A Level Mathematics 9709 Paper 43, May/June 2024

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9709/43 · May/June 2024

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Two particles $P$ and $Q$ of masses $0.2 kg$ and $0.5 kg$ respectively are at rest on a smooth horizontal plane. Particle $P$ is projected with a speed $6 m s^{- 1}$ directly towards $Q$ . After $P$ and $Q$ collide, $P$ moves with a speed of $1 m s^{- 1}$ . Find the two possible speeds of $Q$ after the collision.

[3]

A particle of mass $0.2 kg$ is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point on a vertical wall. The particle is held in equilibrium by a force of magnitude $X N$ , perpendicular to the string, with the string taut and making an angle of $30^{\circ}$ with the wall (see diagram). Find the tension in the string and the value of $X$ .

[3]

A car travels along a straight road with constant acceleration $a m s^{- 2}$ , where $a > 0$ . The car passes through points $A$ , $B$ and $C$ in that order. The speed of the car at $A$ is $u m s^{- 1}$ in the direction $A B$ . The distance $B C$ is twice the distance $A B$ . The car takes $8 seconds$ to travel from $A$ to $B$ and $10 seconds$ to travel from $B$ to $C$ .

Find $u$ in terms of $a$ .

[4]

Find the speed of the car at $C$ in terms of $a$ .

[2]

A particle travels in a straight line. The velocity of the particle at time $t s$ after leaving a point $O$ is $v m s^{- 1}$ , where $v = k t^{2} - 4 t + 3.$ The distance travelled by the particle in the first $2 s$ of its motion is $6 m$ . You may assume that $v > 0$ in the first $2 s$ of its motion.

Find the value of $k$ .

[4]

Find the value of the minimum velocity of the particle. You do not need to show that this velocity is a minimum.

[3]

A van of mass $4500 kg$ is towing a trailer of mass $750 kg$ down a straight hill inclined at an angle of $θ$ to the horizontal where $\sin θ = 0.05$ . The van and the trailer are connected by a light rigid tow-bar which is parallel to the road. There are constant resistance forces of $2500 N$ on the van and $300 N$ on the trailer.

It is given that the tension in the tow-bar is $450 N$ . Find the acceleration of the trailer and the driving force of the van’s engine.

[4]

b(i)

On another occasion, the van and trailer ascend a straight hill inclined at an angle of $α$ to the horizontal where $\sin α = 0.09$ . The driving force of the van’s engine is now $9100 N$ , and the speed of the van at the bottom of the hill is $20 m s^{- 1}$ . The resistances to motion are unchanged. Find the acceleration of the van and the tension in the tow-bar.

[5]

b(ii)

Find the speed of the van when it has travelled a distance of $375 m$ up the hill.

[2]

A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and her bicycle is $80 kg$ . There is a constant resistance force of magnitude $32 N$ to the cyclist’s motion. At an instant when she is travelling at $7 m s^{- 1}$ , her acceleration is $0.1 m s^{- 2}$ .

Find the power output of the cyclist.

[3]

Find the steady speed that the cyclist can maintain if her power output and the resistance force are both unchanged.

[2]

The cyclist later descends a straight hill of length $32.2 m$ , inclined at an angle of $\sin^{- 1} (\frac{1}{20})$ to the horizontal. Her power output is now $120 W$ , and the resistance force now has variable magnitude such that the work done against this force in descending the hill is $1128 J$ . The time taken to descend the hill is $4 s$ . Given that the speed of the cyclist at the top of the hill is $7.5 m s^{- 1}$ , find her speed at the bottom of the hill.

[6]

The diagram shows a track $A B C D$ which lies in a vertical plane. The section $A B$ is a straight line inclined at an angle of $30^{\circ}$ to the horizontal and is smooth. The section $B C$ is a horizontal straight line and is rough. The section $C D$ is a straight line inclined at an angle of $30^{\circ}$ to the horizontal and is rough. The lengths $A B$ , $B C$ and $C D$ are each $2 m$ . A particle is released from rest at $A$ . The coefficient of friction between the particle and both $B C$ and $C D$ is $μ$ . There is no change in the speed of the particle when it passes through either of the points $B$ or $C$ .

It is given that $μ = 0.1$ . Find the distance which the particle has moved up the section $C D$ when its speed is $1 m s^{- 1}$ .

[5]

It is given instead that with a different value of $μ$ the particle travels $1 m$ up the track from $C$ before it comes instantaneously to rest. Find the value of $μ$ and the speed of the particle at the instant that it passes $C$ for the second time.

[4]

Mark scheme

9709/43 · May/June 2024

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1 · 3 marks

M1
For attempt at use of conservation of momentum in at least one case. Must have three non-zero terms. Allow sign errors. Must have correct masses with relevant velocities. Their $v$ may be in opposite direction.
A1
Speed $= 2 {ms}^{- 1}$
A1
Speed $= 2.8 [0] {ms}^{- 1}$ or $\frac{14}{5} {ms}^{- 1}$ or $2 \frac{4}{5} {ms}^{- 1}$

2 · 3 marks

M1
$X \cos 30 - T \sin 30 = 0$
M1
$X \sin 30 + T \cos 30 - 0.2 g = 0$
A1
$X = 1$ , Tension $= 1.73 N [1.7320..]$ or $\sqrt{3} N$

3a · 4 marks

B1
For any one correct expression
B1
For two correct expressions
M1
Attempt to solve simultaneously to obtain an equation in $u$ and $a$ only.
A1
$u = 11 a$

3b · 2 marks

M1
$v = 11 a + a \times 18$
A1
Speed $= 29 a$

4a · 4 marks

M1
For attempt at integration
A1
$\frac{1}{2 + 1} k t^{2 + 1} - \frac{4}{2} t^{1 + 1} + 3 t [= \frac{1}{3} k t^{3} - 2 t^{2} + 3 t] [+ c]$
M1
$\frac{1}{3} k \times 2^{3} - 2 \times 2^{2} + 3 \times 2 [- 0] = 6$
A1
$k = 3$

4b · 3 marks

M1
$2 \times 3 t - 4$ or at min value $t = \frac{- b}{2 a} = \frac{4}{2 \times 3}$
A1
$[2 \times 3 t - 4 = 0 \Rightarrow] t = \frac{2}{3}$
A1
$[v = 3 \times {(\frac{2}{3})}^{2} - 4 \times \frac{2}{3} + 3 =] \frac{5}{3} {ms}^{- 1}$

5a · 4 marks

M1
Use of Newton’s second law for van or trailer or system
A1
Trailer: $450 + 750 g \times 0.05 - 300 = 750 a$ $[525 = 750 a]$ Van: $D + 4500 g \times 0.05 - 2500 - 450 = 4500 a$ $[D - 700 = 4500 a]$ System: $D + 4500 g \times 0.05 + 750 g \times 0.05 - 2500 - 300 = (4500 + 750) a$ $[D - 175 = 5250 a]$
M1
For attempt to solve for $a$ or $D$
A1
$a = 0.7 {ms}^{- 2}$ and $D = 3850 N$

5b(i) · 5 marks

M1
Use of Newton’s second law for van or trailer or system
A1
Trailer: $T - 300 - 750 g \times 0.09 = 750 a$ $[T - 975 = 750 a]$ Van: $9100 - 2500 - 4500 g \times 0.09 - T = 4500 a$ $[2550 - T = 4500 a]$ System: $9100 - 2500 - 300 - (4500 + 750) g \times 0.09 = (4500 + 750) a$ $[1575 = 5250 a]$
A1
Second correct equation
M1
For attempt to solve for $a$ or $T$
A1
$T = 1200 N$ and $a = 0.3 {ms}^{- 2}$

5b(ii) · 2 marks

M1
$v^{2} = 20^{2} + 2 \times their 0.3 \times 375$
A1
$v = 25 {ms}^{- 1}$

6a · 3 marks

B1
$D F = \frac{P}{7}$
M1
$D - 32 = 80 \times 0.1$
A1
$[Power =] 280 W$

6b · 2 marks

M1
$[At steady speed driving force =] 32 = \frac{280}{v}$
A1
Steady speed $= 8.75 {ms}^{- 1}$ or $\frac{35}{4} {ms}^{- 1}$

6c · 6 marks

B1
$120 \times 4$ $[480]$
B1
$\frac{1}{2} \times 80 v^{2}$ $[40 v^{2}]$ or $\frac{1}{2} \times 80 \times {7.5}^{2}$ $[2250]$
B1
$80 g \times \frac{1}{20} \times 32.2$ $[1288]$ $[80 \times 10 \times 1.61]$
M1
Attempt at work-energy equation
A1
$120 \times 4 + 80 g \times \frac{1}{20} \times 32.2 - 1128 = \frac{1}{2} \times 80 (v^{2} - {7.5}^{2})$ $[480 + 1288 - 1128 = 40 v^{2} - 2250]$
A1
Speed $= 8.5 [0] {ms}^{- 1}$ or $\frac{17}{2}$

7a · 5 marks

B1
For CD $R = m g \cos 30$
M1
Use of $F = 0.1 R$ for either BC or CD $F_{B C} = 0.1 m g [= m]$ OR $F_{C D} = 0.1 m g \cos 30 [= \frac{\sqrt{3}}{2} m]$
A1
$0.1 m g \cos 30 d + m g d \sin 30 [(\frac{\sqrt{3}}{2} + 5) m d]$
M1
$m g \times 2 \sin 30 = 0.1 m g \times 2 + 0.1 m g \cos 30 d + m g d \sin 30 + \frac{1}{2} m \times 1^{2}$ $[10 m = 2 m + m \cos 30 d + 5 m d + \frac{1}{2} m]$
A1
$d = 1.28 m$ or $\frac{15 (10 - \sqrt{3})}{97}$ $[1.27854 \dots]$

7b · 4 marks

M1
$m g \times 2 \sin 30 = 2 μ m g + 1 \times μ m g \cos 30 + m g \times 1 \sin 30$ $[10 m = 20 m μ + 10 m \cos 30 μ + 5 m OR 10 m = 20 m μ + m 5 \sqrt{3} μ + 5 m]$
A1
$μ = 0.174$ or $\frac{4 - \sqrt{3}}{13}$ $[0.174457 \dots]$
M1
$m g \times 1 \sin 30 - 1 \times μ m g \cos 30$ $[5 m - 5 \sqrt{3} m μ]$
A1
$m g \times 1 \sin 30 - 1 \times μ m g \cos 30 = \frac{1}{2} m v^{2}$ $[5 m - 5 \sqrt{3} m μ = \frac{1}{2} m v^{2}]$ Speed $= 2.64 {ms}^{- 1}$ $[2.64164 \dots]$

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