Mathematics 9709/42 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A cyclist and bicycle have a total mass of 72 kg. The cyclist rides along a horizontal road against a total resistance force of 28 N.
Find the total work done by the cyclist to increase his speed from 8 m s -1 to 16 m s -1 while travelling a distance of 100 metres. [3]
Mark scheme
1 1 B1 Correct expression for either KE or correct expression for
Initial KE = × 72 × 82 [ = 2304] work done against resistance.
2
1
OR Final KE = × 72 × 16 2 [ = 9216]
1
( )
For reference, × 72 × 16 2 − 82 = 6912.
2
2
OR Work done against resistance = 28 ×100[ = 2800]
Attempt at work-energy equation M1 4 terms; allow sign errors; dimensionally correct.
1 2 1 2
2 × 72 × 8 + WD = 2 × 72 × 16 + 28 × 100
WD = 9712 J A1 OE. Condone 9710 J.
Do not ISW.
Alternative method for Question 1:
162 − 82 (B1) 192
162 = 82 + 2 × 100 × a a = = 0.96 OE, e.g. a = .
2 × 100 200
Use of suvat in a complete method to find an expression
for a. Must be of the form ' a = ' .
Attempt at Newton’s second law (M1) Three terms; dimensionally correct; allow sign errors; must
DF − 28 = 72 × ( their 0.96 ) be using their value of a.
WD[ = 97.12 ×100] = 9712 J (A1) OE. Condone 9710 J.
Do not ISW.
3
A particle P moves in a straight line. At time t s after leaving a point O on the line, P has velocity v m s -1 , where v = 44t - 6t 2 - 36 .
(a) Find the set of values of t for which the acceleration of the particle is positive. [2]
(b) Find the two values of t at which P returns to O. [3]
Mark scheme
2(a) Attempt to differentiate given v M1 Decrease power by 1 and a change in coefficient in at least
one term (which must be the same term); allow
unsimplified.
v
Use of a = scores M0.
t
11 A1 44 2
[ 44 − 12t > 0 ] t < OE, e.g. , 3 , 3.67 or better.
3 12 3
11
Do not allow t .
3
11
May solve 44 − 12t = 0 , but final answer must be t < .
3
If a lower limit included it must be 0. Allow t > 0 or t 0
11 11
. Allow 0, or 0, .
3 3
Alternative Method for Question 2(a):
11 2 (M1) OE
Use completing the square to get −6 t − + +
3
11 (A1) CWO
t< If a lower limit included it must be 0. Allow t > 0 or t 0
3
11 11
. Allow 0, or 0, .
3 3
2(a) Alternative Method 2 for Question 2(a):
t +t (M1) Complete method for finding the value of t at maximum,
Solving 44t − 6t 2 − 36 = 0 and find 1 2 , or equivalent.
2 b
or use − with correct a and b.
2a
11 ± √ 67
For reference, t = .
3
11 (A1) If a lower limit included it must be 0. Allow t > 0 or
t<
3 11 11
t 0. Allow 0, or 0, .
3 3
2
2(b) Attempt to integrate given v M1 Increase power by 1 and a change in coefficient in at least
one term (which must be the same term). Use of s = vt is
M0.
44 1+1 6 2 +1 A1 Allow unsimplified.
(s =) t − t − 36t ( + c ) = 22t 2 − 2t 3 − 36t ( + c )
1+1 2 +1
22t 2 − 2t 3 − 36t = 0 t = 2, 9 ( and 0) ONLY A1 CWO
Ignore t = 0 if not rejected.
3
PN
40° 25°
2N i°
10 N
16 N
Four coplanar forces of magnitude P N, 10 N, 16 N and 2 N act at a point in the directions shown in the diagram. It is given that the forces are in equilibrium.
Find the values of θ and P. [6]
Mark scheme
3 Resolving either direction to get an equation *M1 Correct number of relevant terms; allow sign errors; allow
sin/cos mix.
10cos 25 = 2cos 40 + 16sin θ A1
9.06307787 = 1.532088886 + 16sin θ
7.530988984 = 16sin θ
sin θ = 0.4706868115
P = 10sin 25 + 16cosθ + 2sin 40 A1 This may be with their θ .
P = 4.226182617 + 16cosθ + 1.285575219
P = 5.511757837 + 16cos θ
10cos 25 − 2cos 40 DM1 From equation(s) with correct number of relevant terms.
Attempt to solve for θ = sin −1 Must be a numerical expression for θ .
16
Attempt to solve for P = 10sin 25 + 16cos ( their θ ) + 2sin 40 DM1 From equation(s) with correct number of relevant terms.
Using their θ .
Must be a numerical expression for P.
θ = 28.1 AND P = 19.6 A1 28.07888819 and 19.6285636.
AWRT 28.1 and AWRT 19.6 from correct work.
6
A car has mass 1400 kg. When the speed of the car is v m s -1 the magnitude of the resistance to motion is kv 2 N where k is a constant.
(a) The car moves at a constant speed of 24 m s -1 up a hill inclined at an angle of a to the horizontal where sin a = 0.12 . At this speed the magnitude of the resistance to motion is 480 N.
(i) Find the value of k. [1]
(ii) Find the power of the car’s engine. [3]
(b) The car now moves at a constant speed on a straight level road.
Given that its engine is working at 54 kW , find this speed. [3]
Mark scheme
4(a)(i) 5 B1 480
k × 242 = 480 k = OE, e.g. , 0.833 or better.
6 576
1
4(a)(ii) Attempt at Newton’s second law *M1 3 terms; allow sign errors; allow sin/cos mix.
[ DF = 480 + 1400g × 0.12 DF = 2160] Allow DF = 480 + 1400 g × sin 6.9 or better.
5
May see DF = their 242 + 1400 g × 0.12.
6
Power = ( their 2160 ) × 24 DB1 For using P = DF x v, where DF is numerical.
51840 W A1 Allow W missing, but if given in kW units must be
present.
Allow 51.84 kW.
Allow 51800, 51.8 kW.
3
4(b) 54000 5 *B1FT 5
DF = and DF = their v 2 FT their > 0.
v 6 6
Get an equation of the form av3 = b and attempt to solve for v to get a DM1 a and b must both be positive or both negative.
positive value Must get to a value for v; if cubic not seen, the cubic may
be implied by the correct answer for their equation.
Speed = 40.2 m s–1 A1 40.165977.
AWRT 40.2 from correct work.
3
TN
35°
0.8 kg
28°
A particle of mass 0.8 kg lies on a rough plane which is inclined at an angle of 28° to the horizontal.
The particle is kept in equilibrium by a force of magnitude T N. This force acts at an angle of 35° above a line of greatest slope of the plane (see diagram). The coefficient of friction between the particle and the plane is 0.2 .
Find the least and greatest possible values of T. [8]
Mark scheme
5 Attempt at resolving perpendicular to plane to get an equation *M1 3 relevant terms; allow sign errors; allow sin/cos mix;
allow g missing; m = 0.8 must be used; correct angles
must be used.
R + T sin 35 = 0.8 g cos 28 R + ( 0.57357 ) T = 7.06358 A1
Attempt to resolve parallel to plane for one of the possible cases to get an *M1 3 relevant terms; allow sign errors; allow sin/cos mix;
equation allow g missing; m = 0.8 must be used; correct angles
must be used.
T cos35 = 0.8 g sin 28 + F ( 0.81915) T = 3.75577 + F A1 May use their F.
T cos35 = 0.8 g sin 28 − F ( 0.81915) T = 3.75577 − F A1 May use their F.
Use of F = 0.2R to get an equation in T only DM1 Dependent on previous 2 M marks.
May be implied by correct T value.
Allow g missing.
If resolved equations incorrect and no working seen, then
this mark is implied by the correct T value for their
equations.
Solve to get T = 5.53 A1 5.534499898
AWRT 5.53 from correct work.
Allow 5.54 from correct work.
Solve to get T = 3.33 A1 3.326141531
AWRT 3.33 from correct work.
Allow 3.32 from correct work.
8
Three particles A, B and C of masses 5 kg, 1 kg and 2 kg respectively lie at rest in that order on a straight smooth horizontal track XYZ. Initially A is at X, B is at Y and C is at Z. Particle A is projected towards
B with a speed of 6 m s -1 and at the same instant C is projected towards B with a speed of v m s -1 . In the subsequent motion, A collides and coalesces with B to form particle D. Particle D then collides and coalesces with C to form particle E and E moves towards Z.
15 - v
(a) Show that after the second collision the speed of E is m s -1 . [3]
4
(b) The total loss of kinetic energy of the system due to the two collisions is 63 J.
Use the result from (a) to show that v = 3. [3]
(c) It is given that the distance XY is 36 m and the distance YZ is 98 m.
(i) Find the time between the two collisions. [4]
(ii) Find the time between the instant that A is projected from X and the instant that E reaches Z.
[1]
Mark scheme
6(a) Attempt at conservation of momentum for the 1st collision *M1 3 non-zero terms; allow sign errors; using correct masses.
5 × 6 = ( 5 + 1) vD For reference vD = 5.
If mgv used, allow M1 M1 A0 max.
Attempt at conservation of momentum for the 2nd collision DM1 6 non-zero terms; allow sign errors; using correct masses;
( 5 + 1)( their vD ) − 2v = ( 5 + 1 + 2 ) vE allow their numerical vD .
Allow v = vE for this mark.
Note: 5 × 6 − 2v = ( 5 + 1 + 2 ) vE is M2.
If mgv used, allow M1 M1 A0 max.
15 − v A1 AG
( vE = ) Must in terms of v, as v is given in the question or
4
explicitly defined their letter used as v.
Do not allow v = vE for this mark.
Any error seen is A0 but condone saying ‘divide by 2’ or
equivalent.
If mgv used, allow M1 M1 A0 max.
3
6(b) 1 1 B1 For either KEinitial or KE final correct.
KE initial = × 5 × 6 2 + × 2 × v 2 = 90 + v 2
2 2
2
1 15 − v
KE final = ( 5 + 1 + 2 )
2 4
Attempt difference in KE is 63 to get an equation M1 Using ( sum of two initial KEs ) − KE final = ±63.
1 1 1 15 − v
2
Correct number of relevant terms – correct masses, must
× 5 × 6 + × 2 × v − (5 + 1 + 2)
2 2
= 63 be adding 2 KE terms for KEinitial .
2 2 2 4
( sum of two initial KEs ) and KE final coming from use of
correct formula and of the correct form.
Solve algebraically 3v 2 + 30v − 117 = 0 OE to get v = 3 ONLY A1 AG
Any error seen is A0.
Allow solving correct quadratic expression, rather than
correct quadratic equation, for full marks.
If v = −13 seen it must be discarded.
Must see solving for this mark. A quadratic equation
followed by the answer is insufficient.
6(b) Alternative Method for Question 6(b): Using loss of KE in second collision
1 1 (B1) For either KE or KE final correct.
KE after 1st collision = × 6 × 52 + × 2 × v 2 = 75 + v 2 after 1st collision
2 2
2
15 − v
( 5 + 1 + 2 )
1
KE final =
2 4
1 1 (M1) Using KE − KE final = ± ( 63 − their 15 ) .
Attempt difference in KE is 63 − × 5 × 62 − × 6 × 52 [ = 63 − 15 = 48] to after 1st collision
2 2 Correct number of relevant terms.
get an equation KE after 1st collision , KE final and their 15 coming from use of
1 1 1 15 − v
2
correct formula and of the correct form.
× 6 × 5 + × 2 × v − (5 + 1 + 2)
2 2
= 63 − 15
2 2 2 4
Solve algebraically 3v 2 + 30v − 117 = 0 OE to get v = 3 ONLY (A1) AG
Any error seen is A0.
If v = −13 seen it must be discarded.
Must see solving for this mark. A quadratic equation
followed by the answer is insufficient.
Alternative Method 2 for Question 6(b): Verifying that v = 3
1 1 (B1)
KE initial = × 5 × 6 2 + × 2 × 32 = 99
2 2
1 15 − 3
2 (B1)
KE final = ( 5 + 1 + 2 ) = 36
2 4
KEinitial − KE final = 63 , hence loss in KE is 63 J (B1) Must have a conclusion for this mark.
3
6(c)(i) Time A to B = 6 s B1
Distance BC = 98 − 3 × ( their 6 ) [ = 80] *B1FT FT their 6 which MUST come from 6t = 36.
Use sum of distance moved by D and distance moved by C is 80 m DM1 Using their vD from part (a).
( their 5 ) t + 3t = their 80 vD ≠ 6 or 3 and their 80 ≠ 98.
OR
use distance moved by C divided by relative velocity
80
( their 5 ) + 3
Time = 10 s A1 Do not ISW.
4
6(c)(ii) 3 × 10 + 3 × 6 B1
6 + 10 + 3
= 32 s
1
P
2.5 kg
2m
Q
0.5 kg
30°
Two particles P and Q of masses 2.5 kg and 0.5 kg respectively are connected by a light inextensible string that passes over a small smooth pulley fixed at the top of a plane inclined at an angle of 30° to the horizontal. Particle P is on the plane and Q hangs below the pulley such that the level of Q is 2 m below the level of P (see diagram).
Particle P is released from rest with the string taut and slides down the plane. The plane is rough with coefficient of friction 0.2 between the plane and P.
(a) Find the acceleration of P. [5]
(b) Use an energy method to find the speed of the particles at the instant when they are at the same vertical height. [5]
Additional page
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Mark scheme
7(a) 25 3 B1 5 3
R = 2.5 g cos30 = = 21.65063509 Note: F = 0.5 g cos30 = = 4.330127019.
2 2
Attempt at Newton’s second law *M1 Correct number of dimensionally correct/relevant terms;
allow sign errors; allow sin/cos mix.
Using this twice to get equations for P and Q; allow
different T’s (equations with 0.5 and 2.5).
Using once to get a system equation (equation with
0.5 + 2.5).
EITHER: T − 0.5 g = 0.5a AND 2.5 g sin 30 − F − T = 2.5a A1 EITHER: Both correct; allow their F; must be the same T.
OR: 2.5g sin30 − F − 0.5g = ( 2.5 + 0.5) a OR: correct system equation; allow their F.
Use F = 0.2R to get an equation in a only DM1 Where R is a component of weight of P only; from
equation(s) with the correct number of dimensionally
2.5 g sin 30 − 0.2 × 2.5 g cos 30 − 0.5 g = ( 2.5 + 0.5 ) a correct/relevant terms.
a = 1.06 m s–2 A1 15 − 5 3
Allow .
6
1.05662433.
AWRT 1.06 from correct work.
5
7(b) 4 *B1 Where x is the distance P moves down the plane or Q
[ x sin 30 = 2 − x ] x = moves vertically upwards, or where y is the vertical
3
2 distance of Q below P’s starting point.
OR ( 2 − y ) sin 30 = y y = Allow x = 1.3 or better; allow y = 0.67 or better.
3
1 1
OR × 1.056× t 2 + × 1.056× t 2 × sin 30 = 2 t = 1.5886
2 2
1 2
x = × 1.056× (1.5886..) sin 30 =
2
2 3
1 2 4
OR y = × 1.056× (1.5886..) =
2 3
3 B1 Using their x ( ≠ 2 or 1 or 0 ) , 0 < x < 2;
Change in PE = ± ( 2.5 g ( their x ) sin 30 − 0.5 g ( their x ) ) = ± g ( their x )
4 or their y ( ≠ 2 or 1 or 0 ) , 0 < y < 2.
OR = ± ( 2.5 g ( 2 − their y ) sin 30 − 0.5 g ( 2 − their y ) )
= ± ( 3 ( their y ) − 1) g
WD against friction = 0.2 × 2.5 g cos 30 × ( their x ) = 4.33 ( their x ) B1 Using their x ( ≠ 2 or 1 or 0 ) , 0 < x < 2;
OR WD against friction = 0.2 × 2.5 g cos 30 × ( 2 − ( their y ) ) or their y ( ≠ 2 or 1 or 0 ) , 0 < y < 2.
1 1 DM1 Attempt at work energy equation; dimensionally correct; 5
× 2.5 × v 2 + × 0.5 × v 2 = relevant terms; allow sign errors; allow sin/cos mix.
2 2
Must be using correct values of x or y.
4 4 4
2.5 g sin 30 − 0.5 g − 0.2 × 2.5 g cos30 ×
3 3 3
1 1
OR × 2.5 × v 2 + × 0.5 × v 2 =
2 2
2 2 2
2.5 g 2 − sin 30 − 0.5 g 2 − − 0.2 × 2.5 g cos30 × 2 −
3 3 3
v = 1.68 A1 1.67859014
AWRT 1.68 from correct work.
7(b) Alternative Method for Question 7(b): Considering energy on Q only
Must be using tension and mass 0.5 kg only to be awarded the last 4 marks
4 (*B1) Where x is the distance P moves down the plane or Q
[ x sin 30 = 2 − x ] x = moves vertically upwards, or where y is the vertical
3
2 distance of Q below P’s starting point.
OR ( 2 − y ) sin 30 = y y = Allow x = 1.3 or better; allow y = 0.67 or better.
3
1 1
OR × 1.056× t 2 + × 1.056× t 2 × sin 30 = 2 t = 1.5886.
2 2
1 2
x = × 1.056× (1.5886..) sin 30 =
2
2 3
1 2 4
OR y = × 1.056× (1.5886..) =
2 3
Change in PE = ±0.5 g × ( their x ) (B1) Using their x ( ≠ 2 or 1 or 0 ) , 0 < x < 2;
OR Change in PE = ±0.5 g × ( 2 − ( their y ) ) or their y ( ≠ 2 or 1 or 0 ) , 0 < y < 2.
WD by tension = ( their 5.528312164 ) × ( their x ) (B1) Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
OR WD by tension = ( their 5.528312164 ) × ( 2 − ( their y ) )
Using their x ( ≠ 2 or 1 or 0 ) , 0 < x < 2;
or their y ( ≠ 2 or 1 or 0 ) , 0 < y < 2.
4 1 4 (DM1) Using their tension from 7(a) from equation(s) with the
0.5 g + × 0.5 × v 2 = ( their 5.528312164 ) × correct number of dimensionally correct/relevant terms.
3 2 3
Attempt at work energy equation; dimensionally correct; 3
2 1 2
OR 0.5 g 2 − + × 0.5 × v 2 = ( their 5.528312164 ) × 2 − relevant terms; allow sign errors.
3 2 3 Must be using correct values of x or y.
v = 1.68 (A1) 1.67859014
AWRT 1.68 from correct work.
7(b) Alternative Method 2 for Question 7(b): Considering energy on P only
Note: must be using tension and mass 2.5 kg only to be awarded the last 4 marks
4 2 (*B1) Where x is the distance P moves down the plane or Q
[ x sin 30 = 2 − x ] x =
OR ( 2 − y ) sin 30 = y y = moves vertically upwards, or where y is the vertical
3 3
distance of Q below P’s starting point.
1 1
OR × 1.056× t 2 + × 1.056× t 2 × sin 30 = 2 t = 1.5886. Allow x = 1.3 or better; allow y = 0.67 or better.
2 2
1 2
x = × 1.056× (1.5886..) sin 30 =
2
2 3
1 2 4
OR y = × 1.056× (1.5886..) =
2 3
Change in PE = ±2.5 g ( their x ) sin 30 OR = ±2.5g ( their y ) (B1) Using their x ( ≠ 2 or 1 or 0 ) , 0 < x < 2;
or their y ( ≠ 2 or 1 or 0 ) , 0 < y < 2.
WD by tension = ( their 5.528312164 ) × ( their x ) (B1) Using their tension from 7(a) from equation(s) with the
correct number of dimensionally correct/relevant terms.
OR WD against friction = 0.2 × 2.5g cos30 × ( their x )
Using their x ( ≠ 2 or 1 or 0 ) , 0 < x < 2;
or their y ( ≠ 2 or 1 or 0 ) , 0 < y < 2.
4 1 (DM1) Using their tension from 7(a) from equation(s) with the
2.5 g sin 30 = × 2.5 × v 2 + correct number of dimensionally correct/relevant terms.
3 2
Attempt at work energy equation; dimensionally correct; 4
4 4
0.2 × 2.5 g cos30 × + ( their 5.528312164 ) × relevant terms; allow sign errors; allow sin/cos mix.
3 3 Must be using correct values of x or y.
v = 1.68 (A1) 1.67859014
AWRT 1.68 from correct work.
7(b) Special Case for using constant acceleration: Maximum 2 marks
4 (B1) Where x is the distance P moves down the plane or Q
[ x sin 30 = 2 − x ] x = moves vertically upwards, or where y is the vertical
3
2 distance of Q below P’s starting point.
OR ( 2 − y ) sin 30 = y y = Allow x = 1.3 or better; allow y = 0.67 or better.
3
1 1
OR × 1.056× t 2 + × 1.056× t 2 × sin 30 = 2 t = 1.5886.
2 2
1 2
x = × 1.056× (1.5886..) sin 30 =
2
2 3
1 2 4
OR y = × 1.056× (1.5886..) =
2 3
2 4 (B1) 1.67859014
v = 2 × 1.06 × 3 v = 1.68 AWRT 1.68 from correct work.
5
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