Mathematics 9709/43 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.
Two particles P and Q, of masses 0.3 kg and 0.2 kg respectively, are at rest on a smooth horizontal plane. P is projected at a speed of 4 m s−1 directly towards Q. After P and Q collide, Q begins to move with a speed of 3 m s−1 .
(a) Find the speed of P after the collision. [2]
After the collision, Q moves directly towards a third particle R, of mass m kg, which is at rest on the plane. The two particles Q and R coalesce on impact and move with a speed of 2 m s−1 .
(b) Find m. [2]
Mark scheme
1(a) 0.3 × 4 + 0 = 0.3v + 0.2 × 3 M1 For attempt at use of conservation of momentum
Speed = 2 ms−1 A1
2
1(b) 0.2 × 3 + 0 = ( 0.2 + m ) × 2 M1 For attempt at use of conservation of momentum
m = 0.1 A1
2
A particle P is projected vertically upwards from horizontal ground. P reaches a maximum height of 45 m. After reaching the ground, P comes to rest without rebounding.
(a) Find the speed at which P was projected. [2]
(b) Find the total time for which the speed of P is at least 10 m s−1 . [3]
Mark scheme
2(a) 0 = u 2 − 2 g × 45 M1 For use of v 2 = u 2 + 2as
OE complete method that would lead to finding u
Speed = 30 ms−1 A1
2
2(b) 10 = 30 − gt leading to t = 2 M1 For use of v = u + at to find time to 10 ms–1
or use of ‘suvat’ to find time for one stage of motion
2× 2 s M1 2 × time to 10 ms–1 OE
Total time = 4 s A1
3
s (m)
240
200
D
160 C
120
80
B
40
0
A E t (s)
0 5 10 15 20
The displacement of a particle moving in a straight line is s metres at time t seconds after leaving a fixed point O. The particle starts from rest and passes through points P, Q and R, at times t = 5, t = 10 and t = 15 respectively, and returns to O at time t = 20. The distances OP, OQ and OR are 50 m,
150 m and 200 m respectively.
The diagram shows a displacement-time graph which models the motion of the particle from t = 0 to t = 20. The graph consists of two curved segments AB and CD and two straight line segments BC and DE .
(a) Find the speed of the particle between t = 5 and t = 10. [1]
(b) Find the acceleration of the particle between t = 0 and t = 5, given that it is constant. [2]
(c) Find the average speed of the particle during its motion. [2]
Mark scheme
3(a) Speed = 20 ms−1 B1
1
3(b) 20 = 0 + a × 5 M1 Use of v = u + at OE
a = 4 ms−2 A1
2
3(c) 50 + 100 + 50 + 200 M1 total distance
Use of OE
20 total time
Average speed = 20 ms−1 A1
2
The diagram shows a block of mass 10 kg suspended below a horizontal ceiling by two strings AC and
BC, of lengths 0.8 m and 0.6 m respectively, attached to fixed points on the ceiling. Angle ACB = 90Å.
There is a horizontal force of magnitude F N acting on the block. The block is in equilibrium.
(a) In the case where F = 20, find the tensions in each of the strings. [5]
(b) Find the greatest value of F for which the block remains in equilibrium in the position shown.
[3]
Mark scheme
4(a) TA × 0.8 − TB × 0.6 – 20 = 0 or TA × 0.6 + TB × 0.8 − 10g = 0 M1 Resolving horizontally or vertically
TA × 0.8 − TB × 0.6 – 20 = 0 A1
TA × 0.6 + TB × 0.8 − 10g = 0 A1
0.6 (10 g − 0.6TA ) M1 Attempt to solve simultaneously
0.8TA − = 20 → TA =
0.8
TA = 76 N, TB = 68 N A1
5
4(b) 500 B1 From using TB = 0
TA × 0.6 − 10g = 0 ⇒ TA =
3
TA × 0.8 − F = 0 M1
400 A1 Allow F = 133 to 3 s.f.
F=
3
3
A cyclist is riding along a straight horizontal road. The total mass of the cyclist and her bicycle is 70 kg. At an instant when the cyclist’s speed is 4 m s−1 , her acceleration is 0.3 m s−2 . There is a constant resistance to motion of magnitude 30 N.
(a) Find the power developed by the cyclist. [3]
The cyclist comes to the top of a hill inclined at 5Å to the horizontal. The cyclist stops pedalling and freewheels down the hill (so that the cyclist is no longer supplying any power). The magnitude of the resistance force remains at 30 N. Over a distance of d m, the speed of the cyclist increases from
6 m s−1 to 12 m s−1 .
(b) Find the change in kinetic energy. [2]
(c) Use an energy method to find d . [3]
Mark scheme
5(a) F – 30 = 70 × 0.3 M1 Use of Newton’s Second law
P = 4F B1 Using P = Fv
[= 51 × 4] = 204 W A1
3
5(b) 1 1 M1
Change in KE = × 70 × 12 2 − × 70 × 6 2
2 2
3780 J A1
2
5(c) For work energy equation M1
70 g × d sin 5 − 30d = 3780 A1 FT FT change in kinetic energy from (b)
d = 122 A1
3
Two particles P and Q, of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley at B which is attached to two inclined planes. P lies on a smooth plane AB which is inclined at 60Å to the horizontal. Q lies on a plane BC which is inclined at 30Å to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes (see diagram).
(a) It is given that the plane BC is smooth and that the particles are released from rest.
Find the tension in the string and the magnitude of the acceleration of the particles. [5]
(b) It is given instead that the plane BC is rough. A force of magnitude 3 N is applied to Q directly up the plane along a line of greatest slope of the plane.
Find the least value of the coefficient of friction between Q and the plane BC for which the particles remain at rest. [5]
Mark scheme
6(a) Attempt to use Newton’s Second law M1
For P: 0.3g sin 60 − T = 0.3a A1 For any one equation
For Q: T + 0.2 g sin 30 = 0.2a
A1 For any second equation
System: 0.3g sin 60 + 0.2 g sin 30 = 0.5a
0.3 g sin60 − T 0.2 g sin30 + T
=
0.3 0.2
0.3g sin 60 + 0.2 g sin 30 = 0.5a a = M1 For solving for a or T
Magnitude of acceleration = 7.20 ms−2 Tension = 0.439 N A1
5
6(b) R = 0.2g cos 30 B1
3 3 B1 Equilibrium for P
[0.3g sin 60 − T = 0] T= or T = 2.598...
2
T + 0.2 g sin 30 − F − 3 = 0 M1 Equilibrium for Q on the point of moving down
3 3 M1 Use of F = μ R
+ 0.2 g sin 30 − μ ( 0.2 gcos30 ) − 3 = 0
2
μ = 0.345 A1
5
A particle P moves in a straight line through a point O. The velocity v m s−1 of P, at time t s after passing O, is given by
9 b v= + − ct2 ,
4 t + 12 where b and c are positive constants. At t = 5, the velocity of P is zero and its acceleration is − 13
12 m s−2 .
(a) Show that b = 9 and find the value of c. [5]
(b) Given that the velocity of P is zero only at t = 5, find the distance travelled in the first 10 seconds of motion. [5]
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) 9 b B1 Use of v ( 5) = 0 to form equation in b and c
0= + − c × 52
4 ( 5 + 1)2
b M1 dv 13
a = −2 − 2ct For use of a = and a ( 5 ) = −
( t + 1) 3
dt 12
13 b A1
− = −2 − 2c × 5
12 ( 5 + 1)3
b 9 b 13 M1 Attempts to solve simultaneous equations
− 25c = − and − − 10c = − leading to b = … or c = …
36 4 108 12
b = 9 and c = 0.1 A1 b = 9 (AG)
5
7(b) 9 M1 For use of s = v dt
9
+ − 0.1t 2 dt = …
4 ( t + 1) 2
9 9 1 A1 FT FT their value of c from (a)
= t− − t3 [+ K ]
4 ( t + 1) 30
9 9 1
10
9 9 1 9 9 1 M1 For evaluation from 0 to 5 or from 5 to 10
t− − t 3 = × 10 − − × 103 − × 5 − − × 53
4 ( t + 1) 30 5 4 (10 + 1) 30 4 ( 5 + 1) 30
= M1 For evaluation from 0 to 5 and from 5 to 10 to find
9 9 1 9 9 1 9 9 1 distance travelled
× 5 − − × 53 − ( −9) − ×10 − − ×103 − × 5 − − × 53
4 ( 5 + 1) 30 4 (10 + 1) 30 4 ( 5 + 1) 30
175 2275
= 5.583 + 9 + 11.651 + 5.583 or +
12 132
= 31.8 m A1 350
or
11
5
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Mathematics 9709/43 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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