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4762 Mark Scheme June 2005 Q 1 mark Sub
(a) (i) 240 i N s → B1 1 (ii) M1 Equating to their 240 i in this part (A) 240 i = 70 i +50 v so v = 3.4 i m s –1 A1 FT 240 i (B) 240 i = 70u i – 50u i M1 Must have u in both RHS terms and opposite signs u = 12 so v = − 12 i m s –1 A1 FT 240 i (C) 240 i = 280( i + j ) +50vB B M1 FT 240 i Must have all terms present so vB = (-0.8 i – 5.6 j) m s B
–1 A1 cao 6 (b) (i)
NEL 5.0
4212 −=
−−− vv
M1 NEL
so 312 =− vv A1 Any form PCLM 1 28 6 2 3v v− = + M1 PCLM A1 Any form Solving v2 = 1.6 so 1.6 m s –1 → A1 Direction must be clear (accept diagram) v1 = 1.4 so 1.4 m s− –1 ← A1 Direction must be clear (accept diagram). [Award A1 A0 if v1 & v2 correct but directions not clear] 6 (ii) 1.6 m s –1 B1 FT their 1.6 at 60° to the wall (glancing angles both 60°) B1 No change in the velocity component parallel E1 Must give reason to the wall as no impulse No change in the velocity component E1 Must give reason perpendicular to the wall as perfectly elastic 4 total 17
2 kg 3 kg
before
after
4 m s -1
v1 v2
+ ve
2 m s -1
4762 Mark Scheme June 2005 Q 2 mark Sub
(i)
We need mght
= 850 9.8 6020
× × = 24 990 M1 Use of mght
so approx 25 kW E1 Shown 2 (ii) Driving force – resistance = 0 B1 May be implied v80025000 = M1 Use of P = Fv so v = 31.25 and speed is 31.25 m s –1 A1 3 (iii)
Force is 25000 250010
= N B1
N2L in direction of motion 2500 800 850a− = M1 Use of N2L with all terms a = 2 so 2 m s –2 A1 3 (iv) 22 158505.0208505.0 ××=×× M1 W-E equation with KE and power term B1 One KE term correct 90.625000 ×+ B1 Use of Pt .Accept wrong sign x800− B1 WD against resistance. Accept wrong sign A1 All correct x = 122.6562… so 123 m (3 s. f.) A1 cao 6 (v) either 22 208505.08505.0 ××=×× v M1 W-E equation inc KE, GPE and WD
20
1058.9850 ××− M1 GPE term with attempt at resolution
A1 Correct. Accept expression. Condone wrong sign. 105800 ×− B1 WD term. Neglect sign. ...452.992 =v so 9.97 m s –1 A1 cao or N2L + ve up plane (800 850 0.05) 850g a− + × = M1 N2L. All terms present. Allow sign errors. a = -1.43117… A1 Accept ± ( )2 220 2 1.43117... 105v = + × − × M1 Appropriate uvast. Neglect signs. A1 All correct including consistent signs. Need not follow sign of a above. ...452.992 =v so 9.97 m s –1 A1 cao 5 19
4762 Mark Scheme June 2005 Q 3 mark Sub
(i)
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
52
261
250
2
25
216
205
222
1628yx
M1 B1 B1
Complete method Total mass correct 3 c. m. correct (or 4 x- or y-values correct)
5.25.2
==
yx
A1 A1
[Allow A0 A1 if only error is in total mass]
[If yx = claimed by symmetry and only one component worked replace final A1, A1 by B1 explicit claim of symmetry A1 for the 2.5]
5 (ii) yx = B1 Or by direct calculation M1 Dealing with ‘folded’ parts for x or for z 2212024621628 ×+×+×+×+×=x A1 At least 3 terms correct for x
1431
=x (2.21428…) A1
74
28)2(4)1(8
−=−×+−×
=z (- 0.57142…) A1 All terms correct allowing sign errors
A1
Distance is 222
74
1431
1431
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ M1 Use of Pythagoras in 3D on their c.m.
= 3.18318.. so 3.18 m (3 s. f.) F1 8 (iii) M1 c.m. clearly directly below A
B1 Diagram showing α and known lengths (or equivalent). FT their values. Award if final answer follows their values.
4sin / 3.18318..7
α = M1 Appropriate expression for α . FT their values.
so α = 10.3415… so 10.3° (3 s. f.) A1 cao 4 total 17
C
A
centre of mass
α 3.18318…
4/7
4762 Mark Scheme June 2005 Q 4 mark Sub
(a) Moments c.w. about A (i) 2R = 5L so R = 2.5L E1 Resolve → U = 0 E1 Resolve ↑ V + R = L M1 Resolve vertically or take moments about B (or C) so LV 5.1−= E1 4 (ii)
M1 Equilibrium at a pin-joint
For equilibrium at A M1 Attempt at equilibrium at A or C including resolution with correct angle 05.145cos =+↑ LTAB
so TAB = 223 L
− so 223 L N (C) in AB A1 (2.12L (3 s. f.))
045cos =+→ ABAC TT
so TAC = 2
3L so 2
3L N (T) in AC F1 (1.5L)
At C ↓ BCcos = 0L T θ+ M1 Must include attempt at angle tan 3/ 2 cos 2 / 13θ θ= ⇒ = B1
so BC13 13 so 2 2
L LT = − N (C) in BC A1 (1.80 L (3 s. f.))
F1 Award for T/C correct from their internal forces. Do not award without calcs 8 (b) (i)
B1 All forces present with arrows and labels. Angles and distances not required.
1 (ii) c.w.moments about B R 3 – W × 1 cosθ = 0 × M1 If moments about other than B, then need to resolve perp to plank as well A1 Correct
so θcos31 WR = A1
3 (iii) Resolve parallel to plank θsinWF = B1
θθ
θμ tan3cos
31
sin===
W
WRF
M1 Use of RF μ= and their F and R
A1 Accept any form. 3 total 19
A
1.5 L
TAC
TAB
45°
A G
B
R
S
W
F
θ
4762 Mark Scheme January 2006
Q 1 mark Sub
(i) 16 0.4v= M1 Use of I mv= Δ so 40 m s –1 A1 2 (ii) PCLM ↑ + ve P0.4 32 0.6 0.4 0.6 4u v× − = + × M1 Use of PCLM A1 Any form NEL ↑ +ve P4
0.132v
u−
= −− −
M1 Use of NEL. Allow sign errors.
A1 Any form Solving u = 18 E1 Must be obtained from a pair of correct equations. If given u = 18 used then vP = -
1
must be obtained from 1 equation and both values tested in the second equation
P 1v = − so 1 m s –1 A1 cao. Accept use of given u = 18 downwards A1 cao 7 (iii)
Considering the momenta involved M1 PCLM applied. May be implied.
D
3.6 30.5 0.2 0.3
5.2 4−⎛ ⎞ ⎛ ⎞
= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
v B1 LHS
B1 momentum of C correct A1 Complete equation. Accept sign error.
D
86−⎛ ⎞
= ⎜ ⎟⎝ ⎠
v so a = 8− and b = 6 A1 A1
cao cao
Gradients of the lines are 43
and 68−
M1 Any method for the angle
Since 4 6 13 8× = −−
, they are at 90° E1 Clearly shown
8 17
4762 Mark Scheme January 2006 Q 2 mark Sub
(i) Moments about C
D240 2 3R× = M1 Moments about C or equivalent. Allow 1 force omitted
D 160R = so 160 N A1 Resolve vertically
C D 240R R+ = M1 Resolve vertically or moments about D or equivalent.
All forces present. C 80R = so 80 N F1 FT from their DR only 4 (ii) (A) Moments about D 240 1 4 sin 40T× = M1 Moments about D or equivalent M1 Attempt at resolution for RHS A1 RHS correct T = 93.343… so 93.3 N (3 s. f.) A1 4 (ii) (B) In equilibrium so horizontal force
needed
to balance cpt of T . This must be friction Need reference to horizontal force that
must
and cannot be at C. come from friction at D. 1 (iii)
(A) Moments about B
3 240 cos30 6P× × = M1 All terms present, no extras. Any resolution required attempted.
60 3P = (103.92…..) E1 Accept decimal equivalent
P inclined at 30° to vertical B1 Seen or equivalent or implied in (iii) (A) or (B).
Resolve horizontally. Friction force F sin 30F P= M1 Resolve horizontally. Any resolution
required
attempted so 30 3F = (51.961…) A1 Any form 5
4762 Mark Scheme January 2006 (iii)
(B) Resolve vertically. Normal reaction R
cos30 240P R+ = M1 Resolve vertically. All terms present.and resolution attempted
A1 Using F Rμ= M1 30 3
3240 60 32
μ =− ×
A1 Substitute their expressions for F and R
30 3 3240 90 5
= =−
= 0.34641 so 0.346 (3
s. f.) A1 cao. Any form. Accept 2 s. f. or better
5 19
4762 Mark Scheme January 2006 Q 3 mark Sub
(a) (i) 6 1 11
80 48 12 202 3 9
xy
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
M1 Correct method for c.m.
52080
240xy
⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ B1 Total mass correct
B1 One c.m. on RHS correct [If separate components considered, B1
for 2 correct]
6.5x = E1 3y = A1 cao 5 (ii) Consider x coordinate 520 76 6.4 4x= × + M1 Using additive principle o. e. on x cpts B1 Areas correct. Allow FT from masses from
(i)
so 8.4x = A1 cao 3 (iii)
y coordinate is 1 so we need B1 Position of centre of square 240 76 4 1y= + × and 3.10526...y = M1 so 3.11 (3 s. f.) A1 cao 3 (b) (i) Moments about C 4 120 3 120 2R = × + × M1 Moments equation. All terms present so 4 R = 600 and R = 150 E1 2 (ii)
B1
A↑ AE150 cos30 0T+ = M1 Equilibrium at a pin-joint AE 100 3T = − so 100 3 N (C) A1 Any form. Sign correct. Neglect (C) E↓ AE EB120 cos30 cos30 0T T+ + = M1 Equilibrium at E, all terms present
EB 20 3T = so 20 3 N (T) F1 Any form. Sign follows working. Neglect (T).
F1 T/C consistent with answers
120 N 120 N
150 N
A B
E D
TAB
TAE
TED
TEB TDB TDC
TBC
4762 Mark Scheme January 2006 6 (iii)
Consider → at E, using (ii) gives ED as thrust E1 Clearly explained. Accept ‘thrust’
correctly
deduced from wrong answers to (ii). 1 20
4762 Mark Scheme January 2006 Q 4 mark Sub
(i) 2 20.5 20 8 0.5 20 5 510
6× × − × × + M1 Use of P = WD/t
B1 ΔKE. Accept 390± soi A1 All correct including signs = 150 W A1 4 (ii) (A) 320 5
5g x gx× − M1 Use of mgh on both terms
B1 Either term (neglecting signs) 7gx (68.6x) gain A1 7gx± in any form. A1 cao 4 (B) 11gx B1 1 (C) 20.5 25 4 7 11 18gx gx gx× × = + = M1 Use of work-energy equation. Allow 1
RHS term omitted.
B1 KE term correct x = 1.13378… so 1.13 m (3 s. f.) A1 cao. Except follow wrong sign for 7gx
only.
3 (iii)
either 20.5 35 0.5 35 16v× × − × × M1 Use of work-energy. KE, GPE and WD against friction terms present. 15 0.5 11 0.5 12 0.5g g g= × − × − × B1 ΔGPE correct inc sign (1.5g J loss) A1 All correct 2 13.76v = so v = 3.70944… so 3.71 m s –1 (3 s. f.) A1 cao or 15 15g T a− = 12 11 20T g g a− − = M1 N2L in 1 or 2 equations. All terms
present
so 2.24a = − A1 cao 2 24 2 ( 2.24) 0.5v = + × − × M1 Use of appropriate (sequence of) uvast so 3.71 m s –1 (3 s. f.) A1 cao 4 16
4762 Mark Scheme June 2006
Q 1 mark Sub (a) (i) PCLM → +ve (A) 2 4 6 2 8v× − × = M1 Use of PCLM and correct mass on RHS A1 Any form v = 0.5− so 0.5 m s –1 in opposite
direction to A1 Direction must be negative and consistent or clear.
initial motion of P Accept use of a diagram. 3 (B) ( )22 20.5 2 4 0.5 6 2 0.5 8 0.5× × + × × − × × − M1 Use of KE. Must sum initial terms. Must have correct masses = 27 J A1 FT their (A) only 2 (ii) (A) PCLM → +ve
P Q2 4 6 2 2 6v v× − × = + M1 Use of PCLM P Q3 2v v+ = − A1 Any form NEL → +ve Q 2
2 4 3Pv v−
= −− −
M1 NEL
Q P 4v v− = A1 Any form Q 0.5v = so 0.5 m s –1 in orig direction of P A1 cao. Direction need not be made clear. P 3.5v = − so 3.5 m s –1 in opp to orig dir of
P A1 cao. Direction must be negative and consistent or
clear
(e.g diag) 6 (B) → +ve 2 3.5 2 4 15× − − × = − N s M1 Use of change in momentum with correct mass. so 15 N s in opp to orig direction A1 FT (A). Dir must be clear (e.g. diag) 2 (b) Let arcsin(12 13) and =arcsin(3 5)α β= Parallel: 26cos cosuα β= M1 PCLM parallel to plane attempted. At least one resolution correct A1
so 5 42613 5
u× = × and u = 12.5 A1
Perp: sin
26sinue β
α= M1 NEL on normal components attempted.
F1 FT their u 312.5 55
12 162613
×= =
× F1 FT their u
6 19
4762 Mark Scheme June 2006
Q 2 mark Sub
(i) Diagrams B1 Internal force at B must be shown cw moments about A B2 90 3 0R× − = M1 1st moments equation attempted for either force. B 60R = so 60 N upwards A1 Accept direction not specified cw moments about R: T ↓ 75 1 3 60 0.5 0T× + − × = M1 2nd moments equation for other force. All forces present. No extra forces. A1 Allow only sign errors 15T = − so 15 N upwards A1 Direction must be clear (accept diag) 6 (ii) cw moments about A 90 2cos30 3cos30 3cos 60 0V U× − × − × = M1 Moments equation with resolution. Accept terms missing A1 All correct. Allow only sign errors. giving 60 3 3U V= + E1 Clearly shown 3 (iii) Diagram B1 U and V correct with labels and arrows 1 (iv) ac moments about C 75 2cos30 3.5 cos30 3.5 cos 60 0V U× + − = M1 Moments equation with resolution. Accept term missing B1 At least two terms correct (condone wrong signs) 300 3 3
7U V= − A1 Accept any form
Solving for U and V M1 Any method to eliminate one variable 360 3
7U = ( = 89.0768…) A1 Accept any form and any reasonable accuracy
607
V = ( = 8.571428…) F1 Accept any form and any reasonable accuracy
[Either of U and V is cao. FT the other] Resolve → on BC F = U M1
so frictional force is 360 37
N F1
( = 89.1 N (3 s. f.)) 8 18
4762 Mark Scheme June 2006
Q 3 mark Sub (a) ( )20000 900 0.1 16R g= + × × M1 Use of P = Fv, may be implied. B1 Correct weight term A1 All correct R = 368 so 368 N A1 4 (b) (i)
max cosF mgμ α= B1 Correct expression for maxF or wt cpt down slope
(may be implied and in any form) Force down slope is weight cpt sinmg α B1 Identifying 5sin as 13α or equivalent Require cos sinmg mgμ α α≥
so 5tan12
μ α≥ = E1 Proper use of F Rμ≤ or equivalent.
[ tanμ α= used WW; SC1] 3 (ii) either 20.5 11 v× × M1 Use of work energy with at least three required
terms
attempted 5 1211 1.5 0.2 11 1.5 9
13 13g g= × × + × × × + B1 Any term RHS. Condone sign error.
B1 Another term RHS. Condone sign error. A1 All correct . Allow if trig consistent but wrong 2 18.3717...v = v = 4.2862… so 4.29 m s –1 (3 s. f.) A1 cao or 5 + ve up the slope 5 1211 0.2 11 6 11
13 13g g a− × − × × − = M1 Use of N2L
B1 Any correct term on LHS a = - 6.1239 m s-2 A1 v2 = - 3a M1 use of appropriate uvast v = 4.286 m s-1 A1 c.a.o. (iii) continued overleaf
4762 Mark Scheme June 2006
3 continued (iii) either Extra GPE balances WD against
resistances M1 Or equivalent
sinmgx α B1 6( 3) 0.2 11 cos ( 3)x g xα= + + × × + B1 One of 1st three terms on RHS correct B1 Another of 1st 3 terms on RHS correct A1 All correct. FT their v if used. x = 4.99386… so 4.99 m (3 s. f.) A1 cao. 6 or M1 Allow 1 term missing 0.5 11 18.3717...× × B1 KE. FT their v 5(1.5 ) 11 6(1.5 )
13x g x= + × × − + B1 Use of 1.5 + x (may be below)
12(1.5 ) 0.2 1113
x g− + × × × B1 WD against friction
A1 All correct x = 4.99386… so 4.99 m (3 s. f.) A1 cao. or + ve down the slope 5 1211 0.2 11 6 1113 13g g a× − × × − = M1 N2L with all terms present A1 all correct except condone sign errors 1.4145...a = m s-2 A1 4.2862 = 2a(1.5+x) M1 use of appropriate uvast B1 for (1.5 + x) (may be implied) x = 4.99 A1 c.a.o. 18
4762 Mark Scheme June 2006
Q 4 mark Sub (i) 5 10 20 25
100 10 30 30 300 15 15 30
xy
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
M1 Correct method for c.m.
B1 Total mass correct B1 One c.m. on RHS correct [If separate components considered, B1 for 2
correct]
1700
1001800
xy
⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
17x = A1 cao 18y = A1 cao. [Allow SC 4/5 for 18x = and 17y = ] 5 (ii) ( )17,18, 20 B1 x- and y- coordinates. FT from (i). B1 z coordinate 2 (iii) cw moments about horizontal edge thro’
D M1 Or equivalent with all forces present
x component ( )20 60 20 17 0P × − × − = B1 One moment correct (accept use of mass or
length)
B1 correct use of their x in a distance P = 9 A1 FT only their x 4 (iv) Diagram B1 Normal reaction must be indicated acting vertically upwards at edge on Oz and weight be in approximately the correct place. 1 (v) On point of toppling ac moments about edge along Oz M1 Or equivalent with all forces present 30 60 17 0Q× − × = B1 Any moment correct (accept use of mass or
length)
Q = 34 F1 FT only their x Resolving horizontally F = Q B1 As 34 > 30, slips first B1 FT their Q correctly argued. 5 17
4762 Mark Scheme Jan 2007
52
Q 1 mark sub
(i)
2 110 0.5 0.5 29.5v v× = + M1 PCLM and two terms on RHS A1 All correct. Any form. 1 2 0.8
0 10v v−
= −−
M1 NEL
A1 Any form v1 = 0.3 so V1 = 0.3 A1 v2 = 7.7− so V2 = 7.7 m s –1 A1 Speed. Accept ± . in opposite to original direction F1 Must be correct interpretation of clear working 7 (ii) (A) 10 0.5 30V× = M1 PCLM and coalescence A1 All correct. Any form.
so 16
V = A1 Clearly shown. Accept decimal equivalence. Accept
no direction. 3 (B) Same velocity E1 Accept speed No force on sledge in direction of motion E1 2 (iii)
B1
2 40 0.5 39.5u V× = + M1 PCLM, masses correct A1 Any form u – V = 10 B1 May be seen on the diagram. Hence V = 1.875 A1 Accept no reference to direction. 5 17
39.5 kg 0.5 kg
u V
before
after
2 m s –1 2 m s –1
before
after
0.5 kg 29.5 kg
10 m s –1
v2 m s –1 v1 m s –1
0
4762 Mark Scheme Jan 2007
53
Q 2 mark comment sub
(i) cos30X R= (1) B1 sin 30Y R L+ = (2) M1 Attempt at resolution A1 3 (ii) ac moments about A R – 2L = 0 B1 Subst in (1) and (2) M1 Subst their R = 2L into their (1) or (2) 32 so 3
2X L X L= = E1 Clearly shown
122
Y L L+ × = so Y + L = L and Y = 0 E1 Clearly shown
4 (iii) (Below all are taken as tensions e. g. ABT in B1 Attempt at all forces (allow one omitted) AB) B1 Correct. Accept internal forces set as tensions or thrusts or a mix 2 (iv) ↓ A AD cos30 ( ) 0T Y− = M1 Vert equilibrium at A attempted. Y = 0 need not be explicit so AD 0T = E1 2 (v) Consider the equilibrium at pin-joints M1 At least one relevant equilib attempted A → AB AB0 so 3T X T L− = = (T) B1 (T) not required C ↓ CE cos30 0L T+ = B1 Or equiv from their diagram
so CE2 2L 2 3 so
33 3L LT
⎛ ⎞−= =⎜ ⎟
⎝ ⎠ (C) B1 Accept any form following from their
equation. (C) not required. C ← BC CE cos60 0T T+ = B1 Or equiv from their diagram
so BC2 3 1 3
3 2 3L LT
⎛ ⎞= − − × =⎜ ⎟
⎝ ⎠ (T) B1
FT their CET or equiv but do not condone inconsistent signs even if right answer obtained. (T) not required.
F1 T and C consistent with their answers and their diagram 7 (vi) ↓ B BD BEcos30 cos30 0T T+ = M1 Resolve vert at B so BD BET T= − so mag equal and opp sense E1 A statement required 2 20
4762 Mark Scheme Jan 2007
54
Q 3 mark sub
(i) (10, 2, 2.5) B1 1 (ii) By symmetry
x = 10, B1
2y = B1 (240 80) 80 0 240 2.5z+ = × + × B1 Total mass correct M1 Method for c.m. so 1.875z = A1 Clearly shown 5 (iii) 10x = by symmetry E1 Could be derived 10 10
(320 80) 320 2 80 41.875 3
xyz
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
M1 Method for c.m.
B1 y coord c.m. of lid B1 z coord c.m. of lid 2.4y = E1 shown 2.1z = E1 shown 6 (iv)
B1 Award for correct use of dimensions 2.1 and 2.4 or equivalent
c.w moments about X 40 0.024cos30 40 0.021sin 30× − × B1 1st term o.e. (allow use of 2.4 and 2.1) B1 2nd term o.e. (allow use of 2.4 and 2.1) = 0.41138… so 0.411 N m (3 s. f.) E1 Shown [Perpendicular method: M1 Complete method: A1 Correct lengths and angles E1 Shown] 4 (v) 0.41138… - 0.05P = 0 M1 Allow use of 5 P = 8.22768…… so 8.23 (3 s. f.) A1 Allow if cm used consistently 2 18
30°
P
40 N
5 cm
2.1 cm 2.4 cm
X
4762 Mark Scheme Jan 2007
55
Q 4 mark sub
(i) maxF Rμ= M1 Must have attempt at R with mg resolved R = 2g cos 30 B1 so Fmax = 0.75 2 9.8 cos30 12.730...× × × = so 12.7 N (3 s. f.) A1 [Award 2/3 retrospectively for limiting friction seen below] either Weight cpt down plane is 2gsin 30 = 9.8 N B1 so no as 9.8 < 12.7 E1 The inequality must be properly justified or Slides if tan 30μ < B1 But 0.75 > 0.577… so no E1 The inequality must be properly justified 5 (ii) (A) Increase in GPE is 2 9.8 (6 4sin 30) 156.8× × + = J M1 Use of mgh B1 6 + 4 sin 30 A1 3 (B) WD against friction is 4 0.75 2 9.8 cos30 50.9222...× × × × = J M1 Use of WD = Fd A1 2 (C) Power is 10 × (156.8 + 50.9222…)/60 M1 Use P = WD/t = 34.620… so 34.6 W (3 s. f.) A1 2 (iii) 20.5 2 9× × M1 Equating KE to GPE and WD term. Allow sign errors and one KE term omitted. Allow ‘old’ friction as well. = 2 9.8 (6 sin 30)x× × + + 20.5 2 4× × B1 Both KE terms. Allow wrong signs. 90− A1 All correct but allow sign errors A1 All correct, including signs. so x = 3.8163…. so 3.82 (3 s. f.) A1 cao 5 17
4762 Mark Scheme June 2007 Q 1
(a) (i) Impulse has magnitude N s 2 9 18× = B1
speed is 18 36
= m s –1 . B1
2 (ii) PCLM → 3 6 1 2 8v× − × = M1 Use of PCLM + combined mass RHS A1 All correct v = 2 so 2 m s –1 in orig direction of A E1 Must justify direction (diag etc) 3 (iii) 2 2 2 1 6→ × − × − = N s M1 Attempted use of mv - mu A1 for 6 N s dir specified (accept diag) 2 (iv) (A)
B1 Accept masses not shown
1 (B) PCLM → 2 8 10 1.8 8 10 1.9v× + × = + × M1 PCLM. All terms present A1 Allow sign errors only v = 1.875 A1 3 (C)
NEL 1.9 1.875
1.8 2e−
= −−
M1 Use of NEL with their v
A1 Any form. FT their v so e = 0.125 F1 FT their v (only for 0 1e< ≤ ) 3 (b) Using 2 2 2v u a= + s 2 10 9.8 14v = × × = B1 Allow 14±
rebounds at 414 7
×
M1 Using their vertical component
= 8 m s –1 F1 FT from their 14. Allow ± No change to the horizontal component B1 Need not be explicitly stated Since both horiz and vert components are 8 m s –1 the angle is 45° A1 cao 5 19
AB C
2 ms -1
1.9 m s –1
1.8 m s –1
v ms –1
58
4762 Mark Scheme June 2007 Q 2
(i)
2πθ = B1
8sin 162gives CG
2
π
π π= = E1
16 ,8π
⎛ −⎜⎝ ⎠
⎞⎟ justified E1
3 (ii) 16 36
(8 72) 8 7208
xy
π π π⎛ ⎞−⎛ ⎞ ⎛ ⎞⎜ ⎟+ = +⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
M1 Method for c.m.
B1 Correct mass of 8�or equivalent A1 1st RHS term correct A1 2nd RHS term correct 25.3673... 25.37
2.06997... 2.07xy
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (4 s. f.) E1
E1
[If separate cpts award the A1s for x- and y- cpts correct on RHS] 6 (iii)
B1
General position and angle (lengths need not be shown)
13.93
13.93tan25.37
α = M1 Angle or complement attempted. arctan or equivalent.
M1 Attempt to get 16 – 2.0699… A1 Obtaining 13.93… cao Accept use of 2.0699… but not 16. α = 28.7700… so 28.8° (3 s. f.) A1 cao 5 (iv) [FT use of 2.0699…] c. w. moments about A M1 Moments about any point, all forces present 12 13.93 16 0F× − = A1 so F = 10.4475… A1 (1.5525... if 2.0699… used) 3 17
α 25.37
G
A
59
4762 Mark Scheme June 2007 Q 3
(i) Moments c.w. about B A200 0.6 0.8 0R× − = M1 Accept about any point. Allow sign errors. RA = 150 so 150 N A1 Resolve or moments M1 RB = 50 so 50 N B F1 4 (ii) Moments c.w. about D
C0.8 1.2 200 0R− + × = M1 Or equiv. Accept about any point. All terms present. No extra terms. Allow sign errors.
RC = 300 ↑ A1 Neglect direction Resolve or moments M1 Or equiv. All terms present. No extra terms.
Allow sign errors.
RD = 100 ↓ A1 Neglect direction E1 Both directions clearly shown (on diag) 5 (iii) Moments c.w. about P Q0.4 200cos 0.8 0Rα× − = M1 Or equiv. Must have some resolution. All terms present. No extra terms. Allow sign errors. A1 Correct
RQ = 96 so 96 N A1 [No direction required but no sign errors in working]
resolve perp to plank P Q200cosR Rα= + M1 Or equiv. Must have some resolution. All terms present. No extra terms. Allow sign errors. A1 Correct RP = 288 so 288 N A1 [No direction required but no sign errors in
working]
6 (iv) Need one with greatest normal reaction So at P B1 FT their reactions Resolve parallel to the plank 200sinF α= so F = 56 B1 F
Rμ = M1 Must use their F and R
56 7288 36
= = (= 0.194 (3 s. f.)) A1 cao
4 19
60
4762 Mark Scheme June 2007 Q 4
(i) either 20.5 20 0.5 20 9.8 4× × + × × M1 KE or GPE terms B1 KE term B1 GPE term = 786.5 J A1 cao or 1
32a = B1 120 20 32T g− = × M1 N2L. All terms present. T = 196.625 A1 WD is 4T = 786.5 so 786.5 J A1 cao 4 (ii) 20 0.5 10g g× = so 98 W M1 Use of P = Fv or WD
tΔ
Δ A1 All correct A1 3 (iii) GPE lost is J 35 9.8 3 1029× × = B1 KE gained is J 2 20.5 35 (3 1 ) 140× × − = M1 Δ KE A1 The 140 J need not be evaluated so WE gives WD against friction is M1 Use of WE equation 1029 – 140 = 889 J A1 cao 5 (iv) either
20.5 35 3 35 9.8 0.1 150x x× × + × × = M1 WE equation. Allow 1 missing term. No extra terms.
B1 One term correct (neglect sign) B1 Another term correct (neglect sign) A1 All correct except allow sign errors x = 1.36127… so 1.36 m (3 S. F.) A1 cao or
35 0.1 150 35g a× − = M1 Use of N2L. Must have attempt at weight component. No extra terms.
A1 Allow sign errors, otherwise correct a = – 3.3057… A1 cao 0 = 9 – 2ax M1 Use of appropriate uvast or sequence
x = 1.36127… so 1.36 m (3 S. F.) A1 cao 5 17
61
4762 Mark Scheme January 2008
37
4762 Mechanics 2
Q1 Mark Comment Sub(a) (i) either In direction of the force I = Ft = mv M1 Use of Ft = mv so 1500 8 4000v× = A1 giving v = 3 so 3 m s -1 A1 or N2L gives 1500
4000a = M1 Appropriate use of N2L and uvast
v = 0 + 1500 8
4000× A1
giving v = 3 so 3 m s -1 A1 3 (ii)
PCLM R S12000 4000 500V V= + M1 Appropriate use of PCLM so R S24 8V V= + A1 Any form
NEL S R 0.20 3
V V−= −
− M1 Appropriate use of NEL
so S R 0.6V V− = A1 Any form Solving R S2.6, 3.2V V= = A1 Either value so ram 2.6 m s -1 and stone 3.2 m s -1 F1 6 (iii) 2 2 20.5 4000 3 0.5 4000 2.6 0.5 500 3.2× × − × × − × × M1 Change in KE. Accept two terms B1 Any relevant KE term correct (FT their
speeds)
= 1920 J A1 cao 3 (b) see over
500 4000
4000
3 m s -1
V Rm s -1V S m s -1
before
after 500
0
4762 Mark Scheme January 2008
38
Q 2 Mark Comment Sub(i) (A) 20.5 80 3 360× × = J M1 Use of KE A1 2 (B) 360 12F= × M1 W = Fd attempted so F = 30 so 30 N F1 FT their WD 2 (ii) Using the WE equation M1 Attempt to use the WE equation. Condone
one missing term
2 20.5 80 10 0.5 80 4× × − × × M1 Δ KE attempted 80 9.8 1600h= × × − B1 1600 with correct sign A1 All terms present and correct (neglect signs) h = 6.32653… so 6.33 (3 s. f.) A1 cao 5 (iii) (A) We have driving force F = 40 B1 May be implied so 200 = 40v M1 Use of P = Fv and v = 5 so 5 m s -1 A1 3 (B) From N2L, force required to give accn
is M1 Use of N2L with all terms present (neglect signs)
40 80 2F − = × A1 All terms correct so F = 200 A1 P = 200 0.5× = 100 so 100 W M1 correct use of P = Fv A1 cao 5 17
1 Mark Comment Sub(b) (i) 72i N s B1 Neglect units but must include direction 8(9cos60 9sin 60 )+i j
= ( 36 36 3+i j ) N s E1 Evidence of use of 8 kg , 9 m s -1 and 60°
2 (ii) 72 (36 36 3 ) 12( )u v+ + = +i i j i j M1 PCLM. Must be momenta both sides Equating components M1 72 + 36 = 12u so u = 9
36 3 12 so 3 3v v= = A1 Both
3 (iii) either 4 18 8 9× = × so equal momenta so
60/2 = 30° M1 Must be clear statements
A1 cao or ( ) ( )3 3 1
9 3arctan arctan 30= = ° M1 FT their u and v. A1 cao 2 19
4762 Mark Scheme January 2008
39
Q 3 Mark Comment Sub (i) For z M1 Method for c.m. ( )2 20 100 2 50 120 z× × + × × B1 Total mass of 16000 (or equivalent) 2 2000 50 2 6000 60= × × + × × B1 At least one term correct so z = 57.5 A1 NB This result is given below. and y = 0 B1 NB This result is given below. Statement (or
proof) required.
N.B. If incorrect axes specified, award max 4/5
5 (ii) y and z are not changed with the
folding E1 A statement, calculation or diagram required.
For x 100 120 0 2 20 100 10 16000x× × + × × × = M1 Method for the c.m. with the folding B1 Use of the 10 so 40000 2.5
16000x = = E1 Clearly shown
4 (iii) Moments about AH. M1 Normal reaction acts through this
line B1 May be implied by diagram or statement
c.w. ( )120 72 20 2.5 0P × − × − = B1 20 2.5− or equivalent A1 All correct so P = 10.5 A1 cao 5 (iv) maxF Rμ= M1 Allow F Rμ= so max 72F μ= A1 Must have clear indication that this is max F For slipping before tipping we
require
72 10.5μ < M1 Accept ≤ . Accept their maxF and R. so ( )7
480.1458333... μ < A1 cao 4 18
4762 Mark Scheme January 2008
40
Q 4 Mark Comment Sub(i) Centre of CE is 0.5 m from D B1 Used below correctly a.c. moment about D M1 Use of their 0.5 2200 0.5 1100× = so 1100 N m E1 0.5 must be clearly established. c.w moments about D 2.75 1100 0R × − = M1 Use of moments about D in an equation B1 Use of 1100 and 2.75 or equiv R = 400 so 400 N A1 6 (ii) c.w moments about D 1.5 1100 440 2.75 0W × − − × = M1 Moments of all relevant forces attempted A1 All correct so W = 1540 E1 Some working shown 3 (iii) (A) c.w. moments about D M1 Moments equation. Allow one missing term;
there must be some attempt at resolution.
1.5 1540cos20 1.751100cos 20 400 2.75cos20 0
T× −− − × =
M1 At least one res attempt with correct length Allow sin ↔ cos
A1 Any two of the terms have cos 20 correctly
used (or equiv)
B1 1.75 T A1 All correct T = 59.0663… so 59.1 N (3 s. f.) A1 cao Accept no direction given 6 (iii) (B) either Angle required is at 70° to the
normal to CE B1
so 1 cos70 59.0663...T = M1 so 1 172.698...T = so 173 N (3 s.f.) A1 FT (iii) (A)
or 400cos 20 2.75 1100cos20× + M1 Moments attempted with all terms present 1540cos 20 1.5 sin 20 1.75T= × − × A1 All correct (neglect signs) T = 172.698… so 173 N (3s.f.) A1 FT(iii)(A) 3 18
4762 Mark Scheme June 2008
47
4762 Mechanics 2 Q 1 mark comment sub
(a) (i) In i direction: 6 12 18u − = M1 Use of I-M so u = 5 i.e. 5i m s -1 E1 Accept 6 12 18u − = as total working. Accept 5
instead of 5i.
either In i direction: 0.5 12 0.5 11v + = × M1 Use of I-M B1 Use of + 12i or equivalent 13v = − so 13− i m s -1 A1 Accept direction indicated by any means or 6 × 5 + 0.5 v = 6 × 3 + 0.5 × 11 M1 PCLM v = – 13 A1 Allow only sign errors so – 13i m s-1 A1 Accept direction indicated by any means 5 (ii)
Using NEL: 11 313 5
e− = −− −
M1 Use of NEL. Condone sign errors but not reciprocal expression
F1 FT only their 13− (even if +ve) 4 (0.4)9e = F1 FT only their 13− and only if –ve (allow 1 s.f. accuracy) 3 (iii) In i direction: 2 7 0.5 0.5 11v− × = − × M1 Use of I = Ft M1 Use of I = m(v –u ) v = 17− so 17− i m s -1 A1 For 17± A1 cao. Direction (indicated by any means) or – 2 i = 0.5 a M1 Use of F = ma so a = – 4 i m s-2 A1 For 4± v = 11i – 4i × 7 M1 Use of uvast v = 17− so 17− i m s -1 A1 cao. Direction (indicated by any means) 4 (b) u i + ev j B1 For u B1 For ev
tan vu
α = , tan evu
β = M1 Use of tan. Accept reciprocal argument. Accept use of their components
B1 Both correct. Ignore signs.
tan tanve eu
β α⎛ ⎞= =⎜ ⎟⎝ ⎠
E1 Shown. Accept signs not clearly dealt with.
5 17
4762 Mark Scheme June 2008
48
Q 2 m a r k comment sub
(i)
( ) 3 6 3 02 3 6 6 6 6 2
0 3 6 7xy
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ × = + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
M1 Method for c.m.
B1 Total mass correct 18 36 18 72
2018 36 14 68
xy
+ +⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ B1 For any of the 1st 3 RHS terms
B1 For the 4th RHS term 3.6x = E1 3.4y = A1 cao [If separate cpts, award the 2nd B1 for 2 x- terms
correct and 3rd B1 for 2 7× in y term]
6
(ii) B1 Diagram showing G vertically below D B1 3.6 and their 3.4 correctly placed (may be
implied)
3.6 3.6arctan arctan2 (6 3.4) 4.6
⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟+ − ⎝ ⎠⎝ ⎠ M1 Use of arctan on their lengths. Allow reciprocal
of argument.
Some attempt to calculate correct lengths needed
B1 2 + (6 – their 3.4) seen so 38.047… so 38.0° (3 s. f.) A1 cao 5 (iii) moments about D M1 moments about D. No extra forces BP5 3.6 6 T× = × so tension in BP is 3 N F1 FT their values if calc 2nd Resolve vert: DQ3 5T+ = M1 Resolve vertically or moments about B. so tension in DQ is 2 N F1 FT their values if calc 2nd 4 (iv) We require x-cpt of c.m. to be zero M1 A method to achieve this with all cpts either ( ) 21
220 20 3.6L x L+ = × − or ( ) 22 6 0.5 6 6 6 0.5 0L× × × + × − × = B1 For the 20.5 L× A1 All correct L = 12 A1 4 19
vertical
2
3.6 2.6
3.4
. D
G
O
6
4762 Mark Scheme June 2008
49
Q 3 mark comment sub
(a) (i)
B1 B1
Internal forces all present and labelled All forces correct with labels and arrows (Allow the internal forces set as tensions, thrusts or a mixture)
2
(ii) M1 Equilibrium equation at a pin-joint attempted A1 1st ans. Accept + or –. A ↑ M1 Second equation attempted TAD sin 30 – L = 0 so TAD = 2L so 2L N
(T) F1 2nd ans. FT any previous answer(s) used.
A → TAB + TAD cos 30 = 0 so TAB = 3L− so 3L N (C) M1 Third equation attempted B ↑ TBD sin 60 – 3L = 0 A1 3rd ans. FT any previous answer(s) used. so TBD = 2 3L so 2 3L N (T) B → M1 Fourth equation attempted TBC + TBD cos 60 –TAB = 0 F1 4th ans. FT any previous answer(s) used. so TBC = 2 3L− so 2 3L N (C) E1 All T/C consistent [SC 1 all T/C correct WWW] 9 Accept only 1 leg considered (and without
comment)
(b) Leg QR with frictional force F ← moments c.w. about R 2 sin 60 cos60 0U l Wl× − = M1 Suitable moments equation. Allow 1 force
omitted
A1 a.c. moments A1 c.w. moments Horiz equilibrium for QR F = U M1 A second correct equation for horizontal or
vertical equilibrium to eliminate a force
(U or reaction at foot) [Award if correct moments equation containing
only W and F]
E1 * This second equation explicitly derived
Hence 1 32
W F= M1 Correct use of 2nd equation with the moments equation
and so 3
6F W= E1 Shown. CWO but do not penalise * again.
7 18
A B C
D
L N 3L N
30° 60°
TAD
TBD
TAB TBC
4762 Mark Scheme June 2008
50
Q 4 mark comment sub
(a) (i) Tension is perp to the motion of the
sphere E1
(so WD, cosFd θ = 0) 1 (ii) Distance dropped is 2 2cos 40− =
0.467911.. M1 Attempt at distance with resolution used. Accept
sin cos↔ E1 Accept seeing 2 2cos40− GPE is mgh so M1 0.15 9.8 0.467911... 0.687829...× × = J B1 Any reasonable accuracy 4 (iii) 20.5 0.15 0.687829...v× × = M1 Using KE + GPE constant so v = 3.02837… so 3.03 m s -1 (3 s. f.) F1 FT their GPE 2 (iv) ( )2 21 0.15 2.5
2v× − M1 Use of W-E equation (allow 1 KE term or GPE
term omitted)
B1 KE terms correct 400.687829... 0.6 2 2
360π= − × × × M1 WD against friction
A1 WD against friction correct (allow sign error) v = 2.06178… so 2.06 m s -1 (3 s. f.) A1 cao 5 (b)
N2L down slope: 13 sin 30 38
g F g− = × M1 Must have attempt at weight component
A1 Allow sign errors.
so 98gF = (= 11.025) A1
332
R g= × (= 25.4611…) B1
34
FR
μ = = (= 0.43301…) M1 Use of F Rμ=
E1 Must be worked precisely 6 18
4762 Mark Scheme January 2009
38
4762 Mechanics 2
Q 1 Mark Sub (i) either M1 Use of I = Ft 2 5m u F× = A1 so F = 0.4mu in direction of the velocity A1 Must have reference to direction. Accept diagram. or M1 Use of suvat and N2L
a = 25u A1 May be implied
so F = 0.4mu in direction of the velocity A1 Must have reference to direction. Accept diagram. 3 (ii) M1 For 2 equns considering PCLM, NEL or Energy PCLM → 2 3 3P Qum um mv mv+ = + NEL → 2Q Pv v u u u− = − = Energy 2 21 1
2 2(2 ) (3 )m u m u× + × = 2 21 1
P Q2 2 (3 )m v m v× + × A1 One correct equation A1 Second correct equation Solving to get both velocities M1 Dep on 1st M1. Solving pair of equations. 3
2Quv = E1 If Energy equation used, allow 2nd root discarded
without comment.
2Puv = A1
[If AG subst in one equation to find other velocity, and no more, max SC3] 6 (iii) either
After collision with barrier Q3 2euv = ← B1 Accept no direction indicated
so 3 3 42 2 4u eu um m m→ − = − M1 PCLM
A1 LHS Allow sign errors. Allow use of Q3mv . A1 RHS Allow sign errors
so 13
e = A1
At the barrier the impulse on Q is given by 3 1 3 3
2 3 2u um ⎛ ⎞→ − × −⎜ ⎟
⎝ ⎠ M1 Impulse is m(v – u)
F1
3 12 3u
± ×
so impulse on Q is 6 mu− → F1 Allow ± and direction not clear. FT only e. so impulse on the barrier is 6 mu → A1 cao. Direction must be clear. Units not required. 9 18
4762 Mark Scheme January 2009
39
Q 1 continued mark sub (iii) or
After collision with barrier Q3 2euv = ← B1
Impulse – momentum overall for Q
→ 2 3 44umu mu I m+ + = − × M1 All terms present
A1 All correct except for sign errors 6I mu= − A1 so impulse of 6mu on the barrier → A1 Direction must be clear. Units not required. Consider impact of Q with the barrier to give speed Qv after impact
→ Q3 3 6 32u m mu mv× − = M1 Attempt to use I - M
F1
so Q 2uv = − F1
3 12 2 3u ue = ÷ = A1 cao
9
4762 Mark Scheme January 2009
40
Q 2 Mark Sub (i) 80 cosR g θ= or 784cosθ B1 Seen maxF Rμ= M1 so 32 cos or 313.6cosg θ θ N A1 3 (ii)
Distance is 1.25sinθ
B1
WD is Fmax d M1
so 1.2532 cossin
g θθ
× E1 Award for this or equivalent seen
= 392
tanθ
3 (iii) Δ GPE is mgh M1 so 80 9.8 1.25 980× × = J A1 Accept 100g J 2 (iv) either P = Fv M1 so (80 sin 35 32 cos35) 1.5g g+ × B1 Weight term A1 All correct = 1059.85… so 1060 W (3 s. f.) A1 cao or WDP
t=
Δ M1
so
392980tan 35
1.25 1.5sin 35
+
⎛ ⎞ ÷⎜ ⎟⎝ ⎠
B1 B1
Numerator FT their GPE Denominator
= 1059.85… so 1060 W (3 s. f.) A1 cao 4 (v) either Using the W-E equation M1 Attempt speed at ground or dist to reach required speed. Allow only init KE omitted 2
2 1 3920.5 80 0.5 80 9802 tan 35
v ⎛ ⎞× × − × × = −⎜ ⎟⎝ ⎠
B1 KE terms. Allow sign errors. FT from (iv).
B1 Both WD against friction and GPE terms. Allow sign errors. FT from parts above. A1 All correct v = 3.2793.. so yes A1 CWO or N2L down slope M1 All forces present a = 2.409973… A1 distance slid, using uvast is 1.815372… A1 vertical distance is 1.815372…× sin35 M1 valid comparison = 1.0412… < 1.25 so yes A1 CWO 5 17
4762 Mark Scheme January 2009
41
Q 3 Mark Sub (i) M1 Correct method for y or z B1 Total mass correct M1 15cosα or 15sinα attempted either part 30: 250 4 125 8 cos 375
2y yα⎛ ⎞× + + =⎜ ⎟
⎝ ⎠ B1
308 cos2
α⎛ ⎞+⎜ ⎟⎝ ⎠
B1 250 4× 28 19
3 3y = = E1 Accept any form
30: (250 0 ) 125 sin 3752
z zα× + × = B1 LHS
z = 3 E1 8 (ii) Yes. Take moments about CD. E1 c.w moment from weight; no a.c moment
from
table E1 [Award E1 for 19 8
3> seen or ‘the line of action
of the weight is outside the base] 2 (iii) c.m. new part is at (0, 8 + 20, 15) M1 Either y or z coordinate correct M1 Attempt to ‘add’ to (i) or start again. Allow mass
error.
28375 125 28 5003
y× + × = so 14y = E1
375 3 125 15 500z× + × = so 6z = E1 4 (iv) Diagram B1 Roughly correct diagram B1 Angle identified (may be implied)
Angle is 6arctan14
M1 Use of tan. Allow use of 14/6 or equivalent.
= 23.1985… so 23.2° (3 s. f.) A1 cao 4 18
4762 Mark Scheme January 2009
42
Q 4 mark sub (a) Let the ↑ forces at P and Q be P Q and R R (i) c.w. moments about P M1 Moments taken about a named point. 2 600 3 0QR× − = so force of 400 N ↑ at
Q A1
a.c. moments about Q or resolve M1
P 200R = so force of 200 N ↑ at P A1 4 (ii) P 0R = B1 Clearly recognised or used. c.w. moments about Q M1 Moments attempted with all forces. Dep on P 0R = or PR not evaluated. 2 1 600 0L − × = so L = 300 A1 3 (b) (i) 15 8 8
17 17 15cos or sin or tanα α α= = = B1 Seen here or below or implied by use. c.w moments about A M1 Moments. All forces must be present and appropriate resolution attempted. 16 340cos 8 0Rα× − = A1 so R = 600 E1 Evidence of evaluation. 4 (ii) Diagram B1 Must have 600 (or R) and 340 N and reactions at
A.
B1 All internal forces clearly marked as tension or thrust.
Allow mixture. (Solution below assumes all internal forces [Max of B1 if extra forces present] set as tensions) 2 (iii) M1 Equilibrium at a pin-joint B ↓ BC340cos cos 0Tα α+ = so BC 340T = − (Thrust of) 340 N in BC A1 C → BC ACsin sin 0T Tα α− = so AC 340T = − (Thrust of) 340 N in AC F1 B ← AB BC sin 340sin 0T T α α+ − = M1 Method for ABT so AB 320T = (Tension of) 320 N in AB A1 Tension/ Thrust all consistent with
working F1
[Award a max of 4/6 if working inconsistent with diagram] 6 19
4762 Mark Scheme June 2009
56
4762 Mechanics 2
Q 1 mark comment sub (a) (i) B1 1
(ii)
3umu kmu mv km− = + M1 PCLM applied
A1 Either side correct (or equiv) 41
3kv u⎛ ⎞= −⎜ ⎟
⎝ ⎠ E1 Must at least show terms grouped
3 (iii)
Need v < 0 E1 Accept 4 13k > without reason
so 34
k > B1
[SC1: v = 0 used and inequality stated without reason]
2 (iv)
132
u v
u u
−= −
− − M1 Use of NEL
A1 so 2
3uv = − E1
2 413 3u ku ⎛ ⎞− = −⎜ ⎟
⎝ ⎠ M1
so k = 1.25 A1 cao 5 (b) (i) 1 3
9 5 82 2
⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠V M1 Use of PCLM
B1 Use of mass 8 in coalescence M1 Use of I = Ft 3
1⎛ ⎞= ⎜ ⎟−⎝ ⎠
V E1
4 (ii) i cpt 13 3
2→ − × M1 Allow wrong sign
P m kg
Q km kg
u u
v 3
u
before
after
4762 Mark Scheme June 2009
57
j cpt unchanged B1 May be implied
new velocity 1.51
−⎛ ⎞⎜ ⎟−⎝ ⎠
m s -1 A1 cao [Award 2/3 if barrier taken as
10⎛ ⎞⎜ ⎟⎝ ⎠
]
3 18 Q 2 mark comment sub (a) (i) (A) Yes. Only WD is against
conservative E1 Accept only WD is against gravity or no work done
forces. against friction. (B) Block has no displacement in
that direction E1
2 (ii) 20.5 50 1.5 20 5gx gx× × = − M1 Use of WE with KE. Allow m =
25.
B1 Use of 50 M1 At least 1 GPE term A1 GPE terms correct signs x = 0.38265… so 0.383 m (3 s.
f.) A1 cao
5 (iii) 2 20.5 50 0.5 50 1.5V× × − × × M1 WE equation with WD term. Allow
GPE terms missing
B1 Both KE terms. Accept use of 25. 2 20 2 5 180g g= × − × − B1 Either GPE term B1 180 with correct sign V = 2.6095… so 2.61 m s -1 A1 cao 5 (b) Force down the slope is 2000 450 sin 20g+ M1 Both terms. Allow mass not
weight
B1 Weight term correct Using P = Fv M1 ( )2000 450 sin 20 2.5P g= + × F1 FT their weight term P = 8770.77… so 8770 W (3 s.
f.) A1 cao
5 17
4762 Mark Scheme June 2009
58
Q 3 mark comment sub (i) c.w. moments about A M1 Moments equation. B5 3 85 0R − × = so B 51R = giving
51 N ↑ A1 Accept no direction given
Either a.c. moments about B or resolve ↑ M1
A 34R = so 34 N ↑ F1 Accept no direction given 4 (ii) c.w. moments about A M1 Moments with attempt to resolve
at least
one force. Allow s c↔ . 85 3cos 27.2 5sin 0α α× − × = B1 Weight term B1 horiz force term so 3 85 15tan
27.2 5 8α ×= =
× E1 Must see some arrangement of
terms
or equiv 4 (iii)
B1 All forces present and labelled
a.c. moments about B M1 Moments with attempt to resolve forces
and all relevant forces present 85 2 cos 34 2.5 5 sin 0Sα α× × + × − × = B1 34 2.5× A1 All other terms correct. Allow sign
errors.
S = 37.4 A1 All correct Resolving horizontally and
vertically M1 Either attempted
34sin 0S F α→ − − = so F = 7.4 E1 85 34cos 0R α↑ − − = A1 R = 101 need not be evaluated
here
[Allow A1 for the two expressions if
correct other than s c↔ ] Using F Rμ= M1 7.4
101μ = = 0.07326… so 0.0733
(3 s. f.) A1 cao
10 18
S
R
F
A
B 85 N
34 N
α
4762 Mark Scheme June 2009
59
Q 4 mark comment sub (i) Taking a y-axis vert downwards
from O Allow areas used as masses
22 8 4 2 82kkπσ πσ× × + × × × M1 Method for c.m.
B1 ‘4’ used B1 16 kπ B1 k/2 used ( )22 8 2 8k yπσ πσ= × + × B1 Masses correct
so 264
16 2kyk
+=+
E1 Must see some evidence of simplification
Need no reference to axis of symmetry
6 (ii) k = 12 gives OG as 5.2 and
mass as 320πσ B1 Allow for either. Allow 1σ =
2320 5.2 8 12πσ πσ× + × × M1 Method for c.m. combining with (i) or starting again
B1 One term correct B1 Second term correct ( )320 64 yπσ πσ= + 16
3y = E1 Some simplification shown
5 (iii)
B1 B1 B1
G above edge of base 1 23 312 6 5− = seen here or below
8 seen here or below
23
8tan5
θ = M1 Accept
235
8 or attempts based on
136 and 8.
θ = 54.6887… so 54.7° (3 s. f.) A1 cao 5 (iv) Slips when tanμ θ= M1 Or ….
23
8 1.4117...5
= B1
< 1.5 so does not slip A1 There must be a reason 3 19
O
G
8 12
θ
1 23 312 6 5− =
4762 Mark Scheme January 2010
38
4762 Mechanics 2 1 (a) (i) Let vel of Q be v → 6 1 4 2 4v× = + × M1 Use of PCLM A1 Any form v = – 0.5 so 0.5 m s -1 A1 in opposite direction to R A1 Direction must be made clear. Accept – 0.5 only
if + ve direction clearly shown
4 (ii) Let velocities after be R: vR → ; S: vS
→
PCLM +ve → R S4 2 1 3 2 3v v× − × = + M1 PCLM R S2 3 5v v+ = A1 Any form NEL +ve → S R
1 4 0.1v v−− − = − M1 NEL
so vS – vR = 0.5 A1 Any form Solving gives vR = 0.7 → A1 Direction not required vS = 1.2 → A1 Direction not required Award cao for 1 vel and FT second 6 (iii) R and S separate at 0.5 m s -1 M1 FT their result above. Either from NEL or from difference in final velocities Time to drop T given by 20.5 9.8 0.4T× = so 2
7T = (0.28571…) B1 so distance is 2 1
7 70.5× = m (0.142857…m)
A1 cao
3 (b)
u u→ B1 ( )v ev→ − B1 Accept v ev→ KE loss is ( ) ( )2 2 2 2 21 1
2 2m u v m u e v+ − + M1 Attempt at difference of KEs = 2 2 2 2 21 1 1 1
2 2 2 2mu mv mu me v+ − − E1 Clear expansion and simplification = ( )2 21
2 1mv e− of correct expression 4 17
u u
ev v before after
4762 Mark Scheme January 2010
39
2(i) GPE is 1200 × 9.8 × 60 = 705 600 B1 Need not be evaluated Power is (705 600 + 1 800 000) ÷ 120 M1 power is WD ÷ time B1 120 s = 20 880 W = 20 900 W (3 s. f.) A1 cao 4 (ii) Using P = Fv. Let resistance be R N M1 Use of P = Fv. 13500 = 18F so F = 750 A1 As v const, a = 0 so F – R = 0 Hence resistance is 750 N E1 Needs some justification We require 750 × 200 = 150 000 J
(= 150 kJ) M1 Use of WD = Fd or Pt
F1 FT their F 5 (iii) M1 Use of W-E equation with ‘x’ ( )2 21
2 1200 9 18× × − B1 2 KE terms present 1200 9.8 sin 5 1500x x= × × − M1 GPE term with resolution A1 GPE term correct A1 All correct Hence 145800 = 475.04846…x so x = 306.91… so 307 m (3 s, f,) A1 cao 6 (iv) P = Fv B1 and N2L gives F – R = 1200a B1 Substituting gives P = (R + 1200a)v E1 Shown If 0a ≠ , v is not constant. But P and R
are constant so a cannot be constant. E1
4 19 3 (i) Let force be P (A) a.c. moments about C P × 0.125 – 340 × 0.5 = 0 M1 Moments about C. All forces present. No extra
forces.
A1 Distances correct P = 1360 so 1360 N A1 cao 3 (i) (B) Let force be P c.w. moments about E P × 2.125 – 340 × (2 – 0.5) = 0 M1 Moments about E. All forces present. No extra
forces.
A1 Distances correct P = 240 so 240 N A1 cao 3
4762 Mark Scheme January 2010
40
(ii) sin 2.125 cos 0.9Q Qθ θ× + × M1 Moments expression. Accept s c↔ . B1 Correct trig ratios or lengths 25.5 4.5
13 13Q Q= +
= 3013
Q so 3013
Q N m E1 Shown 3 (iii) We need 30
13 340 1.5Q = × M1 Moments equn with all relevant forces so Q = 221 E1 Shown Let friction be F and normal reaction R Resolve → 221cos 0Fθ − = M1 so F = 85 A1 Resolve ↑ 221 sin θ + R = 340 M1 so R = 136 A1 F Rμ< as not on point of sliding M1 Accept ≤ or = so 85 136μ< A1 Accept ≤ . FT their F and R so 5
8μ > E1 9 18 4 (i) 30 50
4000 4800 80040 20
xy
= −
M1 Any complete method for c.m.
A1 Either one RHS term correct or one component of both RHS terms correct so 26x = E1 44y = A1 [SC 2 for correct y seen if M 0] 4 (ii)
250xy
M1 Any complete method for c.m.
0 20 40 50 60110 40 40 20 40
55 0 20 40 60 = + + + +
B1
Any 2 edges correct mass and c.m. or any 4 edges correct with mass and x or y c.m. coordinate correct.
B1 At most one consistent error 23.2x = E1 40.2y = A1 5
4762 Mark Scheme January 2010
41
(iii)
B1 Indicating c.m. vertically below Q
B1 Clearly identifying correct angle (may be implied) and lengths
Angle is 23.2arctan110 40.2 −
M1
Award for ( )arctan ba where b = 23.2 and a = 69.8
or 40.2 or where b = 69.8 or 40.2 and a = 23.2. Allow use of their value for y only.
= 18.3856…. so 18.4° (3 s. f.) A1 cao 4 (iv) 26 23.2
10 2 1.5 744 40.2
xy
= × × +
M1 Combining the parts using masses
B1 Using both ends A1 All correct 24.04x = so 24.0 (3 s.f.) A1 cao 41.34y = so 41.3 (3 s.f.) F1 FT their y values only. 5 18
G
Q
O
N
23.2 40.2
110 – 40.2
Oxford Cambridge and RSA Examinations
GCE
Mathematics (MEI) Advanced GCE 4762
Mechanics 2
Mark Scheme for June 2010
4762 Mark Scheme June 2010
1
Q 1 mark sub
(i) For P
P200 5 250 200v× + = M1 Award for I-M
P 6.25v = so 6.25i m s -1 E1 Accept no i and no units For Q
Q250 5 250 250v× − = M1 Must have impulse in opposite sense
Q 4v = so 4i m s -1 A1 Must indicate direction. Accept no units. 4 (ii) i direction positive PCLM: Q2250 200 4.5 250w= × + M1 PCLM used. Allow error in LHS FT from (i) F1 Any form. FT only from (i)
Q 5.4w = so 5.4i m s -1 E1 NEL: Q 4.5
4 6.25
we
−− = − M1 NEL . Allow sign errors
A1 Signs correct. FT only from (i) e = 0.4 A1 cao 6 (iii) i direction positive Suppose absolute vel of object is – Vi
200 × 4.5 = – 20V + 180 × 5.5 M1 Applying PCLM. All terms present. Allow sign errors.
B1 Correct masses A1 All correct (including signs) V = 4.5 A1 speed of separation is 5.5 + 4.5 = 10 m s -1 F1 FT their V. 5 (iv) 180 5.5 250 5.4 430W× + × = M1 Using correct masses and velocities W = 5.4418… so 5.44 i m s -1 (3 s. f.) A1 cao 2 17
4762 Mark Scheme June 2010
2
Q 2 mark sub
(i) 20 0 25
20 15 3 20 100 200
x
y
= + +
M1 Method to obtain at least 1 coordinate
B1 ‘100’ or ‘25’ correct
A1 Either one RHS term correct or one component of two
RHS terms correct 17.5x = A1 35y = A1 5 (ii)
350 40
25 5700 200
x
y
= +
M1 Using (i) or starting again
so 22x = , 68y = E1 Clearly shown. 2 (iii) We need the edge that the x position is
nearest M1 This may be implied
x = 22; distances are 22 to PQ, 18 to SR B1 One distance correct 15 to QR B1 All distances correct so edge QR A1 4 (iv) Moments about RS M1 Moments about RS attempted B1 Use of weight not mass below. FT mass from here In sense xOy sin50 200 cos50 40T T× − × M1 Attempt to find moment of T about RS, including
attempt at resolution. May try to find perp dist from G
to line of action of the force. A1 20 (40 17.5) 0g− × − = B1 40 – 17.5 A1 All correct allowing sign errors T = 34.5889… so 34.6 N (3 s. f.) A1 cao (except for use of mass) 7 18
4762 Mark Scheme June 2010
3
Q 3 mark sub
(i) a.c. moments about A 1 × T – 2 × 300 = 0 so T = 600 E1 Resolving → X = 0 B1 Justified ↑ T – Y = 300 M1 so Y = 300 A1 4 (ii) Diagram B1 All external forces marked consistent with (i) The working below sets all internal forces
as tensions; candidates need not do this.
B1 All internal forces with arrows and labels 2 (iii) Let angle DAB be θ .
312 2cos , sinθ θ= =
B1 Or equivalent seen
M1 Attempt at equilibrium at pin-joints M1 1 equilib correct, allowing sign errors
ABA 300 sin 0T θ↑ − − = so AB 200 3 so force is 200 3T = − (C) A1
AD ABA cos 0T T θ→ + =
so AD 100 3 so force is 100 3T = (T) F1
CDC sin 300 0T θ↑ − =
so CD 200 3 so force is 200 3T = (T) F1
BC CDC cos 0T T θ← + =
so BC 100 3 so force is 100 3T = − (C) F1
AB BDB sin 0T Tθ↑ + =
so BD 300 so force is 300T = (T) F1 F1 All T/C consistent with their calculations and diagrams 9 (iv) AD, AB, BC, CD B1 300 N, X and Y not changed. Equilibrium equations at A and C are not altered E1
AB BDB sin ' 600 0T Tθ↑ + + = M1 so BD' 300 so force is 300T = − (C) A1 C not needed.
[If 300 N (C) given WWW, award SC1 (NB it must be made clear that this is a compression)]
4 19
4762 Mark Scheme June 2010
4
Q 4 mark sub
(i) Let friction be F N and
normal reaction R N
max 58cos35F = B1 Need not be explicit
16 58sin 35R g= + M1 Both terms required. A1
maxF Rμ= M1 so μ = 0.249968… about 0.25 E1 5 (ii) WD is 70cos35 3 210cos35 × = M1 Use of WD = Fd . Accept cos35 omitted. so 172.0219… = 172 J (3 s. f.) A1 Average power is WD/time M1 so 34.4043…. = 34.4 W (3 s. f.) A1 cao 4 (iii) Using the constant acceleration result ( )1
2s u v t= + with s = 3, u = 0, v = 1.5 M1 Attempt to substitute in suvat (sequence)
and t = 5 we see that
( )123 0 1.5 5 3.75≠ + × = E1 Conclusion clear
2 (iv) 172.0219… M1 Using W-E equn, allow 1 missing term = 21
2 16 1.5× × M1 KE term attempted A1 correct ( )0.25 16 70sin 35 3g+ × + × M1 Attempt at using new F in maxF Rμ=
+ WD A1 A1 All correct so WD by S is 6.30916…
so 6.31 J (3 s. f.) A1 cao
7 18
GCE
Oxford Cambridge and RSA Examinations
Mathematics (MEI) Advanced GCE
Unit 4762: Mechanics 2
Mark Scheme for January 2011
4762 Mark Scheme January 2011
Q 1 m a r k notes
(i) Let normal reaction be R 2sin 1 0.8 0.6 B1 Accept any form and implied M1 Use of maxF R 2.5 9.8 0.8R B1 Expression for R; may be implied max 0.85F R 16.66 F1 FT their R Wt cpt down slope is 2.5 × 9.8 × 0.6 = 14.7 B1 16.66 > 14.7 so at rest E1 FT if their F and weight component show given result
If g omitted, allow B1M1B0F1B0E1, so 4/6 [Award as follows for use of tan :
B1 3tan
4
E1 tan shown] 6 (ii) Let the speeds down the plane be vA and vB . PCLM down the plane A B1.5 16 2.5 1.5v v M1 PCLM so 48A B5 3v v A1 Any form NEL +ve down the plane A B 0.4
0 16v v
M1 NEL. Allow sign errors
A B 6.4v v A1 Any form A 8.4v so 8.4 m s -1 down plane E1 Condone direction not clear if +8.4 seen
B 2v so 2 m s -1 down plane F1 Condone direction not clear if +2 seen. SC1 if 2 equations obtained and 8.4 substituted into one to obtain answer 2 (instead of E1F1)
6 (iii) 1.5 × (2 – 16) down plane M1 Use of m(v – u) If impulse on A found, treat as MR unless final answer relates this to
impulse on B = – 21 N s down the plane A1 ± 21 N s so 21 Ns up the plane A1 Direction explicitly commented on 3
6
4762 Mark Scheme January 2011
Q 1 m a r k notes
(iv) either max(2.5 9.8 0.6 ) 2.5(0 8.4)F t M1 Using Impulse-momentum (must use 8.4) . sufficient to consider one term on LHS B1 Either side correct A1 Allow only sign errors so t = 10.7142… 10.7 s (3 s. f.) A1 cao or Using N2L down the plane M1 Using N2L ; sufficient to consider one force term a = – 0.784 A1 Allow sign errors M1 Using appropriate suvat must use a or-a found by use of N2L and u = 8.4 using v = u + at, t = 10.7142… 10.7 s (3 s. f.) A1 cao or 20.5 2.5 8.4 (14.7 16.66) 0x M1 Use energy with 8.4, sufficient to consider one non-KE term x= 45 A1 M1 Using appropriate suvat T = 10.7142….. 10.7 (3 s. f.) A1 cao 4 19
7
4762 Mark Scheme January 2011
Q 2 m a r k notes
(a)
Energy: 2 21 12 20.004 0.060 0.8v V M1 Use of KE in two terms in an equation.
2 215 400v V A1 Any form PCLM in i direction: 0.06 0.004 0V v M1 PCLM. Accept sign errors. v = 15V A1 Any form Solving M1 Valid method for elimination of v or V from a linear and a quadratic 2 2(15 ) 15 400V V so 2 400 5
240 3V and 53V i A1 Accept 1.29099…i Accept no direction
5315 v i (= 375 i ) F1 Accept – 19.3649…i Accept no direction
Second answer follows from first A1 (Relative) directions indicated - accept diagram. Both speeds correct. 8 (b) (i) W is work done by resistances on car 2 21
2 800 12 30 800 9.8 20 W M1 Use of WE. Must have KE, W and GPE. Allow -W B1 Both KE terms. Accept sign error A1 All correct with W or -W W = – 145 600 so 145 600 J done by car against resistances A1 cao 4
-1
C 0.004 kg B 0.060 kg
v m s V m s -1 i
8
4762 Mark Scheme January 2011
Q 2 m a r k notes
(ii) either The slope is 18 × 25 = 450 m long B1 800 9.8 20 750 450
25 M1 Use of P = (Work done) / (elapsed time) used for at least one work done term
M1 WD is force × distance used for at least one force A1 Allow only sign errors both terms = 19 772 W A1 cao. or The angle of the slope is arcsin (1/22.5) B1 1800 9.8 750 18
22.5
M1 Use of P = Fv used for at least one term
M1 Attempt at weight component A1 Allow only sign errors both terms = 19 772 W A1 cao. 5 17
9
4762 Mark Scheme January 2011
Q 3 m a r k notes
(i) Horizontal X – 50 = 0 B1 Any form Vertical: R – Y – 45 = 0 B1 Any form 2 (ii) a. c. moments about A 1 × R = 3 × 45 M1 so R = 135 E1 Clearly shown so 135 – Y – 45 = 0 and Y = 90 E1 Shown 3 (iii) In analysis below all internal forces are taken B1 Correct arrow pairs for all internal forces as tensions B1 Correct labels 2
10
4762 Mark Scheme January 2011
Q 3 m a r k notes
(iv) M1 Equilibrium attempted at a pin-joint M1 Equilibrium attempted at a 2nd pin-joint M1 Either Equilibrium equation for 2nd direction at a pin-joint or 3rd pin-joint considered At C B1 At least 3 equations of resolution correct or follow through CD cos30 45 0T so CD 30 3T and force in CD is 30 3 N (T) A1 BC CD cos 60 0T T so BC 15 3T and force in BC is 15 3 N (C) F1 At D BD CDcos30 cos30 0T T so BD 30 3T and force in BD is 30 3 N (C) F1 AD BD CDcos 60 cos 60 50 0T T T so AD 50 30 3T and the force in AD is 50 30 3 N (T) F1 At A AB cos30 90 0T so AB 60 3T and the force in AB is 60 3 N (C) F1 B1 At least 4 T/C correct 10 (v) The equilibria at C depend only on the Resolve in two directions at C and obtain same results as in (iv) M1A1 framework geometry and the 45 N. E1 These are not changed so forces in CB and CD are not changed E1 2 19
11
4762 Mark Scheme January 2011
Q 4 m a r k notes
(i) (2, 2.5) B1 Condone writing as a vector 1 (ii) By symmetry, 2.5y B1 Some justification needed
For x : 1 15 5 6 5 5 6 22 2 2
hh x h
M1 These next 4 marks may be obtained from correct FT of their “2” from (i)
A1 1st term RHS correct (allow sign error) A1 Either other term correct A1 All correct so 25 15 2.5 30h x h so 25 3 2.5(12 )h x h
and 212
2( 3)hx
h
E1 Clearly shown, including signs.
6 (iii) Need 0x M1 Allow 0x or = 0
So 212 0
2( 3)h
h
Hence 12 – h² > 0 Since h > 0, 0 2 3h B1 2√3 or - 2√3 oe seen A1 Accept only +ve root mentioned. WWW for signs Accept 2 3h as answer strict inequality for final A mark 3
12
4762 Mark Scheme January 2011
13
Q 4 m a r k notes
Q4 continued (iv) When h = 3, 0.25x B1 Could be scored in (v) Let mag of vert force be T N a.c moments about axis thro’ O T × 6 – 15×0.25 = 0 M1 If moments about another point need all relevant forces. Allow sign errors. Condone use of 15g so T = 0.625 so 0.625 N A1 cao 3 (v) Let magnitude of force be U N a.c. moments about axis thro’ D U cos30 × 5 – 15 × (3 + 0.25) = 0 M1 Each term must be a moment. If moments about another point need all relevant forces.
Condone use of 15g . B1 moment of U (5Ucos30 or …) oe A1 (3 + 0.25) oe U = 11.25833… so 11.3 N (3 s. f.) A1 cao 4 17
GCE
Oxford Cambridge and RSA Examinations
Mathematics (MEI) Advanced GCE
Unit 4762: Mechanics 2
Mark Scheme for June 2011
4762 Mark Scheme June 2011
Q 1 mark notes
(a) (i) 13T = 10(4.75 – (– 1.75)) M1 Use of I = Ft . Allow sign errors A1 Signs correct on RHS so T = 5. So 5 s. A1 cao OR: 13 = 10a
4.75 ( 1.75)5
1.3T
B1 M1 A1
N2L Use of suvat cao
3 (ii) PCLM: P+Q10 4.75 15 0.5 25v M1 PCLM with combined mass. Allow sign errors P+Q 1.6v so 1.6 m s -1 in +ve direction A1 No need for reference to direction 2 (iii) PCLM: Q10 4.75 15 0.5 10 1 15v M1 PCLM with all correct terms. Allow sign errors A1 Any form Hence and Q has velocity 2 m sQ 2v -1 A1 Accept no direct reference to direction
NEL: Q 10.5 4.75
ve
M1 NEL. Accept their and any sign errors. Fraction must be correct way up Qv
A1 Any form. FT their . Qv so e = 0.19047… so 0.190 (3 s. f.) A1 cao accept 0.19 , 4/21 accept 0.2 only if 0.19 seen earlier 6
1
4762 Mark Scheme June 2011
(b) Initial vert cpt is 14sin30 = 7 B1 1st hits ground at v given by 2 27 2 9.8 3.125v M1 Appropriate suvat. Allow ±9.8 etc Condone u =14 v = 10.5 A1 Vert cpt after 2nd bounce
10.5 × 0.6² M1 their for n = 1, 2 or 3 Condone use of their initial vertical component. Do not award if horiz component is also multiplied by 0.6
10.5 0.6n
B1 use of × 0.6² or attempt at two bounces with 0.6 used each time Horiz cpt is unchanged throughout
(14cos30 ) B1 Award even if value wrong or not given
Angle is
210.5 0.6arctan 17.31586...14cos30
M1 FT their horiz and vert components. oe. Fraction must be for correct angle.
so 17.3° (3 s. f.) A1 cao SC answer of 11.7 will usually earn 5/8 8 19
2
4762 Mark Scheme June 2011
Q 2 mark notes
Penalise answers to fewer than 4sf only once (i) cw moments about A Let force be S 600 0.8 2 0S M1 Moments. All forces. No extras A1 S = 240 so 240 N vertically upwards A1 Need statement of direction or diagram 3 (ii) cw moments about A Let tension be T M1 Moments. All forces. No extras. Attempt at moment of T (need not be resolved) Note that mmts about B needs
forces at hinge. 600 0.8 sin 50 0.3 0T M1 Correct method for moment of T. Allow length errors and s c A1 Moment of T correct (allow sign error) A1 All correct T = 2088.65… ( 1600
sin 50 ) so 2089 N (4 s. f.) A1 cao 5 (iii) Resolve X – Tcos50 = 0 M1 Resolving horiz. Allow sign error. T must be resolved, allow s c so X = 1342.55…. = 1343 (4 s. f.) F1 FT their T only. Allow 1600cot50 Resolve Y – Tsin50 + 600 = 0 M1 NB other methods possible so Y = 1000 F1 FT their T only Method for either R or M1 M dependent on attempts at X and Y using moments/resolution 2 2 2 = 1674.05.. 1600 cot 50 1000R so 1674 (4 s. f.) F1 FT their X and Y Numerical value only 1000arctan
1600cot 50
= 36.6804… so 36.68° (4 s. f.) F1 FT their X and Y Numerical value only Accept 36.67 7 (iv) Angle GAP is above so 36.68° (4 s. f.) B1 Weight, T and R are the only forces acting on E1 Must be clear the beam which is in equilibrium. Hence they are concurrent. Or geometrical calculation 2 17
3
4762 Mark Scheme June 2011
Q 3 mark notes
(i) 1 11 1
2 22 21 12 2
1 210 4 2 3
3 22 3xy
M1 Correct method clearly indicated for x or y component.
B1 If 2D method, at least 1 mass + cm correct for a region. If separate cpts, at least 2 mass + cm correct for one of the cpts
=
1 12 2
1 12 2
2 1 1 7 88 6 3 7 25
E1 Working shown. Either expression shown oe
so
0.82.5
xy
and c.m. is (0.8, 2.5)
E1 Both
4 (ii) c.w. moments about J H3.2 1.8 4 0T B1 Use of 1.8 oe M1 A moments equation with all relevant forces. Allow use of 10 instead of 3.2 so and the force at H is 1.44 N H 1.44T A1 Resolving M1 Or moments again force at J is 3.2 – 1.44 = 1.76 N F1 Only FT if positive final answer 5 (iii) below
4
4762 Mark Scheme June 2011
(iii) 1 1
2 212
12
0 2 210 4 2 2 3 2 3 2 3
0 0 1
xy
z
M1 Dealing with 3D
B1 Dealing correctly with one folded part B1 Dealing with the other folded part
= 0 1 4 5 108 6 7 6 272 0 0 2 0
E1 Working shown. Either expression shown oe
so 1
2.70
xyz
and c.m. is (1, 2.7, 0)
E1 All three components
5 (iv) B1 Recognising that cm is vertically below O (may be implied)
Let angle IOG be B1 Correctly identifying the angle 1
2.7tan M1 Accept 2.71tan oe
so angle is 20.323... so 20.3° (3 s. f.) A1 Do NOT isw 4 18
O
G
I1
2.7
5
4762 Mark Scheme June 2011
6
Q 4 mark notes
(a) 2 21
2 80 6 V M1 WE equation. Allow GPE OR init KE term omitted or wrong. Allow sign errors. There must be 3 terms one of which is the WD term
= 80 × 9.8 × 1600 – 1300000 B1 KE terms correct (accept 40 × (V² – 6² )) B1 GPE term. Allow sign error A1 All terms present. Accept only sign errors, but not the 1300000 and 80x9.8x1600 terms with same sign so V = 34.29285… so 34.3 m s -1, (3 s. f.) A1 Cao accept 14 6 5 (b) (i) N2L up the slope. Driving force is S N S – 1150 – 800 × 9.8 × 0.1 = 800 × 0.25 M1 N2L. Allow either resistance or weight cpt omitted. Allow weight not resolved and sign errors. B1 RHS correct M1 Attempt at weight cpt (800gsinθ is sufficient) Allow missing g A1 Weight cpt correct (numerical) May be implied S = 2134 E1 Power is 2134 × 8 M1 Use of P = Fv = 17072 so 17.1 kW (3 s. f.) A1 7 (ii) Let resistance on sledge be F N N2L up slope for sledge 900 – F – 300 × 9.8 × 0.1 = 300 × 0.25 M1 Need non-zero accn, correct mass and 900. Allow weight missing or unresolved and allow sign errors. Do not award
if 2134 included so F = 531 A1 normal reaction is 300gcosθ B1 Use cosθ = 0.99 or cos5.7 B1 In context 531
300 9.8 0.99
M1 Use of F R for any F and R but not F=900
= 0.181522… so 0.182 (3 s. f.) A1 cao 6 18
4762 Mark Scheme June 2012
Question Answer Marks Guidance 1 (a) (i) KE change: 2 21
2 0.6 7.5 5.5 M1 Difference of two KE terms
= 7.8 J A1 GPE change: 0.6 × 9.8 × 1.5 = 8.82 J B1 Allow –8.82J [3] 1 (a) (ii) W is work done against resistance 7.8 = 8.82 – W M1 W–E all terms. Allow sign errors so W = 1.02 J A1 FT (i) only. Also FT only if mod (their KE) < mod (their PE) 1.02 gets M1A0; 16.62 gets M1A0 [2] 1 (a) (iii) Average resistance is F so F × 1.5 = 1.02 M1 Use of WD = Fs OR find a = 8.667 and use F = 0.6g – 0.6 × 8.667 so F = 0.68 A1 May be implied. FT (ii) Power is 0.68 × 5.5 M1 Use of P = Fv any calculated F = 3.74 so 3.74 W A1 cao [4] 1 (b) (i) R = mgcos 40 B1 Seen or implied Fmax = mgsin 40 B1 Seen or implied Fmax = μR M1 Use of F = μR : substitute F and R
so sin 40 tan 40cos 40
mgmg
E1 This is the minimum amount of working needed to earn the E1 Must see explicit evidence of method
[4] Note: g omitted, treat as MR
1 (b) (ii) EITHER
tan 40 × 0.8 × 9.8 × cos 20 × 3 (= 18.545)
B1 M1
Use of Fmax = μR with tan 40 and cos 20 Use of WD = Fs NOTE: This mark may be awarded here or for use in PE term
(+)0.8 × 9.8 × 3 sin 20 (= 8.044)
B1 Use of mgh Allow sin cos interchange
M1 Two relevant terms added = 26.5897… so 26.6 J (3 s.f.) A1 Cao Allow 26.7 Allow 27 Omission of g can get B0M1B1M1A0
5
4762 Mark Scheme June 2012
Question Answer Marks Guidance OR
tan 40 × 0.8 × 9.8 × cos 20 (= 6.182) B1 Use of Fmax = μR with tan 40 and cos 20 (+) 0.8 × 9.8 × sin 20 (= 2.68) B1 Allow sin cos interchange (= 8.8632444…) M1 Two relevant forces added WD is 3 × 8.8632444… M1 Use of WD = Fs (for at least one of forces)
= 26.5897… so 26.6 J (3 s.f.) A1 cao [5] Omission of g can get B0B1M1M1A0 2 (i) a.c. moments about B 10TC – 15× 2 = 0 M1 Moments with all forces present, no extra forces. so TC = 3. Tension at C is 3 N A1 TC + TB – 15 = 0 M1 May take moments again so TB = 12. Tension at B is 12 N F1 [4] 2 (ii) a.c. moments about A 25Tsin 30 – 15 × 17 = 0 M1 Attempt at moments with resolution; allow cos sin error. All forces
present, no extra forces
so T = 20.4 A1 cao At A Let force be Y N Y + T sin30 – 15 = 0 so Y = 4.8 B1 FT (can take moments about C) X = T cos 30 = 17.6669… N B1 FT Need not be evaluated 2 24.8 ( cos30)T M1
= 18.3073755… so 18.3 N (3 s.f.) A1 cao [6] 2 (iii) Let force be P. a.c. moments about D. 8 × 15 – 12 × = 0 P M1 Moments about D with all forces present, no extra forces so = 10 on point of tipping P A1 cao Using maxF R on point of slipping M1 with R = 15 B1 gives maxF = 0.65 × 15 = 9.75 A1 cao so slips first E1 cao and WWW [6]
6
4762 Mark Scheme June 2012
Question Answer Marks Guidance 3 (a) (i) B1 Correctly identifying the position of the c.m of triangle EFH (10, –4) 6 4 10
300 72 192 363 6 4
xy
M1 A systematic method for at least 1 cpt
B1
Either all x or all y values correct or 2 vector terms correct or allow one common error in both components, e.g. one wrong mass, misunderstanding of c.m. of triangle
6961080
xy
so 2.32x A1 3.6y A1 Allow FT for either if only error is common to both [5] 3 (a) (ii)
centre of mass is at G
M1* Identifying correct angle. May be implied B1 At least 1 relevant distance found. FT (i) tan 9.6
14.32α M1dep* Use of arctan 9.614.32 or arctan 14.32
9.6 o.e. so 33.8376...α so 33.8° (3 s.f.) A1 cao or 180 – 33.8 [4] 3 (b) (i) Marking given tension and thrust B1 Each labelled with magnitude and correct direction Marking all other forces internal to rods
acting on A, B and C (as T or C) B1 Need ALL forces at A, B and C. Need pairs of arrows on AB, AC and BC
[2]
12 C B 2.32α
6O
3.6G
7
4762 Mark Scheme June 2012
Question Answer Marks Guidance 3 (b) (ii) Equilibrium at A AB cos30 18 0T M1 Equilibrium at one pin-joint
AB 12 3T . Force in AB: 12 3 N (T) A1 20.8 Sign consistent with tension on their diagram
A AC AB cos60 5 0T T M1
AC (5 6 3)T . –15.39
Force in AC: ( 5 6 3 ) N (C) F1 FT their ABT
At B in direction AB BR ABcos60 0T T
so 24 3BRT M1 Allow FT Other methods are possible, but award this M1 only for a complete method that would lead to BCT
At B in direction BC BC BR cos30 0T T BC 36T . Force in BC: 36 N (T) F1 A1 cao WWW T/C all correct [7] 4 (i) 26t = 3× 13 M1 Use of Ft = m(v – u) or N2L to find a (= 26/3) and use v u at t = 1.5 so 1.5 s A1 cao [2] 4 (ii) PCLM 10 × 0 + 3 × 13 = 10vQ + 3vP M1 Use of PCLM 39 = 10vQ + 3vP A1 Any form NEL Q
0 13Pv v
e
M1 Use of NEL. Allow sign errors but not inversion
Q 13Pv v e A1 Any form M1 Eliminating one of vQ or vP OR allow substitution of given result in one
equation and check both answers in other equation
vQ = 3(1 + e) B1 cao; aef vP = 3 – 10e E1 Properly shown [7]
8
4762 Mark Scheme June 2012
9
Question Answer Marks Guidance 4 (iii) Need vP < 0 so 3 – 10e < 0 M1 Accept Hence 3
10 1e A1 cao (Allow 1e omitted) Correct answer www gets 2/2 [2] 4 (iv) When 3
10e , its speed is 10e – 3 M1 We require (10e – 3) > 3(1 + e) M1 FT their SC1 for (3 – 10e) > Qv 3(1 + e)
A1 FT their Qv
so 7e > 6 and so 67 1e A1 cao. Allow 6
7e (0.857) Correct answer www gets 4/4 [4] 4 (v) Either
vQ = 4.5 and vP = – 2 M1 Substitute e = 0.5; FT their Qv
When they collide the speed of Q is – 4.5 and of P is 2 M1 Change signs of their velocities
PCLM 10 × –4.5 + 3 × 2 = 13V M1 Use of PCLM Allow sign errors so V = –3 and velocity is –3 m s -1 A1 cao; OR 3 m s -1 to the right or use argument about final LM is –ve of original LM [4]
Or 10 3 1 3 10 3 13e e V
M1
Use of PCLM; Allow sign errors ; FT their Qv
M1 Change signs of their velocities 39 13V M1 Simplify so V = – 3 and velocity is – 3 m s -1 A1 cao; OR 3 m s -1 to the right [4] 4 (vi)
3( – 3 – 2) = – 15 N s B1 FT Using 10(–3 +4.5) = 15 gets B0 until it leads to correct answer
3(their( ) 2)v
[1]
4762 Mark Scheme January 2013
6
Question Answer Marks Guidance
1 (a) Take j north and i east
velocity: before 5i - 5√3j (after 3i) B1
Resolving initial velocity (may be implied). Allow 5i +5√3j or 5i - 5√3j
oe
I = m(v – u) M1
May be implied Allow if only one direction considered or both combined
without vectors. Must include an attempt to resolve 10
so I = 120 000 000(– 2i + 5√3j) A1 Accept mass of 120 000
Modulus is 120 000 000 × 8.888194…
= 1.0665… × 10 9
N s
so 1.07 × 10 9
N s (to 3 s. f.) A1 cao
Alternative method using a diagram, cos and sine rules
[4]
1 (b) (i) PCLM
0.4 × 6 = 0.5 V M1 Implied by 4.8 or -4.8
V = 4.8 ms-1
direction is opposite to that
of P A1 Allow -4.8 as the speed
[2]
1 (b) (ii) P travels 6 × 2
3 = 4 m before the collision B1 Or find
13
24
t for time from edge to collision AND 3.25d
so Q travels 4 – 2 × 0.75 = 2.5 m in 23
s B1 3.25 - 0.75 = 2.5
Q(4.8 ) 22.5
2 3
v M1 Using appropriate suvat FT their 2.5
Hence vQ = 2.7 ms-1
E1 Answer given
[4]
4762 Mark Scheme January 2013
7
Question Answer Marks Guidance
1 (b) (iii) Suppose friction on Q is F
– F × 23
= 0.5(2.7 – 4.8) so F = 1.575 B1 Using Ft = m(v – u) or find a = - 3.15 and use F = ma. FT their 2.7
1.575 = × 0.5 × 9.8 M1 F = R
A1 R correct (4.9)
= 0.32142... so 0.321 (3 s. f.) A1 cao
[4]
Note: F and R need not be explicit:
F=ma and R=mg give a
g (M1A1). Find a = -3.15(B1) gives 0.321
(A1)
1 (b) (iv) Let the speeds after be PV and QV .
PCLM
0.4 × 6 + 0.5 × 2.7 = 0.4 PV + 0.5QV M1 PCLM. FT their 2.7 from (ii). Award M1A0 for use of their 4.8 from (i)
instead of 2.7
so 4 PV + 5QV = 37.5 A1 FT their 2.7 from (ii). Accept any form
NEL
Q P 1
2.7 6 8
V V
M1 NEL. FT their 2.7 from (ii). Award M1A0 for use of their 4.8 from (i)
instead of 2.7
so Q P 0.4125V V A1 FT their 2.7 from (ii). Accept any form
QV = 4.35 so 4.35 m s
-1 A1 cao
[5]
4762 Mark Scheme January 2013
8
Question Answer Marks Guidance
2 (i) 455 = 0.01× 80 × 9.8 × cos4 × 12 + WD M1 Use of Fx
B1 rolling friction force correct (7.82) 12 not needed
A1 All correct terms in an equation (allow sign errors)
WD = 361.149… so 361 J (3 s. f.) A1 cao
[4]
SC B1B1 for final answer 30.1 seen
2 (ii) 0.5 × 80 × v² – 0.5 × 80 × 2² M1 Use of W-E equation. Must include GPE, at least one KE and the WD
B1 Either KE term
= 80 × 9.8 × 12 × sin 4 – 455 B1 GPE term (656.27)
A1 All correct terms in an equation (allow sign errors)
v = 3.0052.. so 3.01 m s -1
(3 s. f.) A1 cao
[5]
2 (iii) Using N2L with driving force S M1 N2L with at most one force term missing
S – (15 + 0.01 × 80 × 9.8 × cos 5) B1 Both resistance terms seen (15 and 7.81)
– 80 × 9.8 × sin 5 B1 Condone wrong sign (68.33)
= 80 × 1.5 A1 All correct terms present; allow sign errors
S = 211.1402… A1 May be implicit
405 = Sv M1 Use of Power = Sv with any S calculated using N2L
so v = 1.918… so 1.92 m s -1
(3 s. f.) A1 FT their S
[7]
Note: missing out one term in N2L can earn 4/7 (M1B1B0A0A0M1A1)
4762 Mark Scheme January 2013
9
Question Answer Marks Guidance
3 (i) 1 1.5 4 4.515 2 9 2 2
2.5 1.5 0.5 1
x
y
M1 A systematic method for at least 1 cpt
A1 Either all x or all y values correct or 2 vector terms correct on RHS
A1 Completely correct expressions seen for all components
=
28.5
17.5
Need not be explicit
so 1.9x A1 Accept any form
7
6y A1 Accept any form (1.17, 1.2) but not 1.16
[5]
3 (ii) Referred to Fig 3.1 with c.m. G, G is 2 +
1.9 to the right of K and 3 – 7
6 = 11
6
below K
B1 FT from (i) May be implied
When hanging, G is vertically below K B1 May be implied
Angle is
116arctan
3.9
M1 o.e. FT their values but must be attempting to find the appropriate angle
= 25.1775… so 25.2° (3 s. f.) A1 cao
[4]
3 (iii) New c.m. is at (1.5, 1.5) & mass of
object is 0.3 kg
B1 Do not penalise below if mass of lamina is taken to be 15
For x : M1
Recognising need first to produce an equation in terms of m for the x-
component
(0.3 + m) × 1.5 = 0.3 × 1.9 + m × 0 M1 Must be 0 not x
m = 0.08 A1 FT their 1.9 from (i). If 15 used, accept m = 4
For y : (0.3 + 0.08) × 1.5 = 0.3 × 7
6 +
0.08y M1
so particle should be at (0, 2.75) A1
cao. Condone no reference to x component. Allow obtained using 15.
Allow 2.74, 2.7375 (from 1.17), 2.775 (from 1.16), 2.625 (from 1.2)
[6]
4762 Mark Scheme January 2013
10
Question Answer Marks Guidance
3 (iv) The c.m. must lie inside KFDL as seen in
the plan in Fig. 3.2
E1 Some indication of this is what is required. Accept a closed region with
KF correct and sides parallel to KL and FD.
E1 Correct. Accept freehand.
The c.m. shown to be in this region M1 Recognition that com is at (1.7, their y ) and is related to their critical
region even if region is incorrect
or calculation with at least 1 correct equation ( 3 2 9 and 3 4 6y x y x )
Do NOT award simply for a recalculation of com as (1.7, 7/6)
E1 Properly established including a statement. (i.e. correct region, correct
com marked and statement of stability)
[4]
4762 Mark Scheme January 2013
11
Question Answer Marks Guidance
4 (i) Let vertical force from support be R N
and tension in string T N.
moments about A
30 × 0.5 × 2.4 – R × (2.4 – 0.6) = 0 M1 Use of moments with all relevant moments attempted
R = 20 so force from block is 20 N A1 (FT from T if T found first)
R + T – 30 = 0 M1
T = 10 so tension is 10 N F1 FT from R
[4]
4 (ii) (A)
sin cos 0R F M1
A1 Must be consideration of a force at A
As on the point of slipping F = 0.6R M1 F and R must be identified, e.g. on a diagram
so sin 0.6 cosR R so sin 0.6cos M1
and tan 0.6 E1 Complete argument
[5]
OR sin sinF mg S M1 Resolve parallel and perpendicular to rod
cos cosR mg S A1 Both correct
As on the point of slipping F = 0.6R M1 F and R must be identified, e.g. on a diagram
sin sin
cos cos
mg SF
R mg S
M1 Divide factored expressions with S included
tan 0.6 E1
[5]
R N
S N
30 N
0.9 m 1.2 m
A
B
F N P
4762 Mark Scheme January 2013
12
Question Answer Marks Guidance
4 (ii) (B)
AP is 1.5 gives sin = 0.6 or cos =
0.8 B1 oe. or 36.9
c. w. moments about A
1.5 ' 30 1.2 cos 0R M1 Moments and all terms present. Accept cos or 0.8
R’ = 19.2 so 19.2 N A1 cao
' 'cos 30 0S R
M1
An equilibrium equation with all relevant forces, resolved appropriately,
e.g. cos 30cos sin R S F . Allow sin cos
( S’ = 14.64) A1
Correct equation involving only S’. Numerical answer not required
'sin ' 0R F M1
Second equilibrium equation with all relevant forces, resolved
appropriately.
e.g. cos sin 30sin F S . Allow sin cos
( F‘ = 11.52) A1 Correct equation involving only F’. Numerical answer not required
11.52
14.64 M1 Use of ' 'F S for a calculated F‘ and S’
= 0.78688… so 0.787 (3 s. f.) A1 cao
[9]
R’ N
S’ N
30 N
0.9 m 1.2 m
A
B
F’ N
P
0.3 m
4762 Mark Scheme June 2013
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Question Answer Marks Guidance 1 (a) (i) 3 × 4 + 21 × 2 = 4U M1 Use of PCLM and I = Ft 4U = 54 so U = 13.5 and speed is 13.5 m s–1 A1 [2] OR 21 4 : 5.25a a and 3 2 5.25v M1 Use of F ma and suvat speed is 13.5 m s–1 A1 [2] 1 (a) (ii) Let V be the speed of S in direction PQ 54 – 2 × 3 = (4 + 2)V M1 PCLM for coalescence 6V = 48 so V = 8 and velocity is 8 m s–1 in direction PQ E1 Answer given. Accept no reference to direction. [2] 1 (a) (iii) Let velocities of R be u before and v after, both in the
direction SR
6 × 8 + 4u = 6 × 5 + 4v M1 Use of PCLM. Allow any sign convention. All masses and speeds must be correct.
v – u = 4.5 A1 Any form. 5 1
8 4vu
M1 Use of NEL correct way up; allow sign errors
4v + u = 28 A1 Any form signs consistent with PCLM eqn Solving u = 2 so 2 m s–1 in the direction SR A1 cao NOTE that a sign error in NEL leads to u =-2; this gets A0 v = 6.5 so 6.5 m s–1 in the direction SR A1 cao. Withhold only 1 of the final A marks if the directions not clear. [6] Directions can be inferred from a CLEAR diagram
4762 Mark Scheme June 2013
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Question Answer Marks Guidance
1 (b) (i) Find v, the speed at which particle hits the plane ½ × 0.2 × v² – ½ × 0.2 × 5² = 0.2 × 10 × 10 M1 Use of WE or suvat must use distance of 10 allow g = 9.8 so v² = 225 and v = 15 A1 Answer not required (v = 14.9 if g = 9.8)
cos = 45
, sin 35
B1 Use of either expression or use of 36.9
Let velocity after be at to the plane Parallel to the plane 15cos= 13cos M1 Attempt to conserve velocity component parallel to plane.
Allow use of 5 instead of 15
So cos= 1213
and = 22.61.. so 22.6° (3 s. f.)
A1 ( 23.8 if g =9.8)
Perpendicular to the plane: 13sin = e × 15sin M1 Attempt to use NEL perpendicular to plane: Allow use of 5 instead of 15 or use tan = etan
sin = 5
13
A1 o.e. find tan = 5
12
so 5 313 15
13 5e and 5
9e A1 cao Accept 0.56 (e = 0.589 if g =9.8)
[8]
OR: First three marks as above Parallel to plane, 15cos ( 12)xu and
( 12)x xv u
M1A1B1
M1
Attempt to conserve velocity component parallel to plane. Allow use of 5 instead of 15
12cos13
xvv
22.6 A1
Perpendicular to plane, 15sin ( 9)yu and
( 9 )y yv eu e
M1 Attempt to use NEL perpendicular to plane. Allow use of 5 instead of 15
2 2 213x yv v A1 Use Pythagoras' theorem for velocities after collision in attempt to find e
22 212 9 13e so 2 25 5
81 9e e
A1
[8]
4762 Mark Scheme June 2013
7
Question Answer Marks Guidance 1 (b) (ii) Impulse is perp to plane with mod 0.2(13sin ( 15sin )) = 0.2(5 – ( – 9) ) M1 For use of I = m (v – u) perp to the plane 0.2(5-9) gets M1A0 = 2.8 N s A1 cao [2] 2 (i) WD is 800 × 9.8 × 6 + 400 × 6 J M1 WD as Fd Used in TWO terms = 49 440 E1 Power is 49440 ÷ 12 M1 Power is WD / t = 4120 W A1 cao [4] 2 (ii) Power is (800 × 9.8 + 400) × 0.55 M1 Power as Fv in one term A1 All correct = 4532 W A1 cao [3] 2 (iii) Let speed be v 21
2 800 800 9.8 3 400 3v M1 Use of W-E equation Must include KE and at least one WD term A1 Allow only sign errors A1 All correct 2 55.8v so v = 7.4699…
and speed is 7.47 m s -1 (3 s.f.) A1 SC: Use of N2L and suvat : M1 Complete method A1 7.47 cao
[4] 2 (iv) 2
21 12 2800 800
4v v M1 Use of W-E equation Must include 2 KE terms and a WD term
B1 Final KE term correct. FT their v. = (800 × 9.8 – 400) × 0.8 B1 One correct WD term – WD A1 All terms present. Allow sign errors and FT their v.
WD is 22 692 so 22 700 J (3 s. f.) A1
cao SC Use of N2L and suvat: Award maximum of B1 for 'Average force (28365) x 0.8'
[5]
4762 Mark Scheme June 2013
8
Question Answer Marks Guidance
Condone using cm not m in moments in any part if consistent 3 (i) c.w. moments about A
60cos40 0.3 60sin 40 0.1
M1
oe e.g. 60 0.3 0.1tan 40 sin50 or 160 cos 90 arctan 3 4010
Method of dealing with moment of weight. Allow cos sin A1 Both weight terms correct. Allow wrong overall sign but not both terms
with the same sign = 9.93207… so 9.93 N m (3 s. f.) E1 [3] 3 (ii) cos40 0.2 9.93207... 0P M1 Moments of all relevant forces attempted. No extra terms. Allow
cos sin P = 64.827… so 64.8 (3 s. f.) A1 cao (64.813… if 9.93 used) [2] 3 (iii) a.c. moments about A to find NR, R, at B 0.8 9.93R
or 0.8 60sin 40 0.1 60cos 40 0.3 0R
M1 Attempt to use moments to find R. Moments of all relevant forces attempted. No extra terms. Allow cos sin Note that mmts about B can score M1 only if mmt of horiz compt of force at A is included. If R is taken as vertical, M0
A1 FT their moment of weight from (i) R = 12.4150… Not a required answer Resolve vertically Y – 60 + Rcos 40 = 0 depM1 Note that the second M mark awarded in this part must be for a complete
method to find Y: so Y = 50.489… so 50.5 N (3 s. f.) A1 FT their calculated R [4]
4762 Mark Scheme June 2013
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Question Answer Marks Guidance
3 (iv) resolve perp to plane 60cos40 200sin 40 0R M1 All terms present and no extra terms. Components of 60 and 200;
allow cos sin A1 R = 174.52… Not a required answer N2L up the plane 60200cos 40 60sin 40 1.75
9.8F M1 Use of N2L with all terms present and no extras. Components of 60
and 200; allow cos sin Allow use of 60 for mass B1 Use of mass not weight A1 FT use of weight and/or sign errors F = 103.927… A1 All correct. Not a required answer As friction limiting F R so 103.927...
174.520... M1
FT their F and their R
= 0.59550.. so 0.596 (3 s. f.) A1 cao [8]
4762 Mark Scheme June 2013
10
Question Answer Marks Guidance
4 (a) (i) Write d = 0.8
2.5 1.2 1.3 2.4x
dy
M1 Method for c.m (length is 7.4 m, mass is 5.92 kg)
=
1.2 2.4 1.8 1.22.5 1.2 1.3 2.4
0.35 0.1 0.25 0d d d d
B1
One rod mass and cpts correct or if done by separate x and y equations 2 rod components and masses correct. (Allow length used instead of mass)
OR: 2 0.96 1.04 1.92
xy
=1.2 2.4 1.8 1.2
2 0.96 1.04 1.920.35 0.1 0.25 0
B1 Another rod dealt with correctly or if done by separate x and y equations, the other equation attempted with 2 rod components and masses correct. (Allow length used instead of mass)
3 2.88 2.34 2.88 11.17.4
0.875 0.12 0.325 0 0.67xy
OR: 2.4 2.304 2.304 1.872 8.88
5.920.7 0.096 0.26 0 0.536
xy
1.5x E1 Clearly shown, with at least one intermediate step 0.090540...y = – 0.0905 (3 s. f.) A1 Condone - 0.09 [5]
4762 Mark Scheme June 2013
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Question Answer Marks Guidance 4 (a) (ii) EITHER: New c.m. has 1.2x M1 Identifying and using a suitable condition. (5.92 + m)× 1.2 = 5.92 × 1.5 + m × 0 M1 Complete method m = 1.48 A1 cao [3] OR: Moment about any point is zero M1 Identifying a suitable condition. e.g. about S: 1.2 0.3 5.92mg g M1 Allow g omitted. Correct number of terms must be included m = 1.48 A1 cao [3] 4 (b) (i) Consider the equilibrium at R Resolving horizontally gives TQR = 0 E1 Then resolving vertically gives TOR = 0 E1 [2] 4 (b) (ii) c.w. moments about O 120 1 60 2 3T M1 May also be argued by first considering internal forces so T = 80 A1 Resolve to give X = 80 and Y = 180 A1 FT X = T. Only Y =180 scores 0 [3] 4 (b) (iii)
B1 All correct. Accept T, X and Y labelled but not substituted. Accept mixes of T and C. Require pairs of arrows with label on OQ, OP and PQ.
[1] Forces internal to the rods have been taken to be tensions. 4 (b) (iv) Take angle OPQ as At P OP60 sin 0T M1 Equilibrium at ANY pin-joint (not R) A1 Correct equation(s) that leads directly to finding OPT or QPT 3sin
13 : 56.3
OP
60sin
T
= 20 13 so 20 13 N (C) A1 o.e. Accept 72.1 N
At P QP OP cos 0T T M1 A second equilibrium equation leading to a second internal force so QP 40T so 40 N (T) A1 cao T/C correct for both rods [5]
4762 Mark Scheme June 2014
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1. Annotations and abbreviations
Annotation in scoris Meaning
Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response.
and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme
Meaning
E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph M1 dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working
4762 Mark Scheme June 2014
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2. Subject-specific Marking Instructions for GCE Mathematics (MEI) Mechanics strand a Annotations should be used whenever appropriate during your marking.
The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.
b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.
c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result.
4762 Mark Scheme June 2014
3
Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.
f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.)
We are usually quite flexible about the accuracy to which the final answer is expressed and we do not penalise over-specification.
When a value is given in the paper Only accept an answer correct to at least as many significant figures as the given value. This rule should be applied to each case.
When a value is not given in the paper Accept any answer that agrees with the correct value to 2 s.f.
4762 Mark Scheme June 2014
4
ft should be used so that only one mark is lost for each distinct error made in the accuracy to which working is done or an answer given. Refer cases to your Team Leader where the same type of error (e.g. errors due to premature approximation leading to error) has been made in different questions or parts of questions. There are some mistakes that might be repeated throughout a paper. If a candidate makes such a mistake, (eg uses a calculator in wrong angle mode) then you will need to check the candidate’s script for repetitions of the mistake and consult your Team Leader about what penalty should be given.
There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question.
g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.
If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.
h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.
i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader.
j If in any case the scheme operates with considerable unfairness consult your Team Leader.
4762 Mark Scheme June 2014
5
Question Answer Marks Guidance
1 (a) (i)
B1 Accept V in either direction. Given velocities and masses must be
correct.
PCLM + ve
5 30 5 30
3 2
u uu V M1 PCLM. Allow sign errors only
so 12
uV A1
Award even if direction of V used in PCLM does not match their
diagram, so or 12 12
u uwill get this A1
A1 WWW. Direction of V correct (may be implied from diagram).
2 12
3
u u
eu
u
M1 FT their V : allow sign errors, but must be right way up
=
5
5124 16
3
(= 0.3125) A1 cao
[6]
1 (a) (ii) +ve
30
12 3
u u M1 Allow sign errors.
7.5u
Or: find impulse on P and reverse the sign
A1
M1
A1
155
2 2
uu u and 7.5u cao M0A0 unless sign is reversed
Direction must be given (may be implicit from diagram).
[2]
P Q
before
after
5 kg 30 kg
u m s –1 3u m s –1
2u m s –1 V m s –1
4762 Mark Scheme June 2014
6
Question Answer Marks Guidance
1 (b) (i) Either
As the parts move at 90°, PCLM in final directions M1
For 2kg: 5 6sin 2u M1 PCLM
so 35
5 6 2u A1 Any form
and u = 9 A1
For 3kg: 5 6sin 3v M1 PCLM
so 45
5 6 3v A1 Any form
and v = 8 A1
Or PCLM 5 6 2 sin 3 sinu v M1
PCLM Allow cos instead of sin if error in both terms; allow sign
errors; masses need to be there. Award if embedded in vector method
so 3 45 5
30 2 3u v A1 Any form
PCLM
2 cos 3 cos 0u v M1
PCLM Allow sin instead of cos if error in both terms and cos used in
previous PCLM eqn; allow sign errors; masses need to be there.
Award if embedded in vector method
so 345 5
2 3u v A1 Any form
Solving M1 A complete method involving 2 equations each in u and v
u = 9 A1 cao for one of u or v
v = 8 F1 for the other: FT substitution into their eqn
Note: Award SC5 for v = 6, u = 12 (from cos/sin reversal)
Uses velocity instead of mmtum: M0M0M1A0F1 max 2/7
Uses mass in one eqn only: M1A1M0M1A0F1 max 4/7
[7]
1 (b) (ii) KE is
2 2 21 1 12 2 2
2 9 3 8 5 6 M1 M1 for attempt at difference of KE (3 terms of correct form)
= 87 J A1 cao
[2]
4762 Mark Scheme June 2014
7
Question Answer Marks Guidance
2 (i) Let the mass of each face be m
2
5 4 0amz m m M1 Any complete method. Accept no mention of m oe.
A1
so 25az A1 CLOSED/bottomless box is NOT a MR. Mark as per scheme giving
method mark if appropriate: max 1/3
[3]
2 (ii) By symmetry , 0x or by calculation M1
A1
Can be awarded if closed/bottomless box used
2 2
8 5 2a amy m m ma M1 Any complete method. Accept no mention of m oe.
92a so 9
16ay E1 Shown (answer given)
25
8 5 3amz m ma M1 Any complete method. Accept no mention of m oe.
= 5a so 58az A1 cao CLOSED/bottomless box: max M1A1M1A0M1A0
[6]
Alternative form of solution:
0 0 / 2 0
8 0 / 2 / 2
/ 2 0 / 2 / 2
/ 2 / 2 0 / 2
/ 2 / 2 / 2
/ 2
x a
m y m m a m a m a
z a a a
a a a
m a m a m a m a
a a a a
=
0
9 / 2
5
m a
a
so 0,x 9 /16y a , 5 / 8z a
M1
Each coordinate
A1 cao
4762 Mark Scheme June 2014
8
Question Answer Marks Guidance
2 (iii)
B1
B1
G vertically above bottom edge
Use of their z and a y oe.
M1 Use of tan (or equivalent) with either z or a z and y or a y
716
58
tan 0.7a
a
M1
(or equivalent)
so = 34.992... so 35° (3 s. f.) A1 cao.
[5] 55 as answer can get B1B1M1M0A0: 3/5
2 (iv) Friction F N, normal reaction R N Allow 28a as M throughout
cos30R Mg B1
N2L down plane
sin30 2Mg F M M1 Attempt to use N2L with all terms (allow a missing g). Allow sign
errors
A1
F R M1 Used correctly
so
sin 30 2
cos30
g
g = 0.34169…
so 0.342 (3 s. f.) A1
[5]
G
916
aa
58a
4762 Mark Scheme June 2014
9
Question Answer Marks Guidance
3 (a) (i) Vertical through C intersects AB at X
BX = 1 and XA = 3 B1 May be implied
R – Y – 60 = 0 B1 Must have an correct equation involving Y .
ac moments about A
60 × 4 – R × 3 = 0 so R = 80 B1 AG
Y = R – 60 = 20 and X = 0 B1 Both. Can be awarded independent of previous B1
[4] MR-1 for AB = 2
3 (a) (ii) B1
All (8 forces, with labelled pairs of arrows for internal forces) present
and consistent . R and Y can be used
In the solutions below all internal forces are set as
tensions
[1]
3 (a) (iii) For example: B BC60 cos30 0T M1
Attempt an equation for the equilibrium in any direction at any pin-
joint (all correct (resolved) terms present, allow sign errors, s c
so BC 40 3T (Force of 40 3 N (C)) A1 Ignore T/C ; sign of force must be consistent with their T/C convention
A AC20 sin30 0T M1 2
nd equilibrium equation attempted
so AC 40T Force of 40 N (C) A1 Ignore T/C ; sign of force must be consistent with their T/C convention
A AB AC cos30 0T T M1 3
rd equilibrium equation attempted
so AB 20 3T Force of 20 3 N (T) Ignore T/C
All three internal forces correct, including T/C A1
NOTE: Award first A1 for ANY force correct (need not be first one
calculated) Award second A1 for a second force correct, FT if
dependent on first one. Award third A1 as cao for everything correct,
including T/C.
[6]
4762 Mark Scheme June 2014
10
Question Answer Marks Guidance
3 (b) (i) Take force as P to give + ac moment about B
ac moments about B
817
cos B1 Seen or implied, e.g. in cos61.9
2 68 2 cos 102 4 sin 0P
M1
Moments equation with all terms attempted and no extras. Allow
s c and sign errors
Moments about other points must include all relevant forces
A1 Substitution of sin/cos not required
P = 148 A1 cao
[4]
3 (b) (ii) Take Q
ac moments about B
2 sin 68 2 cos 102 4 sin 0Q
M1
Moments equation with all terms attempted and no extras. Allow
s c and sign errors
Moments about other points must include all relevant forces
so Q = 167.7333… so 168 (3 s. f.) F1 FT errors in 2,4,cos,sin, sign from part(i) in 2nd and 3rd terms
Horiz force at B is 102 + 167.733.. B1 Adding. FT their Q
Magnitude is 2 2269.7333.. 68 M1 FT their horizontal force at B; Must use 68
= 278.172… so 278 N (3 s. f.) A1 cao
Alternative for the B1M1A1: finding compts of
force at B along and perpendicular to the rod:
102sin 68cos sin
102cos 68sin cos
Y Q
X Q
187.06; 206.34X Y B1 FT their Q
Magnitude is
2 2187.06 206.34
= 278.172… so 278 N (3 s. f.)
M1
A1
FT their X and Y
cao
[5]
4762 Mark Scheme June 2014
11
Question Answer Marks Guidance
4 (a) (i) KE at A is ½ × 10 × 16.6 2 = 1377.8 J
GPE at B is 10 × 9.8 × 14 = 1372 J M1
Calculate relevant quantities (KE at A and PE at B or v = 1.08 at B or h
= 14.1)
KE at A > GPE at B so gets beyond B E1 Clear argued comparison (e.g. 1377.8 > 1372)
[2]
4 (a) (ii) Let speed at D be v m s -1
Note: No use of friction can get B1 max
M1 Use of WD = 14x
B1 x = 25
A to D: 2 21 1
2 210 10 16.6v
M1
WE equation with at least one KE, GPE and WD by friction terms, all
of correct form
10 9.8 7 25 14 A1 Allow only sign errors
( 2 68.36v )
so v = 8.2680… so 8.27 (3 s. f.)
A1
cao
OR:
B to D: 21
210 1377.8 1372v M1
WE equation with at least one KE, GPE and WD by friction terms, all
of correct form
10 9.8 7 25 14 M1 Use of WD = 14x
B1 x = 25
A1 Allow only sign errors
( 2 68.36v )
so v = 8.2680… so 8.27 (3 s. f.) A1 cao
[5]
4762 Mark Scheme June 2014
12
Question Answer Marks Guidance
4 (a) (iii) Consider only the vertical motion. Suppose the
object hits the ground at V m s -1
and rises h m
2 9.8 7V (11.7) AND 12
2 9.8V h
(5.86) M1 Use of 2 2v gs oe Must be 7 in V. Using '8.27' as u gives M0
M1 e used appropriately: must use their attempt at a vertical velocity
so h = ¼ × 7 = 1.75 A1 cao
[Award SC 2 if 1.75 seen WWW]
[3]
4 (b) Driving force (D) =
50
P B1
Use of P = force x velocity. May be implied e.g. by sight of 0.8P/50 or
0.2P/50 in N2L
P = 50F (D = F) B1 Accept any form
N2L along the road
0.81500 0.08
50
PF M1
Use of N2L with all terms attempted and consistent with power
reduction. Allow sign errors.
so 0.8P – 50F = – 6000 A1 Accept any form
Solving gives
M1
Attempt to solve 2 equations each involving P and F. Dependent on
N2L equation attempted with 3 terms.
F = 600 P = 30 000 A1 cao both
[Taking 80% reduction in P gives P = 7500 and F = 150 for 5/6]
[6]
Oxford Cambridge and RSA Examinations
GCE
Mathematics (MEI)
Unit 4762: Mechanics 2
Advanced GCE
Mark Scheme for June 2015
OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2015
4762 Mark Scheme June 2015
3
Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme
Meaning
E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph M1 dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working
4762 Mark Scheme June 2015
4
Subject-specific Marking Instructions for GCE Mathematics (MEI) Mechanics strand a Annotations should be used whenever appropriate during your marking.
The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.
b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.
c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result.
4762 Mark Scheme June 2015
5
Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.
f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) We are usually quite flexible about the accuracy to which the final answer is expressed and we do not penalise over-specification. When a value is given in the paper Only accept an answer correct to at least as many significant figures as the given value. This rule should be applied to each case. When a value is not given in the paper Accept any answer that agrees with the correct value to 2 s.f. ft should be used so that only one mark is lost for each distinct error made in the accuracy to which working is done or an answer given. Refer cases to your Team Leader where the same type of error (e.g. errors due to premature approximation leading to error) has been made in different questions or parts of questions.
4762 Mark Scheme June 2015
6
There are some mistakes that might be repeated throughout a paper. If a candidate makes such a mistake, (eg uses a calculator in wrong angle mode) then you will need to check the candidate’s script for repetitions of the mistake and consult your Team Leader about what penalty should be given. There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question.
g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.
h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.
i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader.
j If in any case the scheme operates with considerable unfairness consult your Team Leader.
4762 Mark Scheme June 2015
7
Question Answer Marks Guidance
1 (i) Suppose U N at J and V N at point of contact with
cylinder
Taking moments about point of contact with the cylinder
or J
0.8U – 0.2 × 30 = 0 or 0.8V – 0.6 × 30 = 0 M1 A correct moments equation
so U = 7.5 or V = 22.5 A1
Resolve U + V – 30 = 0 so V = 22.5 or U = 7.5
B1 FT use of 1st answer. (Or use moments again)
3
(ii) Taking moments about point of contact with the cylinder
0.4 0.2 30 0W M1 A correct moments equation and reaction at J = 0
so W = 15 A1 Award SC2 for 15 seen WWW 2
(iii) Taking moments about point of contact
0.9cos 30 0.3cos 0S
M1
A moments equation about point of contact: allow sin instead of cos:
Must be trig fn in both terms. Allow slip in distances if clearly
taking moments about point of contact
so S = 10 A1 Award SC2 for 10 seen WWW
cw moments about J M1
Attempt at moments or resolution, involving F or R but not both,
with all appropriate terms present and no extras.
30 0.6cos 0.9 0R A1 No need to substitute for S, if it is present.
so 20cosR A1 Or (30 )cosR S o.e. FT their S
resolve up the rod
Attempt at another moments or resolution, that will enable F or R to
be found, with all appropriate terms present and no extras
sin 30sin 0S F M1
A1 No need to substitute for S, if it is present.
so 20sinF A1 Or (30 )sinF S o.e. FT their S
SC Resolving vert and horiz leads to simultaneous eqns.
M1
A1
One eqn in F and R
M1
A1
Second eqn in F and R AND attempt to solve simult eqns for F or R
20cosR A1 Or (30 )cosR S o.e. FT their S
20sinF A1
Or (30 )sinF S o.e.FT their S
8
4762 Mark Scheme June 2015
8
Question Answer Marks Guidance
(iv) Need F R B1 Use of F R or F R or F R
20sin 20cos M1 Needs an inequality, using their F and R from (iii)
so tan A1 FT incorrect S (Strict inequality gets 2/3)
3
16
2 (i) The system loses 10gh – 8g2h = 6gh or 58.8h J M1 Difference of GPEs
B1 Use of 2h or sin30h (sight of 39.2h)
A1 Clearly shown: Loss must be stated
3
(ii) (A) 212
18 6 1.2v g (= 70.56) M1 Use of WE equation (a KE = a GPE)
B1 Use of ‘18’.
v = 2.8 so 2.8 m s -1 A1 cao 3
OR
10 8 sin30 18g g a M1 Use N2L AND use suvat
B1 Use of ‘18’.
/ 3a g or 3.26: 2 0.8 7.84v g : 2.8v A1 cao 3
(ii) (B) Q passes through point 0.3 m above floor going down at
1.05 m s -1
2 21 12 2
18 18 1.05 6 0.3V g M1
WE with 2 KE terms and a difference in GPE terms: must be using
18; may consider motion of P and motion of Q separately
A1 Allow FT on their 6g
( 2 3.0625V ) and V = 1.75 so 1.75 m s -1 A1 cao
OR
10 8 sin30 18g g a and suvat eqn M1 Use N2L (both particles considered) AND use suvat: must be using
18
/ 3a g (= 3.26) A1 cao
( 2 3.0625V ) and V = 1.75 so 1.75 m s -1 A1 cao 3
4762 Mark Scheme June 2015
9
Question Answer Marks Guidance
(iii) max 0.9 8 cos30F g M1 Attempt to use F R with weight of 8g
A1 May not be evaluated (61.10675…or 35.28 3 )
21max2
18 2 6gh F h M1 WE. Correct KE term (with 18), a GPE term and Fh. Allow sign
errors
A1 All correct
h = 0.30023… so 0.300 m (3 s. f.) A1 cao 5
OR
max 0.9 8 cos30F g M1 Attempt to use F R with mass of 8
A1 May not be evaluated (61.10675…)
10 8 sin30 18F g g a M1 Use N2L for both particles, or combined system, must be 18a; all
terms present and use suvat
6.66a A1 cao
h = 0.30023… so 0.300 m (3 s. f.) A1 cao
5
(iv) Could P stay at rest in equilibrium?
Force up the plane on P is 10g = 98 N. Force down
M1
Attempt to compare forces up and down the plane including friction
plane on P due to its weight is 8gsin30 = 39.2 N
OR: Force needed for equilibrium = 10 8 sin30 58.8g g and
compare with max frictional force 61.1
39.2 + Fmax 100.3 > 98 so yes, stays at rest E1 A clear argument 58.8 is less than 61.1 so equilibrium possible
2
(v) 2× 61.10675… M1 Attempt to use P = Fv with v = 2 and their frictional force
= 122.21… so 122 W (3 s. f.) F1
2
18
4762 Mark Scheme June 2015
10
Question Answer Marks Guidance
3 (i) either Consider a square side 1 removed from a square side a
x y by symmetry B1 Or by calculation
For x
2 2 1
1 12 2
aa a x M1 Allow omitted
A1 Any one term correct
A1 All correct
Hence
321
12 2
aa x At least one intermediate step shown
and
3
2
1
2 1
ax
a
:
2 1
2 1
a ax
a
E1
5
or Splitting horizontally into 2 rectangles
x y by symmetry B1 Or by calculation
11 1 1 1 1
2 2
aaa a a x a a a
M1 Allow omitted. o.e.
A1 Any one term correct
A1 All correct
…. hence
2 11
2 2
aaa x
Dividing by (a – 1) o.e.
and
2 1
2 1
a ax
a
E1 Some simplification shown
[Or splitting vertically into 2 rectangles or…]
(ii) Need
2 11
2 1
a ax
a
B1
so 2 1 2 2a a a and 2 1 0a a M1 oe
Solving for the +ve root gives
1 1 4 1 5
2 2a
A1
3
4762 Mark Scheme June 2015
11
Question Answer Marks Guidance
OR
Substitute expression for a into expression for x and
attempt to simplify M1 Includes an attempt at squaring out.. Decimals used M0
Obtain 1x A1
This is on perimeter of shape B1
3
(iii) a = 4 gives x y = 2.1 B1
Take the mass of the particle to be m
4 2.14 3
0 2.1
xm m m
y
M1 Allow if a not substituted as 4
A1 At least 1 correct term Allow if a not substituted as 4
x = 2.575 A1
y = 1.575 A1 FT for 1 of the final answer marks if there is a single error
G is vertically below A B1 May be implied
Note: Can use cosine rule in triangle AGB:
AG = 3.537, AB = 5.657, GB = 2.124:
M1 for attempt to find all three lengths
A1 all correct
M1 use cosine rule
A1 1.72 cao
2.5752.425
arctan 2.4252.575
45 arctan M1 Correct angle attempted (may be scored below)
M1
Use of arctan oe to find angle OAG
must be using their 2.575 and 4 - their 1.575
A1 Use of appropriate lengths. FT their values.
= 1.71835… so 1.72º (3 s. f.) A1 cao
10
18
A
B
G
O
2.425
1.575 2.575
45º
4762 Mark Scheme June 2015
12
Question Answer Marks Guidance
4 (a) (i) Let speed of P after collision be u m s -1
21 1 42 2 9
4 4 36u M1 Use of KE; allow use of 59
in an equation involving only u
B1 Correct use of 49
of a KE term
u = 4 E1 Condone 4
Cannot be in the direction of Q as 4 > 3 E1 Can use PCLM and NEL to show 'in direction of Q' gives negative e
4
(ii) Taking +ve
Impulse is 4(4 ( 6)) M1
Allow sign error. Must be 4(final vel - initial vel); condone 4(6 +4)
= 40 N s A1
Note: impulse on Q = 5(3 - - 5) = 40 needs to be related to impulse
on P to score any marks
Let force be F N
15
40 F M1 Use of Impulse = Ft
so F = 200 A1 Sign consistent with sign of impulse
4
(iii) Taking +ve and vel of Q as U m s -1
PCLM 24 5 4 4 5 3U M1 o.e. e.g. using impulse
so 5U i.e. to the left A1 Must state direction but could be implied by diagram or equivalent
4 3 7
6 5 11e
(0.636 to 3 sig figs) M1
F1 FT their U
4
4762 Mark Scheme June 2015
13
Question Answer Marks Guidance
(b) (i) Vert cpt of vel in m s -1 is 1517
5.95 5.25 B1 May be implied
Particle hits the plane with vert speed
25.25 2 9.8 2.5 M1 Must be vertical component
= 8.75 A1 May be implied
Particle leaves plane with vert speed
8.75 0.8 = 7 B1 Award for the ×0.8 on vert cpt (dependent on M1)
Height reached is given by
20 7 2 9.8 h so h = 2.5 E1 Clearly shown cwo
5
(ii) Time taken in seconds is
down
8.75 5.25
9.8
=
5
14= 0.357 :up
7
9.8=
5
7= 0.714: M1 Finding both times and adding
so
15
14 s F1 FT only their 8.75 from (i)
Horiz distance is
15 85.95
14 17 = 3 m A1 cao
3
20
Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, CB1 2EU Registered Company Number: 3484466 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: 01223 552552 Facsimile: 01223 552553 © OCR 2015
OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge CB1 2EU OCR Customer Contact Centre Education and Learning Telephone: 01223 553998 Facsimile: 01223 552627 Email: [email protected] www.ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored
Oxford Cambridge and RSA Examinations
GCE
Mathematics (MEI)
Unit 4762: Mechanics 2 Advanced GCE
Mark Scheme for June 2016
OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2016
4762 Mark Scheme June 2016
3
Annotations and abbreviations
Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme
Meaning
E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph M1 dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working
4762 Mark Scheme June 2016
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Subject-specific Marking Instructions for GCE Mathematics (MEI) Mechanics strand a Annotations should be used whenever appropriate during your marking.
The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.
b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.
c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.
4762 Mark Scheme June 2016
5
E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.
f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.)
We are usually quite flexible about the accuracy to which the final answer is expressed and we do not penalise over-specification.
When a value is given in the paper Only accept an answer correct to at least as many significant figures as the given value. This rule should be applied to each case.
When a value is not given in the paper Accept any answer that agrees with the correct value to 2 s.f.
4762 Mark Scheme June 2016
6
ft should be used so that only one mark is lost for each distinct error made in the accuracy to which working is done or an answer given. Refer cases to your Team Leader where the same type of error (e.g. errors due to premature approximation leading to error) has been made in different questions or parts of questions. There are some mistakes that might be repeated throughout a paper. If a candidate makes such a mistake, (eg uses a calculator in wrong angle mode) then you will need to check the candidate’s script for repetitions of the mistake and consult your Team Leader about what penalty should be given.
There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question.
g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.
If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.
h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.
i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader.
j If in any case the scheme operates with considerable unfairness consult your Team Leader.
4762 Mark Scheme June 2016
7
Question Answer Marks Guidance
1 (a) (i) Take + ve →
PCLM
Q0.5 4 0.75 1 0.5 2 0.75 v M1 Application of PCLM. Allow sign errors
A1 Correct. Any form
so
Q
7
3v A1 Exact or anything that rounds to 2.33 or better
NEL
72
3
1 4e
M1 NEL. Accept sign errors but not approach/separation
so
1
9e (or 0.11) A1ft ft their vQ
[5]
(ii) Suppose direction reversed. Given LM conserved
Q0.5 4 0.75 1 0.5 2 0.75 v B1 Award for the correct LM equation
so Q 5v
NEL gives
5 2
1 4e
so
7
3e
M1 Using their re-calculated vQ (not equal to 7/3 from (i))
e > 1 so not an elastic collision E1 www
OR for last two marks:
KE after collision = 10.375, before collision = 4.375:
Increase in energy not possible, because no work put into
system
[3]
M1
E1
4762 Mark Scheme June 2016
8
Question Answer Marks Guidance
(iii) No (external horizontal) force acts on the truck B1 Force or momentum considered or correct momentum
equation
so no change in momentum of truck (less object)
(So no change in velocity. Still)
7
3 m s
-1.
B1 FT their value from (i): seen
[2]
(iv) Before 0.5 kg at 2 m s -1
→
After 0.05 kg at U m s -1
← and 0.45 kg at V m s -1
→ M1 Allow if (10 − 2) = 8 used instead of U
PCLM
0.5 2 0.05 0.45U V M1 Allow only sign errors
U + V = 10 B1 oe: relative velocity used correctly
Solving, V = 3 so 3 m s -1
A1
[4]
cao
SC1 Using U = 10, giving V = 10/3
(b) Consider the LM parallel to the plane M1 Accept considering the horizontal components of velocity
before and after o.e. and arguing/stating they should be the
same
Before: 10cos60 5m m After: 6cos40 4.6mm A1 Need not include m. Using sine gets 0/3
Not the same. (LM not conserved.) Plane cannot be smooth. E1 Accept arguments from velocity
[3]
4762 Mark Scheme June 2016
9
Question Answer Marks Guidance
2 (a) (i)
WD against resistance = KE lost
M1
210.04 50 0.2
2F
A1
so F = 250 and resistive force is 250 N
OR:
Use suvat: 2 2 2v u as : 2500
0.4a
Use N2L: F = 0.04 a = - 250
Resistive force = 250 N
A1
M1
A1
A1
Accept -250
Complete method: suvat and N2L
Correct a
Correct F
[3]
(ii) PCLM
0.04 50 3.96 0.04 V M1
so V = 0.5 so 0.5 m s-1
A1 cao
[2]
(iii) Energy lost is
2 21 10.04 50 4 0.5
2 2 M1 Correct masses
= 49.5 J A1ft ft their 0.5 from (ii). May be implied
equating WD against resistance to energy lost
250 49.5x M1 ft their 250 x = a difference in non-zero KEs
so x = 0.198 and distance is 0.198 m A1 cao
[4]
4762 Mark Scheme June 2016
10
Question Answer Marks Guidance
(b) (i) Using W-E equation. Friction is F N
2 21
6 7 1 91.5 6 sin30 82
F g M1 o.e. All 5 terms present, no extras. Allow sign errors
B1 KE terms (both)
B1 Resolved weight or GPE term
M1 WD is Force × distance
so F = 44.1
A1 cao
(As slipping) F R M1 Used
36 cos30 6
2R g g B1
Does not need to be evaluated
so
44.1 1.5 3
23 3 3g
(0.8660…) A1 cao Any form. 0.87 or better
[8] Using suvat and N2L: Max possible is B1 for resolved
weight, then last 3 marks.
Award SC(4) if N2L and suvat used, and correct, www
(ii) Power is T × v M1
so 91.5 × 7 = 640.5 W A1 cao accept 640 or 641. Must be their final answer
[2]
4762 Mark Scheme June 2016
11
Question Answer Marks Guidance
3 (i) 0.825 0.825 1.652 0.6 0.4
0.6 0 0.6
xM M M M
y
M1 Complete method
A1 At least 2 RHS vector terms or 3 component terms correct
0.99x A1
0.42y A1
[4]
(ii) c.w moments about R
Q300 0.99 1.65 0Y M1 Or moments about Q to find
RY
so Q 180Y A1 Must be established not using given
RY
Q R 300Y Y so
RY = 120 E1 AG
[3]
(iii) Mark in 120 N and 180 N and all internal forces B1 Accept labelled internal forces marked as T or C
[1]
(iv) c.w moments about A
D120 0.75 180 2.4 2 0X
Or = 0 M1 Appropriate moments considered
so D 261X E1 Convincingly shown
[2]
4762 Mark Scheme June 2016
12
Question Answer Marks Guidance
(v) At B This solution takes all internal forces +ve when T(ensions)
↑
AB sin 180 0T M1 Equilibrium at a pin-joint to find one of required forces
(all relevant forces)
so
AB
13180 468
5T so force in AB is 468 N (T) A1 Do not need T/C here
←
BC AB cos 0T T M1 2
nd equilibrium at the same or another pin-joint to find
another required force (all relevant forces)
so
BC
12468 432
13T so force in BC is 432 N (C) F1 FT their values. Do not need T/C here
At D
→ DC Dcos 0T X
M1 3
rd equilibrium at a pin-joint (complete method to find
third required force)
so
DC
5261 435
3T so force in DC is 435 N (C) F1 FT their values and all T/C correct
↑ D DC sin 0Y T
so
D
4435 348
5Y so 348 N B1 cao for the first of YD or YA found
D A 300Y Y so
A 48Y so 48 N B1 ft for the second of YD or YA : difference = 300
[8]
4762 Mark Scheme June 2016
13
Question Answer Marks Guidance
4 (i) Let centre of mass be at G
G is on CO by symmetry B1
Let CG = Y and curved surface density be Accept taken to be 1 without comment. o. e.
2 20.1 4 2 0.1 0.1 4 2 0.1
2
hh Y h h M1 Complete method
B1 ‘masses’ in correct ratios: 0.04: 0.2h
B1 Correct use of ‘h’ and ‘h/2’
A1 All correct
so 25 2
2 10
h hY
h
E1 Convincingly shown
[6]
(ii) Let the lowest point of contact of the cylinder with the plane be A
On point of tipping G is vertically above A B1 May be shown on a diagram
Angle AGO is α B1
May be shown on a diagram can be implied by
subsequent work
0.1 2tan
3Y M1 Allow reciprocal of RHS
so 22 5
0.3 22 10
h h
h
A1
so 250 5 3 0h h AG A1 cwo and convincingly shown
Either
5 1 10 3 0h h M1 Clear evidence of 2 roots
(only positive root is) h = 0.2. E1 No need to comment on the negative root
or
2
50 0.2 5 0.2 3 2 1 3 0 B1
And this is the only positive root E1 Need statement but no need to show this.
[7]
4762 Mark Scheme June 2016
14
Question Answer Marks Guidance
(iii) (sin β =0.6; cos β =0.8)
c.w. moments about ‘furthest’ point of the base (through which acts the M1 Both forces present in a moments equation
NR)
cos 0.5 sin 0.2 42 0.1 0T T M1
A1
A1
Attempt to find moment of force of magnitude T in
horizontal and vertical components oe
Correct distances oe
Correct equation, numerical values of cos/sin do not
need to be substituted
(so 0.4 0.12 4.2T )
so T = 15 A1 cao
[5]
Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, CB1 2EU Registered Company Number: 3484466 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: 01223 552552 Facsimile: 01223 552553 © OCR 2016
OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge CB1 2EU OCR Customer Contact Centre Education and Learning Telephone: 01223 553998 Facsimile: 01223 552627 Email: [email protected] www.ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored