Mathematics 9709/41 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
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]
Mark scheme
1 Use of Newton’s second law for either particle or system *M1 Correct number of terms; allow sign errors. Dimensionally
correct.
T − 1.2 g = 1.2a A1 For any 2 correct equations.
1.8 g − T = 1.8a
1.8 g − 1.2 g = (1.2 + 1.8) a
For attempt to solve for T DM1 From equations with the correct number of relevant terms.
If a found first, then substituting into an equation with the correct
number of relevant terms and solving.
a = 2ms−2 A1
Both correct.
T = 14.4 N
4
A
12.5 m
B
A particle of mass 7.5 kg, starting from rest at A, slides down an inclined plane AB. The point B is
12.5 metres vertically below the level of A, as shown in the diagram.
(a) Given that the plane is smooth, use an energy method to find the speed of the particle at B. [2]
(b) It is given instead that the plane is rough and the particle reaches B with a speed of 8 m s -1. The plane is 25 m long and the constant frictional force has magnitude F N.
Find the value of F. [3]
Mark scheme
2(a) 1 *B1 Either correct.
KE = × 7.5 × v 2
2
PE = 7.5 × g ×12.5 [= 937.5]
v = 15.8 ms −1 DB1 5 10 .
12.5
SC B1 for v 2 = 02 + 2 ( g sin θ ) v = 15.8 (or with cos).
sin θ
B0 for v 2 = 02 + 2 g × 12.5 v = 15.8 or correct answer with no
working.
2
2(b) KE B = 0.5 × 7.5 × 82 [= 240] B1
7.5 × g × 12.5 = 0.5 × 7.5 × 82 + F × 25 M1 Attempt at work energy equation; 3 terms; dimensionally correct;
allow sign errors.
F = 27.9 A1
ALTERNATIVE FOR 2(b)
82 = 02 + 2a × 25 a = 1.28 B1 Finding the correct acceleration down the plane.
12.5 M1 Newton’s second law parallel to the plane; allow sign errors and
7.5 g × − F = 7.5 × a sin/cos mix on the weight component; dimensionally correct.
25
Allow with their a, or just a .
F = 27.9 A1
3
52 N
i° 39 N
PN
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 i. [4]
Mark scheme
3 Resolving in any direction to get an equation *M1 52
Allow sin/cos mix. only – or for P 2 = 392 + 522 or tan θ = .
39
(allow reciprocal for M mark).
P cosθ = 39 P sinθ = 52 A1 52
Both correct – or for both P 2 = 392 + 522 and tan θ = .
39
P = 392 + 522 DM1 Attempt to solve for either P or θ from equations with the correct
number of relevant terms.
52
θ = tan −1
39 OE using sin/cos with P
P = 65 θ = 53.1 A1 Both correct; 53.13010…
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 m s -2 until it reaches a speed of 16 m s -1. It then travels at this constant speed before decelerating at a constant rate of 0.75 a m s -2 , coming to rest at B. The total time for the journey is 240 s.
(a) Sketch the velocity-time graph for the bus’s journey from A to B. [1]
v (m s−1)
t (s)
(b) Find an expression, in terms of a, for the length of time that the bus is travelling with constant speed. [2]
(c) Given that the distance from A to B is 3000 m, find the value of a. [3]
Mark scheme
4(a) B1 Correct shape, starting at O and finishing on the t-axis.
1
4(b) 16 64 *B1 Attempt at finding either the time for accelerating or for
t1 = , t2 = decelerating – must be in terms of a.
a 3a
16 64 DB1 OE – allow un-simplified.
T = 240 − +
a 3a
2
4(c) 1 *M1 Use distance is area under the graph.
3000 = ×16 × (T + 240) [T = 135]
2
1 16 64 DM1 Get an expression in terms of a ONLY using their T from part
3000 = ×16 × 240 + 240 − + (b) and solve for a – their T must have come from an expression
2 a 3a
k k
16 64 of the form 240 − 1 − 2 where k1 and k 2 are positive
135 = 240 − + a a
a 3a constants.
OE e.g.
1 16 1 64
3000 = 16 × 240 − ×16 × − × 16 × .
2 a 2 3a
16 A1 Allow 0.356 or better.
a=
45
3
A particle, A, is projected vertically upwards from a point O with a speed of 80 m s -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 m s -1. At time T s after the first particle is projected, the two particles collide and coalesce to form a particle C.
(a) Show that T = 3.5 . [4]
(b) Find the height above O at which the particles collide. [1]
(c) Find the time from A being projected until C returns to O. [5]
Mark scheme
5(a) sA = 80T − 12 gT 2 *M1 For use of s = ut + 1 at 2 at least once with a = ± g and u = 80 or
2
sB = 100(T −1) − 12 g(T −1)2 100 – allow t, T, t ± 1 , T ± 1.
Two correct expressions for the displacement of both particles at A1 Allow t for T.
time T
100(T − 1) − 5(T − 1) 2 = 80T − 5T 2 DM1 Equate and attempt to solve for T or t – must not be using the
same time for both expressions (so must be using the equivalent
of T in one and T ± 1 in the other).
Leading to T = 3.5 A1 AG – no errors seen (but allow all working in terms of t).
4
5(b) s = 80 × 3.5 − 12 g × 3.52 = 218.75m B1 OR 100 × 2.5 − 1 ×10 × 2.52 .
2
1
5(c) v A = 80 − g × 3.5 [ = 45] *M1 For use of v = u + at at least once to find the speed at collision
vB = 100 − g × 2.5 [= 75] with a = ± g , u = 80 or 100 – with t = 2.5 or 3.5 only (but
condone 2.5 with v A and 3.5 with v B ).
45m + 75m = 2mv DM1 Use of conservation of momentum, 3 non-zero terms, allow sign
errors. If total momentum before collision not correct then it
must be clear where both terms came from.
v = 60 A1
−218.75 = 60t − 12 g × t 2 DM1 Complete method to find an equation in t using their v, their
height from part (b) and ±g - dependent on both previous M
marks.
t = 14.9 + 3.5 =18.4s A1
5
PN 1.2 kg
i
A particle of mass 1.2 kg is placed on a rough plane which is inclined at an angle i to the horizontal,
7 where sin i = 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]
Mark scheme
6 Attempt at resolving perpendicular to the plane to get an equation *M1 Correct number of relevant terms, allow sign errors, allow sin/cos
mix, allow g missing.
For reference R = 1.2 g × cos16.26... + P × sin16.26... - allow with
an angle of 16 or better.
24 7 A1 288 7
R = 1.2 g × + P× R = 11.52 + 0.28P or R = + P.
25 25 25 25
Attempt at resolving parallel to the plane to get an equation *M1 Correct number of relevant terms, allow sign errors, allow sin/cos
mix, allow g missing.
For reference F + P × cos16.26... = 1.2 g × sin16.26...
allow with an angle of 16 or better.
24 7 A1 24 84
F + P× = 1.2 g × F + 0.96P = 3.36 or F + P= .
25 25 25 25
Use of F = 0.15R to get an equation in P only DM1 Dependent on both previous M marks – where R is initially a
linear combination of a P component and a weight component (or
a mass component).
7 24 24 7
1.2 g × − P × = 0.15 × 1.2 g × + P × .
25 25 25 25
Solve to get P = 1.63 A1 272
Allow ,1.62874...
167
6
A car has mass 1200 kg. When the car is travelling at a speed of v m s -1, there is a resistive force of magnitude kv N. The maximum power of the car’s engine is 92.16 kW.
(a) The car travels along a straight level road.
(i) The car has a greatest possible constant speed of 48 m s -1.
Show that k = 40. [1]
(ii) At an instant when its speed is 45 m s -1, find the greatest possible acceleration of the car. [3]
(b) 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]
Mark scheme
7(a)(i) Power = k × 482 = 92160 k = 40 B1 AG
1
7(a)(ii) 92160 B1 For any use of power = Fv e.g. 45 × DF = 92160 .
[DF =] [= 2048]
45
2048 − 40 × 45 = 1200a M1 Apply N2L using their DF ≠ 92160,92.16,1920.
3 terms; allow sign errors. Dimensionally correct.
248 31 A1 Allow 0.207 or better.
a= = ms−2
1200 150
3
7(b) DF = 40v + 1200 g × 0.15 *M1 Two term expression for the driving force up the hill, allow sign
errors and sin/cos mix – dimensionally correct.
92160 DM1 Set up an equation in v only – must be using DF × v = 92160 .
= 40v + 1200 g × 0.15
v
40v 2 + 1800v − 92160 [ = 0] DM1 Attempt to solve their 3TQ in v – dependent on both previous M
marks.
v = 30.5 ms −1 A1 30.51179…
4
A particle P moves in a straight line, passing through a point O with velocity 4.2 m s -1. At time t s after
P passes O, the acceleration, a m s -2 , of P is given by a = 0.6t - 2.7 .
Find the distance P travels between the times at which it is at instantaneous rest. [7]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
8 v = 0.3t 2 − 2.7t + c [c = 4.2] *M1 Attempt to integrate a – increase power by 1 and a change in
coefficient in at least one term (which must be the same term).
0.3t 2 − 2.7t + 4.2 [= 0] DM1 Set up 3TQ in t with correct constant term.
[(t − 2)(t − 7) = 0 ] t = 2,7 A1 Both correct values of t (method not required).
Attempt to integrate v DM1 Attempt to integrate v – increase power by 1 and a change in
coefficient in at least one term (which must be the same term) –
expression for v must be at least two terms (so may not include a
constant term) so dependent on first M mark only.
s = 0.1t 3 − 1.35t 2 +4.2t [ +c] A1
For use of their positive t limits in their cubic expression for s M1 Dependent on all previous M marks.
Using their two positive t values correctly in their three term
cubic expressions for s (cubic must contain non-zero t n terms
where n = 1,2 and 3).
Total distance = 6.25 m A1 For reference:
(0.1× 23 − 1.35 × 22 + 4.2 × 2)
− (0.1× 73 − 1.35 × 72 + 4.2 × 7)
If integration of v not explicitly shown, then this can score max
*M1 DM1 A1 then SC B1 for correct answer of 6.25 (so 4 marks
max.).
7
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Mathematics 9709/41 for October/November 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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