AQA Maths M2. Topic Questions from Papers. Differential Equations. Answers
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1 AQA Mahs M Topic Quesions from Papers Differenial Equaions Answers PhysicsAndMahsTuor.com
2 Q Soluion Marks Toal Commens M = A Applying Newonís second law wih 0 and. Correc equaion = 0 dm Separaing ariables ln = + c 0 dm inegraing o ge ln erm. = Ae 0 A Correc inegral wih or wihou c = 0, = 0 c= 0 dm Finding consan = 0 e 0 A 7 Correc final resul Toal 7 M PhysicsAndMahsTuor.com Three erm energy equaion (Q, Jan 006) 6(a) 0 = 0 = = = + c = 0, = c= 0 AG 0 = Toal 7 M A dm dm A dm A 7 applying Newonís second law wih d correc differenial equaion separaing ariables inegraing correc inegrals finding he consan of inegraion correc final resul from correc working = 0 B correc ime Toal 8 TOTAL 60 (Q6, June 006)
3 PhysicsAndMahsTuor.com 7(a) Max speed zero acceleraion used M Implied M 7000 = k k = 0 A (i) 0 = ñ00 d = M A see, ± (ii) M M separaing ariables = ñ A Alernaie 0 ln = A ln = ñ (+ c) A [ ] [ ] 0 0 ln0 ñ ln 0 = ñ m A = ln or 7. or ñ ln A 6 = 0, = 0, c = ln0 =, = 0, ln0 = ñ + ln0 = ln or 7. m A A Toal 8(a) M (Q7, Jan 007) Q Soluion Marks Toal Commens 7(a) Using F = ma: ñ λ m = ma = m d M Condone no ë ñ í d AG Noe: no use of m no marks in (a) = λ M ln = ñ λ + c A Needs ë+ cí = C e When = 0, = U C = U M Needs correc working = U e A AG Toal 6 8(a) Q is in equilibrium E Q a res, or no moing (Q7, June 007)
4 8 8(a) Power of engine is 8kW Force exered by engine = 8000 MA M for Power = F Using F = ma: m 8000 k = 600 d 600 d k = 0 A AG PhysicsAndMahsTuor.com (i) When engine is urned off, power is zero: k = 600 d B AG (ii) 600 k = M 600 = k + c A Need + c When = 0, = 0: c = 600 = 0 0 A 600 = k + 0 When = 0, k = 0: M 0 0 = A SC k k Toal 0 (Q8, Jan 008) Q Soluion Marks Toal Commens 66(a) Using F = ma 0.0 = m d d 0.0 B Need o see m erms = 0.0 B ln = c M Need firs erms 0.0 = Ce When = 0, = 0, C = 0 M 0.0 = 0e A fully correc soluions (c) 0.0 When = 0, 0 = 0e M 0.0 e = A = ln 0.0 =.9 A Accep 0 ln Toal 8 (Q6, June 008)
5 PhysicsAndMahsTuor.com 7 8(a) Using F = ma: 0.08 = 0.0a B d B AG; condone sign error in firs B =.6 M =.6 (+ c) A Condone.6 c.6 c When = 0, = c = M.6 = * A 8 = = + A AG; all working lines correc from * Toal 7 (Q8, Jan 009) Toal 0 88(a) Using F = ma: = m d M = λ A AG = λ M = λ+ c A Condone no +c When = 0, = 9 c = M Dep. on correc inegraion A ( accep sign or ' ' error ) = λ + A = λ + 6 = + λ m Needs correc algebra 6 = A 7 AG ( + λ) (c) When =, 6 or ( ) = λ + = λ + = λ MA A Toal + λ ( λ ) = M + = 9 A = A needs saemen why λ + λ (Q8, June 009)
6 Q Soluion Marks Toal Commens 9 (a) Using F = ma, 0. = m d = 0. B AG Mus see equ n conaining m = 0. = 0.+ c Am m for + c M When = 0, = 6 C = 8 A PhysicsAndMahsTuor.com = ( 0.) = A AG =, = 0. M (c) When ( ) 0. =± = 0 or 0 A if use = 8 0. no need o see 0 = 0 A 0 as ball sops when = 0 (d) Inegraing ( ) = 0. : = x 6 0. d = + + M M for firs erms or ( 0 ) 0 When = 0, x= 0 d = 0 A 0.0 x = When speed is ms, = 0 x = m dep on M aboe = 0 A [No d, marks only] Toal (Q, Jan 00)
7 PhysicsAndMahsTuor.com 0 Toal λ = M d = λ m Condone one of +, λd, λ = λ + c AA m = 0, = u c = = u λ u A m for + c = u λ A 7 Toal 7 (Q, June 00) Q Soluion Marks Toal Commens 8(a)(i) F = 6g 60 Accep 60 6g = 6(9.8 ) B AG mus see 6g or 60 (ii) Using F = ma 6 = 6(9.8 ) M Need o see erms in m (condone sign) = (.) A AG. = d. = ln (.) = + c B M A M log side correc + c. = Ce = 0, = C = 7. or e A Or c = ln7. or.8 =. + 7.e e A Toal 8 (Q8, Jan 0)
8 PhysicsAndMahsTuor.com Q Soluion Marks Toal Commens 6(a) Using F = ma m = m d d d = AG B B: Mus see d m = m or d m = ma and correc final answer. = = + c M A M: Two inegrals wih one in he form f () where f () = ± or ±. The oher inegral mus no conain erms. A: Correc expression. Condone lack of + c for his A, bu no subsequen marks if no c. When = 0, = 6 c = = = + = + dm A AG A Toal 6 dm: Using = 0 and = 6 o find c. A: Obaining c =. A: Correc final answer. Mus see = + + = = + or or Or if hey obain = + c 6, 0 6 = = =, condone c = c (no oher roo considered) (Q6, June 0)
9 PhysicsAndMahsTuor.com 6(a) using F = ma 0. = M d = 0 ( 0.) A Needs line aboe hence d 0. = 0 ln( 0.) = 0 + c MA m M for any side inegraed correcly m for + c (and M gained) 0 0. = Ce = 0, = C = 0. A 0 = e A condone = 0. + e 0 (c) when = 0., 0. = e M 0 0 = e A = ln0 0 = 0.0 A Toal 0 subsiue 0. ino C s, afer finding c, possible numerical error (Q6, Jan 0) Q Soluion Marks Toal Commens 7(a) Using F = ma: m = or g 9.8 d M Need o see m or d d or a d =.96 ( ) A Mus see m erms (no a = ) =.96 M And one side inegraed ln ( ) =.96 + c AA Need + c, A each side When = 0, = 7 c = ln A OE ln = e.96 e A CAO Toal 7 (Q7, June 0)
10 PhysicsAndMahsTuor.com Q Soluion Marks Toal Commens (a) Using F = ma: d B d = M condone,incorrec side = + c A condone lack of + c 9 When = 0, = 8 c = 8 MA 9 = A 6 Paricle is a res when = 0 9 The alue of is 8 B Toal 7 6(a) Resole erically: (Q, Jan 0) 66(a) Using F = ma M = = 0 A 0 00 = B 0 00 = M 0ln(00 ) = + c A Condone lack of + c When = 0, = 0 c = 0ln00 MA 0ln(00 ) = 0ln00 00 = 0ln e 0 = = 00 00e or 00( e ) A 6 Toal 8 7 Using power = force elociy (Q6, June 0)
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