Mark Scheme (Results) January 2008
GCE
GCE Mathematics (6679/01)
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January 2008 6679 Mechanics M3 Mark Scheme Scheme
Question Number 1.(a)
λ ×e
T or
l
λ × 0.16 0.4
= mg (even T=m is M1, A0, A0 sp case)
Special case giving
θ
A1
or 5g
A1 (3)
T sin θ = mg
= 30 is M1 A0 A0 unless
R(↑)
there is evidence that they think θ is with horizontal – then M1 A1 A0
49.
M1
= 2g
⇒ λ = 49 N
(b)
Marks
T cos θ = mg or cos θ =
0.32 . cosθ = 19.6 or 4 g.cos θ = 2 g or 2mg.cos θ = mg 0.4 ⇒ cos θ =
1 2
⇒ θ = 60°
( or
mg T
M1
(ft on their λ )
π
A1ft A1 (3)
radians)
3
6 2. m ‘a’ = ±
16 d 2 x dv dv , with acceleration in any form (e.g. , v , or 5x 2 dt 2 dx dt
B1
a) Uses a = v
dv dv 32 to obtain k v = ± k’ 2 dx dx x
Separates variables,
Obtains
1 2
k
∫
v2 = m
v dv = k’
32 (+ C) x
∫
M1 32 dx x2
dM1
or equivalent e.g.
0.1 2
v2 = –
16 (+ C) 5x
Substituting x = 2 if + used earlier or – 2 if – used in d.e. x = 2, v = ± 8 ⇒ 32 = –16 + C ⇒ C = 48 (or value appropriate to their correct equation) v = 0 ⇒
N.B
32 = 48 ⇒ x = x
2 3
m
(N.B. - 23
is not acceptable for final answer )
d 1 16 2 ( 2 mv ) = , is also a valid approach. dx 5x2
Last two method marks are independent of earlier marks and of each other
A1 M1 A1 M1 A1 cao 8
Scheme
Question Number
Marks
3.(a) Large cone Vol. C of M
1 3
small cone
π ( 2r ) 2 ( 2 h ) 1 2
1 3
h,
πr 2 h 5 4
8 3
S 7 3
h
πr 2 h. 12 h –
πr 2 h x
1 3
tan θ =
(or equivalent)
πr 2 h . 54 h = → x =
(b)
(accept ratios 8 : 1 : 7)
11 28
7 3
πr 2 h . x
h
or equivalent
∗
2r 2r 2r 28 = 11 , = 11 = 11 x 28 h 14 r
θ ≈ 68.6 ° or 1.20 radians (Special case – obtains complement by using tan
θ
=
2r giving 21.4° or .374 radians M1A0A0) x
Centres of mass may be measured from another point ( e.g. centre of small circle, or vertex) The Method mark will then require a complete method (Moments and subtraction) to give required value for x ). However B marks can be awarded for correct values if the candidate makes the working clear.
B1 B1, B1
M1 A1 (5) M1, A1 A1 (3) 8
4. (a)
Energy equation with at least three terms, including K.E term
1 mV 2 + .. 2 1 2mg a 2 1 1 2mg 9a 2 . , + mg. a.sin 30, = . . + .. . 2 a 16 2 2 a 16 ⇒ V = (b)
ga 2
Using point where velocity is zero and point where string becomes slack:
1 mw2 = 2
dM1 A1 (6)
M1
A1 (4)
3ag 8
Alternative (using point of projection and point where string becomes slack): 1 2
A1, A1, A1
A1, A1 1 2mg 9a 2 3a . . , − mg . .sin 30 2 a 16 4
⇒ w =
M1
2
mw2 − 12 mV1 , = So w =
mga mga − 16 8 3ag 8
M1,A1 A1 A1 10
In part (a) DM1 requires EE, PE and KE to have been included in the energy equation. If sign errors lead to V 2 = −
ga , the last two marks are M0 A0 2
In parts (a) and (b) A marks need to have the correct signs In part (b) for M1 need one KE term in energy equation of at least 3 terms with distance
3a to indicate first method, and two KE terms in energy equation of at least 4 terms with 4 a distance to indicate second method. 4 SHM approach in part (b). (Condone this method only if SHM is proved)
a 2g and x = ± . 4 a 3ag w = . 8
Using v 2 = ω 2 (a 2 − x 2 ) with ω 2 = Using ‘a’ =
a 2
to give
M1 A1 A1 A1
5.(a)
N
mv 2 = µ N , = µ mg r
µN
M1, A1
mg
µ= (b)
v2 212 = = 0 .6 rg 75 × 9.8
∗
R(↑) R cos α , m0.6 R sin α = mg
A1 (3)
M1, A1, A1
25mg ⎛ 4 3 3⎞ ⇒ R⎜ − . ⎟ = mg ⇒ R = 11 ⎝ 5 5 5⎠ (c) R(Å) R sin α , ±0.6 R cos α =
mv 2 r
v ≈ 32.5 m s–1
A1 (4)
M1, A1, A1 dM1 A1cao (5) 12
In part (b) M1 needs three terms of which one is mg If cos α and sin α are interchanged in equation this is awarded M1 A0 A1 In part (c) M1 needs three terms of which one is
mv 2 or mrω 2 r
If cos α and sin α are interchanged in equation this is also awarded M1 A0 A1 If they resolve along the plane and perpendicular to the plane in part (b), then attempt at
R − mg cos α =
mv 2 mv 2 sin α , and 0.6 R + mg sin α = cos α and attempt to eliminate v r r
Two correct equations Correct work to solve simultaneous equations Answer
M1 A1 A1 A1
In part (c) Substitute R into one of the equations Substitutes into a correct equation (earning accuracy marks in part (b))
M1 A1
Uses R =
25mg 11
Obtain v = 32.5
(or
25mg ) 29
(4)
A1 M1A1
(5)
6.(a)
Energy equation with two terms on RHS,
ga (5 + 4sin θ ) 2 mv 2 R(\\ string) T − mg sin θ = a ⇒ v2 =
(b)
(c)
T = 0 ⇒ sin θ , = − 56
A1 cso (3)
∗ (3 terms)
mg ⇒T = (5 + 6 sin θ ) o.e. 2
Has a solution, so string slack when α ≈ 236(.4)° or 4.13 radians At top of small circle,
(d)
M1, A1
1 2 1 5 ga mv = m. + mga sin θ 2 2 2
1 2 1 5 ga mga mv = m. − 2 2 2 2 2 3 ⇒ v = 2 ga = 14.7a
Resolving and using Force =
3 ga mv 2 , T + mg = m. 21 r 2a
⇒ T = 2mg
(M1 for energy equation with 3 terms)
M1 A1 A1 (3) M1, A1 A1 (3) M1 A1 A1
(M1 needs three terms, but any v)
M1 A1 A1 (6) 15
Use of v 2 = u 2 + 2 gh is M0 in part (a)
7.(a)
2 && x = 2 g − 98( x + 0.2) , and so && x = −49x
(Measuring x from E)
SHM period with ω 2 = 49 so T = (b)
2π 7
d M1 A1cso (5)
Max. acceleration = 49 × max. x = 49 x 0.4 = 19.6 m s–2
(c)
String slack when x = –0.2:
B1 (1)
M1 A1
v2 = 49(0.42 – 0.22) –1
⇒ v ≈ 2.42 m s (d)
M1 A1, A1
A1
7 3 = 5
Uses x = a cos ωt or use x = a sin ω t but not with x = 0 or ± a
M1
Attempt complete method for finding time when string goes slack –0.2 = 0.4 cos 7t ⇒ cos 7t = – 12
dM1 A1 A1
2π t = ≈ 0.299 s 21 Time when string is slack =
(2) × 2.42 2 3 = ≈ 0.495 s g 7
M1 A1ft (2 needed for
A) Total time = 2 x 0.299 + 0.495 ≈ 1.09 s (a)
(b) (c)
DM1 requires the minus sign. Special case 2 && x = 2 g − 98 x is M1A1A0M0A0 2 && x = −98 x is M0A0A0M0A0 & & No use of x , just a is M1 A0,A0 then M1 A0 if otherwise correct. Quoted results are not acceptable. Answer must be positive and evaluated for B1 M1 – Use correct formula with their ω , a and x but not x = 0. A1 Correct values but allow x = +0.2 Alternative It is possible to use energy instead to do this part
(d)
1 2
mv 2 + mg × 0.6 =
λ × 0.62 2l
M1 A1
π this is dM1, A1 if done correctly If they use x = a sin ω t with x = ±0.2 and add π7 or 14 If they use x = a cos ωt with x = -0.2 this is dM1, then A1 (as in scheme)
If they use x = a cos ωt with x = +0.2 this needs their A1
(3)
π
7
minus answer to reach dM1, then
A1 (7) 16