Mathematics 9709/41 May/June 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A car starts from rest and moves in a straight line with constant acceleration for a distance of 200 m, reaching a speed of 25 m s−1 . The car then travels at this speed for 400 m, before decelerating uniformly to rest over a period of 5 s.
(a) Find the time for which the car is accelerating. [2]
(b) Sketch the velocity–time graph for the motion of the car, showing the key points. [2]
(c) Find the average speed of the car during its motion. [2]
Mark scheme
1(a)
200 =
( 0 + 25 ) t M1
For use of s =
( u + v ) t or other complete method to find t
2 2
e.g. v 2 = u 2 + 2as followed by v = u + at
N.B. a = 1.5625
t = 16 s A1
2
1(b) Trapezium B1 Through (0, 0) and positive t-axis
All correct through (0, 0), (16, 25), (32, 25), (37, 0) B1 FT their value of t from part (a) (t +16 + 5)
2
1(c) 1 M1 1
Total distance = 200 + 400 + × 25 × 5 [= 662.5] Or trapezium × 25 (16 + 37 ) [ = 662.5]
2 2
25 + 0
Or 200 + 400 + ×5
2
Allow their value of t from part (a) in calculating 200 and their
time from the constant speed section from part (b) in calculating
400
Allow their a from part (a) if used in calculating 200
662.5 A1 1325 67
Average speed = = 17.9 m s−1 (3 s.f.) Allow = 17
37 74 74
2
Two particles P and Q, of masses 0.5 kg and 0.3 kg respectively, are connected by a light inextensible string. The string is taut and P is vertically above Q. A force of magnitude 10 N is applied to P vertically upwards.
Find the acceleration of the particles and the tension in the string connecting them. [5]
Mark scheme
2 Use of Newton’s second law for P or Q or system M1 Allow sign errors. Do not allow g missing
For P: 10 – 0.5g – T = 0.5a, For Q: T – 0.3g = 0.3a, A1 For any two correct
For system: 10 – 0.5g – 0.3g = 0.8a
For attempt to solve for either a or T M1 From equation(s) with no missing or extra terms. Allow g missing.
a = 2.5 ms–2 A1
T = 3.75 N A1
5
A crate of mass 300 kg is at rest on rough horizontal ground. The coefficient of friction between the crate and the ground is 0.5. A force of magnitude X N, acting at an angle ! above the horizontal, is applied to the crate, where sin ! = 0.28.
Find the greatest value of X for which the crate remains at rest. [5]
Mark scheme
3 For attempt at resolving horizontally or vertically M1 Allow sin/cos mix. Allow sign error. Allow g missing. Correct
number of terms.
R = 300g – 0.28X or R = 300 g − X sin16.3 A1 α = 16.26…
0.96X – F = 0 or 0.96 X − 0.5(300 g − X sin α ) = 0 A1 Or using their F
Or X cos 16.3 – F = 0 or X cos 16.3 −0.5(300 g − X sin α ) = 0
Use of F = 0.5R M1 Use to get an equation in X only. Allow sin/cos mix. Allow sign
error. Allow g missing.
Must be from 2 term R, which is a linear combination of 300(g)
and a component of X
X = 1360 [1363.63…] A1
5
Three coplanar forces of magnitudes 20 N, 100 N and F N act at a point. The directions of these forces are shown in the diagram.
Given that the three forces are in equilibrium, find F and !. [6]
Mark scheme
4 Attempt to resolve in any direction M1 For resolving. Allow sin/cos mix. Allow sign error.
Correct number of terms.
F cos α – 20 cos 40 – 100 sin 20 = 0 A1
[F cos α = 15.320…+ 34.202… = 49.5229…]
F sin α + 20 sin 40 – 100 cos 20 = 0 A1
[F sin α = 93.969… − 12.855… = 81.1135…]
M1 OE; Attempt to solve for F; one term missing in total
F = ( 49.5229)2 + ( 81.1135)2
81.1135 M1 OE; Attempt to solve for α; one term missing in total
α = tan −1
49.5229
F = 95(.0), α = 58.6 A1 F = 95.0364… and α = 58.5943…
Alternative mark scheme for question 4: For candidates who use cosine and/or sine rule
Attempt at cosine rule from triangle of forces M1 Must use lengths 100 and 20 with a suitable angle
F 2 = 100 2 + 20 2 − 2 × 100 × 20 cos 70 A1 Correct
F = 95[.0] A1
95.0364 20 95.0364 100 M1 Attempt at sin rule
= OR =
sin 70 sin β sin 70 sin γ
A1 where β = ( 70 − α ) where γ = ( 40 + α )
α = 58.6 A1 α = 58.5943…
4 Alternative mark scheme for question 4: For candidates who resolve in other directions
Attempt to resolve (e.g. parallel or perpendicular to 100 N) M1 For resolving. Allow sin/cos mix. Allow sign error.
Correct number of terms.
F sin (α + 20 ) + 20sin 20 − 100 = 0 F sin (α + 20 ) = 93.159 A1
F cos (α + 20 ) − 20cos 20 = 0 F cos (α + 20 ) = 18.793 A1
F = 93.1592 + 18.7932 M1 OE; Attempt to solve for F; one term missing in total
93.159 M1 OE; Attempt to solve for α; one term missing in total
α = tan −1 − 20
18.793
F = 95[.0], α = 58.6 A1 F = 95.0364… and α = 58.5943…
6
Two racing cars A and B are at rest alongside each other at a point O on a straight horizontal test track. The mass of A is 1200 kg. The engine of A produces a constant driving force of 4500 N. When
A arrives at a point P its speed is 25 m s−1 . The distance OP is d m. The work done against the resistance force experienced by A between O and P is 75 000 J.
(a) Show that d = 100. [3]
Car B starts off at the same instant as car A. The two cars arrive at P simultaneously and with the same speed. The engine of B produces a driving force of 3200 N and the car experiences a constant resistance to motion of 1200 N.
(b) Find the mass of B. [3]
(c) Find the steady speed which B can maintain when its engine is working at the same rate as it is at P. [3]
Mark scheme
5(a) For attempt at work energy equation M1 3 terms. Allow sign errors.
M0 for (constant) acceleration method
1 A1 Correct equation
4500d – 75 000 = × 1200 × 252 [= 375 000]
2
d = 100 A1 AG
Accept verification with d substituted in above line to show LHS =
375 000 or LHS −RHS = 0
1
If no marks scored allow SCB1 for × 1200 × 252
2
3
5(b) 252 = 0 + 2a × 100 [leading to a = 3.125] B1 Allow B1 if acceleration found in part (a) as 3.125 and used or
stated here
3200 – 1200 = m × 3.125 M1 Newton’s second law with 3 terms. Allow sign errors and their a.
Mass of car B = 640 kg A1
Alternative mark scheme for question 5(b)
For attempt at work energy equation M1 3 terms. Allow sign errors.
1 A1 Correct equation
(3200 – 1200) × 100 = × m × 252
2
Mass of car B = 640 kg A1
3
5(c) At P power = 3200 × 25 [= 80 000] B1 For use of power = Fv
80000 M1 Attempt Newton’s second law for car B with a = 0
− 1200 = 0 Allow their 80 000 (dimensionally correct)
v
Steady speed = 66.7 m s−1 A1 200 2
Allow = 66
3 3
3
A particle starts from a point O and moves in a straight line. The velocity v m s−1 of the particle at time t s after leaving O is given by v = k 3t2 − 2t3 , where k is a constant.
(a) Verify that the particle returns to O when t = 2. [4]
(b) It is given that the acceleration of the particle is −13.5 m s−2 for the positive value of t at which v = 0.
Find k and hence find the total distance travelled in the first two seconds of motion. [6]
Mark scheme
6(a) For attempt at integration *M1 The power of t must increase by 1 with a change of coefficient in at
least 1 term.
Allow if k is omitted.
1 A1 Allow unsimplified
k t 3 − t 4 [ +C ]
2
1 DM1 For use of limits 0 and 2 or substituting t = 2
k 8 − × 16 [ −0] OR equate to 0, then solve a quartic equation in t.
2
Allow if k is omitted.
0 A1 AG, CWO including stating C = 0 if not using limits
4
6(b) ( )
v = 0 when k 3t 2 − 2t 3 = 0 M1 For solving for t
Leading to t = 1.5 [or t = 0] A1 A0 for other solutions not discarded
(
a = k 6t − 6t 2 ) M1 For differentiation, the power of t must decrease by 1 with a
change of coefficient in at least 1 term.
Allow if k is omitted
( )
[ −13.5 = k 6 × 1.5 − 6 × 1.52 ⇒ −13.5 = k ( 9 − 13.5) ⇒] k = 3 A1
Distance from t = 0 to t = 1.5 is DM1 1.5 2
3 1 4
1.5
3 1 4
For use of limits. 2 × or 2 × or both integrals. Dependent
3 t − 2 t = 3 1.5 − 2 × 1.5 − 0 0 15
0 on first M in part (a), unless they restart
[= 2.53125] So total distance = 2 × 2.53125 = 5.06 m A1 81 1
Allow distance = =5
16 16
If DM0 then SCB1 for 5.06 without working
6
Two particles A and B, of masses 0.4 kg and 0.2 kg respectively, are moving down the same line of greatest slope of a smooth plane. The plane is inclined at 30Å to the horizontal, and A is higher up the plane than B. When the particles collide, the speeds of A and B are 3 m s−1 and 2 m s−1 respectively.
In the collision between the particles, the speed of A is reduced to 2.5 m s−1 .
(a) Find the speed of B immediately after the collision. [2]
After the collision, when B has moved 1.6 m down the plane from the point of collision, it hits a barrier and returns back up the same line of greatest slope. B hits the barrier 0.4 s after the collision, and when it hits the barrier, its speed is reduced by 90%. The two particles collide again 0.44 s after their previous collision, and they then coalesce on impact.
(b) Show that the speed of B immediately after it hits the barrier is 0.5 m s−1 . Hence find the speed of the combined particle immediately after the second collision between A and B. [7]
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) 0.4 × 3 + 0.2 × 2 = 0.4 × 2.5 + 0.2v M1 Use of conservation of momentum with 4 terms. Allow sign errors.
v = 3 ms−1 A1 Allow M1A0 if g included with the masses.
2
7(b) For A ±0.4 g sin 30° = 0.4a or for B ±0.2 g sin 30° = 0.2a M1 For either. Allow sin/cos mix.
or ± mg sin 30° = ma
a = ± 5 or ± g sin 30° A1 Allow g sin 30° without working for M1A1
For B when hits barrier v 2 = 32 + 2 × 5 × 1.6 [⇒ v = 5] M1 Using their a ≠ ± g and their v from part (a)
OR v = u + at v = 3 + 5 × 0.4 [ v = 5] u+v 3+v
× 0.4 [ v = 5]
OR: use of s = t 1.6 =
2 2
1 1
OR × 0.2 × v 2 − × 0.2 × 32 = 0.2 × 1.6 × g sin 30
2 2
Speed after hitting barrier = 0.1 × 5 = 0.5 A1 AG
vA = 2.5 + 5 × 0.44 [= 4.7] vB = −0.5 + 5 × 0.04 [= −0.3] *M1 Use of v = u + at for either with correct t-value, with initial speeds
or vB = 0.5 + (–5) × 0.04 [= 0.3] ±2.5 or ± 0.5 their ± a ≠ ± g
0.4 × 4.7 + 0.2 × (−0.3) = 0.6 vcomb DM1 Use of v = u + at for BOTH with correct t-values, initial speeds
±2.5, ±0.5 and ± their acceleration (same for both) and use of
conservation of momentum with correct number of terms. Allow
sign errors.
vcomb = 3.03 ms−1 A1 91 1
Allow v = =3
30 30
Allow DM1A0 if g included with the masses.
7
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Mathematics 9709/41 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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