Mathematics 9709/52 May/June 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
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 vertically upwards displacement of the ball from O is 14t − kt2 m, where k is a constant.
(i) State the value of k. [1]
(ii) Show that the initial speed of the ball is 28 m s−1 . [2]
(iii) Find the horizontal displacement of the ball from O when t = 3. [2]
2
B C D
0.3 m
F
0.7 m E
A G
A uniform lamina ABCEFG is formed from a square ABDG by removing a smaller square CDFE from one corner. AB = 0.7 m and DF = 0.3 m (see diagram). Find the distance of the centre of mass of the lamina from A. [4]
Mark scheme
1(i) g B1 Use the trajectory equation from the formula sheet
k= =5
2
1
1(ii) Vsin30 = 14 M1 Use the trajectory equation from the formula sheet
V = 28 ms–1 AG A1
2
1(iii) x = 28cos30 × 3 M1 Use horizontal motion. Allow their V for M1
x = 72.7 m A1
2
2
Original square: Area = 0.72, CoM = ( 0.35 + 0.35 ) and
2 2 B1 0.49, 0.495 from A
0.09, 0.21213... from D or E
Smaller square: Area = 0.3 , CoM = ( 0.15 + 0.15 )
2 2 2
AX ( 0.49 − 0.09 ) + 0.09 ( 0.98 − 0.045 ) = 0.49 × 0.495 M1A1 Attempt to take moments about A
AX = 0.431 m A1
4
Alternative method for question 2
(0.49 × 0.35) = (0.09 × 0.55) + 0.4X → X = 0.305 M1 Take moments about AG or AB
X = Y = 0.305 B1
2
AX = ( 0.305 + 0.305 )
2 2 M1 Use Pythagoras’s theorem
AX = 0.431 A1
4
A particle P of mass 0.4 kg is attached to a fixed point A by a light inextensible string of length 0.5 m.
The point A is 0.3 m above a smooth horizontal surface. The particle P moves in a horizontal circle on the surface with constant angular speed 5 rad s−1 .
(i) Calculate the tension in the string. [3]
(ii) Find the magnitude of the force exerted by the surface on P. [2]
Mark scheme
3(i) r = 0.4 m B1 Use Pythagoras’s theorem
T cosθ = 0.4 × 52 × 0.4 M1 Use Newton’s Second Law
0.4 A1
T× = 4, T = 5 N
0.5
3
3(ii) R = 0.4 g − T sin θ M1 Resolve vertically. Allow for their T for M1
R = 1N A1
2
A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 16 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.8 m vertically below O. When the extension of the string is x m, the downwards velocity of P is v m s−1 and a force of magnitude 25x2 N opposes the motion of P.
dv
(i) Show that, when P is moving downwards, v = 10 − 40x − 50x2 . [2]
dx
(ii) For the instant when P has its greatest downwards speed, find the kinetic energy of P and the elastic potential energy stored in the string. [6]
Mark scheme
4(i) dv 16 x M1 Use Newton’s Second Law vertically
0.5v = 0.5 g − − 25 x 2
dx 0.8
dv A1
v = 10 − 40 x − 50 x 2 AG
dx
2
4(ii)
∫ vdv = ∫ (10 − 40 x − 50 x )dx M1 Attempt to integrate
2
v2 50 x3 A1
= 10 x − 20 x 2 − ( +c)
2 3
0 = 10 – 40x – 50x2 M1 Put the acceleration equal to zero
x = 0.2 (Ignore x = –1 if seen) A1
0.5v 2 8 B1 mv 2
= = 0.533J Use KE =
2 15 2
0.22 B1 λ x2
16 × = 0.4 J Use EE =
( 2 × 0.8) ( 2l )
6
A light elastic string has natural length a m and modulus of elasticity , N. When the length of the string is 1.6 m the tension is 4 N. When the length of the string is 2 m the tension is 6 N.
(i) Find the values of a and ,. [5]
One end of the string is attached to a fixed point O on a smooth horizontal surface. The other end of the string is attached to a particle P of mass 0.2 kg. The particle P moves with constant speed on the surface in a circle with centre O and radius 1.9 m.
(ii) Find the speed of P. [3]
Mark scheme
5(i) λ (1.6 − a ) B1 λx
4= Use T = twice
a l
λ (2 − a) B1
6=
a
1.5 =
(2 − a) M1 Attempt to solve the simultaneous equations
(1.6 − a )
5(i) 0.4 = 0.5(a), a = 0.8 A1
λ=4 A1
5
5(ii) 1.1 B1 λx
T = 4× ( = 5.5) FT Use T = , ft candidates λ and a
0.8 L
0.2v 2 M1 Use Newton’s Second Law horizontally
5.5 =
1.9
v = 7.23 ms–1 A1
3
A particle is projected with speed 15 m s−1 at an angle of 1Å above the horizontal. At the instant 4 s after projection the speed of the particle is 30 m s−1 .
(i) Find 1. [4]
(ii) Show that at the instant 4 s after projection the particle is 33.75 m below the level of the point of projection and find the direction of motion at this instant. [4]
Mark scheme
6(i) 15cosθ = vH and 15sinθ – 4g = vV B1 Use horizontal and vertical motion
(15cosθ)2 + (15sinθ – 4g)2 = 302 M1 Use Pythagoras’s theorem
[225 – 1200sinθ + 1600 = 900] M1 Attempt to solve for θ
θ = 50.4° A1
4
6(i) Alternative Method
g ( 4) B1
2
h = (15sin θ ) × 4 −
2
m (15 ) m ( 30 ) M1 Allow h not replaced
2 2
= + mgh
2 2
M1 Attempt to eliminate h and attempt to solve for θ
θ = 50.4° A1
4
6(ii) 1 M1 Use vertical motion. Allow their θ for first M1
s = 15sin 50.4 × 4 − × g × 42
2
s = 33.75 m AG A1
15cos50.4 M1 Use trigonometry of a right angled triangle
cos α =
30
α = 71.4° below the horizontal A1
4 If g = 9.8 or 9.81 used then M1A0M1A0
rm
A B C
Fig. 1
Fig. 1 shows an object made from a uniform wire of length 0.8 m. The object consists of a straight part AB, and a semicircular part BC such that A, B and C lie in the same straight line. The radius of the semicircle is r m and the centre of mass of the object is 0.1 m from line ABC.
(i) Show that r = 0.2. [3]
A
B
C 7N
Fig. 2
The object is freely suspended at A and a horizontal force of magnitude 7 N is applied to the object at
C so that the object is in equilibrium with ABC vertical (see Fig. 2).
(ii) Calculate the weight of the object. [3]
[Question 7(iii) is printed on the next page.]
The 7 N force is removed and the object hangs in equilibrium with ABC at an angle of 1Å with the vertical.
(iii) Find 1. [6]
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) 2r B1 X = distance of centre of mass of the arc from ABC
X=
π
2r M1 Take moments about ABC
0.8 × 0.1 = π r ×
π
r = 0.2 A1
3
7(ii) AC = 0.8 + 2 × 0.2 – 0.2π (= 0.57168…) B1
0.1W = 7AC M1 AC must be a numerical value. Take moments about A
W = 40(0.) N A1
3
7(iii) (0.8 – 0.2π + 0.2) [= 0.37168…] B1
( 0.8 − 0.2π ) M1A1
0.8Y = ( 0.8 − 0.2π ) × + ( 0.2π ) × ( 0.8 − 0.2π + 0.2 )
2
Y = 0.310(338) A1
0.1 M1
tan θ =
0.310338
θ = 17.9 A1 Allow 17.8
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Mathematics 9709/52 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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