Mark Scheme

Paper 4761 Name Mechanics 1 Session Jan Year 005 Solutions and scheme Q 1 (ii) Differentiate M1 At least 1 cpt correct v = t i + (5 4t) j Award for RHS seen Differentiate M1 Do not award if i and j lost in v. At least 1 cpt correct. FT a = i 4 j F1 FT from their component v F + 1 j = 4( i 4 j) M1 NL. Allow F = mg a. No extra forces. Allow 1j omitted Allow wrong signs otherwise correct with their vector a. F = 8 i 8 j cao total 7 4 3 Q (A) the pulleys are smooth and the string is light E1 Accept only the pulley is smooth. (B) the string is inextensible E1 (ii) Diagrams B1 All forces present with labels and arrows. Acc not reqd. 1 For X, NL upwards T g = a M1 NL. Allow F = mga. All forces present Award for equation for X or Y or combined Any form For Y, NL downwards 4 g T = 4a Any form Solve for a and T g a = 3 ( 3.7 (3 s. f.)) 8 T = g (6.1 (3 s. f.)) F1 FT second answer 3 total 8 5

Paper 4761 Name Mechanics 1 Session Jan Year 005 Solutions and scheme Q 3 (ii) x 4 5 0 7 y 4 0 + + = z 5 7 0 Equating components gives x = -9, y = 3, z = 1 We need 5 4 ( 7) M1 + + M1 = 90 or 9.48683 so 9.49 (3 s. f.) Any reasonable accuracy total 6 [Allow SC /4 if 9, -3, -1 obtained] 4 Q 4 Height reached by first particle is given by 0 = 1 9. 8 s M1 Other methods must be complete. Allow g =± 9.8, ± 10 so s =.5 so.5 m Accept with consistent signs Sol (1) (ii) t seconds after second particle projected its M1 Allow g =± 9.8, ± 10 height is 15t 4.9t and the first particle has height.5 4.9t M1 Allow g =± 9.8, ± 10 ( or 1t 4.9t ) Award only if used correctly either Sub t = 1.5 to show both have same value E1 (or sub t = 3.64 into 1t 4.9t for 1 st & t = 1.5 for nd ) State height as 11.475 m cao. Accept any reasonable accuracy. Don t award if only one correctly used equation obtained. or 15t 4.9t =.5 4.9t M1 giving t = 1.5 and height as 11.475 m Both. t shown. Ht cao (to any reasonable accuracy) Sol () t seconds after second particle projected its M1 Allow g =± 9.8, ± 10 height is 15t 4.9t and the first particle has fallen 4.9t B1 Collide when 15T 4.9T + 4.9T =.5 M1 Or other correct method so T = 1.5 E1 H =.5 4.9 1.5 = 11. 475 m cao. Accept any reasonable accuracy. Don t award if only one correctly used equation obtained. total 8 6

Paper 4761 Name Mechanics 1 Session Jan Year 005 Solutions and scheme Q 5 (ii) T BA 400 N B1 Different labels. All forces present with arrows in correct directions. Condone no angles. Using triangle of forces M1 Attempt at triangle of forces. Ignore angles and arrows. Accept 90, 60, 30 triangle. T BC T BC 1 T BA 30 10 30 400 N B1 Triangle, arrows, labels and angles correct (iii) Triangle isosceles so tension in BC is 400 N cao Tension in BA is 400 cos 30 = 400 3 N F1 FT BC only (693 N, (3 s. f.)) [If resolution used, M1 for 1 equn; M1 for nd equn + attempt to elim; ; F1. For M s all forces present but allow s c and sign errors. No extra forces. If Lami used: M1 first pair of equations in correct format, condone wrong angles.. M1 second pair in correct format, with correct angles.f1 FT their first answer if necessary.] Resolve at B perpendicular to the line ABC E1 Attempt to argue unbalanced force Weight has unbalanced component in this E1 Complete, convincing argument. direction [or Resolve horiz and establish tensions equal E1 Resolve vert to show inconsistency. E1] total 7 4

Paper 4761 Name Mechanics 1 Session Jan Year 005 Solutions and scheme Q 6 Area under curve M1 Attempt to find any area under curve or use const accn results 0.5 0 + 0.5 (0 + 10) 4 + 0.5 10 1 B1 Any area correct (Accept 0 or 60 or 5 without explanation) = 85 m cao (ii) (iii) (iv) (v) (vi) 0 10 M1 v/ t =.5 4 accept ±. 5 upwards B1 Accept.5 downwards (allow direction specified by diagram etc). Accept opposite direction to motion. v =. 5t + c M1 Allow their a in the form v =± at+ cor v=± at ( ) + c v = 0 when t = M1 v =.5t + 5 cao [Allow v= 0.5( t ) ] [Allow /3 for different variable to t used, e.g. x. Allow any variable name for speed] 3 Falling with negligible resistance E1 Accept zero resistance, or no resistance seen. 1.5 4+ 9.5 + 7 = 0 E1 One of the results shown 1.5 36 + 9.5 6 + 7 = 10 1.5 49 + 9.5 7+ 7 = 0 E1 All three shown. Be generous about the show. 7 ( 1.5t + 9.5t + 7) dt M1 Limits not required 3 =[ 0.5t + 4.75t + 7t] 7 for each term. Limits not required. Condone + c = + + 49 ( 4 + 19 + 14) 343 19 49 4 M1 Attempt to use both limits on an integrated expression Correct substitution in their expression including subtraction ( may be left as an expression). = 81.5 m cao. total 19 3 3 1 7

Paper 4761 Name Mechanics 1 Session Jan Year 005 Solutions and scheme Q 7 Horiz ( 40 cos 50) t B1 (ii) Vert Need ( 40sin 50) t 4.9t M1 Use of s = ut + 0.5at with a = ±9.8or ± 10. Allow u = 40. Condone s c. Any form ( 40sin 50) t 4.9t = 0 M1 Equating their y to zero. Allow quadratic y only 40sin 50 so t = 4.9 M1 Dep on 1 st M1. Attempt to solve. = 6.534 so 6.53 s (3 d. p.) E1 Clearly shown [or M1 (allow u = 40 and s c) time to greatest height; E1] Range is ( 40 cos 50) 6.534... M1 Use of their horiz expression = 160.78 so 161 m (3 s. f.) Any reasonable accuracy 3 5 (iii) Time AB is given by M1 Equating their linear x to 30. ( 40 cos 50) T = 30 so T = 1.16679 so 1.17 s then either By symmetry, time AC is time AD time AB M1 Symmetry need not be explicit. Method may be implied. Any valid method using symmetry. 30 so time AC is 6.534 40 cos 50 = 5.086. so 5.09 s (3 s. f.) cao or height is ( 40 sin 50) T 4.9T and we need ( 40sin 50) t 4.9t = ( 40sin 50) T 4.9T M1 Complete method to find time to second occasion at that solved for larger root height (iv) i.e. solve 4.9t (40 sin 50) t + 9.0871... = 0 for larger root giving 5.086 cao x& = 40 cos 50 B1 Must be part of a method using velocities. y& = 40 sin 50 9.8 5.086... M1 Use of vert cpt of vel Allow only sign error. FT use of their 5.086.. y& Need arctan M1 May be implied. Accept arctan x & x& y& y& 0. So 36.761 Accept ± 36.8 or equivalent. Condone direction not so 36.8 below horizontal (3 s.f.) clear. total 17 4 5