Mathematics 9709/42 May/June 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
40 N
17 N
55Å
1Å 20Å
32 N
PN
Coplanar forces of magnitudes 40 N, 32 N, P N and 17 N act at a point in the directions shown in the diagram. The system is in equilibrium. Find the values of P and 1. [6]
Mark scheme
1 [P cos θ = 32 cos 20 – 17 sin 55] M1 Resolve forces horizontally or vertically
[P sin θ = 40 + 17 cos 55 – 32 sin 20] 3 terms horizontally, 4 terms vertically
A1 One correct
A1 Both correct
[P sin θ = 38.8062 P cos θ = 16.1446]
M1 Either use Pythagoras to find P
P= (17 cos55 − 32sin 20 + 40 )2 + ( 32cos 20 − 17 cos35)2 or use their value of θ to find P
(17 cos55 − 32sin 20 + 40) M1 Either use trigonometry to find θ
θ = tan −1 or use their value of P to find θ
(32cos 20 − 17cos35)
[tan θ = 2.4037]
P = 42(.0) and θ = 67.4 A1
6
A car moves in a straight line with initial speed u m s−1 and constant acceleration a m s−2 . The car takes 5 s to travel the first 80 m and it takes 8 s to travel the first 160 m. Find a and u. [6]
Mark scheme
2 Possible equations include: M1 Use the equation
t = 0 to t = 5 → 80 = 5u + 12.5a s = ut + ½at 2 to set up one equation in u and a
t = 0 to t = 8 → 160 = 8u + 32a
t = 5 to t = 8 → 80 = 3(u + 5a) + 4.5a or
i.e. 80 = 3u + 19.5a using speeds as u (at t = 0), u + 5a (at t = 5), u + 8a (at t = 8)
and then apply s = ½ × (u + v) × t
80 = 5u + ½ × a × 52 → 5u + 12.5a = 80 A1 One correct equation in a and u
160 = 8u + 0.5a × 82 → 8u + 32a = 160 A1 Second correct equation in a and u
M1 Attempt to solve a pair of valid simultaneous equations for a
or u
8 A1 Allow a = 2.67
a=
3
28 A1 Allow u = 9.33
u=
3
6
5.
A particle of mass 13 kg is on a rough plane inclined at an angle of 1 to the horizontal, where tan 1 = 12
The coefficient of friction between the particle and the plane is 0.3. A force of magnitude T N, acting parallel to a line of greatest slope, moves the particle a distance of 2.5 m up the plane at a constant speed. Find the work done by this force. [5]
Mark scheme
3 R = 13g cos 22.6 = 13g × (12/13), [R = 120] B1 Resolve perpendicular to the plane
F = 0.3 × 13g cos 22.6 [F = 36] M1 Using F = µR
T = F + 13g sin 22.6 = F + 13g × (5/13), [T = 86] M1 Apply Newton’s second law parallel to the plane with a = 0
WD = T × 2.5 [= 86 × 2.5] M1 WD = T × d
WD = 215 J A1
Alternative method for question 3
R = 13g cos 22.6 = 13g × (12/13), [R = 120] B1 Resolve perpendicular to the plane
F = 0.3 × 13g cos 22.6 [F = 36] M1 Using F = µR
PE gain = 13 × g × 2.5 × (5/13) [= 125] M1 Attempt PE gain. Allow sin 22.6 for 5/13
[WD by T = 13 × g × 2.5 × (5/13) + F × 2.5] M1 Using WD by T = PE gain + WD against F
WD by T = 215 J A1
5
A constant resistance to motion of magnitude 350 N acts on a car of mass 1250 kg. The engine of the car exerts a constant driving force of 1200 N. The car travels along a road inclined at an angle of 1 to the horizontal, where sin 1 = 0.05. Find the speed of the car when it has moved 100 m from rest in each of the following cases.
• The car is moving up the hill.
• The car is moving down the hill. [7]
Mark scheme
4 [1200 – 350 – 1250 × 10 × 0.05 = 1250a] M1 Apply Newton’s second law for motion up the hill
[a = 225/1250 = 0.18] A1 Correct Newton’s law for motion up the hill
[1200 – 350 + 1250 × 10 × 0.05 = 1250a] M1 Apply Newton’s second law for motion down the hill
[a = 1475/1250 = 1.18] A1 Correct Newton’s law for motion down the hill
Up the hill: v 2 = 0 + 2 × 0.18 × 100 M1 Use their a in the constant acceleration equations either to
Down the hill: v 2 = 0 + 2 × 1.18 × 100 find v going up or going down the hill
Up the hill: v = 6 ms–1 A1
Down the hill: v = 15.4 ms–1 A1 Allow v = 2√59
Alternative method for question 4
[1200 × 100 = 350 × 100 + 1250g × 100 × 0.05 + ½ × 1250 × v 2] M1 Attempt the work-energy equation for motion up the hill
A1 Correct work-energy equation for motion up the hill
[1200 × 100 + 1250g × 100 × 0.05 = 350 × 100 + ½ × 1250 × v 2] M1 Attempt work-energy equation for motion down the hill
A1 Correct work-energy equation for motion down the hill
M1 Attempt to solve either energy equation to find
either v going up the hill or v going down the hill
Up the hill: v = 6 ms–1 A1
Down the hill: v = 15.4 ms–1 A1 Allow v = 2√59
7
A B
0.4 kg 0.2 kg
0.5 m
Two particles A and B, of masses 0.4 kg and 0.2 kg respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. Both A and B are 0.5 m above the ground. The particles hang vertically (see diagram). The particles are released from rest. In the subsequent motion
B does not reach the pulley and A remains at rest after reaching the ground.
(i) For the motion before A reaches the ground, show that the magnitude of the acceleration of each particle is 10
3 m s−2 and find the tension in the string. [4]
(ii) Find the maximum height of B above the ground. [4]
Mark scheme
5(i) A: 4 – T = 0.4a M1 Apply Newton’ second law to particle A (3 terms)
B: T – 2 = 0.2a or to particle B (3 terms)
System: 4 – 2 = (0.4 + 0.2)a or to the system (4 terms implied)
A1 Two correct equations
M1 Either solve the system equation for a
or solve two simultaneous equations for a or T
or verify the given value of a by finding the same T value in
both equations
10 8 A1 Both correct AG
a= ,T=
3 3
4
5(ii) M1 Apply v 2 = u 2 +2as to particle A or particle B with a = 10/3
v2 = 0 + 2 × 10/3 × 0.5 A1 [v = 1.83 but not needed specifically]
1 M1 Apply v 2 = u 2 + 2as to particle B to find s, the distance
0 = 10/3 – 2 × 10 × s [s = ]
6 travelled by B after A has hit the ground
7 A1 Maximum height = 1/2 + 1/2 + 1/6 = 7/6 = 1.17
Maximum height = = 1.17 m
6
4
A car has mass 1000 kg. When the car is travelling at a steady speed of v m s−1 , where v > 2, the resistance to motion of the car is Av + B N, where A and B are constants. The car can travel along a horizontal road at a steady speed of 18 m s−1 when its engine is working at 36 kW. The car can travel up a hill inclined at an angle of 1 to the horizontal, where sin 1 = 0.05, at a steady speed of 12 m s−1 when its engine is working at 21 kW. Find A and B. [7]
Mark scheme
6 Case 1: DF = 36000/18 B1 DF = P/v in either case
or
Case 2: DF = 21000/12
18A + B = DF M1 Use DF = resistance (case 1)
[36000/18 = 18A + B = 2000]
18A + B = 2000 oe A1 Correct equation, unsimplified
12A + B = DF + weight component M1 Use DF = resistance + weight component (case 2)
[21000/12 = 12A + B + 1000 g × 1/20]
12A + B = 1250 oe A1 Correct equation, unsimplified
DM1 Solve two simultaneous equations in A and B only for A or B
Dependent on both previous M1’s
A = 125, B = –250 A1 Both correct
7
Particles P and Q leave a fixed point A at the same time and travel in the same straight line. The velocity of P after t seconds is 6t t − 3 m s−1 and the velocity of Q after t seconds is 10 − 2t m s−1 .
(i) Sketch, on the same axes, velocity-time graphs for P and Q for 0 ≤ t ≤ 5. [3]
(ii) Verify that P and Q meet after 5 seconds. [4]
(iii) Find the greatest distance between P and Q for 0 ≤ t ≤ 5. [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) Straight line, reaching positive v-axis and positive t-axis (negative B1
gradient)
Quadratic (U shape, through (0,0) and cutting t-axis at t < 5) B1
Fully correct graphs with correct labelling with B1
t = 3, t = 5, v = 10, v = 60 seen
3
7(ii) s = ∫ (10 − 2t ) dt = 10t − t 2 (+ c) B1 Use either integration to find s for Q or use a correct formula
to find the area under the relevant triangle
or use area of a triangle ½ × 10 × 5 [= 25]
M1 Use integration to find the displacement for P
( )
s = ∫ 6t 2 − 18t dt = 2t 3 − 9t 2 ( + c ) A1 Correct integration for P (unsimplified)
5 B1 Either
s ( P ) = 2t 3 − 9t 2 = 25
0 evaluation of s(P) at t = 5 and show that at t = 5, s(P) = s(Q)
or solve = 25
10t − t 2 = 2t 3 − 9t 2 or show that t = 5 is a solution of the cubic by solving
or verify t = 5 is a solution of the cubic by substitution.
4
7(iii) Distance PQ = |sP – sQ| = ±(2t 3 – 8t 2 – 10t) M1 Find the distance between P and Q Allow either sign
sP and sQ must have been found by integration
Maximum s if 6t 2 – 16t – 10 = 0 M1 Differentiate to obtain an equation in t and attempt to solve
t = 3.19 A1
Maximum Distance PQ = (–)48.4 m A1
Alternative method for question 7(iii)
6t 2 – 18t = 10 – 2t M1 State that greatest distance between P and Q occurs when vP
= vQ
6t 2 – 16t – 10 = 0 M1 Rearrange and attempt to solve for t
t = 3.19 A1
Maximum Distance PQ = (–)48.4 m A1
4
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