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# Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London wc1V 7BH

 tarix 25.06.2016 ölçüsü 311 Kb.  Mark Scheme (Final)

Summer 2007

Edexcel Limited. Registered in England and Wales No. 4496750

Registered Office: One90 High Holborn, London WC1V 7BH  GCE Mathematics (6680/01)

June 2007

## Mark Scheme

General:

For M marks, correct number of terms, dimensionally correct, all terms that need resolving are resolved.

Omission of g from a resolution is an accuracy error, not a method error.

Omission of mass from a resolution is a method error.

Omission of a length from a moments equation is a method error.

Where there is only one method mark for a question or part of a question, this is for a complete method.

Omission of units is not (usually) counted as an error.

 Question Number Scheme Marks 1(a)   KE lost = Fraction of KE lost = = or at least 3sf ending in 7 or M1A1 A1 M1 DM1 A1 (6) (b) = =  M1A1 DM1 A1 (4) a) b) M1 Resolve parallel to the wall Alt: reasonable attempt at equation connecting two variables equation correct A1 u in terms of v or v.v. - not necessarily simplified. or ration of the two variables correct M1 expression for KE lost DM1 expression in one variable for fraction of KE lost – could be u/v as above A1 cao M1 Use NIL perpendicular to the wall and form equation in e A1 Correct unsimplified expression as above or or equivalent DM1 Substitute values for trig functions or use relationship from (a) and rearrange to e = ….. A1 cao accept decimals to at least 3sf The first three marks can be awarded in (b) if not seen in (a) The first two marks can be awarded in (a) 2(a) (b)  R(,  *            Hence B1 M1 A1 (3) M1A1 DM1 A1 M1 M1 A1 (7) a) b) B1 Correct expression involving the driving force. M1 Use of F = ma to form a differential equation. Condone sign errors. a must be expressed as a derivative, but could be any valid form. A1 Rearrange to given form. M1 Separate the variables A1 Separation correct (limits not necessarily seen at this stage) DM1 Attempt a complete integration process M1 Correct use of both limits – substitute and subtract. Condone wrong order. M1 Simplify to find k from an expression involving a logarithm A1 Answer as given, or exact equivalent. Need to see k = lnA + B
 Question Number Scheme Marks 3. (a) (b) (c)  = (+const) * = 0    = 0.32(1)c or 18.4o accept awrt  Hence, when  = 0.32c, i.e. stable M1A1A1 A1 (4) M1A1 M1 A1 (4) M1A1 M1 A1 (4) a) b) c) M1 Expression for the potential energy of the two rods. Condone trig errors. Condone sign errors. BC term in two parts A1 correct expression for AB A1 correct expression for BC A1 Answer as given . M1 Attempt to differentiate V. Condone errors in signs and in constants. A1 Derivative correct M1 Set derivative = 0 and rearrange to a single trig function in  A1 Solve for  or M1A1 find the position of the center of mass M1A1 form and solve trig equation for  M1 Differentiate to obtain the second derivative A1 Derivative correct M1 Determine the sign of the second derivative A1 Correct conclusion. cso Or: M1 Find the value of on both sides of the minimum point A1 signs correct M1 Use the results to determine the nature of the turning point A1 Correct conclusion, cso. These 4 marks are dependent on the use of derivatives 4 (a) (b) (c)  Fix A =11.5 km h-1 (3 s.f.) or: triangle without the right angle identified and  minimum value for M1 As above for A1A1 Ambiguous Sine Rule: 2 possible solutions for   = 62,1o (or 118o) (smaller value gives larger relative velocity) either Or Time = = 1.272…… hrs Earliest time is 13.16hrs or 13.17 hrs accept 1.16 (pm) or 1.17 (pm) M1A1 A1 (3) B1B1 (2) M1A1 A1 M1A1 M1 A1 M1 A1 A1 (8) a) b) c) M1 Velocity of B relative to A is in the direction of the line joining AB. Minimum V requires a right angled triangle. Convincing attempt to find the correct side. A1 15 x sin(their  ) A1 Q specifies 3sf, so 11.5 only B1B1 Convincing argument M1 Use of Sine Rule A1 Correct expression A1 (2 possible values,) pick the correct value. M1 Use trig. to form an equation in v A1 correct equation M1 A1ft correct expression with their v (not necessarily evaluated) A1 correct time in hours & minutes Or: M1 Use of cosine rule A1 A1 (Award after the next two marks) 15.72 or awrt 15.72 M1 Attempt to solve the equation for v A1 (15.72 or 3.562) Finish as above 5. (a) (b) (c) CLM: NIL: v2 = 1, v1 = 3 Dependent on both M’s above Horizontal components unchanged (i.e. 2 & 3) Independent of all other marks vA = 3i + j; vB = 2i - 3j For B: I = m(1-(-3)) = 4m (Or For A: -I = 2m(-1 – 1) I = 4m)   = 37o Alternative:  where   required angle is 2 M1A1 M1A1 M1A1 DM1 A1 A1 (7) M1A1 (2) M1A1 M1 A1 (4) a) b) c) M1 Conservation of momentum along the line of centres. Condone sign errors A1 equation correct M1 Impact law along the line of centres. e must be used correctly, but condone sign errors. A1 equation correct. The signs need to be consistent between the two equations M1 Solve the simultaneous equations for their v1 and v2. A1 i components correct – independent mark A1 vA & vB correct M1 Impulse = change in momentum for one sphere. Condone order of subtraction. A1 Magnitude correct. M1 Any complete method to find the trig ratio of a relevant angle. A1 , , … Or M1 find angle of approach to the line of centres and angle after collision. A1 values correct. (both 71.56 …..) M1 solve for  A1 370 (Q specifies nearest degree) Special case: candidates who act as if the line of centres is in the direction of i: CLM u+2v = 8 NIL v-u = 2 u=4/3, v=10/3 4/3i + j ; 10/3i – j Impulse 2m-4/3m = 2/3m 1.700 Work is equivalent, so treat as a MR: M1A0M1A0M1A1A1 M1A1 M1A1M1A1 6 (a) At E, e=0.2   * M1 A1 B1 (3) (b) (c) (d) R()  ( )  * CF is Hence GS is t = 0, y = o: 0 = A so,   t = 0,       k=1 9t = 2 (or 5t=2  accept 0.698s, 0.70s. M*1 M1 DM*1A1 A1 cso (5) M1 A1 B1 M1 A1 (5) B1 M1 M1 A1 (4) a) b) c) d) M1 Hooke’s law to find extension at equilibrium A1 cao B1 Q specifies reference to a diagram. Correct reasoning leading to given answer. M1 Use of F=ma. Weight, tension and acceleration. Condone sign errors. M1 Substitute for tension in terms of x M1 Use given result to substitute for x in terms of y A1 Correct unsimplified equation A1 Rearrange to given form cso. M1 Correct form for CF B1 Deduce coefficient of cos= 0 M1 Differentiate their y and substitue t=0, A1 y in terms of t. Any exact equivalent. B1 correct M1 set M1 solve for general solution for t: or: A1 Select smallest value

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