Mathematics 9709/41 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A block of mass 12 kg is being pulled by a rope up a rough plane. The plane is inclined at an angle of 20° above the horizontal. The rope pulling the block is parallel to a line of greatest slope of the plane. The coefficient of friction between the block and the plane is 0.4. The acceleration of the block is 2 m s -2 .
Find the tension in the rope. [4]
Mark scheme
1 R = 12 g cos 20 B1 112.7631145.
T − F − 12 g sin 20 = 12 × 2 *M1 For use of N2L with 4 terms. Allow sign errors. Allow sin/cos
mix. Allow g missing.
Allow with their Friction or just F.
T − 0.4 × 12 g cos 20 − 12 g sin 20 = 12 × 2 DM1 Use of F = 0.4R where R is a component of weight, to get an
[T − 45.1052− 41.0424. = 24] equation in T only.
Tension = 110 N [110.147…] A1 AWRT 110 CWO.
4
5N
PN
60°
30° O
10 N
Three coplanar forces of magnitudes P N, 5 N and 10 N act at a point O, as shown in the diagram. The resultant of the three forces has magnitude Q N and acts in a direction perpendicular to the force of magnitude P N.
Find the value of P and the value of Q. [4]
Mark scheme
2 P + 5sin60 −10sin30 = 0 M1 For resolving in direction of P.
3 terms; allow sign errors and allow sin/cos mix.
3 A1 Must be positive.
P =5−5 or 0.670 [0.66987…]
2
Q = ± (10cos30 − 5cos60 ) M1 For resolving in direction of Q.
3 terms; allow sign errors and allow sin/cos mix.
3 5 A1 Must be positive.
Q = 10 − or 6.16 [6.16025…]
2 2
4
v (m s–1)
3
0 t (s)
0 10 40
The diagram shows the velocity-time graph of the motion of a cyclist. The graph consists of three straight line segments. The cyclist passes a point O with speed 3 m s –1 and then accelerates for 10 s with constant acceleration 0.5 m s –2. He then travels at constant speed for 30 s before decelerating, coming to rest at point P, covering a distance of 80 m whilst decelerating.
(a) Find the total time taken for the journey from O to P. [3]
(b) On the given axes, sketch a displacement-time graph for the cyclist’s journey from O to P, showing on your graph the distances travelled after 10 s and 40 s. [4]
s (m)
0 t (s)
0 10 40
Mark scheme
3(a) Velocity at t = 10 is 8 [m s–1] B1
1 M1 Attempt to find time for deceleration.
× 8 (T − 40 ) = 80 Using their 8.
2
1
OR × 8 × t = 80 [ t = 20] and T = their t + 40
2
Total time = 60 s A1
3
3(b) Distance after 10 s = 55 m B1FT FT their 8 15 + 5v .
Distance after 40 s = 295 m B1FT FT their 8 15 + 35v .
B1 For at least one quadratic the right way up or a straight line with
positive gradient between t = 10 and t = 40 .
B1 For fully correct, smooth and continuous at t = 10 and t = 40 .
Ignore values on s axis.
29 Ignore curve for t >6 0 if shown on t-axis.
5
5
5
10 40
4
A lorry of mass 18 000 kg is travelling along a straight road.
(a) On a horizontal section of the road, the power of the lorry’s engine is constant. There is a constant resistance to motion of 1600 N.
(i) The steady speed which the lorry can maintain with the engine working at power P W is 30 m s –1.
Find the value of P. [1]
(ii) At an instant when the speed of the lorry is 16 m s –1, its engine is working at a power of 40 kW.
Find the acceleration of the lorry at this instant. [2]
(b) When the lorry has reached a speed of 20 m s –1, it begins to ascend a section of road inclined at an angle a° to the horizontal. The engine now works at a power of 120 kW. There is no change in the lorry’s speed as it ascends the hill. The constant resistance to motion remains 1600 N.
Find the value of a. [3]
Mark scheme
4(a)(i) P B1
– 1600 = 0 P = 48000 only
30
1
4(a)(ii) 40000 M1 For use of N2L with 3 dimensionally correct relevant terms. Allow
− 1600 = 18000a sign errors.
16
40
Allow .
16
Acceleration = 0.05 m s–2 A1
2
4(b) For attempt to resolve up the hill to form an equation M1 3 relevant terms, allow sign errors, allow sin/cos mix, allow g
missing.
120
Allow .
20
120000 A1
− 1600 − 18000 g sin α = 0
20
α = 1.40 [1.4007] A1
3
When a particle P of mass m kg has speed u m s –1, its momentum is 4 N s and its kinetic energy is 16 J.
(a) Find the value of m and the value of u. [3]
P is now projected on a smooth horizontal surface with speed v m s –1 directly towards a particle Q of mass 1.25 kg which is stationary. After P and Q collide, the velocity of P is w m s –1 and the velocity of
Q is 2w m s –1. The loss of kinetic energy in the collision is 25 J.
(b) Find the value of v and the value of w. [4]
Mark scheme
5(a) 1 2 B1
mu = 16
2
mu = 4 B1
u = 8, m = 0.5 B1
3
5(b) 0.5 × v = 0.5w + 1.25 × 2w [v = 6w] *M1 Using their value of m . Correct number of terms, allow sign
errors.
1 1 1 2 *M1 Using their value of m
× 0.5 × v 2 − × 0.5 × w2 + × 1.25 × ( 2 w ) = ±25
2 2 2
1 25 DM1 For equation in w or v only.
9 w2 − w2 − 2.5w2 − 25 = 0 w2 = 25 Dependent on the two previous M marks
4 4
w = 2, v = 12 A1 For both.
Use of mg in momentum A0
4
B
P Q
0.3 kg 0.6 kg
θ° 30°
A C
Two particles, P and Q, of masses 0.3 kg and 0.6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth pulley fixed at a point B where the inclined planes
AB and BC meet. P lies on the smooth plane AB which is inclined at an angle i° to the horizontal where sin i° = 0.4 . Q lies on the plane BC which is inclined at 30° to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes (see diagram). The particles are released from rest.
(a) It is given that the plane BC is smooth.
Find the tension in the string and the acceleration of Q. [5]
(b) It is given instead that the plane BC is rough. The work done against the frictional force when Q moves 2 m down the plane is 1.8 J. You should assume that P does not reach the pulley and that Q does not reach C.
Use an energy method to find the speed of Q when it has moved 2 m down the plane. [4]
Mark scheme
6(a) Use of Newton’s second law for P or Q or system *M1 Allow g missing. Correct number of terms. Allow sin/cos mix.
Allow sign errors.
Allow using θ = 24 ° or better.
0.6 g sin 30 − T = 0.6a A2 A1 For any one correct equation.
T − 0.3g × 0.4 = 0.3a A2 For any two correct equations.
0.6 g sin 30 − 0.3g × 0.4 = ( 0.3 + 0.6 ) a
For attempt to solve for T or a DM1 Must get to ‘T =’ or ‘a =’.
From equations with the correct number of relevant terms.
a = 2 m s-2 A1 Allow ±2 .
T = 1.8 N
5
6(b) PE change for P = ±0.3g × 2 × 0.4 [ = ±2.4] B1 For either.
PE change for Q = ±0.6 g × 2sin 30 [ = ±6]
1 B1
KE change = × ( 0.3 + 0.6 ) v 2
2
1 M1 Attempt at work-energy equation. 4 terms; dimensionally correct.
× 0.9v 2 = 0.6 g × 2sin 30 − 0.3 g × 2 × 0.4 − 1.8 Allow sign errors. Do not allow missing g.
2
Speed = 2 m s-1 A1
Special Case for use of N2L
0.6 g sin 30 − 0.3g sin − 0.9 = 0.9a → a = 1 B1
v2 = 02 + 2x1x2 → v = 2 B1
4
A particle X moves along a straight track, starting from a point O at time t = 0 . The displacement of X
3 from O at time t s is s m, where s = 3t 2 - 6t .
(a) Find the time at which X is instantaneously at rest, and hence find the total distance travelled by X between t = 0 and t = 16. [6]
A second particle Y moves along another straight track, starting from a point P at time t = 0 . The acceleration of Y at time t s is a m s –2, where a = 0.8 - 0.6 t. The velocity of Y when it leaves P is
7.5 m s –1.
(b) When the velocity of Y is –9.6 m s –1, show that the displacement of X from O is equal to the displacement of Y from P. [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
7(a) For an attempt at differentiation *M1 The power of t must decrease by 1 in at least 1 term with a change
of coefficient in the same term.
9
1 A1 Allow un-simplified.
v = t2 −6
2
16 M1 For attempt to solve for t.
v =0t = Condone sign errors and transposing fractions.
9
Correct squaring of √𝑡.
DM1
16 2
3 16 16
3
16 896 Use of limits 0 and their or their and 16 correctly.
3 × 16 − 6 × 16 − 3 × − 6 × =
2 9 9
9 9 9
3
16 2 16 32
Or 3 × − 6 × − 0 = −
9 9 9
16 16 DM1
For both i.e. F (16 ) − F their − F their − F ( 0 )
9 9
16 16
OR F (16 ) − F their + F their − F ( 0 )
9 9
16
OR F (16 ) − 2 × F their
9
928 A1 AWRT 103 m
m
9
6
7(b) For integration *M1 The power of t must increase by 1 in at least 1 term with a change
of coefficient in the same term.
v = 0.8t − 0.3t 2 + 7.5 A1
v = −9.6 0.3t 2 − 0.8t − 17.1 = 0 DM1 For using v = −9.6 where v = pt + qt 2 + r where p ≠ 0 and q ≠ 0
(no restriction for r) and attempt to solve for t.
19 A1
t = 9 or −
3
s = 0.4t 2 − 0.1t 3 + 7.5t ( +c ) B1FT FT their v which is a 3 term quadratic in t.
3 B1
At t = 9 s X = 3 × 9 2 − 6 × 9 = 27
At t = 9 sY = 0.4 × 92 − 0.1× 93 + 7.5 × 9 = 27 B1
7
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