Mathematics 9709/52 February/March 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Mechanics, February/March 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A particle is projected with speed 24 m s−1 at an angle of 30Å above the horizontal. Find the speed and direction of motion of the particle at the instant 4 s after projection. [5]
Mark scheme
1 x′ = 24cos30 ( = 12 3 ) B1 Use horizontal motion
y′ = 24sin30 – 4g ( = –28) B1 Use vertical motion
V 2 = ( 24cos30 ) + ( 24sin30 − 4 g ) = (12 3 )2 + (–28)2 M1 Where V is the required speed
2 2
and α is the angle below the horizontal
OR tanα = (24sin30 – 4g)/(24cos30) = –28/(12 3 )2
V = 34.9 m s–1 A1
α = 53.4° below the horizontal A1
5
1m
2m
3m
A uniform object is made by joining together three solid cubes with edges 3 m, 2 m and 1 m. The object has an axis of symmetry, with the cubes stacked vertically and the cube of edge 2 m between the other two cubes (see diagram).
(i) Calculate the distance of the centre of mass of the object above the base of the largest cube. [3]
The smallest cube is now removed from the object. It is replaced by a heavier uniform cube with 1 m edges which is made of a different material. The centre of mass of the object is now at the base of the
2 m cube.
(ii) Find the ratio of the masses of the two cubes of edge 1 m. [3]
Mark scheme
2(i) Total volume (= 27 + 8 + 1) = 36 B1
36x = 27×1.5 + 8×4 + 1×5.5 M1 Take moments about base of largest cube
x ( = 13/6 ) = 2.17 m A1
3
2(ii) Mass of new cube = 35 + m B1 Where m is the mass of the new cube
(35 + m) × 3 =27×1.5 + 8×4 + 5.5m (leads to m = 13) M1 Take moments about base of largest cube
13:1 or 1:13 A1 Accept 13
3
A small ball is projected 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, where x = 4t and y = 6t − 5t2 .
(i) Find the equation of the trajectory of the ball. [2]
(ii) Hence or otherwise calculate the angle of projection of the ball and its initial speed. [4]
Mark scheme
3(i) x = 4t and y = 6t – 5 t 2 M1 Use horizontal and vertical motion and attempt to eliminate t
y [= 6x/4 – 5 ( x / 4)2 ] = 1.5x –5 x 2 /16 or 1.5x – 0.3125 x 2 A1
2
3(ii) tanθ = 1.5 M1 Use the trajectory equation from the formula sheet
θ = 56.3° A1
V 2cos 2 56.3 =16 M1 Again use the trajectory equation
V = 7.21 m s–1 A1
OR
Vcosθ = 4 and Vsinθ =6 M1 Initial horizontal and vertical velocities
V 2cos 2 θ + V 2 sin2θ = 42 + 62 OR tanθ = 6/4 M1 Use Pythagoras's theorem or trigonometry of a right angled
triangle
V = 7.21 m s–1 A1
θ = 56.3° A1
4
A
60Å
O P
A particle P of mass 0.3 kg is attached to a fixed point A by a light elastic string of natural length
0.8 m and modulus of elasticity 16 N. The particle P moves in a horizontal circle which has centre O.
It is given that AO is vertical and that angle OAP is 60Å (see diagram). Calculate the speed of P. [6]
Mark scheme
4 Tcos60 = 0.3g M1 Resolve vertically
T=6N A1
T = 16e/0.8 ( = 6 ) leads to e = 0.3 M1 Use T =λx/L
r = (0.8 + 0.3)sin60 ( = 1.1sin60) A1
Tsin60 = 0.3 v 2 /(1.1sin60) M1 Use N2L horizontally
v = 4.06 m s–1 A1
6
A particle P 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 24 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.4 m vertically below O.
(i) Find the greatest speed of P. [5]
(ii) Calculate the greatest distance of P below O. [3]
Mark scheme
5(i) 0.3g = 24e/0.6 M1 Note greatest speed occurs at the equilibrium position. Use T=λx/L
e = 0.075 m A1 Fall = 0.275 m
PE Change = 0.3g × 0.275 B1
0.3 v 2 /2 = 0.3g × 0.275 – 24 × 0.0752 /(2 × 0.6) M1 Set up a 3 term energy equation
v = 2.18 m s–1 A1
5
5(ii) 0.3g(0.2 + E) = 24 E 2 /(2 × 0.6) M1 Set up an energy equation. Note v = 0 at the greatest distance
20 E 2 – 3E – 0.6 = 0 M1 Attempt to solve a 3 term quadratic equation
E = 0.264 and so greatest distance is 0.864 m A1
3
2r
5r
2r
Fig. 1
Fig. 1 shows the cross-section of a solid cylinder through which a cylindrical hole has been drilled to make a uniform prism. The radius of the cylinder is 5r and the radius of the hole is r . The centre of the hole is a distance 2r from the centre of the cylinder.
(i) Find, in terms of r, the distance of the centre of mass of the prism from the centre of the cylinder.
[4]
PN
30Å
Fig. 2
The prism has weight W N and is placed with its curved surface on a rough horizontal plane. The axis of symmetry of the cross-section makes an angle of 30Å with the vertical. A horizontal force of magnitude P N acting in the plane of the cross-section through the centre of mass is applied to the cylinder at the highest point of this cross-section (see Fig. 2). The prism rests in limiting equilibrium.
(ii) Find the coefficient of friction between the prism and the plane. [4]
Mark scheme
6(i) Area of hole = π r 2 and Area of original circle = 25π r 2 M1
Area of cross-section = 24π r 2 A1
π r 2 (2r) = 24π r 2 (d) M1 Take moments about the centre of the cylinder
d = r/12 ( = 0.083333....r) A1
4
6(ii) P(2 × 5r) = W(r/12)cos60 M1 Take moments about the point of contact with the plane
P = Wcos60/120 = W/240 = 0.00417W ( = F ) A1
µ = (Wcos60/120)/W M1 Use F = µR Note R = W by resolving vertically
µ = 1/240 = 0.00417 A1
4
A particle P is projected horizontally from a point O on a rough horizontal surface. The coefficient of friction between the particle and the surface is 0.2. A horizontal force of magnitude 0.06t N directed away from O acts on P, where t s is the time after projection. P comes to rest when t = 4.
(i) The particle begins to move again when t = 8. Show that the mass of P is 0.24 kg. [2]
dv
(ii) Show that, for 0 ≤ t ≤ 4, = 0.25t − 2, and find the speed of projection of P. [5]
dt
(iii) Find the distance from O at which P comes to rest. [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) 0.2mg = 0.06 × 8 M1 Resolve along the plane
m = 0.24 kg AG A1
2
7(ii) ୢ௩ ୢ௩ M1 Use N2L along the plane
m = 0.06t – 0.2mg or 0.24 = 0.06t – 0.2 × 0.24g
ୢ௧ ୢ௧
ୢ௩ A1
= 0.25t – 2 AG
ୢ௧
∫ dv = ∫ ( 0.25t − 2 ) dt M1 Attempt to integrate
v = 0.25 t 2 / 2 – 2t + c , Put v = 0 and t = 4 ( leads to c = 6 ) M1 Attempt to find c
Initial velocity = 6 m s–1 A1
5
7(iii) x = ∫ ( 0.25 t 2 /2 – 2t + 6)dt M1 Attempt to integrate
x = 0.25 t 3 /6 – t 2 + 6t ( + k ) A1ft ft candidates c from part (ii)
Finds or assumes k = 0 and substitutes t = 4 OR uses limits of 0 M1
and 4
2 A1
OP = 32/3 = 10 = 10.7 m
3
4
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Mathematics 9709/52 for February/March 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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