or Capu Learning Aitance Service at UCSB

14 3) The Atwood achine yte hown i copried of a block of a M attached by a ale rope to block of a M. The rope pae over a olid cylindrical pulley of a M and radiu R, and the rope doe not lip on the pulley. ind the acceleration of the heavier block. Ue g for gravitational acceleration. a) /7 g b) /5 g c) / g d) /3 g We can et up force forula for the ae, and a torque forula for the pulley. T M poitive torque R T Mg T Ma T Mg Ma T ( M ) g ( M ) a T Ma Mg T R T R MR a ( ) R T T Ma T M M T ( Mg Ma) ( Ma Mg) Ma a g 7 Mg Mg Note that acceleration cae out negative due to our choice of direction for poitive torque. Make ure the ign in all forula atch up with your choice for poitive, or you will get the wrong anwer. or Capu Learning Aitance Service at UCSB

15 4) A light unifor ladder of length 5 i leaning againt a wall o that the top of the ladder i 4 above the ground and the botto of the ladder i 3 fro the wall, a hown. How high can a peron of a 50 kg walk up the ladder before the ladder lip? Aue the coefficient of tatic friction between the ladder and the ground i 0.6 and that the wall i frictionle. a).0 b).5 c) 3.0 d) wall orce on the ladder are hown in blue (the weight of the ladder i negligible). N d g 4 We need to find the ditance d. f 3 forula we can write down force forula and one torque (ue the ground a the pivot point). l 3 Σ x Σ y 0 f 0 N g 0 N 470N Στ 0 0 3d ( )( 4) ( g)( ) wall wall µ g 0 3 ( 88N)( 4) ( 470N)( d) 0 d 4 5 wall 5 ( 0.6)( 470N) 88N ue iilar triangle to find the lever ar for the weight: l d l 3 5 3d 5 or Capu Learning Aitance Service at UCSB

16 5) A 50c X 5 c cafeteria tray with a 0. kg i loaded with the following ite: A carton of ilk with a 0.5 kg i placed in the upper left corner. An apple with a 0.3 kg i placed in the iddle of the right hand ide. A plate with a unifor layer of ahed potatoe (a kg) i placed in the center. A candy bar with a 0.05 kg i placed in the botto right corner. ind the location of the center of a of the tray. a) x -4.4 c, y.3 c b) x -. c, y.8 c c) x -. c, y.3 c d) x 7.0 c, y -.8 c y x x y We need to ue eparate forula for x c and y c : c c ( 0.kg)( 0) + ( 0.5kg)( 5c) + ( 0.3kg)( 5c) + ( kg)( 0) + ( 0.05kg)( 5c) 0.kg+ 0.5kg+ 0.3kg+ kg+ 0.05kg ( 0.kg)( 0) + ( 0.5kg)(.5c) + ( 0.3kg)( 0) + ( kg)( 0) + ( 0.05kg)(.5c) 0.kg+ 0.5kg+ 0.3kg+ kg+ 0.05kg.c.8c or Capu Learning Aitance Service at UCSB

17 6) Two ball are rolled down a hill. Ball A i a olid phere with a M and radiu R. Ball B i a hollow phere with a M and radiu R. Copare the peed of the ball when they reach the botto of the incline. a) V A 0.6 V B b) V A V B c) V A. V B d) V A.7 V B Moent of Inertia for each ball: I I A B 5 3 MR ( ) M R Ue conervation of energy for each ball: MV + I ω Mgh MV + MR V 0 ( )( ) Mgh VA gh 5 R 7 ( VB M( R) )( ) Mgh VB A A A A A V V MV A B B I B ω B gh. gh Mgh MV B + 3 R 6 5 gh or Capu Learning Aitance Service at UCSB

18 7) A box of a M tart fro ret at the top of a frictionle incline of height h. It lide down the hill and acro a horizontal urface, which i alo frictionle, except for a rough patch of length h, with coefficient of kinetic friction 0.5. The box coe into contact with a pring (pring contant k), copreing it. The pring then unload, ending the box back in the oppoite direction. h E initial gh When the box lide acro the rough patch, energy i lot to friction. W friction f h µ g h 0.5gh (each trip) k k The pring jut change the direction of the box energy i conerved while in contact with the pring. So the total energy lot due to friction i 0.5gh. h E final 0.5gh gh final h final 0.5h or Capu Learning Aitance Service at UCSB

19 8) A diver tuck her body in id-flight, reducing her oent of inertia by a factor of. What happen to her angular peed and kinetic energy? a) Both angular peed and kinetic energy reain the ae. b) Angular peed i doubled, and kinetic energy reain the ae. c) Angular peed i doubled, and kinetic energy i increaed by a factor of 4. d) Both angular peed and kinetic energy are doubled. Ue conervation of angular oentu. I ( I 0) ωf ωf ω0 0 ω0 If ωf I0 ω0 Angular peed i doubled Plug thi in to the kinetic energy forula. K rot,init. I 0 ω 0 K rot,final ( I ) ( ω ) 0 0 I ω K rot,init. Kinetic Energy i doubled or Capu Learning Aitance Service at UCSB

20 9) A unifor arble roll without lipping down the path hown, tarting fro ret. ind the iniu height required for the arble to ake it acro without falling into the pit. a) 3 b) 4 c) 30 d) 35 Thi i a -tage proble. When the arble roll down the hill, we can ue conervation of energy. Then it projectile otion a it flie acro the gap. E top E gh gh gh h h v g 7 0 botto v ( 8 ) 9.8 v v + + Iω ( r ) 5 3 v r We need to find the peed v h Projectile otion initial peed i horizontal, and the arble drop 0. ind tie: y gt 0 ( 9.8 ) t t ec Horizontal ditance i 36: ( ec) v 8 36 v or Capu Learning Aitance Service at UCSB

21 0) A chool yard teeter-totter with a total length of 5. and a a of 36 kg i pivoted at it center. A child of a 8-kg it on one end of the teeter-totter. Where hould the parent puh downward with a force of 0 N to balance the teeter totter? a) 0.5 fro the center b). fro the center 5. c).3 fro the center d d).9 fro the center The torque ut balance out. Meauring fro the center, we have: ( 76.4N)(.6) ( 0N)( d) 0 d N The weight of the teeter-totter i at the center, o it produce no torque N 0 N or Capu Learning Aitance Service at UCSB