A Level Mathematics 9709 Paper 43, May/June 2025

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9709/43 · May/June 2025

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Two particles $P$ and $Q$ , of masses $0.1 kg$ and $0.3 kg$ respectively, are at rest on a smooth horizontal plane. $P$ is projected directly towards $Q$ with speed $4 u {ms}^{- 1}$ . At the same instant, $Q$ is projected directly towards $P$ with speed $u {ms}^{- 1}$ . After $P$ and $Q$ collide, $P$ moves with speed $2 {ms}^{- 1}$ and $Q$ moves with speed $4 {ms}^{- 1}$ .

Find the two possible values of $u$ .

[3]

Find the largest possible loss of kinetic energy in the collision.

[2]

A van of mass $4500 kg$ is towing a trailer of mass $350 kg$ along a straight horizontal road. The van and trailer are connected by a light rigid tow-bar which is parallel to the road. There are resistance forces of $X N$ on the van and $120 N$ on the trailer. The driving force produced by the van's engine is $2500 N$ . The tension in the tow-bar is $T N$ , and the acceleration of the van is $0.4 {ms}^{- 2}$ . Find the value of $X$ and the value of $T$ .

[4]

The diagram shows a velocity-time graph which models the motion of a particle. The graph consists of 3 straight line segments. The velocity of the particle at time $t s$ after passing a fixed point $O$ is $v {ms}^{- 1}$ . The particle leaves $O$ with a velocity of $5 {ms}^{- 1}$ and accelerates at $0.75 {ms}^{- 2}$ for $20 s$ . The particle then decelerates for the next $30 s$ . At $t = 40$ , the velocity of the particle is zero. After $t = 40$ , the particle starts to travel back to $O$ , coming to rest at $O$ at time $T s$ .

Find the value of $T$ .

[5]

Find the acceleration of the particle from $t = 50$ to $t = T$ .

[2]

Three blocks $P$ , $Q$ and $R$ , of masses $25 kg$ , $20 kg$ and $m kg$ respectively, are held in equilibrium by three light inextensible strings $O P$ , $O Q$ and $O R$ . The strings $O P$ and $O R$ both pass over small fixed smooth pulleys $A$ and $B$ respectively, with $P$ and $R$ hanging vertically below the pulleys. The block $Q$ hangs vertically below the point $O$ . The angle between $O A$ and the vertical is $30^{\circ}$ and the angle $B O Q = α^{\circ}$ (see diagram). Find the value of $m$ and the value of $α$ .

[6]

A van of mass $2500 kg$ travelling at speed $v {ms}^{- 1}$ experiences a resistance force of $k v^{2} N$ . The constant power of the van's engine is $62.5 kW$ .

The steady speed that the van could maintain when moving along a straight horizontal road is $50 {ms}^{- 1}$ . Show that $k = 0.5$ , and find the acceleration of the van when its speed is $25 {ms}^{- 1}$ on this straight horizontal road.

[4]

The van begins to ascend a hill inclined at an angle $θ^{\circ}$ to the horizontal. The van travels along a line of greatest slope of the hill. The speed of the van at the start of the hill is $20 {ms}^{- 1}$ , and its acceleration is $5 a {ms}^{- 2}$ . Later, on the same hill, the speed of the van is $30 {ms}^{- 1}$ , and its acceleration is $a {ms}^{- 2}$ . The power of the van's engine remains at $62.5 kW$ , and the resistance force remains at $0.5 v^{2} N$ . Find the value of $a$ and the value of $θ$ .

[5]

The diagram shows the vertical cross-section $A B C$ of a rough waterslide. The section $A B$ is a straight line of length $5 m$ inclined at an angle of $θ$ to the horizontal, where $\sin θ = 0.8$ . The point $B$ is $2.5 m$ above the level of $C$ . A man of mass $80 kg$ , modelled as a particle, slides down the waterslide, starting from rest at $A$ . The coefficient of friction between the man and the straight section of the waterslide is $0.1$ .

Find the speed of the man at $B$ .

[5]

It is given that there is no change in the speed of the man when passing through $B$ and that his speed at $C$ is $11 {ms}^{- 1}$ . Find the work done against the resistance force as the man moves from $B$ to $C$ .

[4]

A particle $X$ moves in a straight line. The displacement of $X$ from $O$ at time $t s$ after leaving $O$ is $s m$ , where $s = 0.3 t^{2} + 0.6 t$ for $0 ⩽ t ⩽ 4$ .

Find the velocity of $X$ at $t = 4$ .

[2]

b(i)

For $t > 4$ , the acceleration of $X$ at time $t s$ after leaving $O$ is $a {ms}^{- 2}$ , where $a = 0.3 t^{\frac{1}{2}}$ . There is no change in the velocity of $X$ at $t = 4$ . The velocity of $X$ at $t = T$ is $14.2 {ms}^{- 1}$ . Find the value of $T$ .

[4]

b(ii)

Find the total distance travelled by $X$ between $t = 0$ and $t = T$ .

[4]

Mark scheme

9709/43 · May/June 2025

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1a · 3 marks

M1
For use of conservation of momentum once. Must have correct number of terms. Allow g included with all 4 masses and sign errors only.
A1
u = 14
A1
u = 10

1b · 2 marks

M1
For expression or equivalent difference. Allow sign errors only. Using their 14 which is the larger of the 2 values found in part (a). If only one value of u found in part (a) then M0. If no value for u substituted, then M0.
A1
Largest loss = 183.6 J

2 · 4 marks

M1
Attempt at Newton’s Second law on either van, trailer or the system
A1
Any 2 correct equations: $T - 120 = 350 \times 0.4$ , $2500 - X - T = 4500 \times 0.4$ , $2500 - X - 120 = (4500 + 350) \times 0.4$
1 mark
Attempt to solve for either T or X
A1
T = 260, X = 440

3a · 5 marks

B1
Velocity at t = 20 is [5 + 3/4 * 20 =] 20 m s-1
B1
Speed at t = 50 is 10 ms-1
M1
Displacement at t = 40 is 1/2(5 + their 20) * 20 + 1/2 * 20 * their 20 [= 450] OR 1/2((their 20 - 5) * 20) + 20 * 5 + 1/2 * 20 * their 20 [= 450] OR Displacement at t = 50 is 1/2(5 + their 20) * 20 + 1/2 * (their 20 - 10) * 30 [= 400]
1 mark
1/2 * (T - 40) * their 10 = their 450 OR 1/2 * (T - 50) * their 10 + 1/2 * 10 * their 10 = their 450 OR 1/2 * (T - 50) * their 10 = their 400
A1
T = 130

3b · 2 marks

M1
Acceleration = their 10 / (their 130 - 50) OR 0 = (their(-10)) + a * (their 130 - 50)
A1
Acceleration = 1/8 m s-2 or 0.125 m s-2

4 · 6 marks

M1
For resolving in either direction to form an equation – diagram for reference
A1
25g cos30 - 20g - mg cos alpha = 0 [216.506... - 200 - 10m cos alpha = 0]
A1
25g sin30 - mg sin alpha = 0 [125 - 10m sin alpha = 0]
1 mark
alpha = tan^-1 (25g sin30 / (25g cos30 - 20g)) OR alpha = sin^-1 (25g sin30 / (their m)g)
1 mark
mg = sqrt((25g sin30)^2 + (25g cos30 - 20g)^2) OR m = 25g sin30 / (g sin(their alpha))
A1
alpha = 82.5 and m = 12.6

5a · 4 marks

M1
62500 = k * 50^2 * 50 or 1250 = k * 50^2
A1
k = 0.5
M1
62500/25 - 0.5 * 25^2 = 2500a [2500 - 312.5 = 2500a]
A1
Acceleration = 0.875 m s-2

5b · 5 marks

M1
Attempt at Newton’s second law at least once to form an equation
A2
62500/30 - 0.5 * 30^2 - 2500g sin theta = 2500a and 62500/20 - 0.5 * 20^2 - 2500g sin theta = 2500 * 5a
1 mark
62500/20 - 0.5 * 20^2 - (62500/30 - 0.5 * 30^2) = 10000a
A1
theta = 3[.00] and a = 0.129

6a · 5 marks

B1
R = 80g * 0.6 [= 480]
M1
F = 0.1 * 80g * 0.6 [= 48]
M1
80g * 0.8 - F = 80a [=> a = 7.4]
1 mark
v^2 = (0+) 2 * (their a) * 5
A1
Velocity = 8.60 m s-1

6b · 4 marks

B1
Change in KE = +/- (1/2 * 80 * 11^2 - 1/2 * 80 * (their v^2))
B1
Change in PE = +/- 80g * 2.5 [= +/- 2000]
M1
80g * 2.5 - W = 1/2 * 80 * 11^2 - 1/2 * 80 * (their v^2) [2000 - W = 4840 - 40 * (their v^2)]
A1
Work done = 120 J

7a · 2 marks

B1
v = 0.6t + 0.6
B1
Velocity at t = 4 is 3 m s-1

7b(i) · 4 marks

M1
For attempt at integration of 0.3t^(1/2) to kt^(3/2) (+c)
1 mark
Their 3 = 0.2(4)^(3/2) + c leading to c = ... [=> v = 0.2t^(3/2) + 1.4]
1 mark
14.2 = 0.2T^(3/2) + their 1.4
A1
T = 16

7b(ii) · 4 marks

B1
Distance travelled in first 4 seconds = 7.2 m
B1
s = 0.08t^(5/2) + 1.4t
1 mark
(0.08(16)^(5/2) + 1.4 * 16) - (0.08(4)^(5/2) + 1.4 * 4) [= 81.92 + 22.4 - 2.56 - 5.6 = 104.32 - 8.16 = 96.16]
A1
Distance = 103.36 m

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