Mathematics 9709/52 October/November 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A particle of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 9 N. The other end of the string is attached to a fixed point O on a smooth horizontal surface. The particle is projected horizontally from O with speed 4 m s−1 . Find the greatest distance of the particle from O. [3]
Mark scheme
1 0.3 × 4 2 9e 2 M1 Set up an energy equation.
= Note the final velocity is zero.
2 2 × 0.6
2 2 A1
e = 0.566 or
5
Distance = 1.17 m A1
3
A small ball is projected from a point O on horizontal ground at an angle of 30Å above the horizontal.
At time t s after projection the horizontal and vertically upwards displacements of the ball from O are x m and y m respectively. It is given that x = 40t.
(i) Calculate the initial speed of the ball, and express y in terms of t. [3]
(ii) Hence find the equation of the trajectory of the ball. [2]
Mark scheme
2(i) Vcos30 = 40 M1 Note V is the velocity of projection
V = 46.2 m s −1 A1 80 80 3
Allow or
3 3
y = 23.1t – 5 t 2 B1FT at 2
Use s = ut + vertically. FT candidates half V but not V = 40
2
used
3
2(ii) 23.1x 5 x 2 M1 x
y= − Attempt to eliminate t by substituting t = into answer to part
40 1600 40
(i)
x2 A1
y = 0.577 x − or y = 0.577 x − 0.003125 x 2
320
2
A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point O. The particle P is projected vertically downwards with speed 2 m s−1 from the point 0.5 m vertically below O. For an instant when the acceleration of P is 4 m s−2 downwards, find the extension of the string and the speed of P. [6]
Mark scheme
3 0.5 × 4 = 0.5g – T M1 Use Newton’s Second Law vertically
12e M1 λx
T= Use T =
0.6 l
3 × 0.6 A1
e = = 0.15
12
12 × 0.152 B1ft
EPE = and distance fallen = 0.6 – 0.5 + 0.15
2 × 0.6
0.5v 2 0.5 × 2 2 12 × 0.152 M1 Set up a 4 term energy equation
= + 0.5 g (0.6 − 0.5 + 0.15) −
2 2 2 × 0.6
v = 2.85 m s −1 A1
6
A particle is projected from a point O on horizontal ground with speed V m s−1 at an angle of 60Å above the horizontal. At the instant 3 s after projection the direction of motion of the particle is 30Å below the horizontal.
(i) Find V . [3]
(ii) Calculate the distance of the particle from O at the instant 3 s after projection. [3]
Mark scheme
4(i) Velocity component vertically = ± (Vsin60 – 3g) B1 Use v = u + at
30 − V sin 60 M1 Use trigonometry of a right angled triangle
tan 30 =
V cos 60
V = 15 3 = 26(.0) m s −1 A1
3
4(ii) g × 32 B1FT at 2
y = 26sin 60 × 3 − Use s = ut + vertically. Their V from part (i)
2 2
D 2 = (26sin60 × 3 – g × 32 ) 2 + ( 26cos 60 × 3) M1 Use Pythagoras’s Theorem
2
D = 45(.0) m A1
3
A
30Å
0.5 m
O P
70Å
B
A and B are two fixed points on a vertical axis with A above B. A particle P of mass 0.4 kg is attached to A by a light inextensible string of length 0.5 m. The particle P is attached to B by another light inextensible string. P moves with constant speed in a horizontal circle with centre O between A and B.
Angle BAP = 30Å and angle ABP = 70Å (see diagram).
(i) Given that the tensions in the two strings are equal, find the speed of P. [5]
(ii) Given instead that the angular speed of P is 12 rad s−1 , find the tensions in the strings. [5]
Mark scheme
5(i) r = 0.5sin30 (= 0.25 m) B1
Tcos30 – Tcos70 = 0.4g M1 Resolve vertically
T = 7.6335.. A1
7.6335sin30 + 7.6335sin70 = 0.4 v 2 / 0.25 M1 v2
Use Newton’s Second Law with a =
r
v = 2.62 m s −1 A1
5
5(ii) Acos30 – Bcos70 = 0.4g and Asin30 + Bsin70 = 0.4 × 122 M1 Resolves vertically and uses Newton’s Second Law with
× 0.5sin30 a = rω2
A1 Both correct
M1 Attempt to solve for A or B
A = 8.82 N A1
B = 10.6 N A1
5
A particle P of mass 0.2 kg is projected horizontally from a fixed point O on a smooth horizontal surface. When the displacement of P from O is x m the velocity of P is v m s−1 . A horizontal force
of variable magnitude 0.09 x N directed away from O acts on P. An additional force of constant magnitude 0.3 N directed towards O acts on P.
dv
(i) Show that v = 0.45 x − 1.5. [2]
dx
(ii) Find the value of x for which the acceleration of P is zero. [2]
(iii) Given that the minimum value of v is positive, find the set of possible values for the speed of projection. [5]
Mark scheme
6(i) dv M1 Use Newton’s Second Law horizontally
0.2v = 0.09 x − 0.3
dx
dv A1 AG
v = 0.45 x − 1.5
dx
2
6(ii) 1 M1 Equate acceleration to zero
0 = 0.45 x 2 – 1.5
100 A1
x=
9
2
6(iii) 1 M1 Attempt to integrate
∫ v dv = ∫ (0.45 x 2 − 1.5)dx
3
3
A1
v 2 0.45 2
= − 1.5 x(+c) = 0.3 x 2 − 1.5 x(+ c)
2 3
2
3 M1
100 2 100
0.3 − 1.5 +c=0
9 9
50 A1
c=
9
v 2 50 10 A1
x = 0, > so v >
2 9 3
5
B 1.2 m C
1.8 m
G
A 2.4 m D
ABCD is a uniform lamina in the shape of a trapezium which has centre of mass G. The sides AD and BC are parallel and 1.8 m apart, with AD = 2.4 m and BC = 1.2 m (see diagram).
(i) Show that the distance of G from AD is 0.8 m. [4]
The lamina is freely suspended at A and hangs in equilibrium with AD making an angle of 30Å with the vertical.
(ii) Calculate the distance AG. [2]
With the lamina still freely suspended at A a horizontal force of magnitude 7 N acting in the plane of the lamina is applied at D. The lamina is in equilibrium with AG making an angle of 10Å with the downward vertical.
(iii) Find the two possible values for the weight of the lamina. [5]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(i) 1.8 B1
Rectangle: Area = 1.2 × 1.8 = 2.16, y = = 0.9
2
1.8 1.8 B1
Triangle(s): Area = 1.2 × = 1.08 , y = = 0.6
2 3
(2.16 + 1.08)Y = 2.16 × 0.9 + 1.08 × 0.6 M1 Take moments about AD
Y = 0.8 m A1 AG
4
7(ii) AGsin30 = 0.8 M1 Use Trigonometry of a right angled triangle
AG = 1.6 m A1
2
7(iii) AD makes an angle of 40° or 20° with the vertical B1
W × AGsin10 = 7 × 2.4cos40 M1 Take moments about A
W = 46.3 N A1
W × AGsin10 = 7 × 2.4cos20 M1 Take moments about A
W = 56.8 N A1
5
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/52, October/November 2019 (9709_w19_ms_52), and the question paper (9709_w19_qp_52), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/52 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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