A Level Mathematics 9709 Paper 42, May/June 2023

Question paper and mark scheme50 marks7 questions

Solve it on ExamPal

Practise

Untimed. Write on the paper and check as you go.

Start

Exam

Timed, with a clock you set, like sitting the real paper.

Start

Opens full screen.

Papers open in

Download PDF Mark scheme

100%

Opening the question paper

Read as textThe questions and the mark scheme as plain text. Easier to read on a phone.

Question paper

9709/42 · May/June 2023

Download PDF

A particle of mass $1.6 kg$ is dropped from a height of $9 m$ above horizontal ground. The speed of the particle at the instant before hitting the ground is $12 m s^{- 1}$ . Find the work done against air resistance.

[3]

Two particles $A$ and $B$ , of masses $3.2 kg$ and $2.4 kg$ respectively, lie on a smooth horizontal table. $A$ moves towards $B$ with a speed of $v m s^{- 1}$ and collides with $B$ , which is moving towards $A$ with a speed of $6 m s^{- 1}$ . In the collision the two particles come to rest.

Find the value of $v$ .

[2]

Find the loss of kinetic energy of the system due to the collision.

[2]

Coplanar forces of magnitudes $30 N$ , $15 N$ , $33 N$ and $P N$ act at a point in the directions shown in the diagram, where $\tan α = \frac{4}{3}$ . The system is in equilibrium.

Show that ${(\frac{14.4}{30 - P})}^{2} + {(\frac{28.8}{P + 30})}^{2} = 1$ .

[4]

Verify that $P = 6$ satisfies this equation and find the value of $θ$ .

[2]

An athlete of mass $84 kg$ is running along a straight road.

Initially the road is horizontal and he runs at a constant speed of $3 m s^{- 1}$ . The athlete produces a constant power of $60 W$ . Find the resistive force which acts on the athlete.

[1]

The athlete then runs up a $150 m$ section of the road which is inclined at ${0.8}^{\circ}$ to the horizontal. The speed of the athlete at the start of this section of road is $3 m s^{- 1}$ and he now produces a constant driving force of $24 N$ . The total resistive force which acts on the athlete along this section of road has constant magnitude $13 N$ . Use an energy method to find the speed of the athlete at the end of the $150 m$ section of road.

[6]

A particle of mass $0.6 kg$ is placed on a rough plane which is inclined at an angle of $35^{\circ}$ to the horizontal. The particle is kept in equilibrium by a horizontal force of magnitude $P N$ acting in a vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the particle and plane is $0.4$ . Find the least possible value of $P$ .

[6]

A particle $P$ starts at rest and moves in a straight line from a point $O$ . At time $t s$ after leaving $O$ , the velocity of $P$ , $v m s^{- 1}$ , is given by $v = b t + c t^{\frac{3}{2}}$ , where $b$ and $c$ are constants. $P$ has velocity $8 m s^{- 1}$ when $t = 4$ and has velocity $13.5 m s^{- 1}$ when $t = 9$ .

Show that $b = 3$ and $c = - 0.5$ .

[1]

Find the acceleration of $P$ when $t = 1$ .

[2]

Find the positive value of $t$ when $P$ is at instantaneous rest and find the distance of $P$ from $O$ at this instant.

[5]

Find the speed of $P$ at the instant it returns to $O$ .

[3]

Two particles $P$ and $Q$ , of masses $2 kg$ and $0.25 kg$ respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. Particle $P$ is on an inclined plane at an angle of $30^{\circ}$ to the horizontal. Particle $Q$ hangs below the pulley. Three points $A$ , $B$ and $C$ lie on a line of greatest slope of the plane with $A B = 0.8 m$ and $B C = 1.2 m$ (see diagram). Particle $P$ is released from rest at $A$ with the string taut and slides down the plane. During the motion of $P$ from $A$ to $C$ , $Q$ does not reach the pulley. The part of the plane from $A$ to $B$ is rough, with coefficient of friction $0.3$ between the plane and $P$ . The part of the plane from $B$ to $C$ is smooth.

a(i)

Find the acceleration of $P$ between $A$ and $B$ .

[4]

a(ii)

Hence, find the speed of $P$ at $C$ .

[5]

Find the time taken for $P$ to travel from $A$ to $C$ .

[4]

Mark scheme

9709/42 · May/June 2023

Download PDF

1 · 3 marks

M1
12² = 2 × 9 × a . OR a = 8 .
M1
1.6g − R = 1.6a . [may see R = 3.2]
A1
WD [= 3.2 × 9] = 28.8 J
B1
(KE =) ½ × 1.6 × 12² OR 115.2
B1
(Loss of PE =) ±1.6g × 9 OR ±144
B1
WD = 28.8 J

2a · 2 marks

M1
±[3.2v + 2.4 × (−6)] = 0
A1
v = 4.5

2b · 2 marks

M1
KE = ± ½ × 3.2 × (their 4.5)² OR ± ½ × 2.4 × 6²
A1
KE_loss = 75.6 J

3a · 4 marks

M1
Resolving either direction.
A1
(33 + 15) × ⅗ = P cos θ + 30 cos θ OR (33 + 15)cos(tan⁻¹(4/3)) = P cos θ + 30 cos θ OR 19.8 + 9 = P cos θ + 30 cos θ
A1
15 × ⅘ + 30 sin θ = 33 × ⅘ + P sin θ OR 15 sin(tan⁻¹(4/3)) + 30 sin θ = 33 sin(tan⁻¹(4/3)) + P sin θ OR 12 + 30 sin θ = 26.4 + P sin θ
A1
[Use cos² θ + sin² θ = 1 with] cos θ = 28.8 / (P + 30) and sin θ = 14.4 / (30 − P) to get (14.4 / (30 − P))² + (28.8 / (P + 30))² = 1

3b · 2 marks

B1
Sub P = 6 into ((14.4)/(30 − P))² + ((28.8)/(P + 30))² to get [((14.4)/24)² + ((28.8)/36)²] = (⅗)² + (⅘)² = 0.36 + 0.64 = 1
B1
θ = 36.9

4a · 1 mark

B1
(Resistive force = DF = 60/3 =) 20 N

4b · 6 marks

B1
PE = ±84g × 150 sin 0.8
B1
KE change = ±(½ × 84v² − ½ × 84 × 3²)
B1
Work Done = ±(24 × 150 − 13 × 150)
M1
Attempt at work-energy equation.
A1
84g × 150 sin 0.8 + ½ × 84v² − ½ × 84 × 3² = 24 × 150 − 13 × 150 [1759.23… + 42v² − 378 = 3600 − 1950 → 42v² = 268.765…]
A1
[v =] 2.53 ms⁻¹
M1
Resolve parallel to slope and use Newton's second law
A1
24 − 13 − 84g sin 0.8 = ±84a
M1
Use constant acceleration formula to get an equation in v or v²
A1
[v =] 2.53 ms⁻¹

5 · 6 marks

M1
Attempt at resolving parallel or perpendicular to the plane.
A1
R = P sin 35 + 0.6g cos 35 [R = (0.573…)P + 4.914…]
A1
F + P cos 35 = 0.6g sin 35 [F + (0.819…)P = 3.441…]
M1
Use of F = 0.4R
M1
Solve for P.
A1
P = 1.41
M1
Attempt at resolving vertically or horizontally.
A1
R cos 35 + F sin 35 = 0.6g
A1
P + F cos 35 = R sin 35
M1
Use of F = 0.4R
M1
Solve for P
A1
P = 1.41

6a · 1 mark

B1
4b + 4^(3/2)c = 8 [→ 4b + 8c = 8] and 9b + 9^(3/2)c = 13.5 [→ 9b + 27c = 13.5] State b = 3 and c = −0.5 OR 3 × 4 + (−0.5) × 4^(3/2) = 8 AND 3 × 9 + (−0.5) × 9^(3/2) = 13.5

6b · 2 marks

M1
(a = dv/dt =) 3 − 0.5 × 3/2 × t^(1/2) OR b + c × 3/2 × t^(1/2)
A1
acceleration = 2.25 ms⁻²

6c · 5 marks

M1
Equate v to 0 and attempt to solve for t
A1
t = 36 ONLY
M1
Attempt to integrate
A1
s = 3/2 t² − 0.5 × 2/5 × t^(5/2) (+ D) = 3/2 t² − 1/5 t^(5/2) (+ D) OR s = b/2 t² + c × 2/5 × t^(5/2) (+ D)
A1
Sub t = 36 (or use limits 36 and 0) to get distance = 388.8 m ONLY

6d · 3 marks

M1
3/2 t² − 0.5 × 2/5 × t^(5/2) = 0
A1
t = 56.25
A1
Speed = 42.2 ms⁻¹ ONLY

7a(i) · 4 marks

M1
Particle P: 2g sin 30 − F − T = 2a [10 − F − T = 2a] Particle Q: T − 0.25g = 0.25a System: 2g sin 30 − F − 0.25g = (2 + 0.25)a
A1
Both particle equations correct (with the same T) or system equation correct.
M1
F = 0.3R = 0.3 × 2g cos 30 [= 3√3 = 5.1961……]
A1
Acceleration from A to B = 1.02 m s⁻²

7a(ii) · 5 marks

M1
Use of suvat from A to B to get an equation in v² or v .
A1
v² = [(80 − 32√3)/15] = 1.64 OR v = 1.28
M1
Find the acceleration from B to C: 2g sin 30 − 0.25g = (2 + 0.25)a
M1
v² = (their 1.28)² + 2 × (their 10/3) × 1.2
A1
Velocity = 3.1(0) ms⁻¹
M1
Use of suvat from A to B to get an equation in v² or v .
A1
v² = [(80 − 32√3)/15] = 1.64 OR v = 1.28
B1
Change in PE = ±(2g × 1.2 sin 30 − 0.25g × 1.2) OR Change in KE = ±[½(2 + 0.25)v² − ½(2 + 0.25)(their 1.28)²]
M1
½(2 + 0.25)v² − ½(2 + 0.25)(their 1.28)² = 2g × 1.2 sin 30 − 0.25g × 1.2
A1
Velocity = 3.1(0) m s⁻¹
B1
Change in PE = ±(2g × 2 sin 30 − 0.25g × 2)
B1
Work done against friction = 0.3 × 2g cos 30 × 0.8
B1
Change in KE = ½(2 + 0.25)v²
M1
½(2 + 0.25)v² + 0.3 × 2g cos 30 × 0.8 = (2g × 2 sin 30 − 0.25g × 2)
A1
Velocity = 3.1(0) ms⁻¹
M1
½(2 + 0.25)v² + 0.3 × 2g cos 30 × 0.8 = 2g × 0.8 sin 30 − 0.25g × 0.8 OR ½ × 2 × v² + (their 2.7559…) × 0.8 + 0.3 × 2g cos 30 × 0.8 = 2g × 0.8 sin 30 OR ½ × 0.25 × v² + 0.25g × 0.8 = (their 2.7559…) × 0.8
A1
v² = [(80 − 32√3)/15] = 1.64 OR v = 1.28
B1
Change in PE = ±(2g × 1.2 sin 30 − 0.25g × 1.2) OR Change in KE = ±[½(2 + 0.25)v² − ½(2 + 0.25)1.28²]
M1
½(2 + 0.25)v² − ½(2 + 0.25)1.28² = 2g × 1.2 sin 30 − 0.25g × 1.2
A1
Velocity = 3.1(0) m s⁻¹

7b · 4 marks

M1
0.8 = 0 + ½ × (their positive answer to (a)(i)) × t₁² and solve for t₁ OR 0.8 = ½ (0 + their positive 1.28 from (a)(ii)) t₁ and solve for t₁ OR (their positive 1.28 from (a)(ii)) = (their positive answer to (a)(i)) t₁ and solve for t₁
M1
1.2 = (their 1.28 from (a)(ii)) t₂ + ½ × (their 10/3) × t₂² and solve for t₂ OR 1.2 = ½ ((their 1.28 from (a)(ii)) + (their answer to (a)(ii))) t₂ and solve for t₂ OR (their answer to (a)(ii)) = (their 1.28 from (a)(ii)) + (their 10/3) t₂ and solve for t₂
A1
t₁ = 1.25 or t₂ = 0.547
A1
Total time = 1.8(0) s

Questions

1
Question 13 marks
Mark scheme ›
2
Question 24 marks
Mark scheme ›
3
Question 36 marks
Mark scheme ›
4
Question 47 marks
Mark scheme ›
5
Question 56 marks
Mark scheme ›
6
Question 611 marks
Mark scheme ›
7
Question 713 marks
Mark scheme ›