ADVANCED General Certificate of Education 206 Mathematics Assessment Unit M2 assessing Module M2: Mechanics 2 [AMM2] THURSDAY 2 JUNE, AFTERNOON MARK SCHEME 9846.0 F
GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS Introduction The mark scheme normally proides the most popular solution to each question. Other solutions gien by candidates are ealuated and credit gien as appropriate; these alternatie methods are not usually illustrated in the published mark scheme. The marks awarded for each question are shown in the right hand column and they are prefixed by the letters M, W and MW as appropriate. The key to the mark scheme is gien below: M W indicates marks for correct method. indicates marks for working. MW indicates marks for combined method and working. The solution to a question gains marks for correct method and marks for an accurate working based on this method. Where the method is not correct no marks can be gien. A later part of a question may require a candidate to use an answer obtained from an earlier part of the same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel problem from the point at which the error occurred. If such a candidate continues to apply correct method, then the candidate s indiidual working must be followed through from the error. If no further errors are made, then the candidate is penalised only for the initial error. Solutions containing two or more working or transcription errors are treated in the same way. This process is usually referred to as follow-through marking and allows a candidate to gain credit for that part of a solution which follows a working or transcription error. Positie marking: It is our intention to reward candidates for any demonstration of releant knowledge, skills or understanding. For this reason we adopt a policy of following through their answers, that is, haing penalised a candidate for an error, we mark the succeeding parts of the question using the candidate s alue or answers and award marks accordingly. Some common examples of this occur in the following cases: (a) a numerical error in one entry in a table of alues might lead to seeral answers being incorrect, but these might not be essentially separate errors; (b) readings taken from candidates inaccurate graphs may not agree with the answers expected but might be consistent with the graphs drawn. When the candidate misreads a question in such a way as to make the question easier only a proportion of the marks will be aailable (based on the professional judgement of the examining team). 9846.0 F 22 [Turn oer
(i) F = ma ( 2i + 2j 4k) = 4a a = 2 i + 2 j k W = u + at = 0+- ^ i + j-kh 6 2 2 = -3i+ 3j-6k W + 2 (ii) s = ut at 2 2 AB = 0+ ^- i+ j-kh^6 h 2 2 2 = -9i+ 9j-8k MW OB = OA + AB = (0i + 2 k) + (-9i+ 9j-8 k) = i+ 9j-6k W (iii) Direction of motion gien by a,, or F unit ector t = = 3 + 3 + 6 2 2 2 = 3 6 MW t = (- 3i+ 3j- 6 k) 3 6 = (- i+ j- 2 k) W 2 6 9846.0 F 33
2 (i) 3 m aboe ground 2. m aboe projection MW + 2 s = ut at 2 2 2. = 7(t) + 2( - 9.8)t W 4.9t 2 7t + 2. = 0 (7t 3)(t ) = 0 3 7 t = andt = MW time aboe 3 m = 7 4 sec MW (ii) u x = 25 cos θ 24 = 25^ 25 h = 24 MW + 2 s = ut at 2 33 = 24(t) + 0 33 t = 24 W + 2 s = ut at 2 33 = 7^ h+ (-98.) ^ s 24 2 33 2 24h = 0.36094 MW height aboe ground = 0.36 + 0.9 =.26 m MW (iii) include air resistance/wind speed MW spin on the ball 9846.0 F 44 [Turn oer
3 (i) s = y dt s = ^t 3 6thi + ^ 2t 3 hj + c W2 t = 0 at s = 5i + 75j c = 5i + 75j MW OP = ^t 3 6t + 5hi + ^75 2t 3 hj (ii) meets OQ when i displacements equal and j displacements equal t 3-3t = t 3-6t + 5 3t = 5 t = 5 and 3t 2-2t 3 = 75-2t 3 t 2 = 25 t = ±5 meet at t = 5 W MW MW (iii) d = s dt 2 2 V = (3t - 3) i+ (6t- 6 t ) j W Q at t = 5 V = 72i 20j MW Q - 9846.0 F 55
4 (i) T T mg 3g MW2 (ii) Resole For A T = 3g W For B T sin θ = mg MW 2 3g^h 5 = mg m =.2 W (iii) Consider F = mrω 2 T cos θ =.2 (2 cos θ)ω 2 M2 W2 T = 2.4ω 2 3g = 2.4ω 2 7 ω = 2 W 9846.0 F 66 [Turn oer
5 (i) p mg sin θ 200 P = F F = P F = ma 60000-200 -mg ^ m( 0.3) 30 2h = - W3 m = 548 W (ii) max. speed acceleration = 0 forces balance mg sin θ + 200 = 78000 = 3.7 ms MW W W (iii) Because air resistance would normally increase with speed MW 6 mass = density olume mass/second = 000 6A = 6000A MW WD/second = KE + PE 2 + 500 = 2m mh g 2 2 + 500 = (6000 A)(6) (6000 A)(9.8)(8) M2 W2 A = 0.0005 W 7 9846.0 F 77
7 (i) km mg F = ma mg km = m(a) Max. speed a = 0 MW g kv = 0 (ii) V = g W k d = g - k dt separate and integrate y y M2 W d ld g - k = t y V 2 0 d g -k y T = 0 V ld t l 2 9- ln(g -k ) C = T W2 k 0 l V l T = - alna k g - kk -- a ln g k W k 2 k sub V = g k T = l g alng -ln k k 2 T = l ln 2 k W 2 Total 75 9846.0 F 88 [Turn oer