Question Scheme Marks AOs

1 (a) Differentiate v to obtain a M1 3.4
a =( 6t − 12 ) i + (18t − 3) j A1 1.1b

(2)

Integrate v to obtain r M1 3.4

 3t 2 
(b) r =( t 3 − 6t 2 ( +C1 ) ) i +  3t 3 − ( + C2 )  j A1 1.1b
 2 
(2)
Solve a = λ j to obtain t M1 3.1a
6t − 12 = 0 ⇒ t = 2 A1 1.1b

(c) Substitute their t M1 1.1b
r=
−16i + 18 j A1 2.2a

(4)

(8 marks)

Notes:

(a)

M1 Powers going down by 1

A1 Correct only

(b)

M1 Powers going up by 1. Condone missing constants of integration

A1 Correct only

(c)

M1 Set coefficient of i equal to zero and solve for t

A1 Correct only

M1
Substitute their t in an expression of the form r = ( at 3
− bt 2 + c ) i + ( dt 3 − et 2 + f ) j where
abde ≠ 0
A1 Correct only
Question Scheme Marks AOs

2 (a)
D




T
B
C

V 60°

A 40 N
H




Take moments about A: M1 3.3

T × 2 = 40 ×1.5cos 30° A1 1.1b

T = 15 3 ( N ) 26.0 ( N ) A1 1.1b

(3)

(b) Resolve horizontally M1 3.4

 15 3 
= =
H T cos 60 °  = 12.99....  A1 1.1b
 2 
Resolve vertically M1 3.4
V + T cos
= 30° 40=(V 17.5) A1 1.1b

3
Combine components : 17.52 + 225 × M1 3.1b
4
= =
475 5=
19 22(N) A1 2.2a

(6)
(b) Resolve parallel to the beam M1 3.4
alt
= ° ( 20 )
X 40 cos 60= A1 1.1b

Resolve perpendicular to the beam M1 3.4

Y +=
T 40 cos 30° (=
Y 20 3 − 15 =
3 5 3 ) A1 1.1b

Combine components : 202 + 25 × 3 M1 3.1b
 = =
475 5=
19 22(N) A1 2.2a

(6)
(c) The tension will not be constant B1 3.5a
The tension will increase as you go up the rope since it is supporting
more rope B1 2.4


(2)



(11 marks)

Notes:

(a)

M1 Or alternative complete method to form an equation in T

A1 Correct unsimplified equation in T

A1 26 or better

(b)

M1 First equation. Required terms and no extras. Condone sign errors and sin/cos confusion.

A1 Correct unsimplified equation
Second equation. Required terms and no extras. Condone sign errors and sin/cos
M1
confusion.
A1 Correct unsimplified equation

M1 Correct strategy to find the resultant force

A1 22 or better (21.79….)
N.B. There are many approaches to this. Alternative equations include e.g. moments
about C and moments about D
(c)

B1 Correct answer

B1 Correct reasoning
Question Scheme Marks AOs

3 (a) Distance travelled M1 3.1b

=
(15 + 17 ) ×16 × 60 + (10 + 12 ) ×16 × 60 A1 1.1b
2 2
= 16 × 16 × 60 + 11×16 × 60 = 15360 + 10560
A1 1.1b
= 25920 = m 25.92 km

(3)

(b) Both trains have travelled the same distance M1 3.1b


=
⇒ 25920
( M + ( M − 2 ) ) × 24 × 60 A1ft 1.1b
2 A1ft 1.1b

Solve for T M1 1.1b

M − 1 = 18 ⇒ T = 33 − 19 = 14 A1 2.2a

(5)

(c) e.g. the platforms at A, B and C have no length
the trains have no length
B1 3.5b
A, B and C modelled as points
the trains are modelled as particles.
(1)

(9 marks)

Notes:

(a)
Complete method to find the distance AC. Condone confusion between minutes and
M1
seconds.
Unsimplified expression with at most one error (the omission of ×60 in each term is one
A1
error)
A1 26000 m or 26 km or better

(b)

M1 Equate the distance travelled by Y to their answer from (a)
A1ft Unsimplified equation in one unknown with at most one error. Follow their (a)
A1ft Correct unsimplified equation in one unknown. Follow their (a)
M1 Solve for T

A1 Correct answer only

(c)

B1 Any relevant assumption
Question Scheme Marks AOs

R
4 (a)
30 N
F


α 20 N



=
Resolve perpendicular to the plane: R 20 cos α + 30sin α M1 3.4
12 5
R = 20 × + 30 × = 30 (N) A1 1.1b
13 13
(2)

(b) Forces parallel to the plane: M1 3.1b
30 cos α …Fmax + 20sin α A1ft 1.1b
Use of Fmax = µ R M1 1.2
12 5 2
30 µ „ 30 × − 20 × , µ„ A1 2.2a
13 13 3
...