Question 3:
3 | A
5m
3m
2m 3m
B 4m C
Total mass 17m | B1
Moments about AB: 4m2a5m2a3m4a17mx | M1A1
30
x  a
17 | A1
Moments about BC: 3m1.5a5m1.5a 17my | M1A1
12
y  a
17 | A1
x
tan for their x, y
3ay | M1A1ft
37.6, 38 | A1
[10]
Notes
3. | B1 for total mass 17m
First M1 for a ‘moments’ equation about AB or another line, with correct
no. of terms , for wire AND masses
First A1 for a correct equation (allow consistent omission of m’s and/or
a’s and interchange of a’s and m’s)
Second A1 for 30a/17, 1.8a or better oe for their axis (allow omission
of a)
Second M1 for a ‘moments’ equation about AB or another line , with
correct no. of terms, for wire AND masses
Third A1 for a correct equation (allow consistent omission of m’s and/or
a’s and interchange of a’s and m’s)
Fourth A1 for 12a/17, 0.71a or better oe for their axis (allow omission
of a)
x
Third M1 independent for tan or its reciprocal (or their
3ay
equivalents if they have used different axes). Allow omission of a, if
their x and y are numbers and x and y do not need to be substituted
(they may not have an x and y )
Fifth A1 ft on their x and y
Sixth A1 for 38o or better (37.5685..)
N.B. The first two M marks are for a complete method in each case so if
they only find the CM of the wire e.g. (1.5a, a) or only find the CM of
the particle system e.g. (2.4a, 0), it’s M0. However, if they do then
combine, the A1 is for all the equations that they use.
N.B. They may take A as their origin. Then x is 30/17, y = 39/17
and tan= x /y
N.B. If they take the CM of the wire (12m) only, to be at the centroid of
triangle ABC, (4a/3, a), and then combine with the masses, can score
max B1M0M0M1A1ftA0. Beware! Since the CM of the wire only, is
(1.5a, a), they will get a correct answer for y using this incorrect
method.
Similarly, if they take the CM of the wire (12m) to be at some other
point, with no working , often at (2a, 1.5a), and then combine with the
masses, can score max B1M0M0M1A1ftA0.
4a |  12 
Normal reaction 6.5gcos  6.5g 6g 
 13  | B1
1 12
Use of F R F  6.5g 2g
3 13 | M1
6.5
Work-energy principle: 366.5gsind Fd
2 | M1
A2 ft
 5 
Substitute and solve for d: 117dg  6.5 2 
 13  | DM1
117
d  2.7m to 2 s.f. GIVEN ANSWER
4.5g | A1 (7)
4b | 6.5 6.5
62  v2 2Fd
2 2 | M1 A2
v2m s-1, 2.0 , 2.00 | A1 (4)
4balt1 | 6.5
Energy: v2 6.5gsind Fd
2 | M1 A2
v2m s-1, 2.0 , 2.00 | A1 (4)
4balt2 | F=ma & suvat: | M1
6.5gsinF  6.5a | A1
v2 = 2 x (g/13 oe) x d | A1
v2m s-1, 2.0 , 2.00 | A1 (4)
Notes
4a | B1 for 6.5gcos(This could be scored in (b) if not seen in (a))
First M1 for F = 1/3 x their R (This could be scored in (b) if not seen in
(a))
Second M1 for work-energy equation: Need KE, PE (k x dsin) and
WD (n x d) terms
First and second A marks ft on their F , -1 each error
Third M1 dependent on second M1, for solving for d.
Third A1 for 2.7 (2 SF) GIVEN ANSWER. Must finish with 2.7 but
(2SF) may be omitted.
N.B. No marks for a non-energy method.
4b | M1 for work-energy equation: Need 2 KE terms and 2 x WD (n x d)
terms
First A2 -1 each error
Third A1 for 2 m s-1. Penalise inaccurate answers e.g. 2.01 or 2.02
Alt 1 M1 for work-energy equation: Need KE, PE (k x dsin) and WD
(n x d) terms
First A2 -1 each error
Third A1 for 2 m s-1 .Penalise inaccurate answers e.g. 2.01 or 2.02
Alt 2 M1 for a complete method ( F = ma and v2=u2+2as)
First A1 for F = ma, equation in a only
Second A1 for v2 = 2 x (g/13 oe) x d (i.e. a must be correct)
Third A1 for 2 m s-1 Penalise inaccurate answers e.g. 2.01 or 2.02
5a | CLM: 3mu3mv4mw | M1A1
1
Impact law: wv  u
3 | M1A1
4
Solve for w: w u
7 | M1A1
4u 16
Impulse 4m  mu *Given answer*
7 7 | A1
N.B. Given answer, so working needs checking | (7)
5b | 1 72
5my2  mu2
2 245 | B1
16
CLM: mu 4mx5my
7 | M1 A1
4
Impact: yxe u
7 | M1A1
7u 7 7
e  
35 4u 20 | A1
(6)
Notes | [13]
5a | First M1 for CLM, correct no. of terms, allow cancelled m’s
First A1 for a correct equation
Second M1 for Impact Law, correct way up
Second A1 for a correct consistent equation
Third M1 for solving for either v or w
Third A1 for either v or w in terms of u (w 4u or v 5 u)
7 21
Fourth A1 for given answer fully justified.
(I  4m.4u or I  3m(5 uu))
7 21
5b | 72
B1 for 15my2  mu2 their y
2 245
First M1 for CLM, condone sign errors
First A1 for 4m.4u/7 = 4mx + 5my their x and y
Second M1 for Impact Law, correct way up (consistent signs)
Second A1 for e. 4u/7 = - x + y their x and y
Third A1 for e = 7/20 oe
x 1u
7
y 12u
35
N.B. y 16u(1e)
63
e2 2e 329 0
400
(e 47)(e 7 )0
20 20
6a | Smooth peg – no friction, so just the normal reaction | B1 (1)
6b | Moments about A:
3aN W2acoskW4acos | M1 A2
W 10
N  612k W12k *Given Answer*
3 10 5 | A1 (4)
6c | 3
Use of F  R
4 | M1
Resolve horizontally: F  Nsin | M1A1
Resolve vertically: RNcosW1k | M1A1
3
Sub into F  R
4
10 1 3 10 3
W12k  (W(1k) W12k )
5 10 4 5 10 | DM1
2
k  *Given Answer
11 | A1
(7)
Notes
6a | B1 for smooth peg so no friction so just normal reaction
6b | M1 for moments about A (or any other complete method)
First A2 for an equation in N only, -1 each error
Third A1 for the given answer. A0 if they go from decimals to surd
form
6c | First M1 for use of F ≤ ¾ R or F = ¾ R (Allow µ instead of ¾)
Second M1 for resolving horizontally or another moments equation
First A1 for a correct equation
Third M1 for resolving vertically or another moments equation
Second A1 for a correct equation
Fourth M1 dependent on all 3 previous M’s for producing an equation
or inequality in k only.
Third A1 for k ≤ 2/11 given answer (must have worked with an
inequality all the way through to earn this mark)
7a | Vertical speed = 0: v 9gt 0 | M1
9
t  0.92(s) 0.918
g | A1
(2)
7b | 1
x4t,y9t gt2
2 | B1, B1
1
Use y  x( k): 4t 9t gt2
2 | M1
40
k  4.1 (4.08)
g | A1 (4)
7c | Complete method using symmetry of times to find other pt :
Time to A = k/4 = 10/g => time to other pt = 9/g – (10/g - 9/g)= 8/g
So x = 4 x 8/g | M1
x = 4k/ 5 or (72/g – k) or (7.3(5) – k) | A1
4 
r k ikj
 
5  | A1 (3)
7c(alt) | Complete method using symmetry of horiz distances to find other pt :
At max ht , x = 4 x 9/g = 36/g => At other pt, x = 36/g – (k - 36/g) | M1
Other point: : x = 4k/ 5 or (72/g – k) or (7.3(5) – k) | A1
4 
r k ikj
 
5  | A1
(3)
7c(alt) | 40 1
Same height: 9t t2
g 2 | M1
Other point: : x = 4k/ 5 or (72/g – k) or (7.3(5) – k) | A1
4 
r k ikj
 
5  | A1
(3)
32 40
NB: an answer of r 3.3i4.1j or r  i j scores M1A1A0
g g
7d | 4 9 16
4i + kj perpendicular to 4i + 9j:  , k 
k 4 9 | M1A1
16 97
 9gT , T  1.1 (1.10)
9 9g | M1A1
(4)
7d | OR: vert: v = 9 – gT AND combine with the horiz cpt. | M1
v = 4i + (9 – gT)j | A1
4 9

9 –gT 4 | M1
97
T  1.1 (1.10)
9g | A1
(4)
Notes
7a | M1 for 0 = 9 – gt (or any other complete method) condone sign errors
A1 for 9/g or 0.918 or 0.92 (correctly obtained) (45/49 is A0)
7b | First B1 for k = 4t
Second B1 for k = 9t – ½ gt2
M1 for eliminating and solving for k or t
A1 for k = 40/g or 4.08 or 4.1
7c | M1 for a complete method using symmetry on times
First A1 for x = 4/ k
5
Second A1 for 4/ k i + k j
5
OR
M1 for a complete method using symmetry on horiz. distances
First A1 for x = 4/ k
5
Second A1 for 4/ i + k j
5
OR
M1 for k = 9t – ½ gt2
First A1 for x = 4/ k
5
Second A1 for 4/ k i + k j
5
N.B. Correct answers not in terms of k, score M1A1A0
7d | First M1 for attempt to find a vector of form (4i + kj) which is perpar to
4i + 9j (must use reciprocal, but condone missing – sign)
First A1 for k = -16/
9
Second M1 for an equation in T only (-16/ = 9 – gT)
9
Second A1 for T = 97/(9g) or 1.10 or 1.1
OR
First M1 for attempt to find velocity vector at time T
First A1 for 4i + (9 – gT)j – This may not be explicit but they must have
BOTH cpts.
Second M1 for using the perpendicularity with 4i + 9j to form an
equation in T only (must use reciprocal, but condone missing – sign)
Second A1 for T = 97/(9g) or 1.10 or 1.1
N.B. -9 = 9 – gT (T = 1.84) is M0A0M0A0
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