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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

6680 Mechanics M4

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

A1 Correct as above or equivalent



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

A1 Integration correct

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

B1B0 Argument with some merit

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

A1 GS for y correct

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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