A Level Mathematics 9709 Paper 41, May/June 2023

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9709/41 · May/June 2023

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Two particles $P$ and $Q$ , of masses $m kg$ and $0.3 kg$ respectively, are at rest on a smooth horizontal plane. $P$ is projected at a speed of $5 {m s}^{- 1}$ directly towards $Q$ . After $P$ and $Q$ collide, $P$ moves with a speed of $2 {m s}^{- 1}$ in the same direction as it was originally moving.

Find, in terms of $m$ , the speed of $Q$ after the collision.

[2]

After this collision, $Q$ moves directly towards a third particle $R$ , of mass $0.6 kg$ , which is at rest on the plane. $Q$ is brought to rest in the collision with $R$ , and $R$ begins to move with a speed of $1.5 {m s}^{- 1}$ . Find the value of $m$ .

[2]

A particle $P$ of mass $0.4 kg$ is projected vertically upwards from horizontal ground with speed $10 {m s}^{- 1}$ .

Find the greatest height above the ground reached by $P$ .

[2]

When $P$ reaches the ground again, it bounces vertically upwards. At the first instant that it hits the ground, $P$ loses $7.2 J$ of energy. Find the time between the first and second instants at which $P$ hits the ground.

[4]

A particle moves in a straight line starting from rest. The displacement $s m$ of the particle from a fixed point $O$ on the line at time $t s$ is given by $s = t^{\frac{5}{2}} - \frac{15}{4} t^{\frac{3}{2}} + 6.$ Find the value of $s$ when the particle is again at rest.

[4]

The velocity of a particle at time $t s$ after leaving a fixed point $O$ is $v {m s}^{- 1}$ . The diagram shows a velocity-time graph which models the motion of the particle. The graph consists of 5 straight line segments. The particle accelerates to a speed of $0.9 {m s}^{- 1}$ in a period of $3 s$ , then travels at constant speed for $6 s$ , and then comes instantaneously to rest $1 s$ later. The particle then moves back and returns to rest at $O$ at time $T s$ .

Find the distance travelled by the particle in the first $10 s$ of its motion.

[2]

Given that $T = 12$ , find the minimum velocity of the particle.

[2]

Given instead that the greatest speed of the particle is $3 {m s}^{- 1}$ , find the value of $T$ and hence find the average speed of the particle for the whole of the motion.

[4]

Four coplanar forces act at a point. The magnitudes of the forces are $F N$ , $10 N$ , $50 N$ and $40 N$ . The directions of the forces are as shown in the diagram.

Given that the forces are in equilibrium, find the value of $F$ and the value of $θ$ .

[6]

Given instead that $F = 10 \sqrt{2}$ and $θ = 45$ , find the direction and the exact magnitude the resultant force.

[3]

Two particles $P$ and $Q$ , of masses $0.2 kg$ and $0.1 kg$ respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley at $B$ which is attached to two inclined planes. Particle $P$ lies on a smooth plane $A B$ which is inclined at $60^{\circ}$ to the horizontal. Particle $Q$ lies on a plane $B C$ which is inclined at an angle of $θ^{\circ}$ to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes (see diagram).

It is given that $θ = 60$ , the plane $B C$ is rough and the coefficient of friction between $Q$ and the plane $B C$ is $0.7$ . The particles are released from rest. Determine whether the particles move.

[4]

It is given instead that the plane $B C$ is smooth. The particles are released from rest and in the subsequent motion the tension in the string is $(\sqrt{3} - 1) N$ . Find the magnitude of the acceleration of $P$ as it moves on the plane, and find the value of $θ$ .

[4]

A car of mass $1200 kg$ is travelling along a straight horizontal road. The power of the car's engine is constant and is equal to $16 kW$ . There is a constant resistance to motion of magnitude $500 N$ .

Find the acceleration of the car at an instant when its speed is $20 {m s}^{- 1}$ .

[3]

Assuming that the power and the resistance forces remain unchanged, find the steady speed at which the car can travel.

[2]

The car comes to the bottom of a straight hill of length $316 m$ , inclined at an angle to the horizontal of $\sin^{- 1} (\frac{1}{60})$ . The power remains constant at $16 kW$ , but the magnitude of the resistance force is no longer constant and changes such that the work done against the resistance force in ascending the hill is $128 400 J$ . The time taken to ascend the hill is $15 s$ . Given that the car is travelling at a speed of $20 {m s}^{- 1}$ at the bottom of the hill, find its speed at the top of the hill.

[6]

Mark scheme

9709/41 · May/June 2023

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1a · 2 marks

M1
$m \times 5 + 0 = m \times 2 + 0.3 v$
A1
Speed $= 10 {m (ms}^{- 1})$

1b · 2 marks

M1
$0.3 \times 10 m + 0 = 0 + 0.6 \times 1.5 [3 m = 0.9]$
1 mark
$m = 0.3$

2a · 2 marks

M1
$0 = 10^{2} + 2 (- g) s \Rightarrow s = \dots$
A1
Max. height $= 5 (m)$

2b · 4 marks

1 mark
KE before impact $= \frac{1}{2} \times 0.4 \times 10^{2} [= 20]$
1 mark
$\frac{1}{2} \times 0.4 \times v^{2} = 20 - 7.2 [v = 8]$ or $0.4 g h = 20 - 7.2 [h = 3.2]$
1 mark
$- 8 = 8 + (- g) t \Rightarrow t = \dots$ or $0 = 8 t + \frac{1}{2} \times - g \times t^{2} \Rightarrow t = \dots$
A1
Time $= 1.6 (s)$

3 · 4 marks

1 mark
For an attempt at differentiation.
A1
$(v =) \frac{5}{2} t^{\frac{3}{2}} - \frac{45}{8} t^{\frac{1}{2}}$
1 mark
$v = 0 \Rightarrow \frac{1}{8} t^{\frac{1}{2}} (20 t - 25) = 0 \Rightarrow t = \dots$
A1
$s = \frac{15}{16} [= 0.9375]$

4a · 2 marks

M1
Distance $= \frac{1}{2} (6 + 10) \times 0.9$
A1
Distance $= 7.2 (m)$

4b · 2 marks

M1
$\frac{1}{2} (12 - 10) v_{min} = - 7.2$
A1
Minimum velocity $= - 7.2 {(ms}^{- 1})$

4c · 4 marks

1 mark
$\frac{1}{2} \times (T - 10) \times 3 = 7.2 or \frac{1}{2} \times t \times 3 = 7.2$
A1
$T = 14.8$
1 mark
$\frac{14.4}{14.8}$
A1
Average speed $= \frac{36}{37} {(ms}^{- 1})$

5a · 6 marks

M1
Resolving either direction.
A1
Vertical: $F \sin θ + 40 \sin 60 - 50 = 0$
A1
Horizontal: $F \cos θ + 10 - 40 \cos 60 = 0$
M1
$θ = \tan^{- 1} (5 - 2 \sqrt{3})$
M1
$F = \sqrt{15.358.. .^{2} + 10^{2}}$
A1
$θ = 56.9, F = 18.3$

5b · 3 marks

B1
$(Y =) \pm (10 \sqrt{2} \sin 45 + 40 \sin 60 - 50) [= \pm (20 \sqrt{3} - 40)]$
B1
$(X =) \pm (10 \sqrt{2} \cos 45 + 10 - 40 \cos 60) [= 0]$
B1
Resultant force is $40 - 20 \sqrt{3} (N)$ in the same direction as the 50(N) force.

6a · 4 marks

B1
$R = 0.1 g \cos 60 [= 0.5]$
M1
$F = 0.7 \times 0.1 g \cos 60 [= 0.35]$
M1
For whole system: LHS of Newton’s second law: $0.2 g \sin 60 - 0.1 g \sin 60 - F [= 0.866... - F]$ Or separately for $P$ and $Q$ : $0.2 g \sin 60 - T (= 0.2 a)$ and $T - 0.1 g \sin 60 - F (= 0.1 a)$ , and eliminate $T$ to get $0.2 g \sin 60 - 0.1 g \sin 60 - F (= 0.3 a)$
A1
As $\frac{\sqrt{3}}{2} - 0.35 > 0$ the particles do move.

6b · 4 marks

M1
Attempt to use Newton’s second law for $P$ : $0.2 g \sin 60 - (\sqrt{3} - 1) = 0.2 a$
A1
$a = 5 {(ms}^{- 2})$
M1
Newton’s second law for system: $0.2 g \sin 60 - 0.1 g \sin θ = 0.3 (5)$ or Newton’s second law for $Q$ : $(\sqrt{3} - 1) - 0.1 g \sin θ = 0.1 (5)$
A1
$θ = 13.4$

7a · 3 marks

B1
Driving force $F = \frac{16000}{20} [= 800]$
M1
$F - 500 = 1200 a$
A1
$a = 0.25 {(ms}^{- 2})$

7b · 2 marks

M1
$\frac{16000}{v} - 500 = 0$
A1
$v = 32 {(ms}^{- 1})$

7c · 6 marks

B1
Work done by engine $= 16000 \times 15 [= 240000]$
B1
KE change $= \pm (\frac{1}{2} \times 1200 \times v^{2} - \frac{1}{2} \times 1200 \times 20^{2})$
B1
PE change $= \pm (1200 \times g \times 316 \times \frac{1}{60}) [= \pm 63200]$
M1
Attempt at work-energy equation.
A1
$16000 \times 15 - 128400 = \frac{1}{2} \times 1200 \times v^{2} - \frac{1}{2} \times 1200 \times 20^{2} + 1200 g \times 316 \times \frac{1}{60}$ $(240000 - 128400 = 600 v^{2} - 240000 + 63200)$
A1
$v = 21.9 {(ms}^{- 1})$

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