PhysicsAndMathsTutor.com. 1 mark Sub (i)

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1 1 mark Sub x = 14cos60t so x = t y = 14sin 60t 4.9t y = 3t 4.9t ( y = t 4.9t ) (ii) (A) time taken to reach highest point 0= T so 3 f.)) s (1.31. = 1.4 s (3 s. ca Consider motion in x direction. Need not resolve. Allow sin cos. Condone +1 seen. Need not be simplified. Suitable uvast used f y with g = ± 9.8, ±10, ± 9.81 soi Need not resolve. Allow sin cos. Allow + 1omitted. Any fm and s. f. Need not be simplified All crect. +1 need not be justified. Accept any fm and s. f. Need not be simplified. Appropriate uvast. Accept u = 14 and sin cos and u v. Require v = 0 equivalent. g = ± 9.8, ±10, ± 9.81 soi. [If time of flight attempted, do not award if twice interval obtained] (B) distance from base is 3 = 3 m (= so 8.66 m (3 s. f.)) Use of their x = t with their T FT their T only in x = t. Accept values rounding to 8.6 and 8.. (C) Height at this time is 3 3 H = + 1 = 8. Subst in their quadratic y with their T. Crect subst of their T in their y which has attempts at all 3 terms. Do not accept u = 14.

2 clearance is 8. 6 =. m Clearly shown. f height above pt of projection 0= ( 3) s Appropriate uvast. Accept u = 14. g = ± 9.8, ±10, ± 9.81 soi Attempt at vert cpt accept sin cos.accept sign errs but not u = 14. (iii ) s =. so clearance is. =. m See over Clearly shown. 4 (iii ) Elim t between y = 3 = t t 4.9t + 1 and x x x so y = F1 so y = 3x their quadratic y (accept bracket errs) Must see their t = x/ fully substituted in Accept any fm crectly written. FT their x and 3 term quadratic y (n using u = 14) (iv) 0.1x + 1 need 6= 3t 4. 9t so 4.9t 3t+ = 0 t = 3± 1 ( ) ( 3 +1) 3 moves by [( ) ] = m their quadratic y from = 6, equivalent. Dep. Attempt to solve this 3 term quadratic. (Allow u = 14). f root Moves by their root - their (ii)(a) equivalent. Award this f recognition of crect dist (no calc) cao [If new distance to wall found must have larger of +ve roots f 3 rd M and award max 4/ f 13.66] using equation of trajecty with y = 6

3 6= 3x 0.1x Solving x 10 3x+0 = 0 x = 3± 1 ( ) distance is = m Equating their quadratic trajecty equn to 6 Dep. Attempt to solve this 3 term quadratic. (Allow u = 14). f root distance is their root their(ii)(b) Award this f recognition of crect dist (no calc) Cao [If new distance to wall found must have larger of + ve roots f 3 rd M and award max 4/ f 13.66] 0

4 mark Height reached by first particle is given by 0 = s so s =. so. m Sol (1) (ii) t seconds after second particle projected its height is 1t 4.9t and the first particle has height. 4.9t ( 1t 4.9t ) Sub t = 1. to show both have same value State height as 11.4 m 1t 4. 9t =. 4.9 t giving t = 1. and height as 11.4 m Sol () t seconds after second particle projected its height is 1t 4.9t and the first particle has fallen 4.9t Collide when 1T 4. 9T T =. so T = 1. H = = 11.4 m total 8 Other methods must be complete. Accept with consistent signs Award only if used crectly ( sub t = 3.64 into 1t 4.9t f 1 st & t = 1. f cao. Accept any reasonable accuracy. Don t award if only one crectly used equation obtained. Both. t shown. Ht cao (to any reasonable accuracy) Or other crect method cao. Accept any reasonable accuracy. Don t award if only one crectly used equation obtained. nd ) 6

5 3 mark Hiz (40 cos 0)t (ii) Vert (40sin 0)t 4.9t Need (40sin 0)t 4.9t = 0 40sin 0 so t = 4.9 = 6.34 so 6.3 s (3 d. p.) Range is (40 cos 0) = so 161 m (3 s. f.) Use of s = ut + 0.at with a = ±9.8 ±10. Allow u = 40. Condone s c. Any fm Equating their y to zero. Allow quadratic y only Dep on 1 st. Attempt to solve. Clearly shown [ (allow u = 40 and s c) time to greatest height; ] Use of their hiz expression Any reasonable accuracy 3 (iii) Time AB is given by Equating their linear x to 30. (40 cos 0)T = 30 so T = so 1.1 s then By symmetry, time AC is time AD time AB Symmetry need not be explicit. Method may be implied. Any valid method using symmetry. 30 so time AC is cos 0 =.086. so.09 s (3 s. f.) cao height is (40sin 0)T 4.9T and we need (40sin 0) t 4.9t = (40sin 0)T 4.9T solved f larger root Complete method to find time to second occasion at that height (iv) i.e. solve 4.9t (40sin 0)t = 0 f larger root giving.086 x& = 40 cos 0 cao Must be part of a method using velocities. 4 y& = 40sin Use of vert cpt of vel Allow only sign err. FT use of their &y Need arctan x& May be implied. Accept arctan x& y& but not use of y& 0. So Accept ±36.8 equivalent. Condone direction not so 36.8 below hizontal (3 s f.) clear. total 1

The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 72.

The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 72. ADVANCED SUBSIDIARY GCE UNIT 761/01 MATHEMATICS (MEI) Mechanics 1 MONDAY 1 MAY 007 Additional materials: Answer booklet (8 pages) Graph paper MEI Examination Formulae and Tables (MF) Morning Time: 1 hour