Mathematics 9709/41 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A bus moves from rest with constant acceleration for 12 s. It then moves with constant speed for 30 s before decelerating uniformly to rest in a further 6 s. The total distance travelled is 585 m.
(a) Find the constant speed of the bus. [2]
(b) Find the magnitude of the deceleration. [1]
Mark scheme
1(a) 6V + 30V + 3V = 585 M1 Use of constant acceleration equations or a v-t graph.
0.5 ( 30 + 48)V = 585 Complete method to set up an equation in V using constant
acceleration equations or correct area formula in v-t graph.
Speed of the bus = 15 ms–1 A1 Must be positive.
2
1(b) Magnitude of deceleration = 2.5 B1 FT OE. Do not allow a = –2.5.
1
Two small smooth spheres A and B, of equal radii and of masses km kg and m kg respectively, where k > 1, are free to move on a smooth horizontal plane. A is moving towards B with speed 6 m s−1 and B is moving towards A with speed 2 m s−1 . After the collision A and B coalesce and move with speed 4 m s−1 .
(a) Find k. [3]
(b) Find, in terms of m, the loss of kinetic energy due to the collision. [2]
Mark scheme
2(a) Attempt at use of conservation of momentum M1 4 terms implied, i.e. m and km included before and after
collision. Velocity after collision is the same for m and km.
km × 6 − m × 2 = ( km + m ) × 4 A1
k =3 A1
3
2(b) 1 1 M1 Attempt at any of the three possible KE terms, unsimplified.
× km × 62 + × m × ( −2 )
2
KE initial = k need not be substituted here.
2 2
1
KE after = × ( km + m ) × 42
2
Loss of KE = 24m J A1 FT KE loss = 56m – 32m
FT on their k , KE loss = (10k − 6 ) m , k > 0.6 .
2
24 N
!
1 PN
20 N
!
36 N
Coplanar forces of magnitudes 24 N, P N, 20 N and 36 N act at a point in the directions shown in the diagram. The system is in equilibrium.
Given that sin ! = 35 , find the values of P and 1. [6]
Mark scheme
3 Attempt at resolving in any direction M1 Correct number of terms. No substitution for α required.
P cosθ = ( 36 − 24) cos36.9 A1
or
P cosθ = ( 36 − 24) × 0.8
P sin θ + 20 = ( 24 + 36) sin 36.9 = 14.4 + 21.6 A1
or
P sin θ + 20 = 60 × 0.6 = 36
P cosθ = 9.6, P sin θ = 16 P = 162 + 9.62 M1 Correct method for solving equations for P.
OE
16 M1 Correct method for solving equations for θ .
θ = tan −1
9.6 5
OE e.g. θ = tan −1
3
P = 18.7 A1 16 34
θ = 59[.0] Allow P = .
5
Allow P = 18.6 .
6
A particle of mass 12 kg is stationary on a rough plane inclined at an angle of 25Å to the horizontal.
A force of magnitude P N acting parallel to a line of greatest slope of the plane is used to prevent the particle sliding down the plane. The coefficient of friction between the particle and the plane is 0.35.
(a) Draw a sketch showing the forces acting on the particle. [1]
(b) Find the least possible value of P. [5]
Mark scheme
4(a) Correct 4 force diagram B1 Angles shown. F either up/down slope.
1
4(b) Attempt to resolve forces parallel to the plane M1 Three terms, allow sign errors.
P + F = 12 g sin 25 [ = 50.7] A1 Must have correct direction of F here.
R = 12 g cos 25 [= 108.8] B1
F = 120cos 25 × 0.35 [ = 38.1] M1 Attempt to solve for P using equations with the correct
number of terms.
P + 38.1 = 120 sin 25
R must be a single-term component of 12g.
P = 12.6 A1 P = 12.64926… Allow P = 12.7
5
A car of mass 1600 kg travels at constant speed 20 m s−1 up a straight road inclined at an angle of sin−1 0.12 to the horizontal.
(a) Find the change in potential energy of the car in 30 s. [3]
(b) Given that the total work done by the engine of the car in this time is 1960 kJ, find the constant force resisting the motion. [3]
(c) Calculate, in kW, the power developed by the engine of the car. [2]
(d) Given that this power is suddenly decreased by 15%, find the instantaneous deceleration of the car. [3]
Mark scheme
5(a) s = 30 × 20 B1
PE change = 1600 × g × s × 0.12 M1 Attempt change in PE. May use angle = 6.9º.
Allow sin/cos error only.
[ PE change = 1600 × g × 20 × 30 × 0.12]
Change in PE = 1152000 J A1
3
5(b) 1 960 000 = WDres + their PE M1 Using work-energy, allow sign error.
[1960 000 = WDres + 1152 000]
[WDres = 808 000 J ]
R = WDres ÷ 600 B1 Using WDres = R × 600 .
Force resisting motion = R = 1350 N to 3sf A1 4040
Allow R = N. Allow R negative.
3
Alternative method for question 5(b)
DF − R − 1600 g × 0.12 = 0 M1 R is the resisting force.
196000 9800 B1
DF = =
20 × 30 3
4040 A1 Allow R negative.
Force resisting motion = R = = 1350 N to 3sf
3
3
5(c) 4040 M1 For using P = DF × v .
P= + 1600 × g × 0.12 × 20 Allow use of their R.
3
196 000
= 3
P = 65.3 kW A1
Alternative method for question 5(c)
1960 000 M1 For using P = Work done ÷ Time .
P=
30
P = 65.3 kW A1
Alternative method for question 5(c)
9800 M1 For using P = DF × v .
P= × 20 Allow use of their DF.
3
P = 65.3 kW A1
2
5(d) 196 000 B1 FT 8330
0.85 × = DF × 20 P = DF × v DF =
3 3
FT on their P.
DF − R − 1600 g × 0.12 = 1600a M1 Newton’s 2nd law, four terms, allow sin/cos error, their R and
8330 4040 their DF.
3 − 3 − 1920 = 1600a
a = [ − ] 0.306 ms −2 A1
a = [-]
490
= [-]
49
1600 160
Alternative method for question 5(d)
9800 = DF × 20 B1 FT Using the reduction in power as the cause of the
deceleration.
9800 = 0.15 × their P = DF × v
DF = 1600 d M1
9800
20 = 1600d
a = [ − ] 0.306 ms −2 A1
a = [-]
490
= [-]
49
1600 160
3
A particle P moves in a straight line starting from a point O and comes to rest 14 s later. At time t s after leaving O, the velocity v m s−1 of P is given by v = pt2 − qt 0 ≤ t ≤ 6, v = 63 − 4.5t 6 ≤ t ≤ 14, where p and q are positive constants.
The acceleration of P is zero when t = 2.
(a) Given that there are no instantaneous changes in velocity, find p and q. [3]
(b) Sketch the velocity-time graph. [3]
(c) Find the total distance travelled by P during the 14 s. [5]
Mark scheme
6(a) a = 2 pt − q *M1 Attempt to differentiate v.
36 p − 6q = 36 DM1 For attempting to set up 2 equations using a = 0 at t = 2 and
4p − q = 0 matching the velocities at t = 6 and solve for p or q .
p = 3, q = 12 A1 Both correct.
3
6(b) Correct quadratic from t = 0 to t = 6 B1 No labelling necessary for this mark.
or
Correct straight line from 6 to 14
Both quadratic and straight line correct B1 Must join and no labelling needed.
All correct and key points shown B1 All correct, labelled at (4, 0), (6, 36) and (14, 0).
3
6(c) Attempt to integrate v *M1 Allow in terms of p and q.
s = t 3 − 6t 2 A1 FT FT on their p and q values.
4
s ( quadratic ) = t 3 − 6t 2 + t 3 − 6t 2
6 DM1 [ = 32 + 32]
0 4
Using limits correctly for t = 0 to t = 6.
Allow in terms of p and q.
14 B1
s ( triangle ) = 63t − 2.25t 2 = 144
6
or area of triangle = 144
Total distance travelled in 14 s = 208 m A1
5
A 2 kg
3 kg
B 0.45 m
18Å
Two particles A and B of masses 2 kg and 3 kg respectively are connected by a light inextensible string. Particle B is on a smooth fixed plane which is at an angle of 18Å to horizontal ground. The string passes over a fixed smooth pulley at the top of the plane. Particle A hangs vertically below the pulley and is 0.45 m above the ground (see diagram). The system is released from rest with the string taut. When A reaches the ground, the string breaks.
Find the total distance travelled by B before coming to instantaneous rest. You may assume that B does not reach the pulley. [8]
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 Particle A: 2 g − T = 2a M1 Apply Newton’s 2nd law to either particle A, or to particle B
Particle B: T − 3 g sin18 = T − 9.27 = 3a or the system.
Correct number of terms.
System: 2 g − 3g sin18 = 2 g − 9.27 = ( 2 + 3) a
A1 A and B correct or system correct.
a = 2.145898034 M1 Attempt to find a using equations with correct number of
[5a = 10.72949017] terms.
v 2 = 2 × a × 0.45 M1 Use of constant acceleration equations with their a ≠ ± g to
find v2 when A reaches the ground.
v 2 = 2 × 2.145898034 × 0.45 = 1.931308Λ A1 Allow unsimplified.
[v = 1.389715162]
T = 0, ±3 g sin18 = 3a M1 Attempt to find a for the motion of B when string becomes
[ a = ±3.0901699] slack. Allow sin/cos error, no extra terms.
[0 = 1.93 − 2 × 3.09 × s] [ s = 0.312] M1 Use constant acceleration equations, using a new a ≠ ± g , to
find the further distance, s, travelled by B before coming to
rest.
Total distance moved by B = 0.45 + 0.312 = 0.762 m A1
Alternative method for question 7
Attempt PE loss as A reaches the ground M1 Allow sin/cos error.
PE loss = 2 g × 0.45 − 3 g × 0.45sin18 A1 Correct unsimplified.
[ = 4.82827]
1 *M1 Apply work-energy equation as
2 g × 0.45 − 3g × 0.45sin18 = × ( 2 + 3) v 2 PE loss = KE gain,
2
allow sign error, sin/cos error, 4 terms implied.
7 Solve for v2 DM1
v 2 = 1.931308… [v = 1.389715162] A1
PE gain = 3g × s sin18 M1 Attempt PE gain for B after string breaks, allow sign error,
sin/cos mix,
s = extra distance travelled by B along the plane.
1 M1 Work energy equation for B as
3g × s sin18 = × 3 × 1.931308 [ s = 0.312] PE gain = KE loss, 2 terms.
2
Total distance moved by B = 0.45 + 0.312 = 0.762 m A1
8
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Mathematics 9709/41 for October/November 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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