Mathematics 9709/51 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
0.8 m
0.15 m
O P v m s−1
A particle P of mass 0.3 kg is attached to a fixed point A by a light inextensible string of length 0.8 m.
The fixed point O is 0.15 m vertically below A. The particle P moves with constant speed v m s−1 in a horizontal circle with centre O (see diagram).
(i) Show that the tension in the string is 16 N. [2]
(ii) Find the value of v. [3]
2 A particle is projected with speed V m s−1 at an angle of 1Å above the horizontal. At the instant 4 s after projection the speed of the particle is 16 m s−1 and its direction of motion is 30Å above the horizontal.
Find V and 1. [5]
Mark scheme
1(i) 0.15 M1 Resolve vertically. θ is the angle between the string and
T cosθ = T × = 0.3g the vertical
0.8
T = 16 N AG A1
2
1(ii) r 2 = 0.82 − 0.152 B1 r = 0.78581...
0.78581... 0.3v 2 M1 Use Newton’s Second Law horizontally
16sin θ = 16 × =
0.8 0.78581...
v = 6.416 A1
3
2
(
V cosθ = 16cos30 = 8 3 = 13.856... ) B1 Use horizontal motion
V sin θ = 16sin 30 + 4 g ( = 48) B1 Use vertical motion
V 2 = (16cos30 ) + (16sin 30 + 4 g ) OR M1 Use Pythagoras’s theorem or trigonometry of a right
2 2
angled triangle
tan θ =
(16sin 30 + 4 g )
16cos30
V = 50(.0) A1
θ = 73.9° A1
5
0.2 m
A
0.2 m
0.7 m
The diagram shows the cross-section through the centre of mass of a uniform solid object. The object is a cylinder of radius 0.2 m and length 0.7 m, from which a hemisphere of radius 0.2 m has been removed at one end. The point A is the centre of the plane face at the other end of the object. Find the distance of the centre of mass of the object from A. [5]
[The volume of a hemisphere is 23 0r3 .]
Mark scheme
3 Volume of cylinder = π × 0.22 × 0.7 (= 0.028π) AND B1 Both volumes required for B1
0.23
Volume of hemisphere = 2π × ( = 0.0053333π )
3
0.2 B1
Distance of centre of mass from object base = 0.7 − 3 × ( = 0.625 )
8
0.23 0.2 0.23 M1A1 Take moments about the plane face
x π × 0.22 × 0.7 − 2π × +
0.7 − 3 × × 2π × = 0.35 × 0.028π
3 8 3
x = 0.285 m A1
5
A small ball is projected with speed 25 m s−1 at an angle of 30Å above the horizontal from a point O on horizontal ground. At time t s after projection the horizontal and vertically upwards displacements of the ball from O are x m and y m respectively.
(i) Express x and y in terms of t and hence find the equation of the trajectory of the ball. [4]
(ii) Find x for the position of the ball when its path makes an angle of 15Å below the horizontal. [4]
Mark scheme
4(i) x = 25cos30t B1 Use horizontal motion
gt 2 B1 Use vertical motion
y = 25sin 30t −
2
x
2 M1 Eliminate t
g
x 25cos 30
y = 25sin 30 −
25cos 30 2
x 4 x2 A1
y= − or y = 0.577 x − 0.0107 x 2
3 375
4
4(ii) dy 1 8x dy M1A1 Differentiate the equation from part (i) to find the gradient
= − or = 0.577 − 0.0214 x
dx 3 375 dx
1 8x M1 Attempt to solve
− tan15 = − or − tan15 = 0.577 − 0.0214x
3 375
x = 39.6 or x = 39.5 A1
4
Alternative method for question 4(ii)
vy vy M1
tan15 = =
vx 12.5 3
v y = 12.5 3tan15 ( = 5.8 ) downwards A1
–5.8 = 12.5 – 10t leading to t = 1.83 M1 Vertical motion using v = u + at
25 3 A1
X = 1.83 × = 39.6
2
4
A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus of elasticity 6 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.5 + x m vertically below O. The particle P comes to instantaneous rest at O.
(i) Find x. [3]
(ii) Find the greatest speed of P. [5]
Mark scheme
5(i) 6 x2 M1 Set up an energy equation
0.4 g ( 0.5 + x ) =
( 2 × 0.5)
6x2 – 4x – 2 = 0 or 3x2 – 2x – 1 = 0 M1 Attempt to solve a 3 term quadratic equation
1 A1
x = 1 (ignore − if seen)
3
3
5(ii) 6e M1 λx
0.4 g = Use T = to find the extension at the equilibrium
0.5 l
position
1 A1
e=
3
1 B1ft Ft for candidate’s e
PE change = 0.4 g 0.5 +
3
1
2 M1 Set up a three term energy equation
6
0.4V 2 1 3
= 0.4 g 0.5 + −
2 3 ( 2 × 0.5 )
V = 3.65 ms–1 A1
5
A
0.3 m
B C
0.6 m
ABC is a uniform lamina in the form of a triangle with AB = 0.3 m, BC = 0.6 m and a right angle at
B (see diagram).
(i) State the distances of the centre of mass of the lamina from AB and from BC. [2]
Distance from AB ……
Distance from BC ……
The lamina is freely suspended at B and hangs in equilibrium.
(ii) Find the angle between AB and the horizontal. [2]
A force of magnitude 12 N is applied along the edge AC of the lamina in the direction from A towards C. The lamina, still suspended at B, is now in equilibrium with AB vertical.
(iii) Calculate the weight of the lamina. [3]
Mark scheme
6(i) From AB = 0.2 B1
From BC = 0.1 B1
2
6(ii) 0.1 M1 θ is the angle between AB and the horizontal
tan θ =
0.2
θ = 26.6° A1
2
6(iii) 12cos26.6 × 0.3 = W × 0.2 M1A1 Take moments about B. (W is the weight of the lamina)
W = 16.1 N A1
3
A particle P of mass 0.5 kg is attached to a fixed point O by a light elastic string of natural length
1 m and modulus of elasticity 16 N. The particle P is projected vertically upwards from O with speed
6 m s−1 . A resisting force of magnitude 0.1x2 N acts on P when P has displacement x m above O.
After projection the upwards velocity of P is v m s−1 .
dv
(i) Show that, before the string becomes taut, v = −10 − 0.2x2 . [2]
dx
(ii) Find the velocity of P at the instant the string becomes taut. [4]
(iii) Find an expression for the acceleration of P while it is moving upwards after the string becomes taut. [2]
(iv) Verify that P comes to instantaneous rest before the extension of the string is 0.5 m. [4]
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) dv M1 Use Newton’s Second Law vertically
0.5v = −0.5 g − 0.1x 2
dx
dv A1
v = −10 − 0.2 x 2 AG
dx
2
7(ii)
∫ vdv = ∫ ( −10 − 0.2 x ) dx M1 Attempt to integrate the expression in part (i)
2
v2 −0.2 x3 A1
− 10 × +c
2 3
v2 0.2 M1 Either use limits or find c and put x = 1
2 = −10 − 3 + 18
v = 3.98 (329…) ms–1
7(iii) dv 16 ( x − 1) M1 Use Newton’s Second Law vertically when string
0.5v = −0.5 g − 0.1x2 − becomes taut
dx 1
dv A1
v = −10 − 0.2 x 2 − 32 x + 32 = 22 − 32 x − 0.2 x 2
dx
2
7(iv)
∫ vdv = ∫ ( 22 − 32 x − 0.2 x ) dx M1 Attempt to integrate after the string becomes taut
2
v2 32 x 2 0.2 x3 A1
= 22 x − − +k
2 2 3
x = 1, v= 3.98 (329…) hence k = 2. Now put x = 1.5 M1 Either use limits or find k and put x = 1.5
1.52 1.53
22 × 1.5 − 32 × − 0.2 × + 2 = −1.225
2 3
v2 A1
As cannot be negative, P comes to rest before the extension of the
2
string is 0.5.
4
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Mathematics 9709/51 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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