Mathematics 9709/42 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.
A particle of mass 1.6 kg is dropped from a height of 9 m above horizontal ground. The speed of the particle at the instant before hitting the ground is 12 m s−1 .
Find the work done against air resistance. [3]
Mark scheme
1 122 = 2 × 9 × a . OR a = 8 . M1 Use of suvat to get an equation in a.
1.6 g − R = 1.6a . M1 Use Newton’s second law with 3 terms, allow sign errors.
[may see R = 3.2 ] Allow their a ≠ g. Allow a if it isn’t subsequently replaced
with g.
WD [ = 3.2 × 9] = 28.8 J A1
Alternative method for Question 1
1 B1 Allow for the expression for KE.
(KE =) × 1.6 × 122 OR 115.2
2
(Loss of PE =) ±1.6 g × 9 OR ±144 B1 Allow for the expression for PE.
WD = 28.8 J B1 Allow if get –28.8 and then say 28.8 without explanation.
Do not allow –28.8 as final answer to working, so if get
28.8 and state –28.8 then ISW.
3
Two particles A and B, of masses 3.2 kg and 2.4 kg respectively, lie on a smooth horizontal table.
A moves towards B with a speed of v m s−1 and collides with B, which is moving towards A with a speed of 6 m s−1 . In the collision the two particles come to rest.
(a) Find the value of v. [2]
(b) Find the loss of kinetic energy of the system due to the collision. [2]
Mark scheme
2(a) ± 3.2v + 2.4 × ( −6 ) = 0 M1 Attempt at conservation of momentum; 2 non-zero terms;
allow sign errors.
v = 4.5 A1 M1A0 for use of mgv.
v = −4.5 is A0.
2
2(b) 1 1 M1 Attempt at either KE term, using their v.
KE = ± × 3.2 × ( their 4.5 ) OR ± × 2.4 × 62
2
2 2 1
Do not allow × 3.2 × ( their 4.5 ± 6 ) ,
2
2
1
or × 2.4 × ( their 4.5 ± 6 ) ,
2
2
1
or × ( 3.2 + 2.4 ) × ( their 4.5 ± 6 ) ,
2
2
1
or × 3.2 × ( their 4.5 − 0 ) ,
2
2
1
or × 2.4 × ( 6 − 0 ) .
2
2
KEloss = 75.6 J A1 Allow –75.6.
1
Note × ( 3.2 + 2.4 ) × 62
2
1
or × ( 3.2 + 2.4 ) × ( their 4.5) is M1A0.
2
2
2
30 N 15 N
1 !
1 !
PN
33 N
Coplanar forces of magnitudes 30 N, 15 N, 33 N and P N act at a point in the directions shown in the diagram, where tan ! = 43 . The system is in equilibrium.
14.4 2 28.8 2
@ A @ A
(a) Show that + = 1. [4]
30 − P P + 30
(b) Verify that P = 6 satisfies this equation and find the value of 1. [2]
Mark scheme
3(a) Resolving either direction. M1 Correct number of terms, allow sign errors, allow sin/cos
mix.
Do not allow with just sin α and cos α.
3 A1 OE, but see note for final A1.
( 33 + 15) × = P cosθ + 30cosθ Allow:
5
28.8 = ( P + 30 ) cosθ
4
OR ( 33 + 15 ) cos tan −1 = P cosθ + 30cosθ ( 33 + 15) cos53 (.1) = P cosθ + 30cosθ
3 19.81 + 9.01 = P cos θ + 30 cos θ
19.86 + 9.03 = P cos θ + 30cos θ .
OR 19.8 + 9 = P cos θ + 30 cos θ
4 4 A1 OE, but see note for final A1.
15 × + 30sin θ = 33 × + P sin θ Allow:
5 5
14.4 = ( 30 − P ) sin θ
4 4 15sin 53 (.1) + 30sin θ = 33sin 53 (.1) + P sin θ
OR 15sin tan −1 + 30sin θ = 33sin tan −1 + P sin θ
3 3 12.00 + 30sin θ = 26.39 + P sin θ
11.98 + 30sin θ = 26.35 + P sin θ .
OR 12 + 30sin θ = 26.4 + P sin θ
28.8 14.4 A1 AG. Must have evidence of where 28.8 and 14.4 come
[Use cos 2 θ + sin 2 θ = 1 with] cosθ = and sin θ = to get from.
P + 30 30 − P
A0 for any error seen.
2 2 A0 if use of inexact angles seen.
14.4 28.8 Any inexact decimals seen for force components, i.e. if
+ =1
30 − P P + 30 14.4 and/or 28.8 have come from rounding to 3sf,
scores M1A1A1A0 max 3/4.
If exact values of sin α and cos α not shown (e.g.
28.8 = ( P + 30 ) cosθ or 14.4 = ( 30 − P ) sin θ
from no working), this scores M1A1A1A0 max 3/4
marks.
4
3(b) 2
14.4 28.8
2 B1 2
3 4
2
Sub P = 6 into + Must see either + = 1 or 0.36 + 0.64 = 1 as
30 − P P + 30 5 5
14.4 2 28.8 2 3 2 4 2 minimum working.
to get + = + = 0.36 + 0.64 = 1
24 36 5 5
θ = 36.9 B1 AWRT 36.9 .
2
An athlete of mass 84 kg is running along a straight road.
(a) Initially the road is horizontal and he runs at a constant speed of 3 m s−1 . The athlete produces a constant power of 60 W.
Find the resistive force which acts on the athlete. [1]
(b) The athlete then runs up a 150 m section of the road which is inclined at 0.8Å to the horizontal.
The speed of the athlete at the start of this section of road is 3 m s−1 and he now produces a constant driving force of 24 N. The total resistive force which acts on the athlete along this section of road has constant magnitude 13 N.
Use an energy method to find the speed of the athlete at the end of the 150 m section of road. [6]
Mark scheme
4(a) 60 B1
Resistive force = DF = = 20 N
3
1
4(b) PE = ±84 g × 150sin 0.8 B1 ± 1759.23…
1 1 B1 1
KE change = ± × 84v 2 − × 84 × 32 ± × 84v 2 − 378
2 2 2
Work Done = ± ( 24 × 150 − 13 × 150 ) B1 ± ( 3600 − 1950 ) = ±1650
Attempt at work-energy equation. M1 5 terms, dimensionally correct, allow sign errors, sin/cos
mix on PE term, PE must include sin 0.8 or cos 0.8 .
1 1 A1
84 g × 150sin 0.8 + × 84v 2 − × 84 × 32 = 24 × 150 − 13 × 150
2 2
1759.23 + 42v 2 − 378 = 3600 − 1950 → 42v 2 = 268.765
[v =] 2.53 m s–1 A1 AWRT 2.53; 2.5296…
6
Special case for use of constant acceleration: Maximum 4 marks
Resolve parallel to slope and use Newton’s second law *M1 Four terms, allow sign errors, allow sin/cos mix.
24 − 13 − 84 g sin 0.8 = ±84a A1 For reference a = ±0.008669
Use constant acceleration formula to get an equation in v o r v 2 DM1 E.g. v 2 = 32 + 2 × ( their a ) × 150.
[v =] 2.53 m s–1 A1 AWRT 2.53; 2.5296…
4
PN 0.6 kg
35Å
A particle of mass 0.6 kg is placed on a rough plane which is inclined at an angle of 35Å to the horizontal. 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 plane is 0.4.
Find the least possible value of P. [6]
Mark scheme
5 Attempt at resolving parallel or perpendicular to the plane. *M1 3 terms, allow sign errors, allow sin/cos mix, allow g
missing.
Forces that need resolving should be resolved, forces that
do not need resolving should not be resolved.
R = P sin 35 + 0.6 g cos35 R = ( 0.573) P + 4.914 A1
F + P cos35 = 0.6 g sin 35 F + ( 0.819) P = 3.441 A1 Their F.
Use of F = 0.4 R *M1 Where R is initially a linear combination of a P
component and a weight component (or a mass
component).
Solve for P. DM1 From equations with the correct number of relevant
resolved terms.
0.6 g
R= = 5.7222 .
cos35 + 0.4sin 35
0.6 g sin 35 − 0.4 × 0.6 g cos35
Must get to P = …, e.g. P =
cos35 + 0.4sin 35
If no working seen, allow this mark if correct solution for
their equations.
If F ⩽ 0.4R used, it should be used correctly.
e.g. 0.6g sin 35 – P cos 35 ⩽ 0.4(P sin 35 + 0.6g cos 35).
P = 1.41 A1 AWRT 1.41 .
If P ⩾ 1.41 seen, must then state the least value explicitly
for A1.
5 Alternative for Question 5: Resolving vertically and horizontally
Attempt at resolving vertically or horizontally. *M1 3 terms, allow sign errors, allow sin/cos mix, allow g
missing.
Forces that need resolving should be resolved, forces that
do not need resolving should not be resolved.
R cos35 + F sin 35 = 0.6 g A1 Their F or R.
P + F cos 35 = R sin 35 A1 Their F or R.
Use of F = 0.4 R *M1 To get 2 equations, one in R (or F) and the other in P and
R (or P and F) from resolved equations with correct
number of relevant terms. Allow g missing.
Solve for P DM1 From equations with the correct number of relevant
resolved terms.
0.6 g
May see R = = 5.7222 ;
cos35 + 0.4sin 35
0.6 g sin 35 − 0.4 × 0.6 g cos35
Must get to P = …, e.g. P = .
cos35 + 0.4sin 35
If no working seen, allow this mark if correct solution for
their equations.
P = 1.41 A1 AWRT 1.41 .
6
A particle P starts at rest and moves in a straight line from a point O. At time t s after leaving O, the
3 velocity of P, v m s−1 , is given by v = bt + ct 2 , where b and c are constants. P has velocity 8 m s−1 when t = 4 and has velocity 13.5 m s−1 when t = 9.
(a) Show that b = 3 and c = −0.5. [1]
(b) Find the acceleration of P when t = 1. [2]
(c) Find the positive value of t when P is at instantaneous rest and find the distance of P from O at this instant. [5]
(d) Find the speed of P at the instant it returns to O. [3]
Mark scheme
6(a) 3 3 B1 Must have 2 correct equations, which do not have to be
4b + 4 2 c = 8 [ → 4b + 8c = 8] and 9b + 9 2 c = 13.5 [ → 9b + 27c = 13.5] State simplified.
b = 3 and c = −0.5 Allow to just state the values of b and c.
Allow substitution of b = 3 and c = – 0.5 in both equations
OR to verify. No further calculation required.
3 3 B0 if any incorrect work seen.
3 × 4 + ( −0.5 ) × 4 2 = 8 AND 3 × 9 + ( −0.5 ) × 9 2 = 13.5
1
6(b) dv 3
1
3
1 M1 Attempt to differentiate, decrease power by 1 and a
a = = 3 − 0.5 × × t 2 OR b + c × × t 2 change in coefficient in at least one term (which must be
dt 2 2
v
the same term); allow unsimplified; a = is M0.
t
acceleration = 2.25 m s–2 A1 9 1
OE, e.g. or 2 .
4 4
SC B1 for 2.25 if no differentiation seen.
2
6(c) Equate v to 0 and attempt to solve for t M1 Must get to t = … and must be positive.
t = 36 ONLY A1 WWW. Allow if t = 0 seen and not rejected.
Attempt to integrate M1 Increase power by 1 and a change in coefficient in at least
one term (which must be the same term).
s = vt is M0.
3 2
5
3 1
5 A1 Allow unsimplified (including indices).
s = t 2 − 0.5 × × t 2 ( + D ) = t 2 − t 2 ( + D )
2 5 2 5
5
b 2
OR s = t 2 + c × × t 2 ( + D )
2 5
Sub t = 36 (or use limits 36 and 0) to get distance = 388.8 m ONLY A1 Allow 389 m.
If no integration seen for the last 3 marks, allow SC B1
for 388.8 m. Max M1A1B1 for 3/5 marks.
5
6(d) 3 2 2
5 M1 Equate their s (that has come from an integration attempt)
t − 0.5 × × t 2 = 0 to 0 and attempt to solve for t. Must get to t =
2 5
t = 56.25 A1 225
WWW. OE, e.g. . Allow 2sf or better.
4
Allow any correct unsimplified expression equivalent to
152
56.25 e.g. 2 or 7.52 .
2
Speed = 42.2 m s–1 ONLY A1 AWRT 42.2 . Speed = – 42.2 is A0. Allow A1 if negative
sign dropped without justification.
3
P
2 kg
A
0. 8 m B
1. 2 m
C
30Å
Q 0.25 kg
Two particles P and Q, of masses 2 kg and 0.25 kg respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. Particle P is on an inclined plane at an angle of 30Å to the horizontal. Particle Q hangs below the pulley. Three points A, B and C lie on a line of greatest slope of the plane with AB = 0.8 m and BC = 1.2 m (see diagram).
Particle P is released from rest at A with the string taut and slides down the plane. During the motion of P from A to C, Q does not reach the pulley. The part of the plane from A to B is rough, with coefficient of friction 0.3 between the plane and P. The part of the plane from B to C is smooth.
(a) (i) Find the acceleration of P between A and B. [4]
(ii) Hence, find the speed of P at C. [5]
(b) Find the time taken for P to travel from A to C. [4]
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)(i) Particle P: 2 g sin 30 − F − T = 2a [10 − F − T = 2a ] M1 Newton’s second law on either particle or for the system
Particle Q: T − 0.25 g = 0.25a with correct masses; correct number of terms, allow
sin/cos mix, allow sign errors.
Allow with their F.
System: 2 g sin 30 − F − 0.25 g = ( 2 + 0.25 ) a
A1 Both particle equations correct (with the same T) or
system equation correct.
Allow with their F.
If their a direction is different to ours, allow if their a is
consistently used e.g. −2 g sin 30 + F + T = 2a ' and
0.25 g − T = 0.25a ' .
F = 0.3R = 0.3 × 2 g cos30 = 3 3 = 5.1961. M1 Use of F = 0.3R, where R is a component of weight.
Acceleration from A to B = 1.02 m s–2 A1 Solving for the acceleration from A to B.
10 − 4 3
Allow ; AWRT 1.02 .
3
10 − 3
May see T = = 2.7559
3
4
7(a)(ii) Use of suvat from A to B to get an equation in v 2 or v . *M1 Using u = 0 and their a from (a)(i) to get a positive v 2 .
E.g. v 2 = 2 × 0.8 × ( their 1.02 )
or v 2 = 2 × ( −0.8 ) × ( −their 1.02 ) .
80 − 32 3 A1 Not v = 1.29 .
v2 = = 1.64 OR v = 1.28 Allow 2sf or better without wrong work,
15
i.e. v 2 = 1.6 or v = 1.3 .
Find the acceleration from B to C: 2 g sin 30 − 0.25 g = ( 2 + 0.25 ) a *M1 Resolving on both particles and eliminate T (if their a
direction is different to ours, allow if their a is
consistently used) OR for the system to get an equation in
a only. Correct number of relevant terms, allow sin/cos
mix, allow sign errors.
10
For reference a = or 3.33 .
3
10
May see T = .
3
10 DM1 Use of suvat from B to C, allow their positive a ≠ g from
v 2 = ( their 1.28 ) + 2 × their × 1.2
2
3 (a)(ii) not their a from (a)(i) and their 1.28 .
Dependent on previous two marks.
Velocity = 3.1(0) m s–1 A1 AWRT 3.1(0) to 3sf.
7(a)(ii) Alternative Method for Question 7(a)(ii): using suvat in first stage and energy in second stage
Use of suvat from A to B to get an equation in v 2 or v . *M1 Using their positive a from (a)(i)
E.g. v 2 = 2 × 0.8 × ( their 1.02 ) .
80 − 32 3 A1 Allow 2sf or better, i.e. v 2 = 1.6 or v = 1.3 .
2
v = = 1.64 OR v = 1.28
15
Change in PE = ± ( 2 g × 1.2sin 30 − 0.25 g × 1.2 ) B1
1 1 2
OR Change in KE = ± ( 2 + 0.25 ) v 2 − ( 2 + 0.25 )( their 1.28 )
2 2
1 1 DM1 Use of work-energy 6 terms; dimensionally correct. Allow
( 2 + 0.25) v 2 − ( 2 + 0.25)( their 1.28) = 2 g × 1.2sin 30 − 0.25 g × 1.2
2
2 2 sign errors. Allow sin/cos mix on PE.
Dependent on previous M.
Velocity = 3.1(0) m s–1 A1 AWRT 3.1(0) to 3sf.
7(a)(ii) Alternative Method for Question 7(a)(ii): using energy for complete motion
Change in PE = ± ( 2 g × 2sin 30 − 0.25 g × 2 ) B1
Work done against friction = 0.3 × 2 g cos30 × 0.8 B1
1 B1
Change in KE = ( 2 + 0.25) v 2
2
1 M1 Use of work-energy 5 terms; dimensionally correct.
( 2 + 0.25) v 2 + 0.3 × 2 g cos30 × 0.8 = ( 2 g × 2sin 30 − 0.25 g × 2 ) Must be considering both particles.
2
Allow sign errors. Allow sin/cos mix on PE and/or WD
against friction.
Velocity = 3.1(0) m s–1 A1 AWRT 3.1(0) to 3sf.
7(a)(ii) Alternative Method for Question 7(a)(ii): using energy in two stages
1 *M1 Use of work-energy 5 terms; dimensionally correct.
( 2 + 0.25) v 2 + 0.3 × 2 g cos30 × 0.8 = 2 g × 0.8sin 30 − 0.25 g × 0.8 Allow sign errors. Allow sin/cos mix on PE.
2
1 OR Use of work-energy 4 terms; dimensionally correct.
OR × 2 × v 2 + ( their 2.7559) × 0.8 + 0.3 × 2 g cos30 × 0.8 = 2 g × 0.8sin 30 Allow sign errors. Allow sin/cos mix on PE.
2
OR Use of work-energy 3 terms; dimensionally correct.
Allow sign errors.
1
OR × 0.25 × v 2 + 0.25 g × 0.8 = ( their 2.7559) × 0.8
2
80 − 32 3 A1 Allow 2sf or better, i.e. v 2 = 1.6 or v = 1.3 .
v2 = = 1.64 OR v = 1.28
15
Change in PE = ± ( 2 g × 1.2sin 30 − 0.25 g × 1.2 ) B1
1 1
OR Change in KE = ± ( 2 + 0.25 ) v 2 − ( 2 + 0.25 )1.282
2 2
1 1 DM1 Use of work-energy 6 terms; dimensionally correct. Allow
( 2 + 0.25) v 2 − ( 2 + 0.25)1.282 = 2 g × 1.2sin 30 − 0.25 g × 1.2 sign errors. Allow sin/cos mix on PE.
2 2
Dependent on previous 2 marks.
Velocity = 3.1(0) m s–1 A1 AWRT 3.1(0) to 3sf.
5
7(b) 1 M1 Use of suvat from A to B to find t1 , using s = 0.8 and
0.8 = 0 + × (their positive answer to (a)(i)) × t12 and solve for t1
2 their positive a ≠ g from (a)(i).
1 Must get to t1 =
OR 0.8 = (0 + their positive 1.28 from (a)(ii)) t1 and solve for t1
2 OR using their positive 1.28 .
OR (their positive 1.28 from (a)(ii)) = (their positive answer to (a)(i)) t1 and OR using their positive a ≠ g from (a)(i) and positive 1.28 .
solve for t1
1 10 M1 Use of suvat from B to C to find t2 , using s = 1.2 and
1.2 = (their 1.28 from (a)(ii)) t2 + × their × t2 2 and solve for t2
2 3 their positive a ≠ g from (a)(ii) (not their a from (a)(i))
OR and their 1.28 which would lead to a positive t2 value.
1 Must get to t2 =
1.2 = ((their 1.28 from (a)(ii)) + (their answer to (a)(ii))) t2 and solve
2
for t2 OR using their 1.28 and/or their answer to (a)(ii) which
would lead to a positive t2 value.
OR
10 OR using their 1.28 and/or their answer to (a)(ii) and their
(their answer to (a)(ii)) = (their 1.28 from (a)(ii)) + their t2 and solve
3 a ≠ g from (a)(ii) (not their a from (a)(i)) which would
for t2 lead to a positive t2 value.
t1 = 1.25 or t2 = 0.547 A1 These can be seen as an expression.
0.8 1.2
Allow t1 = t2 = OE.
0.64 2.19
Allow 2sf or better, e.g. t1 = 1.3 or t1 = 0.55 .
Total time = 1.8(0) s A1 WWW. AWRT 1.8(0) to 3sf.
4
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