Mathematics 9709/42 February/March 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, February/March 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A lorry of mass 16 000 kg is travelling along a straight horizontal road. The engine of the lorry is working at constant power. The work done by the driving force in 10 s is 750 000 J.
(a) Find the power of the lorry’s engine. [1]
(b) There is a constant resistance force acting on the lorry of magnitude 2400 N.
Find the acceleration of the lorry at an instant when its speed is 25 m s−1 . [3]
Mark scheme
1(a) Power = 750000/10 = 75000 W or 75 kW B1 Power = WD/Time
1
1(b) Driving force DF = 75000/25 B1FT Using P = DF × v
[DF – 2400 = 16000a] M1 Using Newton’s 2nd law
a = 0.0375 ms–2 A1 3
Allow a =
80
3
A particle P of mass 0.4 kg is on a rough horizontal floor. The coefficient of friction between P and the floor is -. A force of magnitude 3 N is applied to P upwards at an angle ! above the horizontal, where tan ! = 34 . The particle is initially at rest and accelerates at 2 m s−2 .
(a) Find the time it takes for P to travel a distance of 1.44 m from its starting point. [2]
(b) Find -. [4]
Mark scheme
2(a) [1.44 = 0 + ½ × 2t2] M1 For using a complete method which would lead to an equation for
finding a value of t such as s = ut + ½ at2 with u = 0, s = 1.44 and
a=2
t = 1.2 s A1
2
3
2(b) R = 0.4g – 3 × 5 = 0.4g – 3 sin 36.9 [= 2.2] B1
4
[3 × 5 – F = 3 cos 36.9 – F = 0.4 × 2] [F = 1.6] M1 Use Newton’s 2nd law, 3 terms, to find F.
3 × 54 − 0.4 × 2 1.6 M1 F
μ = = Use of μ =
0.4 g − 3 × 53 2.2 R
μ = 0.727 A1 8
Allow μ =
11
4
B
hm
A
0.5 m
C
The diagram shows the vertical cross-section of a surface. A, B and C are three points on the cross- section. The level of B is h m above the level of A. The level of C is 0.5 m below the level of A. A particle of mass 0.2 kg is projected up the slope from A with initial speed 5 m s−1 . The particle remains in contact with the surface as it travels from A to C.
(a) Given that the particle reaches B with a speed of 3 m s−1 and that there is no resistance force, find h. [3]
(b) It is given instead that there is a resistance force and that the particle does 3.1 J of work against the resistance force as it travels from A to C.
Find the speed of the particle when it reaches C. [3]
Mark scheme
3(a) Initial KE = ½ × 0.2 × 52 B1
or Final KE = ½ × 0.2 × 32
½ × 0.2 × 52 = 0.2gh + ½ × 0.2 × 32 M1 Use conservation of energy
h = 0.8 A1
3
3(b) Apply work-energy equation from A to C M1
½ × 0.2 × 52 – 3.1 + 0.2g × 0.5 = ½ × 0.2v2 A1 Correct work-energy equation
Speed = 2 ms–1 A1
3
A cyclist travels along a straight road with constant acceleration. He passes through points A, B and
C. The cyclist takes 2 seconds to travel along each of the sections AB and BC and passes through B with speed 4.5 m s−1 . The distance AB is 54 of the distance BC.
(a) Find the acceleration of the cyclist. [5]
(b) Find AC. [2]
Mark scheme
4(a) Use the constant acceleration equations to obtain an expression for M1
either sAB or sBC in terms of a
sAB = 2 × 4.5 – ½ × a × 22 A1 or sAB = ½(vA + vB) × 2 = 9 – 2a
sBC = 2 × 4.5 + ½ × a × 22 A1 or sBC = ½(vB + vC) × 2 = 9 + 2a
4 M1 Use the given information to find a valid equation for a
[2 × 4.5 – ½a × 22 = (2 × 4.5 + ½a × 22)]
5
a = 0.5 ms–2 A1
Alternative method for question 4(a)
[4.5 = u + 2a, sAC = 4u + 8a, sAB = 2u + 2a] M1 Any two relevant equations in u, a, sAB and sAC where u is the
velocity at A
Two correct equations A1
Three correct equations A1
5 M1 Use the given information that BC = 5/4AB to find a valid equation
[2(4.5 – 2a) + 6a = {2(4.5 – 2a) + 2a}]
4 such as the one shown OE involving a only
a = 0.5 ms–2 A1
Alternative method for question 4(a)
[AC = 4.5 × 4] M1 Using AC = vB × 4 since vB is the average velocity over AC
BC = 5/9 × AC or AB = 4/9 × AC M1
BC = 10 or AB = 8 A1
[10 = 4.5 × 2 + 2a or 8 = 4.5 × 2 – 2a] M1 Using s = ut + ½ at2 for BC or s = vt – ½ at2 for AB
a = 0.5 ms–2 A1
5
4(b) sAB = 2 × 4.5 – ½ × 0.5 × 22 = 8 M1 Attempt to find the value of sAB or sBC
OR OR attempt to find sAB directly as
sBC = 2 × 4.5 + ½ × 0.5 × 22 = 10 sAC = 3.5 × 4 + ½ × a × 42 or ½ (4.5 – 2a + 4.5 + 2a) × 4
or add the 2 expressions found in 4(a) for sAB and sBC
sAC = 8 + 5/4 × 8 = 18 m A1
OR
sAC = 10 + 4/5 × 10 = 18 m
2
FN
4N
30Å !Å 3N
P
6N
Coplanar forces, of magnitudes F N, 3 N, 6 N and 4 N, act at a point P, as shown in the diagram.
(a) Given that ! = 60, and that the resultant of the four forces is in the direction of the 3 N force, find F. [3]
(b) Given instead that the four forces are in equilibrium, find the values of F and !. [5]
Mark scheme
5(a) [4 sin 30 + F sin 60 – 6 = 0] M1 Resolve forces vertically and equate to zero
Correct equation A1
F = 4.62 A1 8 8
Allow F = or F = 3
3 3
3
5(b) Resolve forces either vertically or horizontally M1
F sin α + 4 sin 30 – 6 = 0 A1 Both equations correct
and [F sin α = 4]
F cos α + 3 – 4 cos 30 = 0 [F cos α = 0.464102...]
[F2 = 42 + 0.4642] M1 Attempt to solve for F using Pythagoras or from a value found for α
or
4 0.464
F = sin 83.4 = cos83.4
−1 4 M1 Attempt to solve for α using trigonometry or from a value found for
α = tan 0.464 F
or
−1 4 −1 0.464
α = sin 4.03 = cos 4.03
F = 4.03 and α = 83.4 A1 Both correct as shown [F = 4.0268…, α = 83.382…]
5
On a straight horizontal test track, driverless vehicles (with no passengers) are being tested. A car of mass 1600 kg is towing a trailer of mass 700 kg along the track. The brakes are applied, resulting in a deceleration of 12 m s−2 . The braking force acts on the car only. In addition to the braking force there are constant resistance forces of 600 N on the car and of 200 N on the trailer.
(a) Find the magnitude of the force in the tow-bar. [2]
(b) Find the braking force. [2]
(c) At the instant when the brakes are applied, the car has speed 22 m s−1 . At this instant the car is
17.5 m away from a stationary van, which is directly in front of the car.
Show that the car hits the van at a speed of 8 m s−1 . [2]
(d) After the collision, the van starts to move with speed 5 m s−1 and the car and trailer continue moving in the same direction with speed 2 m s−1 .
Find the mass of the van. [3]
Mark scheme
6(a) [T – 200 = 700 × –12] M1 Apply Newton’s 2nd law to the trailer or apply Newton’s 2nd law to
Car: –T – 600 – F = 1600 × –12 the car and to the system and eliminate the braking force, F.
System: –600 – 200 – F = 2300 × –12
Magnitude of T = 8200 N A1
2
6(b) Car [T – F – 600 = 1600 × –12] M1 Apply Newton’s second law either to the car or to the system with
or braking force = F and use of their T from 6(a)
System [–600 – 200 – F = 2300 × –12]
Braking force F = 26800 N A1
2
6(c) [v2 = 222 + 2 × –12 × 17.5] M1 A complete method using constant acceleration equations which
would lead to an equation for finding v, using u = 22, s = 17.5 and
a = –12
v = 8 ms–1 A1 AG
2
6(d) [2300 × 8 + m × 0 = 2300 × 2 + m × 5] M1 For applying the conservation of momentum equation to the system
of car, trailer and van, where m = mass of the van
A1 Correct equation
m = 2760 kg A1
3
A particle moves in a straight line through the point O. The displacement of the particle from O at time t s is s m, where s = t2 − 3t + 2 for 0 ≤ t ≤ 6,
2
24 t s= − + 25 for t ≥ 6. t 4
(a) Find the value of t when the particle is instantaneously at rest during the first 6 seconds of its motion. [2]
At t = 6, the particle hits a barrier at a point P and rebounds.
(b) Find the velocity with which the particle arrives at P and also the velocity with which the particle leaves P. [3]
(c) Find the total distance travelled by the particle in the first 10 seconds of its motion. [5]
Additional Page
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Mark scheme
7(a) [v = 2t – 3] M1 For differentiation of s for 0 ⩽ t ⩽ 6
t = 1.5 A1
2
7(b) Velocity at arrival = 9 ms–1 B1 t = 6 used in v
24 M1 For differentiation of s for t ⩾ 6
v = − 2 − 0.5t
t
Velocity when leaves = –3.67 ms–1 A1 Allow v = –11/3
3
7(c) At t = 0, s = 2 or at t = 6, s = 20 B1 SOI
At t = 1.5, s = –0.25 B1 SOI
At t = 10, s = 2.4 B1 SOI
[Total distance = 2 + 0.25 + 0.25 + 20 + (20 – 2.4)] M1 Evidence of distance rather than displacement involving all three
sections, (0, 1.5), (1.5, 6) and (6, 10)
So total distance travelled = 40.1 m A1
5
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