A Level Mathematics 9709 Paper 41, Oct/Nov 2024

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9709/41 · Oct/Nov 2024

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Two particles, of masses $1.8 kg$ and $1.2 kg$ , are connected by a light inextensible string that passes over a fixed smooth pulley. The particles hang vertically. The system is released from rest. Find the magnitude of the acceleration of the particles and find the tension in the string.

[4]

A particle of mass $7.5 kg$ , starting from rest at $A$ , slides down an inclined plane $A B$ . The point $B$ is $12.5 metres$ vertically below the level of $A$ , as shown in the diagram.

Given that the plane is smooth, use an energy method to find the speed of the particle at $B$ .

[2]

It is given instead that the plane is rough and the particle reaches $B$ with a speed of $8 {ms}^{- 1}$ . The plane is $25 m$ long and the constant frictional force has magnitude $F N$ . Find the value of $F$ .

[3]

Coplanar forces of magnitudes $52 N$ , $39 N$ and $P N$ act at a point in the directions shown in the diagram. The system is in equilibrium. Find the values of $P$ and $θ$ .

[4]

A bus travels between two stops, $A$ and $B$ . The bus starts from rest at $A$ and accelerates at a constant rate of $a {ms}^{- 2}$ until it reaches a speed of $16 {ms}^{- 1}$ . It then travels at this constant speed before decelerating at a constant rate of $0.75 a {ms}^{- 2}$ , coming to rest at $B$ . The total time for the journey is $240 s$ .

Sketch the velocity-time graph for the bus’s journey from $A$ to $B$ .

[1]

Find an expression, in terms of $a$ , for the length of time that the bus is travelling with constant speed.

[2]

Given that the distance from $A$ to $B$ is $3000 m$ , find the value of $a$ .

[3]

A particle, $A$ , is projected vertically upwards from a point $O$ with a speed of $80 {ms}^{- 1}$ . One second later a second particle, $B$ , with the same mass as $A$ , is projected vertically upwards from $O$ with a speed of $100 {ms}^{- 1}$ . At time $T s$ after the first particle is projected, the two particles collide and coalesce to form a particle $C$ .

Show that $T = 3.5$ .

[4]

Find the height above $O$ at which the particles collide.

[1]

Find the time from $A$ being projected until $C$ returns to $O$ .

[5]

A particle of mass $1.2 kg$ is placed on a rough plane which is inclined at an angle $θ$ to the horizontal, where $\sin θ = \frac{7}{25}$ . 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 the plane is $0.15$ . Find the least possible value of $P$ .

[6]

A car has mass $1200 kg$ . When the car is travelling at a speed of $v {ms}^{- 1}$ , there is a resistive force of magnitude $k v N$ . The maximum power of the car’s engine is $92.16 kW$ .

a(i)

The car travels along a straight level road. The car has a greatest possible constant speed of $48 {ms}^{- 1}$ . Show that $k = 40$ .

[1]

a(ii)

At an instant when its speed is $45 {ms}^{- 1}$ , find the greatest possible acceleration of the car.

[3]

The car now travels at a constant speed up a hill inclined at an angle of $\sin^{- 1} 0.15$ to the horizontal. Find the greatest possible speed of the car going up the hill.

[4]

A particle $P$ moves in a straight line, passing through a point $O$ with velocity $4.2 {ms}^{- 1}$ . At time $t s$ after $P$ passes $O$ , the acceleration, $a {ms}^{- 2}$ , of $P$ is given by $a = 0.6 t - 2.7$ . Find the distance $P$ travels between the times at which it is at instantaneous rest.

[7]

Mark scheme

9709/41 · Oct/Nov 2024

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1 · 4 marks

1 mark
Use of Newton’s second law for either particle or system
A1
$T - 1.2 g = 1.2 a$ $1.8 g - T = 1.8 a$ $1.8 g - 1.2 g = (1.2 + 1.8) a$
1 mark
For attempt to solve for $T$
A1
$a = 2 {ms}^{- 2}$ $T = 14.4 N$

2a · 2 marks

1 mark
$KE = \frac{1}{2} \times 7.5 \times v^{2}$ $PE = 7.5 \times g \times 12.5 [= 937.5]$
1 mark
$v = 15.8 {ms}^{- 1}$

2b · 3 marks

B1
${KE}_{B} = 0.5 \times 7.5 \times 8^{2} [= 240]$
M1
$7.5 \times g \times 12.5 = 0.5 \times 7.5 \times 8^{2} + F \times 25$
A1
$F = 27.9$

3 · 4 marks

1 mark
Resolving in any direction to get an equation
A1
$P \cos θ = 39 P \sin θ = 52$
1 mark
$P = \sqrt{39^{2} + 52^{2}}$ $θ = \tan^{- 1} (\frac{52}{39})$
A1
$P = 65 θ = 53.1$

4a · 1 mark

B1
Correct shape, starting at $O$ and finishing on the $t$ -axis.

4b · 2 marks

1 mark
$t_{1} = \frac{16}{a}, t_{2} = \frac{64}{3 a}$
1 mark
$T = 240 - (\frac{16}{a} + \frac{64}{3 a})$

4c · 3 marks

1 mark
$3000 = \frac{1}{2} \times 16 \times (T + 240) [T = 135]$
1 mark
$3000 = \frac{1}{2} \times 16 \times [240 + 240 - (\frac{16}{a} + \frac{64}{3 a})]$ $135 = 240 - (\frac{16}{a} + \frac{64}{3 a})$
A1
$a = \frac{16}{45}$

5a · 4 marks

1 mark
$s_{A} = 80 T - \frac{1}{2} g T^{2}$ $s_{B} = 100 (T - 1) - \frac{1}{2} g (T - 1)^{2}$
A1
Two correct expressions for the displacement of both particles at time $T$
1 mark
$100 (T - 1) - 5 (T - 1)^{2} = 80 T - 5 T^{2}$
A1
Leading to $T = 3.5$

5b · 1 mark

B1
$[s = 80 \times 3.5 - \frac{1}{2} g \times {3.5}^{2}] = 218.75 m$

5c · 5 marks

1 mark
$v_{A} = 80 - g \times 3.5 [= 45]$ $v_{B} = 100 - g \times 2.5 [= 75]$
1 mark
$45 m + 75 m = 2 m v$
A1
$v = 60$
1 mark
$- 218.75 = 60 t - \frac{1}{2} g \times t^{2}$
A1
$t = 14.9 + 3.5 = 18.4 s$

6 · 6 marks

1 mark
Attempt at resolving perpendicular to the plane to get an equation
A1
$R = 1.2 g \times \frac{24}{25} + P \times \frac{7}{25}$
1 mark
Attempt at resolving parallel to the plane to get an equation
A1
$F + P \times \frac{24}{25} = 1.2 g \times \frac{7}{25}$
1 mark
Use of $F = 0.15 R$ to get an equation in $P$ only
A1
Solve to get $P = 1.63$

7a(i) · 1 mark

B1
$Power = k \times 48^{2} = 92160 \Rightarrow k = 40$

7a(ii) · 3 marks

B1
$[D F =] \frac{92160}{45} [= 2048]$
M1
$2048 - 40 \times 45 = 1200 a$
A1
$a = \frac{248}{1200} = \frac{31}{150} {ms}^{- 2}$

7b · 4 marks

1 mark
$D F = 40 v + 1200 g \times 0.15$
1 mark
$\frac{92160}{v} = 40 v + 1200 g \times 0.15$
1 mark
$40 v^{2} + 1800 v - 92160 [= 0]$
A1
$v = 30.5 {ms}^{- 1}$

8 · 7 marks

1 mark
$v = 0.3 t^{2} - 2.7 t + c [c = 4.2]$
1 mark
$0.3 t^{2} - 2.7 t + 4.2 [= 0]$
A1
$[(t - 2) (t - 7) = 0 \Rightarrow] t = 2, 7$
1 mark
Attempt to integrate $v$
A1
$s = 0.1 t^{3} - 1.35 t^{2} + 4.2 t [+ c]$
M1
For use of their positive $t$ limits in their cubic expression for $s$
A1
Total distance $= 6.25 m$

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