Mathematics 9709/42 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.
Small smooth spheres A and B, of equal radii and of masses 5 kg and 3 kg respectively, lie on a smooth horizontal plane. Initially B is at rest and A is moving towards B with speed 8.5 m s−1 . The spheres collide and after the collision A continues to move in the same direction but with a quarter of the speed of B.
(a) Find the speed of B after the collision. [3]
(b) Find the loss of kinetic energy of the system due to the collision. [2]
Mark scheme
1(a) Conservation of momentum M1 1
3 terms; allow M1 if speed of A after collision is × 8.5 .
4
Allow 5 × 8.5 = 5 X + 3Y where X and Y are different
which may be seen by later work. If X and Y are
subsequently used as being equal then M0.
5 × 8.5 = 5 × 0.25v + 3v A1 OE e.g. 5 × 8.5 = 5V + 3 × 4V
Speed of B = 10 ms
−1 A1 Do not award if 10 from using mgv, maximum 2/3
–10 is A0 as speed required not velocity
3
1(b) 1 1 Attempt at any of the 3 terms for KE, using their 10 ms −1
KE before = × 5 × 8.52 [ = 180.625]
2 1 1
Not × ( 5 + 3) × 8.52 , not × ( 5 + 3) × 2.52
2 2
1 1
KE after = × 5 × 2.52 + × 3 ×102 [ = 15.625 + 150 = 165.625] 1
not × ( 5 + 3) ×10 unless X = Y seen
2
2 2 2
KE loss [ = 180.625 − 165.625] = 15 J A1 Accept ‒15, AWRT ±15.0
2
Coplanar forces of magnitudes 60 N, 20 N, 16 N and 14 N act at a point in the directions shown in the diagram.
Find the magnitude and direction of the resultant force. [6]
Mark scheme
2 Resolving either direction M1 3 terms; allow sign errors and allow sin/cos mix
( X = ) ± ( 20 cos 60 − 14 − 16 cos 50 ) [ = μ14.2846] A1
(Y = ) ± ( 60 − 20sin 60 − 16sin 50 ) [ = ±30.42278] A1
M1 Attempt to solve for R ; one missing term in total
(14.2846) + ( 30.42278)
2 2
R=
30.42278 M1 Attempt to solve for 𝜃 or 𝛼; one missing term in total
θ = tan −1 = tan ( 2.1297 )
−1
14.2846
14.2846
OR α = tan −1 = tan ( 0.4596 )
−1
30.42278
R = 33.6 N A1 Both correct.
Direction is 64.8° above the 14 N force OE; allow 64.9, 25.1
or 25.2° above the negative 𝑥-axis Giving an angle only is insufficient.
or 25.2° left of the 60 N force Direction may be seen on a diagram, with minimum of arrow
or bearing 335° on resultant.
or 115° anticlockwise from the positive x -axis Arrows on both components only is A0 as it doesn’t show the
direction of the resultant.
However the direction is stated, it must be able to be drawn
uniquely.
6
Two particles A and B, of masses 2.4 kg and 1.2 kg respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. A is held at a distance of 2.1 m above a horizontal plane and B is 1.5 m above the plane. The particles hang vertically and are released from rest. In the subsequent motion A reaches the plane and does not rebound and B does not reach the pulley.
(a) Show that the tension in the string before A reaches the plane is 16 N and find the magnitude of the acceleration of the particles before A reaches the plane. [4]
(b) Find the greatest height of B above the plane. [3]
Mark scheme
3(a) 2.4 g − T = 2.4a M1 Attempt at Newton’s second law on either particle or the
T − 1.2 g = 1.2a system with correct number of terms; allow sign errors.
2.4 g − 1.2 g = ( 2.4 + 1.2 ) a A1 Any 2 consistent and correct
May have an a in opposite direction to our a
Attempt to solve for a or T M1 From equation(s) with correct number of relevant terms.
If g missing then M0A0M1A0, maximum1/4.
Must get a = or T =
Must not assume T = 16 . May attempt to verify a value of a
using T = 16 in 2 equations
10 A1 Both correct; allow a = 3.33 . AG for T = 16 .
T = 16 N and a = ms-2
3 Assuming T = 16 and only one equation is M1A0M0A0
maximum 1/4. Withhold A mark if T = 15.9... ≈ 16 , but
condone T = 1.2 × 3.33 + 12 = 16 or
T = 24 − 2.4 × 3.33 = 16
4
3(b) 10 M1 Use of suvat or use energy to find v or v 2 , using their
v2 = 2 × × 2.1 [ = 14] v = 14 = 3.741 a ≠ ± g (unless 10 comes from their attempt at a ) from (a),
3
1 s = 2.1
OR × 2.4 × v 2 = 2.4 g × 2.1 − 16 × 2.1
2
1 1
OR × 2.4 × v 2 + × 1.2 × v 2 = 2.4 g × 2.1 − 1.2 g × 2.1
2 2
0 = 14 − 2 × g × s → s = M1 Attempt to use v 2 = u 2 + 2as (or other complete method),
using a = − g , to find additional height after string slack,
1
( )
2
or (1.2 ) 14 = (1.2 ) × g × h → h = 2
2 using their v or v .
s = [1.5 + 2.1 + 0.7 = ] 4.3 m A1 AWRT 4.3(0);
Allow use of a = 3.33 to give s = 4.2993 ≈ 4.3 ( 0 )
Allow use of v = 3.74 to give s = 4.29938 ≈ 4.3 ( 0 )
Alternative for question 3(b) - using energy on particle B
16 × 2.1 = 1.2 gH M1 Apply energy to B, 2 terms
H = 2.8 A1
s = [1.5 + 2.8 = ] 4.3 m A1
3
A particle A, moving along a straight horizontal track with constant speed 8 m s−1 , passes a fixed point O. Four seconds later, another particle B passes O, moving along a parallel track in the same direction as A. Particle B has speed 20 m s−1 when it passes O and has a constant deceleration of
2 m s−2 . B comes to rest when it returns to O.
(a) Find expressions, in terms of t, for the displacement from O of each particle t seconds after B passes O. [3]
(b) Find the values of t when the particles are the same distance from O. [3]
(c) On the given axes, sketch the displacement-time graphs for both particles, for values of t from 0 to 20. [3]
Mark scheme
4(a) Use suvat to find expressions for s A or sB M1 For s A must be using u = 8 and time of t ± 4
1 2
For sB , using s = ut + at with u = 20 and a = ±2
2
( s A = ) 8 ( 4 + t ) [ = 32 + 8t ] A1 Any unsimplified expression; ISW
1 A1 Any unsimplified expression; ISW
( sB = ) 20t − × 2 × t 2 If 0 marks scored then allow
2
SC: B1 for ( s A = ) 8t and
1
B1 for ( sB = ) 20 ( t − 4 ) − × 2 × ( t − 4 ) maximum 2/3
2
2
3
4(b) 1 *M1 Equating their expressions for s A and sB to form an equation
8 ( 4 + t ) = 20t − × 2 × t 2
2 in t
where s A is of the form ±8t ± 32
1
and sB is of the form ±20t ± × 2×t2
2
Attempt to solve a 3-term quadratic to find at least one t value DM1 For reference t 2 − 12t + 32 = 0
Allow if no working seen and have correct real solution(s) to
their 3-term quadratic.
If working shown and if using the formula, it must be using
the correct formula. If factorising must have 3 of the 4 terms
correct of ( t − 4 )( t − 8 )
t = 4 and 8 A1 If 0 marks scored then allow
1
SC: M1 for ±8t = ±20 ( t ± 4 ) ± × 2 × ( t ± 4 ) and
2
2
A1 for t = 8 and 12 maximum 2/3.
3
4(c) Straight line B1 FT Positive gradient, intersecting positive s axis.
Full domain not required.
FT if they get s A = 8t using the SC in (a)
Inverted quadratic, passing through origin. B1 FT Full domain not required but must clearly go beyond the
maximum.
1
FT if they get sB = 20 ( t − 4 ) − × 2 × ( t − 4 ) using the SC
2
2
in (a), with curve though positive t axis before turning point.
All correct, line through (0, 32), quadratic through (20, 0), intersections B1 Intersections must occur before the turning point.
indicated at t = 4 and t = 8 .
3
A block of mass 12 kg is placed on a plane which is inclined at an angle of 24Å to the horizontal. A light string, making an angle of 36Å above a line of greatest slope, is attached to the block. The tension in the string is 65 N (see diagram). The coefficient of friction between the block and plane is -. The block is in limiting equilibrium and is on the point of sliding up the plane.
Find -. [6]
Mark scheme
5 Attempt at resolving parallel to the plane *M1 3 terms. Allow sign errors, sin/cos mix. Allow g missing,
otherwise dimensionally correct.
65 cos 36 = 12 g × sin 24 + F A1 F = 3.777707
Attempt at resolving perpendicular to the plane *M1 3 terms. Allow sign errors, sin/cos mix. Allow g missing,
otherwise dimensionally correct.
12 g × cos 24 = R + 65sin 36 A1 R = 71.419
Use F = μ R DM1 To get an equation in μ only.
65 cos 36 − 12 g × sin 24 52.586 − 48.808 3.777 Dependent on two previous M marks. Allow g missing
μ = 12 g × cos 24 − 65sin 36 = 109.625 − 38.206 = 71.419
μ = 0.0529 A1 Allow AWRT 0.053
Do not accept fractional equivalent.
6
A car of mass 900 kg is moving up a hill inclined at sin−1 0.12 to the horizontal. The initial speed of the car is 11 m s−1 . After 12 s, the car has travelled 150 m up the hill and has speed 16 m s−1 . The engine of the car is working at a constant rate of 24 kW.
(a) Find the work done against the resistive forces during the 12 s. [5]
The car then travels along a straight horizontal road. There is a resistance to the motion of the car of
1520 + 4v N when the speed of the car is v m s−1 . The car travels at a constant speed with the engine working at a constant rate of 32 kW.
(b) Find this speed. [3]
Mark scheme
6(a) (
KE change = ± 0.5 × 900 × 162 − 0.5 × 900 × 112 ) B1
[ = ± (115200 − 54450 ) = ±60750]
PE = 900 g × 150 × 0.12 [ = 162000] B1 Allow 900 g ×150 × sin 6.89° or 900 g ×150 × sin 6.9°
Not from use of constant acceleration/Newton’s second law.
[Work done by car’s engine =] 24000 × 12 [ = 288000] B1
OE e.g. 24000 =
WD
12
Work done against resistive forces M1 Use of work-energy 5 terms; dimensionally correct.
= 24000 × 12 − 900 g × 150 × 0.12 − 0.5 × 900 × 162 + 0.5 × 900 × 112 Work done by car’s engine not from using one of the given
= 288000 − 162000 − 115200 + 54450 speeds.
Allow sign errors.
Work done = 65 250 J A1 or 65.25 kJ
Allow AWRT 65300 J or 65.3 kJ from correct work
5
6(b) 32000 B1 OE e.g. 32000 = DF × v
Driving Force =
v
32000 M1 Apply N2L to the car with a = 0 (3 terms) and attempt to
= 1520 + 4v solve a 3-term quadratic in v .
v
For reference 4v + 1520v − 32000 = 0
2
Allow if no working seen and have correct real solution(s) to
their 3-term quadratic. If working shown and if using the
formula, it must be using the correct formula. If factorising
must have 3 of the 4 terms correct of ( v − 20 )( v + 400 )
Speed = 20 ms
−1 A1 Only.
3
A particle P moves in a straight line. The velocity v m s−1 at time t seconds is given by v = 0.5t for 0 ≤ t ≤ 10, v = 0.25t2 − 8t + 60 for 10 ≤ t ≤ 20.
(a) Show that there is an instantaneous change in the acceleration of the particle at t = 10. [3]
(b) Find the total distance covered by P in the interval 0 ≤ t ≤ 20. [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) dv B1 For acceleration during the first 10 seconds
( a =) = 0.5
dt
dv B1 Allow unsimplified
Differentiate to get ( a = ) = 2 × 0.25t − 8 [ = 0.5t − 8]
dt
a [ = 0.5 × 10 − 8] = −3 B1 CWO. Do not award final B mark if more than 2
accelerations seen and not discarded, 2/3 maximum
Ignore any comments, correct or incorrect
3
7(b) Get distance in first 10 seconds as 25 B1 10
From suvat or from 0.5t dt
0
v = 0 when t = 12 and t = 20 B1 SOI
Attempt to integrate v *M1 For integration, the power of t must increase by 1 in at least
1 term with a change of coefficient in the same term.
( )
s = 0.25t 2 − 8t + 60 dt
s = vt is M0
0.25 3 8 2 1 3 A1 Allow unsimplified
s= t − t + 60t ( + c ) = 12 t − 4t + 60t ( + c )
2
3 2
1 3 DM1 Using the correct limits correctly
Attempt to evaluate their t − 4t 2 + 60t for t = 10 to t = 12
12
and t = 12 to t = 20
850 800 14 64 A1
s = 25 + 288 − − − 288 = 25 + − − = 51 m
3 3 3 3
7(b) Special Case for those who use a calculator to integrate. Maximum 4/6
Get distance in first 10 seconds as 25 B1 10
From suvat or 0.5t dt
0
v = 0 when t = 12 and t = 20 B1 SOI
12
14 B1 20
Either s = ( )
0.25t − 8t + 60 dt =
2
3
= 4.67 Allow 0.25t − 8t + 60 dt = 26
2
10 10
20
64
Or s = ( 0.25t − 8t + 60 ) dt = 3 = 21.3
2
12
14 64 B1 Allow if t = 12 and t = 20 not found for 3 marks
s = 25 + + = 51 m
3 3
6
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Mathematics 9709/42 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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