4762 Mark Scheme June 2005

Transcription

1 76 Mark Scheme June 00 Q mark Sub (a) (i) 0 i N s B M Equating to their 0 i in this part (A) 0 i = 70 i +0 v so v =. i m s A FT 0 i (B) 0 i = 70u i 0u i M Must have u in both RHS terms and opposite signs u = so v = i m s A FT 0 i (C) 0 i = 80( i + j ) +0v B B M FT 0 i Must have all terms present (b) (i) so v B B = (-0.8 i.6 j) m s before m s - m s - A cao 6 kg kg after v v + ve NEL v v = 0. M NEL so v v = A Any form PCLM 8 6= v+ v M PCLM A Any form Solving v =.6 so.6 m s A Direction must be clear (accept diagram) v =. so. m s A Direction must be clear (accept diagram). [Award A A0 if v & v correct but directions not clear] 6.6 m s B FT their.6 at 60 to the wall (glancing angles both 60 ) B No change in the velocity component parallel E Must give reason to the wall as no impulse No change in the velocity component E Must give reason perpendicular to the wall as perfectly elastic total 7

2 76 Mark Scheme June 00 Q mark Sub (i) (iii) We need mgh = = 990 M Use of mgh t 0 t so approx kw E Shown Driving force resistance = 0 B May be implied 000 = 800v M Use of P = Fv so v =. and speed is. m s A Force is 000 = 00 N B 0 (iv) (v) NL in direction of motion = 80a M Use of NL with all terms a = so m s A = M W-E equation with KE and power term B One KE term correct B Use of Pt.Accept wrong sign 800x B WD against resistance. Accept wrong sign A All correct x =.66 so m ( s. f.) A cao either = v M W-E equation inc KE, GPE and WD M GPE term with attempt at resolution A Correct. Accept expression. Condone wrong sign B WD term. Neglect sign. 6 v = so 9.97 m s A cao or NL + ve up plane ( g 0.0) = 80a M NL. All terms present. Allow sign errors. a = -.7 A Accept ± v = 0 + (.7... ) 0 M Appropriate uvast. Neglect signs. A All correct including consistent signs. Need not follow sign of a above. v = so 9.97 m s A cao 9

3 76 Mark Scheme June 00 Q mark Sub (i) x = 6 y x =. y =. M B B A A Complete method Total mass correct c. m. correct (or x- or y-values correct) [Allow A0 A if only error is in total mass] [If x = y claimed by symmetry and only one component worked replace final A, A by B explicit claim of symmetry A for the.] x = y B Or by direct calculation M Dealing with folded parts for x or for z 8 x = A At least terms correct for x x = (.8 ) A 8 ( ) + ( ) z = = (- 0.7 ) 8 7 A All terms correct allowing sign errors A (iii) Distance is M =.88.. so.8 m ( s. f.) F A M Use of Pythagoras in D on their c.m. c.m. clearly directly below A 8 α.88 centre of mass /7 C B Diagram showing α and known lengths (or equivalent). FT their values. Award if final answer follows their values. sin α = / M Appropriate expression for α. FT their values. so α = 0. so 0. ( s. f.) A cao total 7

4 76 Mark Scheme June 00 Q mark Sub (a) Moments c.w. about A (i) R = L so R =.L E Resolve U = 0 E Resolve V + R = L M Resolve vertically or take moments about B (or C) so V =. L E A T AC M Equilibrium at a pin-joint. L For equilibrium at A M Attempt at equilibrium at A or C including resolution with correct angle cos +.L = 0 T AB L L so T AB = so N (C) in AB A (.L ( s. f.)) T + T cos = 0 AC AB T AB (b) (i) L L so T AC = so N (T) in AC F (.L) At C L + T cos θ = 0 BC M Must include attempt at angle tanθ = / cosθ = / B so L L T BC = so N (C) in BC A (.80 L ( s. f.)) F Award for T/C correct from their internal forces. Do not award without calcs 8 F A R W G B S θ B All forces present with arrows and labels. Angles and distances not required. R = A c.w.moments about B R W cosθ = 0 M If moments about other than B, then need to resolve perp to plank as well A Correct so W cosθ (iii) Resolve parallel to plank F = W sinθ B F W sinθ μ = = = tanθ R W cosθ M Use of F = μr and their F and R A Accept any form. total 9

5 76 Mark Scheme January 006 Q mark Sub (i) 6 = 0.v M Use of I = Δ mv so 0 m s A (iii ) PCLM + ve u = 0.vP M Use of PCLM A Any form NEL +ve vp = 0. u M Use of NEL. Allow sign errors. A Any form Solving u = 8 E Must be obtained from a pair of correct equations. If given u = 8 used then v P = - must be obtained from equation and both values tested in the second equation v P = so m s A cao. Accept use of given u = 8 downwards A cao Considering the momenta involved M PCLM applied. May be implied = v D = so a = 8 and b = 6 6 Gradients of the lines are and 6 8 v D B LHS B A A A M momentum of C correct Complete equation. Accept sign error. cao cao Any method for the angle Since 6 =, they are at 90 E Clearly shown 8 8 7

6 76 Mark Scheme January 006 Q (i) Moments about C mark 0 = R D M R D = 60 so 60 N A Resolve vertically RC + RD = 0 M R C = 80 so 80 N F FT from their D Moments about C or equivalent. Allow force omitted Resolve vertically or moments about D or equivalent. All forces present. R only (A) Moments about D 0 = T sin 0 M Moments about D or equivalent M Attempt at resolution for RHS A RHS correct T = 9. so 9. N ( s. f.) A (B) In equilibrium so horizontal force needed to balance cpt of T. This must be friction Need reference to horizontal force that must and cannot be at C. come from friction at D. (iii ) (A) Moments about B 0 cos0= 6P M All terms present, no extras. Any resolution required attempted. Sub P = 60 (0.9..) E Accept decimal equivalent P inclined at 0 to vertical B Seen or equivalent or implied in (iii) (A) or (B). Resolve horizontally. Friction force F F = Psin 0 M so F = 0 (.96 ) A Any form Resolve horizontally. Any resolution required attempted

7 76 Mark Scheme January 006 (iii ) (B) Resolve vertically. Normal reaction R P cos 0 + R = 0 M Resolve vertically. All terms present.and resolution attempted A Using F = μr M μ = = = = 0.6 so 0.6 ( 0 90 s. f.) A A Substitute their expressions for F and R cao. Any form. Accept s. f. or better 9

8 76 Mark Scheme January 006 Q (a) (i) x 6 80 = y 9 x 0 80 = y 0 mark M B B Correct method for c.m. Total mass correct One c.m. on RHS correct [If separate components considered, B for correct] Sub (iii ) (b) (i) x = 6. E y = A cao Consider x coordinate 0 = x M Using additive principle o. e. on x cpts B Areas correct. Allow FT from masses from (i) so x = 8. A cao y coordinate is so we need B Position of centre of square 0 = 76y + and y = M so. ( s. f.) A cao Moments about C R = M Moments equation. All terms present so R = 600 and R = 0 E 0 N E T ED D T AE T EB T DB T DC A T AB B T BC B 0 N 0 N A 0 + TAE cos 0 = 0 M Equilibrium at a pin-joint T AE = 00 so 00 N (C) A Any form. Sign correct. Neglect (C) E 0 + TAE cos 0 + TEB cos 0 = 0 M Equilibrium at E, all terms present Any form. Sign follows working. Neglect T EB = 0 so 0 N (T) F (T). F T/C consistent with answers

9 76 Mark Scheme January (iii ) Consider at E, using gives ED as thrust E Clearly explained. Accept thrust correctly deduced from wrong answers to. 0

10 76 Mark Scheme January 006 Q mark Sub (i) M Use of P = WD/t B Δ KE. Accept ± 90 soi A All correct including signs = 0 W A (A) 0g x gx M Use of mgh on both terms B Either term (neglecting signs) 7gx (68.6x) gain A ± 7gx in any form. A cao (B) gx B (C) = gx+ gx = gx M Use of work-energy equation. Allow RHS term omitted. (iii ) x =.78 so. m ( s. f.) B A KE term correct cao. Except follow wrong sign for 7gx only. either 0. v 0. 6 M Use of work-energy. KE, GPE and WD against friction terms present. = g 0. g 0. g 0. B Δ GPE correct inc sign (.g J loss) A All correct v =.76 so v =.709 so.7 m s ( s. f.) A cao or g T = a T g g = 0a M NL in or equations. All terms present so a =. A cao v = + (.) 0. M Use of appropriate (sequence of) uvast so.7 m s ( s. f.) A cao 6

11 76 Mark Scheme June 006 Q mark Sub (a) (i) PCLM +ve (A) 6 = 8v M Use of PCLM and correct mass on RHS A Any form v = 0. so 0. m s in opposite Direction must be negative and consistent or A direction to clear. initial motion of P Accept use of a diagram. (B) (A) (B) (b) ( ) M Use of KE. Must sum initial terms. Must have correct masses = 7 J A FT their (A) only PCLM +ve 6 = vp + 6vQ M Use of PCLM vp + vq = A Any form NEL +ve vq v P = M NEL vq vp = A Any form v Q = 0. so 0. m s in orig direction of P A cao. Direction need not be made clear. v P =. so. m s in opp to orig dir of P A cao. Direction must be negative and consistent or clear (e.g diag) +ve. = N s M Use of change in momentum with correct mass. so N s in opp to orig direction A FT (A). Dir must be clear (e.g. diag) Let α = arcsin( ) and β=arcsin( ) Parallel: 6 cosα = u cos β M PCLM parallel to plane attempted. At least one resolution correct A so 6 = u and u =. A u sin β Perp: e = 6sinα M NEL on normal components attempted. F FT their u. = = 6 6 F FT their u 6 6 9

12 76 Mark Scheme June 006 Q mark Sub (i) Diagrams B Internal force at B must be shown cw moments about A 90 RB = 0 M st moments equation attempted for either force. R B = 60 so 60 N upwards A Accept direction not specified cw moments about R: T 7 + T = 0 M nd moments equation for other force. All forces present. No extra forces. A Allow only sign errors T = so N upwards A Direction must be clear (accept diag) cw moments about A 90 cos 0 V cos 0 U cos 60 = 0 M Moments equation with resolution. Accept terms missing A All correct. Allow only sign errors. giving 60 = U + V E Clearly shown 6 (iii) Diagram B U and V correct with labels and arrows (iv) ac moments about C 7 cos 0 +.V cos 0.U cos 60 = 0 M Moments equation with resolution. Accept term missing B At least two terms correct (condone wrong signs) 00 = U V A Accept any form 7 Solving for U and V M Any method to eliminate one variable 60 U = ( = ) 7 A Accept any form and any reasonable accuracy 60 V = ( = 8.78 ) 7 F Accept any form and any reasonable accuracy [Either of U and V is cao. FT the other] Resolve on BC F = U M so frictional force is 60 7 N F ( = 89. N ( s. f.)) 8 8

13 76 Mark Scheme June 006 Q mark Sub (a) (b) (i) ( R g ) 0000 = M Use of P = Fv, may be implied. B Correct weight term A All correct R = 68 so 68 N A F max = μmgcosα B Force down slope is weight cpt Correct expression for F max or wt cpt down slope (may be implied and in any form) sin α as or equivalent mg sinα B Identifying Require μmg cosα mg sinα so μ tanα = E Proper use of F μr or equivalent. [ μ = tanα used WW; SC] either 0. v M Use of work energy with at least three required terms attempted = g g. + 9 B Any term RHS. Condone sign error. B Another term RHS. Condone sign error. A All correct. Allow if trig consistent but wrong v = v =.86 so.9 m s ( s. f.) A cao or + ve up the slope g 0. g 6 = a M Use of NL B Any correct term on LHS a = m s - A v = - a M use of appropriate uvast v =.86 m s - A c.a.o. (iii) continued overleaf

14 76 Mark Scheme June 006 continued (iii) either Extra GPE balances WD against resistances M Or equivalent mgxsinα B = 6( x+ ) + 0. g cos α( x+ ) B One of st three terms on RHS correct B Another of st terms on RHS correct A All correct. FT their v if used. x =.9986 so.99 m ( s. f.) A cao. 6 or M Allow term missing B KE. FT their v = (. + x) g 6(. + x) B Use of. + x (may be below) (. + x) 0. g B WD against friction A All correct x =.9986 so.99 m ( s. f.) A cao. or + ve down the slope g 0. g 6 = a M NL with all terms present A all correct except condone sign errors a =... m s - A.86 = a(.+x) M use of appropriate uvast B for (. + x) (may be implied) x =.99 A c.a.o. 8

15 76 Mark Scheme June 006 Q mark Sub (i) x = y 0 0 M Correct method for c.m. B B Total mass correct One c.m. on RHS correct [If separate components considered, B for correct] x = y 800 x = 7 A cao y = 8 A cao. [Allow SC / for x = 8 and y = 7 ] ( 7,8,0 ) B x- and y- coordinates. FT from (i). B z coordinate (iii) cw moments about horizontal edge thro D M Or equivalent with all forces present x component P 0 60 ( 0 7) = 0 One moment correct (accept use of mass or B length) B correct use of their x in a distance P = 9 A FT only their x (iv) Diagram B Normal reaction must be indicated acting vertically upwards at edge on Oz and weight be in approximately the correct place. (v) On point of toppling ac moments about edge along Oz M Or equivalent with all forces present 0 Q 60 7 = 0 Any moment correct (accept use of mass or B length) Q = F FT only their x Resolving horizontally F = Q B As > 0, slips first B FT their Q correctly argued. 7

16 76 Mark Scheme Jan 007 Q mark sub (i) before after 0 m s 0 0. kg 9. kg v m s v m s 0 0. = 0.v + 9.v M PCLM and two terms on RHS A All correct. Any form. v v = M NEL A Any form v = 0. so V = 0. A v = 7.7 so V = 7.7 m s A Speed. Accept ±. in opposite to original direction F Must be correct interpretation of clear working (A) 0 0. = 0V M PCLM and coalescence A All correct. Any form. so V = 6 A Clearly shown. Accept decimal equivalence. Accept no direction. 7 (B) Same velocity E Accept speed No force on sledge in direction of motion E (iii) m s m s before 9. kg 0. kg B after V u 0= 0.u+ 9.V M PCLM, masses correct A Any form u V = 0 B May be seen on the diagram. Hence V =.87 A Accept no reference to direction. 7

17 76 Mark Scheme Jan 007 Q mark comment sub (i) X = Rcos0 () B Y + Rsin 0 = L () M Attempt at resolution A ac moments about A R L = 0 B (iii) (iv) Subst in () and () M Subst their R = L into their () or () X = L so X = L E Clearly shown Y + L = L so Y + L = L and Y = 0 E Clearly shown (Below all are taken as tensions e. g. T AB in B Attempt at all forces (allow one omitted) AB) B Correct. Accept internal forces set as tensions or thrusts or a mix A T cos0 ( 0 AD = M Vert equilibrium at A attempted. Y = 0 need not be explicit so T AD = 0 E (v) Consider the equilibrium at pin-joints M At least one relevant equilib attempted A TAB X = 0 so TAB = L (T) B (T) not required (vi) C L + T cos0 0 CE = B Or equiv from their diagram L L L so TCE = so = (C) B Accept any form following from their equation. (C) not required. C cos60 0 BC CE B Or equiv from their diagram L FT their T L CE or equiv but do not condone so TBC = = (T) B inconsistent signs even if right answer obtained. (T) not required. F T and C consistent with their answers and their diagram B TBD cos0 + TBE cos0 = 0 M Resolve vert at B T = T so mag equal and opp sense E A statement required so BD BE 0 7

18 76 Mark Scheme Jan 007 Q mark sub (i) (0,,.) B By symmetry x = 0, B y = B (0 + 80) z = B Total mass correct M Method for c.m. so z =.87 A Clearly shown (iii) x = 0 by symmetry E Could be derived x 0 0 (0 + 80) y = z.87 M Method for c.m. B y coord c.m. of lid B z coord c.m. of lid y =. E shown z =. E shown 6 (iv) P cm. cm X. cm 0 B Award for correct use of dimensions. and. or equivalent 0 N c.w moments about X cos sin 0 B st term o.e. (allow use of. and.) B nd term o.e. (allow use of. and.) = 0.8 so 0. N m ( s. f.) E Shown [Perpendicular method: M Complete method: A Correct lengths and angles E Shown] (v) P = 0 M Allow use of P = so 8. ( s. f.) A Allow if cm used consistently 8

19 76 Mark Scheme Jan 007 Q mark sub (i) (A) (B) Fmax = μr M Must have attempt at R with mg resolved B R = g cos 0 so F max = cos0 = so.7 N ( s. f.) A [Award / retrospectively for limiting friction seen below] either Weight cpt down plane is gsin 0 = 9.8 N B so no as 9.8 <.7 E The inequality must be properly justified or Slides if μ < tan 0 B But 0.7 > 0.77 so no E The inequality must be properly justified Increase in GPE is 9.8 (6 + sin 0) = 6.8 J M Use of mgh B 6 + sin 0 A WD against friction is cos 0 = J M Use of WD = Fd A (C) Power is 0 ( )/60 M Use P = WD/t =.60 so.6 W ( s. f.) A (iii) 0. 9 M Equating KE to GPE and WD term. Allow sign errors and one KE term omitted. Allow old friction as well. = 9.8 (6+ xsin0) + 0. B Both KE terms. Allow wrong signs. 90 A All correct but allow sign errors A All correct, including signs. so x =.86. so.8 ( s. f.) A cao 7

20 76 Mark Scheme June 007 Q (a) (i) Impulse has magnitude 9= 8 N s B (iii) (iv) (A) speed is 8 6 = m s. B PCLM 6 = 8v M Use of PCLM + combined mass RHS A All correct v = so m s in orig direction of A E Must justify direction (diag etc) = 6 N s M Attempted use of mv - mu A for 6 N s dir specified (accept diag) ms -.8 m s AB C B Accept masses not shown v ms.9 m s (B) PCLM = 8v M PCLM. All terms present A Allow sign errors only v =.87 A (C) (b).9.87 NEL = e.8 M Use of NEL with their v A Any form. FT their v so e = 0. F FT their v (only for 0< e ) Using v = u + as v = = B Allow ± rebounds at 7 M Using their vertical component = 8 m s F FT from their. Allow ± No change to the horizontal component B Need not be explicitly stated Since both horiz and vert components are 8 m s the angle is A cao 9 8

21 76 Mark Scheme June 007 Q (i) π θ = B π 8sin 6 gives CG = π = π E 6,8 π justified E 6 x 6 (8π + 7) 8π π = + 7 y 0 8 x = = ( s. f.) y (iii).9 M Method for c.m. B Correct mass of 8 or equivalent A st RHS term correct A nd RHS term correct E E [If separate cpts award the As for x- and y- cpts correct on RHS] 6 A G α.7 B General position and angle (lengths need not be shown).9 Angle or complement attempted. arctan or tanα = M.7 equivalent. M Attempt to get A Obtaining.9 cao Accept use of.0699 but not 6. α = so 8.8 ( s. f.) A cao (iv) [FT use of.0699 ] c. w. moments about A M Moments about any point, all forces present.9 6F = 0 A so F = 0.7 A (... if.0699 used) 7 9

22 76 Mark Scheme June 007 Q (i) (iii) (iv) Moments c.w. about B RA = 0 M Accept about any point. Allow sign errors. R A = 0 so 0 N A Resolve or moments M R B = 0 so 0 N F Moments c.w. about D 0.8R = 0 M C Or equiv. Accept about any point. All terms present. No extra terms. Allow sign errors. R C = 00 A Neglect direction Resolve or moments M Or equiv. All terms present. No extra terms. Allow sign errors. R D = 00 A Neglect direction E Both directions clearly shown (on diag) Moments c.w. about P 0. 00cosα 0.8RQ = 0 M Or equiv. Must have some resolution. All terms present. No extra terms. Allow sign errors. A Correct R Q = 96 so 96 N A [No direction required but no sign errors in working] resolve perp to plank RP = 00cosα + RQ M Or equiv. Must have some resolution. All terms present. No extra terms. Allow sign errors. A Correct R P = 88 so 88 N A [No direction required but no sign errors in working] Need one with greatest normal reaction So at P B FT their reactions 6 Resolve parallel to the plank F = 00sinα so F = 6 B F μ = R M Must use their F and R 6 7 = = (= 0.9 ( s. f.)) 88 6 A cao 9 60

23 76 Mark Scheme June 007 Q (i) either M KE or GPE terms B KE term B GPE term = 786. J A cao or a = B T 0g = 0 M NL. All terms present. T = 96.6 A WD is T = 786. so 786. J A cao 0g 0. 0g M Use of P = Fv or Δ WD Δ t A All correct A (iii) GPE lost is 9.8 = 09 J B KE gained is 0. ( ) = 0 J M Δ KE A The 0 J need not be evaluated so WE gives WD against friction is M Use of WE equation 09 0 = 889 J A cao (iv) either x = 0x M WE equation. Allow missing term. No extra terms. B One term correct (neglect sign) B Another term correct (neglect sign) A All correct except allow sign errors x =.67 so.6 m ( S. F.) A cao or g 0. 0 = a M Use of NL. Must have attempt at weight component. No extra terms. A Allow sign errors, otherwise correct a =.07 A cao 0 = 9 ax M Use of appropriate uvast or sequence x =.67 so.6 m ( S. F.) A cao 7 6

24 76 Mark Scheme January Mechanics Q Mark Comment Sub (a) (i) either In direction of the force I = Ft = mv M Use of Ft = mv so 00 8 = 000v A giving v = so m s - A or NL gives 00 a = M Appropriate use of NL and uvast 000 v = A giving v = so m s - A 0 m s - before after 00 V S m s V Rm s - PCLM 000 = 000VR + 00VS M Appropriate use of PCLM so = 8VR + VS A Any form VS VR NEL = 0. 0 M Appropriate use of NEL so VS VR = 0.6 A Any form Solving VR =.6, VS =. A Either value so ram.6 m s - and stone. m s - F (iii) M Change in KE. Accept two terms B Any relevant KE term correct (FT their speeds) = 90 J A cao (b) see over 6 7

25 76 Mark Scheme January 008 Mark Comment Sub (b) (i) 7i N s B Neglect units but must include direction 8(9cos60i+ 9sin 60 j ) = ( 6i+ 6 j) N s E Evidence of use of 8 kg, 9 m s - and 60 7 i+ (6i+ 6 j) = ( ui+ vj ) M PCLM. Must be momenta both sides Equating components M = u so u = 9 6 = v so v = A Both (iii) either 8= 8 9 so equal momenta so 60/ = 0 M Must be clear statements A cao or arctan = arctan = 0 M FT their u and v. ( ) ( ) 9 A cao 9 Q Mark Comment Sub (i) (A) = 60 J M Use of KE A (B) 60 = F M W = Fd attempted so F = 0 so 0 N F FT their WD Using the WE equation M Attempt to use the WE equation. Condone one missing term M Δ KE attempted = h 600 B 600 with correct sign A All terms present and correct (neglect signs) h = 6.6 so 6. ( s. f.) A cao (iii) (A) We have driving force F = 0 B May be implied so 00 = 0v M Use of P = Fv and v = so m s - A (B) From NL, force required to give accn Use of NL with all terms present (neglect M is signs) F 0 = 80 A All terms correct so F = 00 A P = = 00 so 00 W M correct use of P = Fv A cao 7 8

26 76 Mark Scheme January 008 Q Mark Comment Sub (i) For z M Method for c.m. ( ) z B Total mass of 6000 (or equivalent) = B At least one term correct so z = 7. A NB This result is given below. and y = 0 NB This result is given below. Statement (or B proof) required. N.B. If incorrect axes specified, award max / y and z are not changed with the folding E A statement, calculation or diagram required. For x = 6000x M Method for the c.m. with the folding B Use of the so x = = E Clearly shown (iii) Moments about AH. M Normal reaction acts through this line B May be implied by diagram or statement (iv) c.w. P 0 7 ( 0.) = 0 B 0. or equivalent A All correct so P = 0. A cao Fmax = μr M Allow F = μr = μ A Must have clear indication that this is max F so Fmax 7 For slipping before tipping we require 7μ < 0. M Accept. Accept their F max and R. so ( 8 ) μ < A cao 8 9

27 76 Mark Scheme January 008 Q Mark Comment Sub (i) Centre of CE is 0. m from D B Used below correctly a.c. moment about D M Use of their = 00 so 00 N m E 0. must be clearly established. c.w moments about D R.7 00 = 0 M Use of moments about D in an equation B Use of 00 and.7 or equiv R = 00 so 00 N A 6 c.w moments about D W = 0 M Moments of all relevant forces attempted A All correct so W = 0 E Some working shown (iii) (A) c.w. moments about D. 0 cos 0.7T 00cos0 00.7cos0 = 0 M M Moments equation. Allow one missing term; there must be some attempt at resolution. At least one res attempt with correct length Allow sin cos (iii) (B) A Any two of the terms have cos 0 correctly used (or equiv) B.7 T A All correct T = so 9. N ( s. f.) A cao Accept no direction given either Angle required is at 70 to the normal to CE B so cos M so T = so 7 N ( s.f.) A FT (iii) (A) or 00 cos cos 0 M Moments attempted with all terms present = 0cos 0. T sin 0.7 A All correct (neglect signs) T = so 7 N (s.f.) A FT(iii)(A) 8 6 0

28 76 Mark Scheme June Mechanics Q mark comment sub (a) (i) In i direction: 6u = 8 M Use of I-M so u = i.e. i m s - Accept 6u = 8as total working. Accept E instead of i. either In i direction: 0.v + = 0. M Use of I-M B Use of + i or equivalent v = so i m s - A Accept direction indicated by any means or v = M PCLM v = A Allow only sign errors so i m s - A Accept direction indicated by any means Using NEL: = e M Use of NEL. Condone sign errors but not reciprocal expression F FT only their (even if +ve) e = (0.) 9 F FT only their and only if ve (allow s.f. accuracy) (iii) In i direction: 7 = 0.v 0. M Use of I = Ft M Use of I = m(v u ) v = 7 so 7 i m s - A For ± 7 A cao. Direction (indicated by any means) or i = 0. a M Use of F = ma so a = i m s - A For ± v = i i 7 M Use of uvast v = 7 so 7 i m s - A cao. Direction (indicated by any means) (b) u i + ev j B For u B For ev v ev Use of tan. Accept reciprocal argument. Accept tanα =, tan β = M u u use of their components v tan β = e = etanα u B E 7 Both correct. Ignore signs. Shown. Accept signs not clearly dealt with. 7

29 76 Mark Scheme June 008 Q m a r k comment sub (i) x 6 0 ( + 6) y = M Method for c.m. B Total mass correct x y = = B For any of the st RHS terms B For the th RHS term x =.6 E y =. A cao [If separate cpts, award the nd B for x- terms correct and rd B for 7in y term] 6.6. D.6. G 6 O vertical B Diagram showing G vertically below D B.6 and their. correctly placed (may be implied).6.6 arctan = arctan Use of arctan on their lengths. Allow reciprocal M + (6.).6 of argument. Some attempt to calculate correct lengths needed B + (6 their.) seen so 8.07 so 8.0 ( s. f.) A cao (iii) moments about D M moments about D. No extra forces.6= 6 T BP so tension in BP is N F FT their values if calc nd Resolve vert: + TDQ = M Resolve vertically or moments about B. so tension in DQ is N F FT their values if calc nd (iv) We require x-cpt of c.m. to be zero M A method to achieve this with all cpts either 0 + L x = 0.6 L ( ) or L = 0 L = ( ) B A A 9 For the 0. L All correct 8

30 76 Mark Scheme June 008 Q mark comment sub (a) D (i) T AD T BD A 0 B 60 C T AB T BC B B Internal forces all present and labelled All forces correct with labels and arrows (Allow the internal forces set as tensions, thrusts or a mixture) L N L N M Equilibrium equation at a pin-joint attempted A st ans. Accept + or. A M Second equation attempted T AD sin 0 L = 0 so T AD = L so L N (T) F nd ans. FT any previous answer(s) used. A T AB + T AD cos 0 = 0 so T AB = L so L N (C) M Third equation attempted B T BD sin 60 L = 0 A rd ans. FT any previous answer(s) used. so T BD = L so L N (T) B M Fourth equation attempted T BC + T BD cos 60 T AB = 0 F th ans. FT any previous answer(s) used. so T BC = L so L N (C) E All T/C consistent [SC all T/C correct WWW] 9 (b) Leg QR with frictional force F moments c.w. about R U sin60 l Wlcos60= 0 M A A Horiz equilibrium for QR F = U M Accept only leg considered (and without comment) Suitable moments equation. Allow force omitted a.c. moments c.w. moments A second correct equation for horizontal or vertical equilibrium to eliminate a force (U or reaction at foot) [Award if correct moments equation containing only W and F] E * This second equation explicitly derived Hence W = F M Correct use of nd equation with the moments equation and so F = W E Shown. CWO but do not penalise * again

31 76 Mark Scheme June 008 Q mark comment sub (a) (i) Tension is perp to the motion of the sphere E (so WD, Fd cosθ = 0) Distance dropped is cos0 = M Attempt at distance with resolution used. Accept sin cos E Accept seeing cos0 GPE is mgh so M = J B Any reasonable accuracy (iii) v = M Using KE + GPE constant so v =.087 so.0 m s - ( s. f.) F FT their GPE (iv) 0. ( v. ) M Use of W-E equation (allow KE term or GPE term omitted) (b) B KE terms correct 0 = π 60 M WD against friction A WD against friction correct (allow sign error) v =.0678 so.06 m s - ( s. f.) A cao NL down slope: g sin0 F = g 8 M Must have attempt at weight component A Allow sign errors. 9g so F = (=.0) 8 A R = g (=.6 ) B μ = F R M Use of F = μ R E Must be worked precisely 8 6 0

32 76 Mark Scheme January Mechanics Q Mark Sub (i) either M Use of I = Ft m u = F A so F = 0.mu in direction of the velocity A Must have reference to direction. Accept diagram. or M Use of suvat and NL a = u A May be implied so F = 0.mu in direction of the velocity A Must have reference to direction. Accept diagram. PCLM um + um = mv + mv NEL v v = u u = u Energy = Q P m ( u) + ( m) u P Q P Q m v + ( m) v A One correct equation M For equns considering PCLM, NEL or Energy (iii) A Second correct equation Solving to get both velocities M Dep on st M. Solving pair of equations. u v Q = E If Energy equation used, allow nd root discarded without comment. u v P = A [If AG subst in one equation to find other velocity, and no more, max SC] either eu After collision with barrier v Q = B Accept no direction indicated so m u m eu = m u M PCLM so e = A A LHS Allow sign errors. Allow use of mv Q. A RHS Allow sign errors At the barrier the impulse on Q is given by u u m M Impulse is m(v u) F u ± so impulse on Q is 6 mu F Allow ± and direction not clear. FT only e. so impulse on the barrier is 6 mu A cao. Direction must be clear. Units not required

33 76 Mark Scheme January 009 Q continued mark sub (iii) or eu After collision with barrier v Q = B Impulse momentum overall for Q u mu + mu + I = m M All terms present A All correct except for sign errors I = 6mu A so impulse of 6mu on the barrier A Direction must be clear. Units not required. Consider impact of Q with the barrier to give speed v Q after impact u m 6mu = mvq M Attempt to use I - M F u so v Q = F u u e = = A cao 9 9

34 76 Mark Scheme January 009 Q Mark Sub (i) (iii) (iv) (v) R = 80gcosθ or 78cosθ B Seen Fmax = μr M so g cos θ or.6cosθ N A Distance is. sinθ B WD is F max d M. so g cosθ sinθ E Award for this or equivalent seen = 9 tanθ Δ GPE is mgh M so = 980 J A Accept 00g J either P = Fv M so (80gsin + gcos). B Weight term A All correct = 09.8 so 060 W ( s. f.) A cao or WD P = Δ t M so tan B Numerator FT their GPE. B Denominator. sin = 09.8 so 060 W ( s. f.) A cao either Using the W-E equation M Attempt speed at ground or dist to reach required speed. Allow only init KE omitted v = 980 tan B KE terms. Allow sign errors. FT from (iv). B Both WD against friction and GPE terms. Allow sign errors. FT from parts above. A All correct v =.79.. so yes A CWO or NL down slope M All forces present a =.0997 A distance slid, using uvast is.87 A vertical distance is.87 sin M valid comparison =.0 <. so yes A CWO 7 0

35 76 Mark Scheme January 009 Q Mark Sub (i) (iii) (iv) M Correct method for y or z B Total mass correct M cosα or sinα attempted either part y: cosα = 7y B 8+ cosα B 0 8 y = = 9 E Accept any form z : 0 (0 0 + ) sinα = 7z B LHS z = E Yes. Take moments about CD. E c.w moment from weight; no a.c moment from table E [Award E for 9 > 8 seen or the line of action of the weight is outside the base] c.m. new part is at (0, 8 + 0, ) M Either y or z coordinate correct M Attempt to add to (i) or start again. Allow mass error = 00y so y = E 7 + = 00z so z = 6 E Diagram B Roughly correct diagram B Angle identified (may be implied) 6 Angle is arctan M Use of tan. Allow use of /6 or equivalent. =.98 so. ( s. f.) A cao 8 8

36 76 Mark Scheme January 009 Q mark sub (a) Let the forces at P and Q be RP and R Q (i) c.w. moments about P M Moments taken about a named point. 600 R Q = 0 so force of 00 N at Q A a.c. moments about Q or resolve M R P = 00 so force of 00 N at P A (b) (i) (iii) R P = 0 B Clearly recognised or used. c.w. moments about Q M Moments attempted with all forces. Dep on R P = 0 or RP not evaluated. L 600= 0 so L = 00 A cos α = or sin α = 7 or tanα = B Seen here or below or implied by use. c.w moments about A M Moments. All forces must be present and appropriate resolution attempted. 6 0cosα 8R = 0 A so R = 600 E Evidence of evaluation. Diagram (Solution below assumes all internal forces set as tensions) B B M B 0cosα + TBC cosα = 0 so T BC = 0 (Thrust of) 0 N in BC A C TBC sinα TAC sinα = 0 so T AC = 0 (Thrust of) 0 N in AC F Must have 600 (or R) and 0 N and reactions at A. All internal forces clearly marked as tension or thrust. Allow mixture. [Max of B if extra forces present] Equilibrium at a pin-joint B AB BC sin 0sin 0 M Method for T AB so T AB = 0 (Tension of) 0 N in AB A Tension/ Thrust all consistent with working F [Award a max of /6 if working inconsistent with diagram] 9 6

37 76 Mark Scheme June Mechanics Q mark comment sub (a) (i) B before u u after P m kg v Q km kg u (iii) (iv) u mu kmu = mv + km M PCLM applied A Either side correct (or equiv) k v = u E Must at least show terms grouped Need v < 0 E Accept k > without reason so k > B [SC: v = 0 used and inequality stated without reason] u v = u u M Use of NEL A u so v = E u k = u M so k =. A cao (b) (i) 9 + = 8 V M Use of PCLM B Use of mass 8 in coalescence M Use of I = Ft V = E i cpt M Allow wrong sign 6

38 76 Mark Scheme June 009 j cpt unchanged B May be implied cao [Award / if barrier taken as. new velocity m s - A 0 ] 8 Q mark comment sub (a) (i) (A) Yes. Only WD is against Accept only WD is against gravity E conservative or no work done forces. against friction. (B) Block has no displacement in that direction E (iii) (b) gx = gx M x = 0.86 so 0.8 m ( s. f.) Use of WE with KE. Allow m =. B Use of 0 M At least GPE term A GPE terms correct signs A V M cao WE equation with WD term. Allow GPE terms missing B Both KE terms. Accept use of. = 0g g 80 B Either GPE term B 80 with correct sign V =.609 so.6 m s - A cao Force down the slope is g sin 0 M Both terms. Allow mass not weight B Weight term correct Using P = Fv M P = ( gsin 0). F FT their weight term P = so 8770 W ( s. f.) A cao 7 7

39 76 Mark Scheme June 009 Q mark comment sub (i) c.w. moments about A M Moments equation. RB 8= 0 so R B = giving N A Accept no direction given Either a.c. moments about B or resolve M R A = so N F Accept no direction given c.w. moments about A M Moments with attempt to resolve at least one force. Allow s c. 8 cosα 7. sinα = 0 B Weight term (iii) 8 so tanα = = 7. 8 A S N R F α 8 N B a.c. moments about B B E B M 8 cosα +. S sinα = 0 B. A S = 7. A All correct horiz force term Must see some arrangement of terms or equiv All forces present and labelled Moments with attempt to resolve forces and all relevant forces present All other terms correct. Allow sign errors. Resolving horizontally and M vertically S F sinα = 0 so F = 7. E R 8 cosα = 0 A Using F = μr M 7. μ = = so A ( s. f.) 8 Either attempted R = 0 need not be evaluated here [Allow A for the two expressions if correct other than s c ] cao 0 8

40 76 Mark Scheme June 009 Q mark comment sub (i) Taking a y-axis vert downwards from O Allow areas used as masses k πσ 8 + πσ 8 k M Method for c.m. B used B 6π k B k/ used = πσ 8 + πσ 8k y B Masses correct so ( ) 6 + k y = 6 + k E Must see some evidence of simplification Need no reference to axis of symmetry 6 (iii) k = gives OG as. and mass as 0πσ πσ + πσ M 0. 8 ( πσ πσ ) = y B Allow for either. Allow σ = B B Method for c.m. combining with (i) or starting again One term correct Second term correct y = 6 E Some simplification shown O G 8 6 = B B B G above edge of base 6 = seen here or below 8 seen here or below θ 8 Accept tanθ = M 8 6 and 8. θ =.6887 so.7 ( s. f.) A cao or attempts based on (iv) Slips when μ = tanθ M Or = B <. so does not slip A There must be a reason 9 9

41 76 Mark Scheme January 00 (a) (i) 76 Mechanics Let vel of Q be v 6 = v + M Use of PCLM A Any form v = 0. so 0. m s - A in opposite direction to R A Direction must be made clear. Accept 0. only if + ve direction clearly shown Let velocities after be R: v R ; S: v S PCLM +ve = vr + vs M PCLM vr + vs = A Any form NEL +ve vs vr = 0. M NEL so v S v R = 0. A Any form Solving gives v R = 0.7 A Direction not required v S =. A Direction not required Award cao for vel and FT second (iii) R and S separate at 0. m s - M FT their result above. Either from NEL or from difference in final velocities Time to drop T given by T = 0. so T = 7 (0.87 ) B so distance is 7 0. = 7 m (0.87 m) A cao 6 (b) before v after ev u u B v ( ) ev B Accept v ev KE loss is m u + v m u + e v M Attempt at difference of KEs ( ) ( ) = u u mu + mv mu me v E Clear expansion and simplification = mv ( e ) of correct expression 7 8

42 76 Mark Scheme January 00 (i) GPE is = B Need not be evaluated Power is ( ) 0 M power is WD time B 0 s = W = W ( s. f.) A cao Using P = Fv. Let resistance be R N M Use of P = Fv. 00 = 8F so F = 70 A As v const, a = 0 so F R = 0 Hence resistance is 70 N E Needs some justification We require = J (= 0 kj) M F Use of WD = Fd or Pt FT their F (iii) M Use of W-E equation with x 00 ( 9 8 ) B KE terms present = x sin 00x M GPE term with resolution A GPE term correct A All correct Hence 800 = x so x = 06.9 so 07 m ( s, f,) A cao 6 (iv) P = Fv B and NL gives F R = 00a B Substituting gives P = (R + 00a)v E Shown If a 0, v is not constant. But P and R are constant so a cannot be constant. E 9 (i) Let force be P (A) a.c. moments about C P = 0 M Moments about C. All forces present. No extra forces. A Distances correct P = 60 so 60 N A cao (i) (B) Let force be P c.w. moments about E P. 0 ( 0.) = 0 M Moments about E. All forces present. No extra forces. A Distances correct P = 0 so 0 N A cao 9

43 76 Mark Scheme January 00 Qsinθ. + Qcosθ 0.9 M Moments expression. Accept s c. B Correct trig ratios or lengths.q.q = + = 0 Q so 0 Q N m E Shown (iii) We need 0Q = 0. M Moments equn with all relevant forces so Q = E Shown Let friction be F and normal reaction R Resolve cosθ F = 0 M so F = 8 A Resolve sin θ + R = 0 M so R = 6 A F < μr as not on point of sliding M Accept or = so 8 < 6μ A Accept. FT their F and R so μ > 8 E (i) x y = 0 0 M A so x = 6 E y = A Any complete method for c.m. Either one RHS term correct or one component of both RHS terms correct [SC for correct y seen if M 0] x y M Any complete method for c.m = B Any edges correct mass and c.m. or any edges correct with mass and x or y c.m. coordinate correct. B At most one consistent error x =. E y = 0. A 0

44 76 Mark Scheme January 00 (iii) Q O G. B Indicating c.m. vertically below Q N. Angle is arctan 0 0. B = so 8. ( s. f.) A cao Clearly identifying correct angle (may be implied) and lengths arctan b a where b =. and a = 69.8 M Award for ( ) or 0. or where b = 69.8 or 0. and a =.. Allow use of their value for y only. (iv) x y = + 0. M Combining the parts using masses B Using both ends A All correct x =.0 so.0 ( s.f.) A cao y =. so. ( s.f.) F FT their y values only. 8

45 GCE Mathematics (MEI) Advanced GCE 76 Mechanics Mark Scheme for June 00 Oxford Cambridge and RSA Examinations

46 76 Mark Scheme June 00 Q mark sub (i) (iii) For P = 00v P M Award for I-M v P = 6. so 6.i m s - E Accept no i and no units For Q 0 0 = 0v Q M Must have impulse in opposite sense v Q = so i m s - A Must indicate direction. Accept no units. i direction positive PCLM: 0 = wQ M PCLM used. Allow error in LHS FT from (i) F Any form. FT only from (i) w Q =. so.i m s - E wq NEL: 6.= e M NEL. Allow sign errors A Signs correct. FT only from (i) e = 0. A cao i direction positive Suppose absolute vel of object is Vi 00. = 0V M Applying PCLM. All terms present. Allow sign errors. B Correct masses A All correct (including signs) V =. A speed of separation is. +. = 0 m s - F FT their V. 6 (iv) = 0W M Using correct masses and velocities W =.8 so. i m s - ( s. f.) A cao 7

47 76 Mark Scheme June 00 Q mark sub (i) x y = M B A x = 7. A y = A Method to obtain at least coordinate 00 or correct Either one RHS term correct or one component of two RHS terms correct x 0 0 = + y M Using (i) or starting again so x =, y = 68 E Clearly shown. (iii) We need the edge that the x position is nearest M This may be implied x = ; distances are to PQ, 8 to SR B One distance correct to QR B All distances correct so edge QR A (iv) Moments about RS M Moments about RS attempted B Use of weight not mass below. FT mass from here In sense xoy T sin 0 00 T cos0 0 M Attempt to find moment of T about RS, including attempt at resolution. May try to find perp dist from G to line of action of the force. A 0 g (0 7.) = 0 B 0 7. A All correct allowing sign errors T =.889 so.6 N ( s. f.) A cao (except for use of mass) 7 8

48 76 Mark Scheme June 00 Q mark sub (i) a.c. moments about A T 00 = 0 so T = 600 E Resolving X = 0 B Justified T Y = 00 M so Y = 00 A Diagram B All external forces marked consistent with (i) The working below sets all internal forces as tensions; candidates need not do this. B All internal forces with arrows and labels (iii) Let angle DAB be θ. cos θ =, sinθ = A 00 T sinθ = 0 AB B M M so T AB = 00 so force is 00 (C) A A T + T cosθ = 0 AD AB so T AD = 00 so force is 00 (T) F C T sinθ 00 = 0 CD so T CD = 00 so force is 00 (T) F C T + T cosθ = 0 BC CD so T BC = 00 so force is 00 (C) F B T sinθ + T = 0 AB BD so T BD = 00 so force is 00 (T) F F Or equivalent seen Attempt at equilibrium at pin-joints equilib correct, allowing sign errors All T/C consistent with their calculations and diagrams 9 (iv) AD, AB, BC, CD B 00 N, X and Y not changed. Equilibrium equations at A and C are not altered E B T sin θ + T' = 0 M AB BD so T ' BD = 00 so force is 00 (C) A C not needed. [If 00 N (C) given WWW, award SC (NB it must be made clear that this is a compression)] 9

49 76 Mark Scheme June 00 Q mark sub (i) Let friction be F N and normal reaction R N F max = 8cos B Need not be explicit R = 6g+ 8sin M Both terms required. A Fmax = μr M so μ = about 0. E WD is 70cos = 0cos M Use of WD = Fd. Accept cos omitted. so 7.09 = 7 J ( s. f.) A (iii) (iv) Average power is WD/time M so.0. =. W ( s. f.) A cao Using the constant acceleration result s = u+ v t with s =, u = 0, v =. M Attempt to substitute in suvat (sequence) ( ) and t = we see that 0 +. =.7 ( ) E Conclusion clear 7.09 M Using W-E equn, allow missing term = 6. M KE term attempted A correct + 0. ( 6g + 70sin ) M Attempt at using new F in F max = μr + WD A A All correct so WD by S is so 6. J ( s. f.) A cao 7 8

50 GCE Mathematics (MEI) Advanced GCE Unit 76: Mechanics Mark Scheme for January 0 Oxford Cambridge and RSA Examinations

51 76 Mark Scheme January 0 Q m a r k notes (i) Let normal reaction be R sin B Accept any form and implied M Use of Fmax R R B Expression for R; may be implied Fmax 0.8 R 6.66 F FT their R Wt cpt down slope is =.7 B 6.66 >.7 so at rest E FT if their F and weight component show given result If g omitted, allow BMB0FB0E, so /6 [Award as follows for use of tan : B tan E tan shown] 6 Let the speeds down the plane be v A and v B. PCLM down the plane. 6.vA.vB M PCLM so va vb 8 A Any form NEL +ve down the plane va vb M NEL. Allow sign errors va vb 6. A Any form va 8. so 8. m s - down plane E Condone direction not clear if +8. seen vb so m s - down plane F Condone direction not clear if + seen. SC if equations obtained and 8. substituted into one to obtain answer (instead of EF) 6 (iii). ( 6) down plane M Use of m(v u) If impulse on A found, treat as MR unless final answer relates this to impulse on B = N s down the plane A ± N s so Ns up the plane A Direction explicitly commented on 6

52 76 Mark Scheme January 0 Q m a r k notes (iv) either ( Fmax ) t.(0 8.) M Using Impulse-momentum (must use 8.). sufficient to consider one term on LHS B Either side correct A Allow only sign errors so t = s ( s. f.) A cao or Using NL down the plane M Using NL ; sufficient to consider one force term a = 0.78 A Allow sign errors M Using appropriate suvat must use a or-a found by use of NL and u = 8. using v = u + at, t = s ( s. f.) A cao or ( ) x 0 M Use energy with 8., sufficient to consider one non-ke term x= A M Using appropriate suvat T = ( s. f.) A cao 9 7

53 76 Mark Scheme January 0 Q m a r k notes (a) - Energy: v ms V ms - i C 0.00 kg 0.00 v V 0.8 M Use of KE in two terms in an equation. v V 00 A Any form PCLM in i direction: 0.06V 0.00v 0 M PCLM. Accept sign errors. v = V A Any form Solving M Valid method for elimination of v or V from a linear and a quadratic ( V) V 00 so V and V i A Accept.9099 i Accept no direction 00 0 B kg v i (= 7i ) F Accept 9.69 i Accept no direction Second answer follows from first A (Relative) directions indicated - accept diagram. Both speeds correct. 8 (b) (i) W is work done by resistances on car W M Use of WE. Must have KE, W and GPE. Allow -W B Both KE terms. Accept sign error W = 600 so 600 J done by car against resistances A cao A All correct with W or -W 8

54 76 Mark Scheme January 0 Q m a r k notes either The slope is 8 = 0 m long B M Use of P = (Work done) / (elapsed time) used for at least one work done term M WD is force distance used for at least one force A Allow only sign errors both terms = 9 77 W A cao. or The angle of the slope is arcsin (/.) B M Use of P = Fv used for at least one term M Attempt at weight component A Allow only sign errors both terms = 9 77 W A cao. 7 9

55 76 Mark Scheme January 0 Q m a r k notes (i) Horizontal X 0 = 0 B Any form Vertical: R Y = 0 B Any form a. c. moments about A R = M so R = E Clearly shown so Y = 0 and Y = 90 E Shown (iii) In analysis below all internal forces are taken B Correct arrow pairs for all internal forces as tensions B Correct labels 0

56 76 Mark Scheme January 0 Q m a r k notes (iv) M Equilibrium attempted at a pin-joint M Equilibrium attempted at a nd pin-joint M Either Equilibrium equation for nd direction at a pin-joint or rd pin-joint considered At C B At least equations of resolution correct or follow through CD cos0 0 TCD 0 and force in CD is 0 N (T) A T T BC CD cos 60 0 so TBC and force in BC is N (C) F At D T cos 0 T cos 0 0 BD so TBD 0 CD and force in BD is 0 N (C) F TAD TBD TCD cos 60 cos so TAD 0 0 and the force in AD is 0 0 N (T) F At A T AB cos so TAB 60 and the force in AB is 60 N (C) F B 0 At least T/C correct (v) The equilibria at C depend only on the Resolve in two directions at C and obtain same results as in (iv) MA framework geometry and the N. E These are not changed so forces in CB and CD are not changed E 9

57 76 Mark Scheme January 0 Q mark notes (i) (,.) B Condone writing as a vector By symmetry, y. B Some justification needed For x : h h 6 x h 6 so h x h. 0 so and h x.( h ) h x ( h ) M A A A E 6 These next marks may be obtained from correct FT of their from (i) st term RHS correct (allow sign error) Either other term correct All correct Clearly shown, including signs. (iii) Need x 0 M Allow x 0 or = 0 h So 0 ( h ) Hence h² > 0 Since h > 0, 0 h B or - oe seen A Accept only +ve root mentioned. WWW for signs Accept h as answer strict inequality for final A mark

58 76 Mark Scheme January 0 Q mark notes Q continued (iv) When h =, x 0. B Could be scored in (v) Let mag of vert force be T N a.c moments about axis thro O T 6 0. = 0 M If moments about another point need all relevant forces. Allow sign errors. Condone use of g so T = 0.6 so 0.6 N A cao (v) Let magnitude of force be U N a.c. moments about axis thro D U cos0 ( + 0.) = 0 B A U =.8 so. N ( s. f.) A cao 7 M Each term must be a moment. If moments about another point need all relevant forces. Condone use of g. moment of U (Ucos0 or ) oe ( + 0.) oe

59 GCE Mathematics (MEI) Advanced GCE Unit 76: Mechanics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

60 76 Mark Scheme June 0 Q mark notes (a) (i) T = 0(.7 (.7)) M Use of I = Ft. Allow sign errors A Signs correct on RHS so T =. So s. A cao OR: = 0a.7 (.7) T. B M A NL Use of suvat cao PCLM: v P+Q M PCLM with combined mass. Allow sign errors vp+q.6 so.6 m s - in +ve direction A No need for reference to direction (iii) PCLM: v Q M PCLM with all correct terms. Allow sign errors A Any form Hence v - and Q has velocity m s A Accept no direct reference to direction Q vq M NEL. Accept their v Q and any sign errors. Fraction must be correct way up NEL: e 0..7 A Any form. FT their v Q. so e = so 0.90 ( s. f.) A cao accept 0.9, / accept 0. only if 0.9 seen earlier 6

61 76 Mark Scheme June 0 (b) Initial vert cpt is sin0 = 7 B st hits ground at v given by v M Appropriate suvat. Allow ±9.8 etc Condone u = v = 0. A Vert cpt after nd bounce ² M their n for n =, or Condone use of their initial vertical component. Do not award if horiz component is also multiplied by 0.6 B use of 0.6² or attempt at two bounces with 0.6 used each time Horiz cpt is unchanged throughout (cos0 ) B Award even if value wrong or not given Angle is arctan cos0 M FT their horiz and vert components. oe. Fraction must be for correct angle. so 7. ( s. f.) A cao SC answer of.7 will usually earn /8 8 9

62 76 Mark Scheme June 0 Q mark notes Penalise answers to fewer than sf only once (i) cw moments about A Let force be S S 0 M Moments. All forces. No extras A S = 0 so 0 N vertically upwards A Need statement of direction or diagram cw moments about A Let tension be T T sin M T = ( 600 sin 0 so 089 N ( s. f.) A cao Moments. All forces. No extras. Attempt at moment of T (need not be resolved) Note that mmts about B needs forces at hinge. M Correct method for moment of T. Allow length errors and s c A Moment of T correct (allow sign error) A All correct (iii) Resolve X Tcos0 = 0 M Resolving horiz. Allow sign error. T must be resolved, allow s c so X =.. = ( s. f.) F FT their T only. Allow 600cot0 Resolve Y Tsin = 0 M NB other methods possible so Y = 000 F FT their T only Method for either R or M M dependent on attempts at X and Y using moments/resolution R 600 cot = so 67 ( s. f.) F FT their X and Y Numerical value only 000 arctan 600cot 0 = so 6.68 ( s. f.) F FT their X and Y Numerical value only Accept (iv) Angle GAP is above so 6.68 ( s. f.) B Weight, T and R are the only forces acting on E Must be clear the beam which is in equilibrium. Hence they are concurrent. Or geometrical calculation 7

63 76 Mark Scheme June 0 Q mark notes (i) x 0 y 7 8 = x 0.8 so and c.m. is (0.8,.) y. M B E E Correct method clearly indicated for x or y component. If D method, at least mass + cm correct for a region. If separate cpts, at least mass + cm correct for one of the cpts Working shown. Either expression shown oe Both c.w. moments about J..8 TH 0 B Use of.8 oe M A moments equation with all relevant forces. Allow use of 0 instead of. so TH. and the force at H is. N A Resolving M Or moments again force at J is.. =.76 N F Only FT if positive final answer (iii) below