Answers to test yourself questions

Size: px

Start display at page:

Download "Answers to test yourself questions"

Cori McCormick
5 years ago
Views:

1 Answes to test youself questions opic. Cicula motion π π a he angula speed is just ω 5. 7 ad s. he linea speed is ω m s.. 4 b he fequency is f. 8 s a f. 45 ( 3. 5). m s. 3 a he aeage acceleation is defined as a t. he elocity ectos at A and and the change in the elocity ae shown below. he magnitude of the elocity ecto is 4. m s and it takes a time of π s to complete a full 4. eolution. Hence a time of s to complete a quate of eolution fom A to. he magnitude of 4 is m s and so the magnitude of the aeage acceleation is m s. his is. 785 diected towads noth-west and if this ecto is made to stat at the midpoint of the ac A it is then diected towads the cente of the cicle. b he centipetal acceleation has magnitude. 8. m s diected towads the cente of the cicle.. π 4 he centipetal acceleation is a f. Hence a f s min. 5 a he centipetal acceleation is 4.. m s. he tension is the foce that poides the centipetal. 4 acceleation and so ma... N. b Fom ma. N we hae a 4.. c... 8 m.. m s and so m s.. 4 With a 9. 8 m s we hae that a s 85 min. physics fo the I Diploma Cambidge niesity Pess 5 ANSWES O ES YOSEF QESIONS

2 π m s 7 a a ( 5. 3 ) b he foces on the pobe ae (i) its weight, mg, and (ii) the nomal eaction foce N fom the suface. Assuming the pobe to stay on the suface the net foce would be mg N m m N mg m g m( ) >. his is positie so the pobe can stay on the suface. π π.5 m s km 35 4 (.99 4 ) a m s b.5 8 a c F ma m N 9 he components of ae: x sin θ, y cos θ We hae that sin θ m cos θ mg Diiding side by side: m sin θ cos θ mg tan θ g his gies a km g tan θ 9.8 tan 35 fiction eaction weight b et the nomal eaction foce fom the wall be N. hen N m mg f s Fo the minimum otation speed the fictional foce must be a maximum i.e. f s µs N. I.e. N m mg µs N ANSWES O ES YOSEF QESIONS physics fo the I Diploma Cambidge niesity Pess 5

3 Combining gies mg m g µ s i.e m s. Fom π f we find µ s. f e s 7 e min. π π 5. a et be the speed on the flat pat of the oad befoe the loop is enteed. At the top the net foce on the cat is its weight and the nomal eaction foce fom the oad, both diected etically downwads. hen, mu N mg N mu + mg whee u is the speed at the top. Fo the cat not to fall off the oad, we must hae N > i.e. u > g. Fom conseation of enegy, m mg( ) + mu and so u 4 g. Hence 4 g > g, i.e. > 5g m s. b Fo just about equal to 5g we get u g m s. he tension in the sting must equal the weight of the hanging mass i.e. Mg. he tension sees as the centipetal foce on the smalle mass and so m. Hence m Mg Mg. m 3 et the tension in the uppe sting be and in the lowe sting. oth stings make an angle θ with the hoizontal. We hae that: sinθ mg + sinθ m + We may ewite these as: sinθ sinθ mg m +. 5 Fom tigonomety, sinθ. 5 θ 3. Futhe, m. heefoe the. equations simplify to. 5 ( ) o.. 8 ( + ) Finally, 3. N, 8. N. 4 a y conseation of enegy, mgh m and so gh m s (with this speed, this amusement pak should not hae a licence to opeate!). b he foces on a passenge ae the weight and the eaction foce both in the etically down diection. hus + mg m m mg. he speed at the top is found fom enegy conseation as mgh m + mg( ) Hence N. 3 5 c sing u as we get 49 a 4 and so a 3 m s (some passenges will be fainting 4 now, assuming they ae still alie!). physics fo the I Diploma Cambidge niesity Pess 5 ANSWES O ES YOSEF QESIONS 3

4 . he law of gaitation 5 a F G Mm ( ) 4 b F G Mm ( ) 3 7 c F G Mm (. ) 7 3 a Zeo since it is being pulled equally fom all diections. b Zeo, by Newton s thid law. c F G m, (d) F G m G Mm G m ( m + M ) g g A ( 9) N N 47. N 8 g A ( ) g 9 Since sta A is 7 times as massie and the density is the same the olume of A must be 7 times as lage. Its adius G7M must theefoe be 3 times as lage. Hence g A ( 3) 3. g / g new ( / ) g old et this point be a distance x fom the cente of the Eath and let d be the cente to cente distance between the eath and the moon. hen G8M x ( d x) 8( d x) x 9( d x) x x 9. 9 d a At point P the gaitational field stength is obiously zeo. b he gaitational field stength at Q fom each of the masses is 3. g. 7. N kg. he net field, taking components, is diected fom Q 9 ( ) to P and has magnitude g cos cos. N kg. 4 ANSWES O ES YOSEF QESIONS physics fo the I Diploma Cambidge niesity Pess 5

5 3 We know that m m. ut π and so we deduce that ( 4 ) a Fom we calculate. 7. ( ) 4 4. m. 3 4 π. heefoe m s b he shuttle speed is m s. he elatie speed of the shuttle and Hubble is. 74 m s and so the distance of km will be coeed in 37 s hs a Gm m n Gm n n+ Gm m Gm n. ut π and so π Gm giing n b Fo this to be consistent with Keple s thid law we need n + 3 n physics fo the I Diploma Cambidge niesity Pess 5 ANSWES O ES YOSEF QESIONS 5

AP Physics 1 - Circular Motion and Gravitation Practice Test (Multiple Choice Section) Answer Section

AP Physics 1 - Circular Motion and Gravitation Practice Test (Multiple Choice Section) Answer Section AP Physics 1 - Cicula Motion and Gaitation Pactice est (Multiple Choice Section) Answe Section MULIPLE CHOICE 1. B he centipetal foce must be fiction since, lacking any fiction, the coin would slip off.