Mark Scheme (Final)
Summer 2007
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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 18.4^{o} accept awrt
Hence, when = 0.32^{c},
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
= 62,1^{o} (or 118^{o})
(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:
v_{2} = 1, v_{1} = 3 Dependent on both M’s above
Horizontal components unchanged (i.e. 2 & 3) Independent of all other marks
v_{A} = 3i + j; v_{B} = 2i  3j
For B: I = m(1(3)) = 4m
(Or For A: I = 2m(1 – 1) I = 4m)
= 37^{o}
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 v_{1} and v_{2}.
A1 i components correct – independent mark
A1 v_{A} & v_{B} 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 37^{0} (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 vu = 2
u=4/3, v=10/3
4/3i + j ; 10/3i – j
Impulse 2m4/3m = 2/3m
1.70^{0}
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

 