Mathematics 9709/43 May/June 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Two particles P and Q, of masses 0.1 kg and 0.4 kg respectively, are free to move on a smooth horizontal plane. Particle P is projected with speed 4 m s−1 towards Q which is stationary. After P and Q collide, the speeds of P and Q are equal.
Find the two possible values of the speed of P after the collision. [3]
Mark scheme
1 For attempt at use of conservation of momentum in one case M1 0.1 × 4 + 0 = 0.4v + 0.1v
or 0.1 × 4 + 0 = 0.4v + 0.1( −v ) OE.
Must have correct number of terms. Allow sign errors.
4 A1 Must be positive.
Speed = 0.8 [m s−1] or Allow Max M1A1A0 if g included with the masses.
5
4 A1 Must be positive.
Speed = [m s−1] Allow 1.33
3
3
A car of mass 1500 kg is towing a trailer of mass m kg along a straight horizontal road. The car and the trailer are connected by a tow-bar which is horizontal, light and rigid. There is a resistance force of F N on the car and a resistance force of 200 N on the trailer. The driving force of the car’s engine is 3200 N, the acceleration of the car is 1.25 m s−2 and the tension in the tow-bar is 300 N.
Find the value of m and the value of F. [4]
Mark scheme
2 Attempt to use Newton’s second law M1 Must have correct number of terms. Allow sign errors. Must use
300 and 1.25, not T and a.
Trailer 300 − 200 = m × 1.25 or A1 Any 2 equations.
Car 3200 − F − 300 = 1500 × 1.25 Third equation could be with their m substituted if found already.
System 3200 − F − 200 = (1500 + m ) × 1.25
Solve for m or F M1 Must get to ‘m =’ or ‘F =’. Must have correct number of terms.
Allow sign errors.
Can be implied by correct answers.
m = 80 and F = 1025 A1
4
A
XN
60Å
0.2 kg
B R
A smooth ring R of mass 0.2 kg is threaded on a light string ARB. The ends of the string are attached to fixed points A and B with A vertically above B. The string is taut and angle ABR = 90Å. The angle between the part AR of the string and the vertical is 60Å. The ring is held in equilibrium by a force of magnitude X N, acting on the ring in a direction perpendicular to AR (see diagram).
Calculate the tension in the string and the value of X . [5]
Mark scheme
3 For attempt to resolve in one direction M1 Must use 0.2 substituted for m if just awarding M1 for vertical
equation.
Must have correct number of relevant terms (forces must have
components as required). Allow sin/cos mix. Allow sign errors.
Allow g missing.
X sin 60 + T sin 30 − 0.2 g = 0 A1 OE. Correct vertical.
X cos 60 − T − T cos30 = 0 A1 OE. Correct horizontal. If the two Ts are different, they can get
max M1A1A0M0A0, unless they subsequently state that the two
T s are equal.
For attempt to solve for tension or X M1 Must have correct number of relevant terms in both equations.
Must get to ‘T =’ or ‘X =’. Allow g missing. Can be implied by
correct answers.
If no working shown their values must follow from their
equations.
X = 2, tension in string = 0.536 [N] A1 Allow exact value of tension = 4 − 2 3. Allow awrt 2.00 for X.
3 Alternative method for Question 3: Resolving parallel and perpendicular to the X N force
For attempt to resolve in one direction, with 0.2 substituted for m M1 Must have correct number of relevant terms (forces must have
components as required). Allow sin/cos mix. Allow sign errors.
Allow g missing.
X − 0.2 g cos 30 − T cos 60 = 0 A1 OE. Correct parallel to X.
T + T cos30 − 0.2 g cos 60 = 0 A1 OE. Correct perp to X. If the two Ts are different, they can get
max M1A1A0M0A0 unless they subsequently state that the two
Ts are equal.
For attempt to solve for the tension or for X M1 Must have correct number of relevant terms in both equations.
Must get to ‘T =’ or ‘X =’. Allow g missing. Can be implied by
correct answers. If no working shown their values must follow
from their equations.
X = 2, Tension in string = 0.536 [N] [0.53589…] A1 Allow exact value of tension = 4 − 2 3. Allow awrt 2.00 for X.
5
A lorry of mass 15 000 kg moves on a straight horizontal road in the direction from A to B. It passes
A and B with speeds 20 m s−1 and 25 m s−1 respectively. The power of the lorry’s engine is constant and there is a constant resistance to motion of magnitude 6000 N. The acceleration of the lorry at B is
0.5 times the acceleration of the lorry at A.
(a) Show that the power of the lorry’s engine is 200 kW, and hence find the acceleration of the lorry when it is travelling at 20 m s−1 . [5]
The lorry begins to ascend a straight hill inclined at 1Å to the horizontal. It is given that the power of the lorry’s engine and the resistance force do not change.
(b) Find the steady speed up the hill that the lorry could maintain. [2]
Mark scheme
4(a) For use of P = Fv B1 P P
P = 20 F or P = 25F OE (e.g. F = or F = ).
20 25
But not with wrong F substituted (e.g. 6000).
Attempt to use Newton’s second law in at least one case M1 Must have 3 terms. Allow sign errors. Allow F.
P P 1 A1 OE for both. Allow 2a’ and a’.
− 6000 = 15000a and − 6000 = 15000 a Must be the same P for both.
20 25 2
For solving simultaneously M1 Dependent on 2 equations of the correct form with the correct
number of relevant terms. Must get to ‘P =’ or ‘a =’, but P =
200 kW or 200 000 W with no attempt at a gets M0. Must be the
same P for both.
4 A1 AG.
Power [= 200 000W] = 200 kW, a = [m s-2] OE awrt 0.267. Do not allow 200 000 [W] as final answer. Must
15
show some working when they find P.
Alternative Method for Question 4(a): Using two expressions for P
For use of P = Fv B1 P P
P = 20 F or P = 25F OE (e.g. F = or F = ).
20 25
But not with wrong F substituted (e.g. 6000).
For one expression for P in terms of a only M1 Allow sign errors. Need 2 term expression.
(15000a + 6000 ) × 20 = (15000 × 0.5a + 6000 ) × 25 A1 Correct equation.
For solving for a M1 Must get to ‘a =’.
4 A1 AG.
Power [= 200 000W] = 200 kW, a = [m s-2] OE awrt 0.267. Do not allow 200 000 [W] as final answer. Must
15
show some working when they find P.
4(a) Alternative Method for Question 4(a): Using the given value of P = 200 kW
For use of P = Fv B1 e.g. 200 000 = 20F or 200 000 = 25F OE.
200000 200000
e.g. F = [ = 10000] or F = [ = 8000].
20 25
Attempt to use Newton’s second law in at least one case M1 Must have 3 terms. Allow sign errors. Allow with F. Allow 200 in
place of 200 000.
200000 200000 1 A1 For both. Allow 2a’ and a’ here.
− 6000 = 15000a and − 6000 = 15000 a
20 25 2
For solving for a in both cases. M1
4 A1 awrt 0.267.
For showing that both equations lead to a = [m s-2]
15
5
4(b) For attempt at resolving up hill M1 200000 200 000
Or − 6000 − 2618 = 0 . May see .
v 8618
200000 Must have correct number of terms. Allow sin/cos mix. Allow
− 6000 − 15000 g sin1 = 0
v sign errors. Allow g missing, but not a different acceleration. Do
not allow F.
Steady speed = 23.2 [m s–1] A1
2
A particle starts from rest from a point O and moves in a straight line. The acceleration of the particle
1 at time t s after leaving O is a m s−2 , where a = kt 2 for 0 ≤ t ≤ 9 and where k is a constant. The velocity of the particle at t = 9 is 1.8 m s−1 .
(a) Show that k = 0.1. [3]
For t > 9, the velocity v m s−1 of the particle is given by v = 0.2 t − 92 + 1.8.
(b) Show that the distance travelled in the first 9 seconds is one tenth of the distance travelled between t = 9 and t = 18. [4]
(c) Find the greatest acceleration of the particle during the first 10 seconds of its motion. [3]
Mark scheme
5(a) For attempt at integration M1 The power of t must increase by 1 with a change of coefficient.
Do not penalise missing c. Use of v = at scores M0.
2
3 A1 1
1 2 +1
v = kt 2 [ + c ] Allow unsimplified e.g. v = kt [ + c ] .
3 1.5
2
3
3 B1 AG.
1.8 = k × 92 k = × 1.8 ÷ 27 = 0.1 Must show values substituted OE (e.g. 1.8 = 18k).
3 2
3
5(b) For attempt at integration of either M1 The power of t or ( t − 9 ) must increase by 1 with a change of
2 3
( ) ( )
kt 2 dt or 0.2 ( t − 9 ) + 1.8 dt or 0.2t 2 − 3.6t + 18 dt
3
2 coefficient in at least one term. Use of s = vt is M0.
4 5
9 18 A1 Allow unsimplified. No need for limits. Could include ‘+c’ with
0.2
t 2 and ( t − 9 )3 + 1.8t or either or both.
150 0 3 9
18
0.2 3 3.6 2
3 t − 2 t + 18t
9
4
5
162 M1 Correct use of limits 0, 9 or limits 9, 18. Can be implied by either
= × 9 2 = 6.48 or answer following integration
150 25
0.2 243
( t − 9) + 1.8t − ,
3
“+c” method is as follows: s =
(18 − 9 )3 + 1.8 × 18 − 1.8 × 9 = 64.8 or
0.2 324
or = 3 25
3 5 0.2 3 3.6 2 1458
0.2 3.6 0.2 3 3.6 2 or s = t − t + 18t − ,
× 183 − × 182 + 18 × 18 − ×9 − × 9 + 18 × 9 3 2 25
3 2 3 2 1782 162
or then substitute t = 18 to get , subtract , so distance =
648 324 324 25 25
= 5 − 5 = 5
324
.
5
1 324 162 A1 AG OE. Check working as can get answer from wrong working
6.48 = × 64.8 or 64.8 = 10 × 6.48 or = 10 × NFWW (not from wrong working).
10 5 25
5(b) Alternative Method for Question 5(b): Special Case for those who use a calculator to integrate. Award max 2/4
9
2 3 162 SC B1
Either kt 2 dt = 6.48 or
0
3
25
18
( 0.2 ( t − 9) + 1.8) dt = 64.8 or 5
2 324
Or
9
1 324 162 SC B1 OE.
6.48 = × 64.8 or 64.8 = 10 × 6.48 or = 10 ×
10 5 25
4
5(c) For differentiation M1 The power of t or ( t − 9 ) must decrease by 1 with a change of
Should get a = 0.4 ( t − 9 ) or a = 0.4t − 3.6 v
coefficient. M0 for a = .
t
0.4 [ms-2] [at t = 10] A1 SC B1 for 0.4 with no differentiation seen.
0.3 seen (from the first phase) and state that 0.4 is final answer B1 No working needed.
If M1A0 or M0A0 scored, then SC B1 for 0.3 without mention of
the maximum acceleration.
3
An elevator is pulled vertically upwards by a cable. The elevator accelerates at 0.4 m s−2 for 5 s, then travels at constant speed for 25 s. The elevator then decelerates at 0.2 m s−2 until it comes to rest.
(a) Find the greatest speed of the elevator and hence draw a velocity-time graph for the motion of the elevator. [3]
(b) Find the total distance travelled by the elevator. [2]
The mass of the elevator is 1200 kg and there is a crate of mass m kg resting on the floor of the elevator.
(c) Given that the tension in the cable when the elevator is decelerating is 12 250 N, find the value of m. [3]
(d) Find the greatest magnitude of the force exerted on the crate by the floor of the elevator, and state its direction. [3]
Mark scheme
6(a) Greatest speed = 2 [m s–1] [ 0.4 × 5] B1 This can be seen on the graph and not stated explicitly.
Trapezium shape B1 Sitting on t-axis, starting at origin.
v B1 All correct including height of 2 and t-values of 5, 30, 40 on the
2
horizontal axis. Labels not needed.
Does not need to be to scale.
t
−5 5 10 15 20 25 30 35 40
3
6(b) 1 M1 Allow M1 for finding total area under their trapezium or
Distance = ( 25 + 5 + 25 + their 10) × their 2 appropriate ‘suvat’ in each phase. If presented as 3 areas, they do
2
not need to be added for M1. Allow one wrong value but must
1 1
or × 5 × their 2 + 25 × their 2 + × their10 × their 2 represent all 3 phases of motion.
2 2
Distance = 65 [m] A1
2
6(c) Attempt at Newton’s second law M1 Must have correct number of terms (5). Allow sign errors. Allow
g missing. Use of a = g is M0A0A0 but condone use of a = 0.4
(from wrong phase).
12250 − 1200 g − mg = (1200 + m ) × ( −0.2 ) A1 Correct equation. Note that taking a = 0.2 and omitting mg gets
M0A0A0.
Or 1200 g + mg − 12250 = (1200 + m ) × 0.2
m = 50 A1
3
6(d) Realise that this is when accelerating and attempt Newton’s M1 Must have correct number of terms (3). Allow sign errors. Allow
second law for the crate only g missing. Must use a = ±0.4, M0A0A0 otherwise.
R − 50 g = 50 × 0.4 or 50 g − R = 50 × ( −0.4 ) A1FT Correct equation using their 50.
Force R = 520 [N], upwards A1 Must include ‘upwards’ OE.
3
X
1.8 m
Y
2m
!
Z
The diagram shows the vertical cross-section XYZ of a rough slide. The section YZ is a straight line of length 2 m inclined at an angle of ! to the horizontal, where sin ! = 0.28. The section YZ is tangential to the curved section XY at Y , and X is 1.8 m above the level of Y . A child of mass 25 kg slides down the slide, starting from rest at X . The work done by the child against the resistance force in moving from X to Y is 50 J.
(a) Find the speed of the child at Y . [4]
It is given that the child comes to rest at Z .
(b) Use an energy method to find the coefficient of friction between the child and YZ , giving your answer as a fraction in its simplest form. [6]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) PE lost = mgh = 25 g × 1.8 [ = 450] B1
For work energy equation M1 Must have correct number of terms. Allow sign errors.
Dimensionally correct. Must use 25, not m.
Candidates who try to use constant acceleration can only score
B1.
1 A1 OE. Must be correct.
25g ×1.8 − 50 = × 25v 2
2
v = 4 2 [m s-1] or 5.66 [5.6568…] A1 Allow 32 .
4
7(b) PE gained/lost = ±25 g × 2 × 0.28 [ = ±140] B1FT For either. FT from their v for KE.
Must have α substituted for PE.
1
( )
2
or KE gained/lost = ± × 25 their 4 2 [KE = ± 400] Allow 25 g × 2sin16.26° or 25 g × 2sin16.3° .
2
For work energy equation *M1 Must have correct number of terms. Allow sign errors.
Dimensionally correct. Allow sin/cos mix
Do not allow with WD instead of F × 2 . Must have substituted α
and v.
( )
1 2 A1FT FT their v 2 or v .
F × 2 = 25 g × 2 × 0.28 + × 25 4 2 [⇒ F = 270]
2
R = 25 g ×0.96 [= 240] B1 Allow 25g cos16.26° or 25g cos16.3°.
Use of F = μ R to form an equation in µ only DM1 Must be from 3 term F, dimensionally correct and single term R.
Allow sin/cos mix but must be different components of weight. F
and R must be numerical expressions.
9 A1 1
μ= CAO. Allow 1 , but no other answer.
8 8
Alternative method 1 for first 3 marks: Using energy from the initial position (use existing scheme for final 3 marks).
PE lost = ±25 g × (1.8 + 2 × 0.28 ) [ = ±590] B1 Allow 25 g × (1.8 + 2sin16.26° )
or 25 g × (1.8 + 2sin16.3° ) .
For work energy equation *M1 Must have correct number of terms. Allow sign errors.
Dimensionally correct. Allow sin/cos mix.
Do not allow with WD instead of F × 2 . Must have substituted α.
F × 2 = 25 g × (1.8 + 2 × 0.28 ) − 50 [⇒ F = 270] A1
7(b) Special Case: Use of constant acceleration. Award max 4/6
( ) + 2 ( ±a ) × 2
2
0= 4 2 Use of v2 = u 2 + 2as.
a = ±8 SC FT their v 2 or v . Note: 8.01 or 8.0089 from use of 5.66.
B1FT
R = 25g ×0.96 SC B1 Allow 25 g cos16.26° or 25 g cos16.3°.
Use of F = μ R and attempt at N2L SC M1 To form an equation in μ only. Using their a.
Allow sign errors. Allow sin/cos mix but must be different
If correct should get components of weight. F and R must be numerical expressions.
25 gsin16.3° − μ × 25 gcos16.3° = 25 × ( −8 ) [ 70 − 240 μ = −200] Must have substituted α.
9 SC A1 1
μ= CAO. Allow 1 , but no other answer.
8 8
6
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