Mk Scheme (Results) Jnuy 00 GCE GCE Mthemtics (6679/0) Edecel Limited. Registeed in Englnd nd Wles No. 4496750 Registeed Office: One90 High Holbon, London WCV 7BH
Jnuy 00 6679 Mechnics M Mk Scheme Question Numbe.() Tsinθ = mg Specil cse giving θ = 0 is A0 A0 unless thee is evidence tht they think θ is with hoizontl then A0 Scheme e T o λ = mg (even T=m is, A0, A0 sp cse) l λ 0.6 = g 0.4 λ = 49 N o 5g R( ) Tcosθ = mg o cosθ = 0. 49..cosθ = 9.6 o 4 g.cosθ = g o mg.cosθ = mg (ft on thei λ ) 0.4 mg T Mks () ft cosθ = θ = 60 ( o π dins) () 6. ) 6 m = ± 5 d, with cceletion in ny fom (e.g., dt v dv d, d v d t o B Uses = dv v to obtin k d v v = ± k d d Septes vibles, k v dv = k d Obtins v = m (+ C) o equivlent e.g. 0. v = 6 (+ C) 5 Substituting = if + used elie o if used in d.e. =, v = ± = 6 + C C = 4 (o vlue ppopite to thei coect eqution) d v = 0 = 4 = m (N.B. - is not cceptble fo finl nswe ) co d N.B d ( m 6 v ) =, is lso vlid ppoch. 5 Lst two method mks e independent of elie mks nd of ech othe
Question Numbe.() Scheme Lge cone smll cone S Vol. π () (h) π h 7 π h (ccept tios : : 7) Mks B C of M h, 5 h 4 (o equivlent) B, B. π h h π h. 5 h = 7 π h. o equivlent 4 = h (5) tn θ =, = h = = 4, θ 6.6 o.0 dins (Specil cse obtins complement by using tn θ = giving.4 o.74 dins A0A0) Centes of mss my be mesued fom nothe point ( e.g. cente of smll cicle, o vete) The Method mk will then equie complete method (Moments nd subtction) to give equied vlue fo ). Howeve B mks cn be wded fo coect vlues if the cndidte mkes the woking cle. ()
4. () Enegy eqution with t lest thee tems, including K.E tem mv +.. 9 +... mg., + mg..sin0, =. mg. 6 6 V = Using point whee velocity is zeo nd point whee sting becomes slck: mw = mg 9.., mg..sin0 6 4 w = g Altentive (using point of pojection nd point whee sting becomes slck): mg mg mw mv, = 6 g So w = g,, d (6), (4), 0 In pt () D equies EE, PE nd KE to hve been included in the enegy eqution. g If sign eos led to V =, the lst two mks e M0 A0 In pts () nd A mks need to hve the coect signs In pt fo need one KE tem in enegy eqution of t lest tems with distnce to indicte fist method, nd two KE tems in enegy eqution of t lest 4 tems with 4 distnce to indicte second method. 4 SHM ppoch in pt. (Condone this method only if SHM is poved) g Using v = ω ( ) with ω = nd = ±. 4 Using = to give w = g.
5.() µ N mg N = µ N, = µ mg v µ = = = 0.6 g 75 9., () R( ) R cos α, m 0.6Rsinα = mg,, 4 5 R. = mg R = mg 5 5 5 (4) R( ) Rsin α, ± 0.6Rcosα =,, v.5 m s d co (5) In pt needs thee tems of which one is mg If cos α nd sin α e intechnged in eqution this is wded A0 In pt needs thee tems of which one is o mω If cos α nd sin α e intechnged in eqution this is lso wded A0 If they esolve long the plne nd pependicul to the plne in pt, then ttempt t R mgcosα = sinα, nd 0.6R+ mgsinα = cosα nd ttempt to eliminte v Two coect equtions Coect wok to solve simultneous equtions Answe In pt Substitute R into one of the equtions Substitutes into coect eqution (ening ccucy mks in pt ) 5mg 5mg Uses R = (o ) 9 Obtin v =.5 (4) (5)
6.() Enegy eqution with two tems on RHS, 5g = + m. mg sin θ, (d) R(\\ sting) T = 0 sin θ, = g T mg sin θ = ( tems) v = (5 + 4sin θ ) 5 6 T = mg (5 + 6sinθ ) o.e. Hs solution, so sting slck when α 6(.4) o 4. dins At top of smll cicle, 5g m. mg = g = 4.7 v = ( fo enegy eqution with tems) cso () (), () Resolving nd using Foce = T = mg, g T + mg = m. ( needs thee tems, but ny v) (6) 5 Use of v u gh = + is M0 in pt ()
7.() (Mesuing fom E) && = g 9( + 0.), nd so && = 49 SHM peiod with ω = 49 so T = π 7 M. cceletion = 49 m. = 49 0.4 = 9.6 m s, d cso (5) B () (d) Sting slck when = 0.: v = 49(0.4 0. ) 7 v.4 m s = 5 Uses = cosωt o use = sinωt but not with = 0 o ± Attempt complete method fo finding time when sting goes slck 0. = 0.4 cos 7t cos 7t = A ) t = π 0.99 s Time when sting is slck = ().4 = 0.495 s ( needed fo g 7 Totl time = 0.99 + 0.495.09 s () d ft (7) 6 () (d) D equies the minus sign. Specil cse && = g 9 is A0M0A0 && = 9 is M0A0A0M0A0 No use of &&, just is A0,A0 then A0 if othewise coect. Quoted esults e not cceptble. Answe must be positive nd evluted fo B Use coect fomul with thei ω, nd but not = 0. Coect vlues but llow = +0. Altentive It is possible to use enegy insted to do this pt λ 0.6 + mg 0.6 = l If they use If they use If they use = sinωt with = ± 0. nd dd π π 7 o 4 this is d, if done coectly = cosωt with = -0. this is d, then (s in scheme) π = cosωt with = +0. this needs thei minus nswe to ech d, then 7