A Level Mathematics 9709 Paper 41, Oct/Nov 2023

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Question paper

9709/41 · Oct/Nov 2023

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1

A particle of mass 1.6 kg is projected with a speed of 20 m s−1 up a line of greatest slope of a smooth plane inclined at α to the horizontal, where tan⁡α=34. Use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest.

[3]
2

A particle of mass 2.4 kg is held in equilibrium by two light inextensible strings, one of which is attached to point A and the other attached to point B. The strings make angles of 35∘ and 40∘ with the horizontal (see diagram). Find the tension in each of the two strings.

[5]
Figure 1 of 1 for question 2
3

The diagram shows the velocity-time graph for the motion of a bus. The bus starts from rest and accelerates uniformly for 8 seconds until it reaches a speed of 12.6 m s−1. The bus maintains this speed for 40 seconds. It then decelerates uniformly in two stages. Between 48 and 62 seconds the bus decelerates at a m s−2 and between 62 and 70 seconds it decelerates at 2a m s−2 until coming to rest.

Figure 1 of 1 for question 3
a

Find the distance covered by the bus in the first 8 seconds.

[1]
b

Find the value of a.

[3]
c

Find the average speed of the bus for the whole journey.

[4]
4

Two particles P and Q, of masses 6 kg and 2 kg respectively, lie at rest 12.5 m apart on a rough horizontal plane. The coefficient of friction between each particle and the plane is 0.4. Particle P is projected towards Q with speed 20 m s−1.

a

Show that the speed of P immediately before the collision with Q is 103 m s−1.

[3]
b

In the collision P and Q coalesce to form particle R. Find the loss of kinetic energy due to the collision.

[4]
c

The coefficient of friction between R and the plane is 0.4. Find the distance travelled by particle R before coming to rest.

[2]
5

The diagram shows a particle A, of mass 1.2 kg, which lies on a plane inclined at an angle of 40∘ to the horizontal and a particle B, of mass 1.6 kg, which lies on a plane inclined at an angle of 50∘ to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley P fixed at the top of the planes. The parts AP and BP of the string are taut and parallel to lines of greatest slope of the respective planes. The two planes are rough, with the same coefficient of friction, μ, between the particles and the planes. Find the value of μ for which the system is in limiting equilibrium.

[7]
Figure 1 of 1 for question 5
6

A car of mass 1300 kg is moving on a straight road.

a(i)

On a horizontal section of the road, the car has a constant speed of 30 m s−1 and there is a constant force of 650 N resisting the motion. Calculate, in kW, the power developed by the engine of the car.

[2]
a(ii)

Given that this power is suddenly increased by 9 kW, find the instantaneous acceleration of the car.

[3]
b

On a section of the road inclined at sin⁡−10.08 to the horizontal, the resistance to the motion of the car is (1000+20v) N when the speed of the car is v m s−1. The car travels downwards along this section of the road at constant speed with the engine working at 11.5 kW. Find this constant speed.

[4]
7

A particle moves in a straight line starting from a point O before coming to instantaneous rest at a point X. At time t s after leaving O, the velocity v m s−1 of the particle is given by v=7.2t20⩽t⩽2,v=30.6−0.9t2⩽t⩽8,v=1600t2+kt8⩽t, where k is a constant. It is given that there is no instantaneous change in velocity at t=8. Find the distance OX.

[9]

Mark scheme

9709/41 · Oct/Nov 2023

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1 · 3 marks

  • B1

    ±1.6g×x×35[=9.6x] or ±12×1.6×202[=320]

  • M1

    12×1.6×202=1.6g×xsin⁡α where sin⁡α=35

  • A1

    x=1003

2 · 5 marks

  • M1

    Attempt to resolve horizontally or vertically to form an equation.

  • A1

    T1cos⁡35=T2cos⁡40

  • A1

    T1sin⁡35+T2sin⁡40=2.4g

  • 1 mark

    Attempt to solve for either tension.

  • A1

    T1=20.4 N and T2=19.0 N

3a · 1 mark

  • B1

    Distance=50.4 m

3b · 3 marks

  • M1

    v1=12.6−(62−48)a

  • M1

    0=v1−2a×(70−62)

  • A1

    a=0.42

3c · 4 marks

  • B1

    Speed at time t=62 is 6.72 ms−1

  • B2

    s2=(48−8)×12.6[=504] s3=0.5×(12.6+their 6.72)×(62−48)[=135.24 or 338125 oe] or their 6.72×(62−48)+0.5×(62−48)×(12.6−their 6.72) s4=0.5×their 6.72×(70−62)[=26.88 or 67225 oe]

  • B1

    Average speed=10.236 ms−1

4a · 3 marks

  • B1

    −0.4×6g=6a

  • 1 mark

    v2=202+2×(−4)×12.5

  • A1

    v2=300⇒v=103

4b · 4 marks

  • M1

    6×103=(6+2)v′

  • A1

    v′=7.53

  • B1

    Initial KE=12×6×(103)2[=900] Final KE=12×8×(7.53)2[=675]

  • A1

    Loss of KE=225 J

4c · 2 marks

  • M1

    0=(their 7.53)2+2×(their −4)×s

  • A1

    [Distance=]21.1 m

5 · 7 marks

  • M1

    Resolving parallel to the slope at A or B to form an equation.

  • A1

    1.6gsin⁡50−T−FB=0

  • A1

    T−FA−1.2gsin⁡40=0

  • B1

    RA=1.2gcos⁡40 or RB=1.6gcos⁡50

  • M1

    FA=1.2gμcos⁡40 or FB=1.6gμcos⁡50

  • 1 mark

    1.6gsin⁡50−1.6gμcos⁡50=1.2gsin⁡40+1.2gμcos⁡40

  • A1

    [μ=1.6gsin⁡50−1.2gsin⁡401.2gcos⁡40+1.6gcos⁡50⇒]μ=0.233

6a(i) · 2 marks

  • B2

    Power=19.5 kW

6a(ii) · 3 marks

  • B1

    their 19500+9000=DF×30

  • M1

    DF−650=1300a

  • A1

    a=313=0.231 ms−2

6b · 4 marks

  • B1

    DF=11500v

  • M1

    Attempt at Newton’s second law.

  • A1

    11500v+1300×g×0.08−(1000+20v)=0

  • A1

    Speed=25 ms−1

7 · 9 marks

  • M1

    30.6−0.9×8=160082+8k

  • A1

    k=−0.2

  • 1 mark

    1600t2+(their k)×t=0⇒t=…

  • M1

    Attempt to integrate v for one of the 3 intervals

  • A1

    s=7.23t3(+c)

  • A1

    s=30.6t−0.92t2(+c)

  • A1

    s=1600−1t−1+k2t2(+c)

  • B1

    Either 19.2 or 156.6 or ±86.4

  • B1

    Distance=19.2+(216−59.4)+(−120−(−206.4))=262.2 m

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