AQA Maths M2. Topic Questions from Papers. Energy, Work and Power. Answers

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1 AQA Maths M Topic Questions from Papers Energy, Work and Power Answers PhysicsAndMathsTutor.com

2 4(a) P (30 4) W AG Finding force Correct answer from P Fv (b)(i) F sin5 + 30v Finding force. Correct force 590 (00 9.8sin vv ) d Using P Fv v + 39sin5 v Correct equation from correct working AG (ii) 39sin 5 ± ( 39sin 5 ) 4 ( 764) v Solving quadratic v 8.3 or v 8.3 ms Correct solution Total 8 (Q4, Jan 006) Q Solution Marks Total Comments 7(a) e Use of Hookes law and equilibrium e 0.39 m Correct length (b) J EPE 0.4 AG Use of EPE formula Correct value from correct working 00 (c)(i) 45 ( x 0.4) + 0v ( x) 0.4 Expression for EPE with ( x ± 0.4) Correct EPE Four term energy equation 45 5( x 0.4) + 5v + 98( x) B Correct GPE Correct equation 5v 98x x + 00x 0 d Solving for v v 39.6x 5x 4.6 AG 7 Correct result from correct working (ii) 39.6x 5x x 39.6x ± Solving quadratic x 5 or Correct solutions x Appropriate value selected SC Only correct answers given, award. Total 4 TOTAL 60 (Q7, Jan 006)

3 Q Solution Marks Total Comments 3(a) 30 use of EPE formula 3 EPE J 0.5 correct EPE PhysicsAndMathsTutor.com (b)(i) v v + 9. v ms - AG d 5 three term energy equation two terms correct all terms correct solving for v correct v from correct working (ii) 50.7 v v ms - 3 two term energy equation correct equation correct velocity (c) v v ms 5 Total 5 finding friction force correct friction force three term energy equation correct equation correct velocity (Q3, June 006) 4(a) 35 v ñ v 4 (ms ) 3 Energy method (b) Air resistance or friction B (c) Energy lost ñ 35 ( 90) F 0 90 m Work done: ( ) 0F 90 F 45.5(N) 4 F > 0 Difference attempted ± Total 8 (Q, Jan 007)

4 58(a) 49 x g 0.5 x 0. 3 (b) 49 ( 0.) EPE (J) (c)(i) 49 x ( 0. + x) 0.5 A3 x + 0.6x ñ All terms attempted for - EE from A3 (ii) 0.6 ± x x 0.04 x 0.04 only identified Total TOTAL 75 (Q8, Jan 007) 6(a) Kinetic energy 5 0 Full method 50 J (b) Using conservation of energy: KE when box hits ground Initial KE + Change in potential energy Could have sign errors g ft 70 J 3 AG; SC g (c) mv 70 V 688 Speed is 6. m s ñ 3 CAO; accept 688 or 4 43 ; SC 6.3 (d) No air resistance E Or no resistance forces Box is a particle E Deduct mark for unacceptable third reason Total 0 (a) Symmetry of the lamina about PQ E Accept ëmirror lineí (Q, June 007)

5 Total 9 λ x 76(a) EPE is l 00(0.5).5 J PhysicsAndMathsTutor.com (b) When string becomes slack, using mv loss in EPE: NB Using 5 to answer (a) and thus (b) U 5 U v.5 Speed is 5m s ñ 3 AG no marks (c) Resolving vertically, R 5g B F µ R 0.4 U 5g g Using change in energy work done: g U 0.5 for force distance ( ) v, first term (or.5) ñ v v.08 Speed is.04 m s ñ 7 Total second term (inc ñ ) (Q6, June 007) Q Solution Marks Total Comments 8(a) Kinetic energy J (b) Using mgh mv : g h h g.5 m 3 (c) When 3 m above ground level: Change in PE is 0.6 g J KE of ball is J Speed of ball is m Dep on.9 m s 4 No KE given: speed.9 SC3 (d) eg ball is a particle, no air resistance, E Accept no spin, no wind weight is the only force acting etc Total 0 (Q, Jan 008)

6 6(a) EPE λ x 9 l 300 (.5) J (b) When string is slack, gain in PE is mgh 6 g.5sin J KE EPE gain in PE m v v AG (c) At A, PE gained above initial position is 6 g 5.5 sin30 Or PE above position string slack is J B KE at A is 77.3 This is more than initial elastic potential B energy particle will not reach A E 3 Or Using v u + as Total 0 a 0.5g s.37 or.366 B B [or.87 above starting point] Hence stops before A E Vertical height above sling slack is Vertical height above starting point is.435 (Q6, Jan 008)

7 Q Solution Marks Total Comments 04(a) Using power force velocity Power (40 50) 50 00,000 watts PhysicsAndMathsTutor.com (b) When speed is 5, max force exerted is N B Accelerating force is 3000N Using F ma a Need 3 terms eg 4000 ± 000 ma or 000 ± 000 ma M0 for 000 ma a ms 3 (c) When van is at maximum speed force against gravity is mgsin 6 (parallel B to slope) Force against gravity and resistance is mg sin v v Speed is maximum when v For 3 terms; and other term v v correct 40 v v CAO Speed is 34.4 ms 6 Total (Q4, June 008)

8 8(a) (b)(i) (ii) Work done e e λx d x l 0 λx l 0 Needs limit of 0 λe l 3 AG Or Area under a straight line λe average force distance l Using T λ x l 5g 50 x 0.6 Extension is 0.96 m EPE λ x l 50 (0.3) J (iii) When x above P, 50 (0.3 x) 50 (... x) EPE 0.6 for 0.6 PE[relative to P] ( )5 g x for 5 g distance KE + EPE [at new point] EPE [at P] gain in PE 4 terms, all signs correct, terms correct mv + 50 (0.3 x) (0.3) 0.6 5gx mv + x 50 ( 0.6 x) 0.6 5gx m Equation involving terms in v, x and x only.5.v + 5 x 75 x 49x v 0.4x 50 x 7 (iv) Particle is at rest when v 0 0.4x 50 x 0 x 0 [not required] Or x m above P. 50 Total 6 (Q8, June 008)

9 (a) Initial KE mv Total 4 6 Allow one of m and v incorrect 43 J (b)(i) When it hits the ground, conservation of energy gives KE Initial KE + loss in PE g 4 Need 6 g 4 or J (3sf) AG (ii) v Speed is 4.9 m s 3 PhysicsAndMathsTutor.com (iii) Stone is a particle B Not g constant No air resistance B No other forces acting Total 9 (Q, Jan 009) 36(a) At maximum speed, tractive force resistance force Using power force velocity: F 40 F N 3 (b) Using force distance work done change in energy: s (40 36 ) Fs change of KE of 3 terms correct all 3 terms correct Distance 456 m 4 Total 7 (Q6, Jan 009)

10 Total 7 49(a) When acceleration is zero, tension gravitational force 784x 80g 6 Both terms correct x 6, x + 6 3m for x 6 Length of cord is 3 m 3 (b)(i) When bungee jumper comes to rest, EPE 784 x 6 49x Change in PE 80 g (6 + x) 49x (6 + x) x 3x + 5 x 3x AG PhysicsAndMathsTutor.com Or 80 g 65 ( 80g[ 6 + x] ) ( or 80g ( 49 x) ) 3 ± (ii) x x Distance below point of jump is m Distance between jumper and ground is m 4 Accept 5.87, 5.3 Total (Q9, Jan 009) (a) KE J (b) Change in PE as slides down: mgh cos 30 Need cos 30 or sin Using Conservation of Energy: KE at end of slide m a J accept Speed of Anne is m m s 6 9 (c) Anne is a particle; no air resistance E Total 9 (Q, June 009)

11 8 65 Force acting against gravity is mg sinθ Or Force acting against gravity and resistance is mg sinθ m mg sinθ g sinθ Using power force velocity F 8330 kw 6 Total 6 λ (Q5, June 009) 76(a) 6 λx EPE l J (b) Using initial EPE KE when string becomes slack: 48 5 v F v m s F 3 ft ' a '.5 (c) Normal reaction is 5g or 49 Frictional force is 5g µ m Work done by frictional force is 5µ g m 0µ g Stops at wall 0µ g 48 m m 0 µ g ' a ' µ accept 4 49 OE Total (Q6, June 009) Q Solution Marks Total Comments 8 Work done Fs cosθ Accept Fs sinθ for 40 5 cos J 3 Total (Q, Jan 00)

12 98(a) When x, KE is 49 v ( ) 078 x EPE is Change in PE is 49 g x for any 078 p Conservation of energy: ( x ) v + 49 g x ( ) v + x 49gx 3 terms (KE, PE, EPE) terms correct all 3 terms correct + ( ) 9.6 SC3 49 v + 49 e 49 g( e + ) v x x 5v 38x 5x 40 6 AG [could use x for e] (b) If x is not greater than, cord is not stretched B (c) At maximum value of x, v 0 5x 38x ± x 5 m dep on above x or for either solution 54.8 E 4 Needs to give a reason for deletion of second root. Both roots must be positive: one above, one below (d)(i) When speed is a maximum, a 0 or tension gravitational force ( d5v ) 38 0x dx 078( x ) 49g 0 at maximum speed 38 0x 0 x 9.8 x AG (ii) From part (a), v v.96 Maximum speed is 3.0ms Total 6 (Q8, Jan 00)

13 0(a) Kinetic energy 3 4 Total 3 4 (J) (b) PE lost is 3 g 5 53g or J Accept 499, 53g (c)(i) KE is g (a) + (b) J (if done (c)(i) in (b) 0 marks; if done (b) and then (c)(i) in (b) only) (ii) Using KE mv v 05.6 Speed of stone is 3.9 ms 4 Accept 3.8 from 50 PhysicsAndMathsTutor.com If use constant acceleration formulae in D, possible 4 marks in (c) BUT no marks if initial speed is treated as being vertical (d) eg Stone is a particle E No air resistance Total 9 Not no resistance; accept no wind resistance (Q, June 00) Q Solution Marks Total Comments 6(a) Using power force velocity Power (30 48) watts AG (b) When speed is 40 m s, max force exerted is N B Accelerating force is N Using F ma: 58 00a m a 0.44 m s 4 (c) Force exerted by engine is 690 v B Force exerted by the engine 30v mg sin3 (Use of cos3 delete, of 3 A terms) 30v (or 00gsin 3) 690 A All terms correct v Two terms correct 30v 65.47v SC3 for 30v v ± v 30 Speed is 59.3 m s 7 Total 3 (Q6, June 00)

14 (a) PE lost is 4 g 5 cos 70 4 g 5 cos or sin 0 or J (b) KE is loss of PE KE is 67.0 J B ft (c) Using KE mv v 33.5 Speed of particle is 5.79 m s (ft from (b)) Total 5 (Q, Jan 0) 5 33(a) PE is 400 g g [or 3 360] B (b) KE is B (c) Work done per minute is 3 60 J Power ( a) + ( b) W CAO Accept 537 from 3400 in (a) Total 4 (Q3, Jan 0)

15 Q Solution Comments 47(a) Work done λx l λe l e 0 e λx d x Condone lack of limits and dx l 0 Must include limits from integral 3 AG (b)(i) λ x Using T, 7g 96 x l 4g x could use 3g or 4g at least side correct (ii) By C of Energy, when next at rest, EPE (initial) PE change (for platform) + EPE (when at rest) x 4 g (0.7 x) + 3 terms (not including ) mv of 3 terms correct all correct x + 5x 00x 80x m (0x 7)(0x ) 0 x 0. 6 [last, must give 0., not 0. and 0.7] Alternative (b)(ii) ( ) X 4gX + () () () 4gX X + 49X (m) X 0, 0.6 () (where X is distance moved upwards) (iii) Max speed when T mg 96x 4g x Or mid-point of values 0. and 0.6 above SC Total 5 (Q7, Jan 0)

16 Q Solution Marks Total Comments (a) KE 58 : Correct fully substituted expression 5 for KE. 6 J : CAO (b) Change in PE: mgh : Expression for PE with 58 and 9.8 or 9.8 with 6 or 7 for the height (or and 4, and 5 or 0 and 4) : Accept 3980 or 3970 or 3978 or 3979 or Accept 398 or 3983 or KE J : Adding their two previous answers J Speed of Kim is PhysicsAndMathsTutor.com d d: Seeing expression for v (not v ), dependent on second.88 m s.9 m s 5 : Accept.88 or.8 or.9 Accept.88 or.8 or.9 or AWRT.89 from g 9.8. Obtaining v u + gh followed by incorrect substitution M0, unless h is 6 or 7, which is.0 (from h 6) 7 v + g 4..9 v 4 + 4g.9 v + g Total 7 (Q, June 0)

17 65(a) km h 90 ms m s AG B B: Must see or (b) Resistance is 5000 N B B: Obtaining Using power force velocity : Using P Fv with 5 and their F. 5 kw 3 : Correct final answer, must be in kw. Total 4 5W or W B 5 B (Q5, June 0)

18 Q Solution Marks Total Comments 9(a) EPE λ x 7 l 800 (4) B B: Extension 4. 6 : Substitution of 6, 800 and their extension into EPE formula. 400 J 3 : Correct EPE (b) 800 ( x) : Equation with EPE and KE terms, both correct. x 4.67 x 6.53 m : Correct extension. Accept or 6.53 or AWRT Distance from O is.5 m 3 : Correct distance. Accept.5 or AWRT.53. (c) Resistance force is 800 N Work done by resistance force is 800 (x + 6) B B: Correct work done by resistance force. C of Energy gives 800 ( x) ( x + 6) x + 800( x+ 6) x + 6x 3 0 or 50x + 800x ± x 3 d : Three energy terms, KE, Work Done and EPE. : EPE correct. : Correct equation. : Correct quadratic equation with no brackets. d: Solving their quadratic equation with correct formula and correct substitution x.5497 : Correct positive solution stated. Accept.54 or.55 or AWRT.55. Distance from O is 7.55 m 8 : Correct distance from O. Accept 7.55 or 7.54 or AWRT OR Use d for distance: 800 d C of Energy gives 800 ( d 6) d d 000d d 0d ± x 3 d 7.55 (B) () () () (d) () PhysicsAndMathsTutor.com B: Correct work done by resistance force. : Three energy terms, KE, Work Done and EPE. : Seeing d 6 in EPE : EPE correct. : Correct equation. : Correct quadratic equation with no brackets. d: Solving their quadratic equation. : Correct distance from O. Accept 7.55 or 7.54 or AWRT Total 4 TOTAL 75 (Q9, June 0)

19 7g 8g (a) KE at P correct F J (b) change in PE as it falls: mgh correct 8330 ISW (c)(i) using Conservation of Energy: KE at ground ft C s (a) and (b) J ft if gained in (a) and (b) ( J to 3sf) (ii) speed of packet is R m s CAO ft C s (c)(i) Total (Q, Jan 0) Total 0 94(a) using power force velocity power (5 4) 4 power is watts (b) when speed is 5 m s, max force exerted is N B resistance force is N accelerating force is N 565 using F ma a m a.7 m s 4 Total 6 (Q4, Jan 0)

20 308(a) λx using EPE, l 3. EPE B B for J 3 (b) by C of Energy, when next at rest, EPE (initial) work done against friction + EPE (when at rest) for work done by friction or 5F F terms; all correct 5F B B 8.8 frictional force is 3.6N 6 3. (c) at B, tension is N B tension > friction hence particle starts to move E (d) when particle is next at rest, work done against friction is EPE at B 3.6 distance 8.8 distance is.76 CAO. m (e) total distance is B ft from in (d) or total distance 3.6 original EPE, m total distance is 7. m Total 4 (Q8, Jan 0)

21 (a) KE All terms correct 9 79 J J (b) Change in PE: mgh J All terms correct J 3 00 J (c)(i) KE when touches down on ground J Their values, one correct 5 88 J J CAO (ii) Speed of Alan is m s 37.3 m s CAO Total 8 (Q, June 0)

22 38(a) Initial EPE x l 0 (0.5) 5 for formula with extension J Initial KE is J When block is at A, mv v Speed is 9.80 m s 4 Accept 4 6 ; condone 9.79 (b)(i) Normal reaction is mg 0.4g Frictional force is 0.4μg N Work done by frictional force is 5.5 (0.4μg) or. g m C of Energy, when at A, gives (0.4μg) 0.4 v 9.. μg 0. v v 96 g Three terms, eg initial energy in (a) (3 or 9.); work done; KE at A. Fully correct 6 Ft v v in (a) g (ii) Speed when rebounding is 96 g Bft Block is stationary at B 0.4 (96 g).μg Three terms 4 Two terms correct with sign 0 (0.5) 5 Third term correct with sign 0.(96 g ).μg μg 3 Or g. g 3 μ Total 6 TOTAL 75 (Q8, June 0)

23 (a) KE J (b) Change in PE: mgh J (c)(i) KE when reached point B J (a) (b).84 J cao (ii) Speed of ball is If added in (c)(i) 0 marks for (c)(i) 4.8 for c(ii) m s 4.80 m s Condone 4.8,4.79 Total 8 dv (Q, Jan 03) 343 Force acting against gravity is mgsin Force acting against gravity and resistance is mgsin g sin N or 8590 N Condone cos or - for M marks Using power force velocity 8588 dep W 89 kw 5 Accept 88.9 or 88 Total 5 4(a) Symmetry E (Q3, Jan 03)

24 e x 358(a) Work done d x SC l 0 x l e l e 0 SC 3 e 0 e de l x dx with no limits l (b)(i) Using T x : l 5g 39 x.6 5g.6 x Extension is 0. m (ii) When extension is 0.6 m, EPE x l B B for (0.6) J 3 (iii) Let y metres be distance particle is above A. C of energy, when particle has speed 0.8 m s, gives 39 (0.6 y) 5 g y (0.6).6 49y +.5(0.6 y) y 47y +.5y y 98y F 4 terms, correct 4 terms, 3 correct A 4 terms correct Ft answer to (b)(ii) y y and if x used instead of 0.6 y, here for x Speed first becomes 0.8 when y E 5 Total 3 (Q8, Jan 03)

25 36(a) KE 5 7 Total 6 74 J 70 J (b) Change in PE: mgh Carol s KE when she reaches the net J J 5350 J 3 (c) Speed of Carol is m s 4.3 m s 3 Total 8 (Q, June 03) Total Using power force velocity F 0 F 000 Accelerating force is N B Using F ma 000a a 0.38 or m s 6 Total 6 (Q7, June 03)

26 Q Solution Marks Total Comments 389(a)(i) λx Using T l 60.5 Tension in string is 3 50 N B Frictional force on A [using F μr] is g 3.36 N B which is less than tension in string Thus particle A moves towards the hole B 3 (ii) Gravitational force on B is 3g 9.4 B which is less than tension in string Thus particle B moves towards the hole B (b) λx EPE l 60 (.5) J (c) Let x be the distance B has moved upwards Work done by friction [on A] is J When B is at rest, extension is.04 x λx EPE l 60 (.04 x) 3 0(.04 x) J B C of Energy, when particle B is at rest, gives 3 g x +0(.04 x) x.4x Or 0x.4x x.555 and 0.45 Particle B is first at rest when it has moved upwards.56 m 7 Accept.55 Total 4 TOTAL 75 (Q9, June 03)