Mathematics 9709/43 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Particles P of mass 0.4 kg and Q of mass 0.5 kg are free to move on a smooth horizontal plane. P and
Q are moving directly towards each other with speeds 2.5 m s−1 and 1.5 m s−1 respectively. After P and Q collide, the speed of Q is twice the speed of P.
Find the two possible values of the speed of P after the collision. [4]
Mark scheme
1 0.4 × 2.5 – 0.5 × 1.5 M1 Attempt momentum before impact.
0.4 × 2.5 – 0.5 × 1.5 = 0.4v + 0.5 × 2v M1 Use of conservation of momentum, either case.
0.4 × 2.5 – 0.5 × 1.5 = 0.4v + 0.5 × 2v A1 One correct equation
or 0.4 × 2.5 – 0.5 × 1.5 = –0.4v + 0.5 × 2v
Speed is 0.179 m s–1 or 0.417 m s–1 A1 Both values
4
A cyclist is travelling along a straight horizontal road. She is working at a constant rate of 150 W. At an instant when her speed is 4 m s−1 , her acceleration is 0.25 m s−2 . The resistance to motion is 20 N.
(a) Find the total mass of the cyclist and her bicycle. [3]
The cyclist comes to a straight hill inclined at an angle 1 above the horizontal. She ascends the hill at constant speed 3 m s−1 . She continues to work at the same rate as before and the resistance force is unchanged.
(b) Find the value of 1. [2]
Mark scheme
2(a) 150 B1 OE. P = Fv used correctly.
Forward force exerted by cyclist = N [= 37.5 N]
4
150 M1 Use of Newton’s second law
− 20 = m × 0.25
4
m = 70 kg A1
3
2(b) 150/3 – 20 – 70gsin θ = 0 M1 For resolving up the plane
θ = 2.5° to 1d.p. A1 FT From 2.456….
FT θ = sin–1 3 from (a)
m
2
20 N
FN
1Å 60Å
"Å !Å
30 N
40 N
Four coplanar forces act at a point. The magnitudes of the forces are 20 N, 30 N, 40 N and F N. The directions of the forces are as shown in the diagram, where sin !Å = 0.28 and sin "Å = 0.6.
Given that the forces are in equilibrium, find F and 1. [6]
Mark scheme
3 F sin θ + 20sin 60 – 30sin α – 40 sin β = 0 M1 For resolving in either direction
Vertical: F sin θ + 20sin 60 – 30 × 0.28 – 40 × 0.6 = 0 [F sinθ = 15.07949…] A1
Horizontal: F cosθ + 40 × 0.8 – 30 × 0.96 – 20cos 60 = 0 [F cosθ = 6.8] A1
15.0794 … M1 For method for finding θ
θ = tan −1
6.8
M1 For method for finding F
F= 15.07949…2 + 6.82
θ = 65.7, F = 16.5 A1
6
A particle is projected vertically upwards with speed u m s−1 from a point on horizontal ground. After
2 seconds, the height of the particle above the ground is 24 m.
(a) Show that u = 22. [2]
(b) The height of the particle above the ground is more than h m for a period of 3.6 s.
Find h. [4]
Mark scheme
4(a) 1 M1 Use of s = ut + ½at2
24 = u × 2 − g × 22
2
u = 22 A1 AG
2
4(b) At maximum height 0 = 222 − 2gs M1 Use of v2 = u2 + 2as to find maximum height.
Maximum height s = 24.2 m A1
Height down = 0.5g × 1.82 (=16.2) M1 Find distance travelled down in 1.8 s.
h=8 A1
Alternative method for Question 4(b)
0 = 22 – 10t M1 Use of v = u – gt with u = 22 and v = 0 to find time to
reach maximum height
t = 2.2 A1
1 M1 Use of s = ut + 1 at2 to find value of h
h = 22 × (2.2 – 1.8) – g × (2.2 – 1.8)2 2
2
h=8 A1
Alternative method for Question 4(b)
1 2 1 M1 Use of s = ut + 1 at2 for times t and t + 3.6 to find time
22t – gt = 22 × (t + 3.6) – g × (t + 3.6)2 2
2 2
taken to reach height h.
t = 0.4 (or t + 3.6 = 4) A1
1 M1 Use s = ut + 1 at2 to find value of h.
h = 22 × 0.4 – g × 0.42 2
2
h=8 A1
4
A car of mass 1400 kg is towing a trailer of mass 500 kg down a straight hill inclined at an angle of 5Å to the horizontal. The car and trailer are connected by a light rigid tow-bar. At the top of the hill the speed of the car and trailer is 20 m s−1 and at the bottom of the hill their speed is 30 m s−1 .
(a) It is given that as the car and trailer descend the hill, the engine of the car does 150 000 J of work, and there are no resistance forces.
Find the length of the hill. [5]
(b) It is given instead that there is a resistance force of 100 N on the trailer, the length of the hill is
200 m, and the acceleration of the car and trailer is constant.
Find the tension in the tow-bar between the car and trailer. [4]
Mark scheme
5(a) Increase in KE = ½ × 1900 × 302 – ½ × 1900 × 202 [= 475000 J] B1 May be implied by energy equation.
Loss of PE = 1900 × g × s sin 5 [= 1655.95s J] B1 May be implied by energy equation.
1900 × g × s sin 5 + 150 000 = ½ × 1900×302 – ½ × 1900 × 202 M1 For attempt at work/energy equation
A1 Correct
s = [Length of hill =] 196 m A1
5
5(b) 302 = 202 + 2a × 200 M1 Use of v2 = u2 + 2as
a = 1.25 m s–2 A1
T – 100 + 500g sin 5 = 500a M1 For applying Newton’s second law to the trailer.
T = 289 N A1
4
A particle moves in a straight line and passes through the point A at time t = 0. The velocity of the particle at time t s after leaving A is v m s−1 , where
v = 2t2 − 5t + 3.
(a) Find the times at which the particle is instantaneously at rest. Hence or otherwise find the minimum velocity of the particle. [4]
(b) Sketch the velocity-time graph for the first 3 seconds of motion. [3]
(c) Find the distance travelled between the two times when the particle is instantaneously at rest.
[3]
Mark scheme
6(a) (2t – 3)(t – 1) = 0 leading to t = ….. M1 Attempt to solve v = 0
t = 1 or t = 1.5 A1
Minimum velocity when t = 1.25 leading to v = ….. M1 Uses roots or dv/dt=0 to find t for vmin and attempts
dv substitution to obtain vmin. Alternatively completes
or = 4t – 5 = 0 t = 1.25 leading to v = ….. square.
dt
2
or v = 2 t − 5 − 25 + 3 leading to v = …..
4 16
Minimum velocity is −0.125 m s–1 A1 1
Allow v = −
8
4
6(b) Quadratic curve (two roots and v(3) > v(0)) B1
Goes through (1.25, –0.125), (0, 3), (1, 0), (1.5, 0), (3,6) B1 3 of the 5 key points shown on axes or as coordinates
All five points shown on a totally correct graph B1
3
6(c) 2 3 5 2 M1 For use of s = ∫v dt
s= t − t + 3t
3 2
2 5 2 3 5 2 M1 Correct use of limits (their 1 and 1.5)
( ) ( ) ( ) – (1) − (1) + 3(1)
3 2
3 1.5 − 1.5 + 3 1.5
2 3 2
Distance = 0.0417 m A1 A0 for –0.0417
3
4N P
0.3 kg
1
A particle P of mass 0.3 kg rests on a rough plane inclined at an angle 1 to the horizontal, where
7 . A horizontal force of magnitude 4 N, acting in the vertical plane containing a line of sin 1 = 25 greatest slope of the plane, is applied to P (see diagram). The particle is on the point of sliding up the plane.
(a) Show that the coefficient of friction between the particle and the plane is 43 . [4]
The force acting horizontally is replaced by a force of magnitude 4 N acting up the plane parallel to a line of greatest slope.
(b) Find the acceleration of P. [3]
(c) Starting with P at rest, the force of 4 N parallel to the plane acts for 3 seconds and is then removed.
Find the total distance travelled until P comes to instantaneous rest. [3]
Additional Page
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Mark scheme
7(a) 24 7 M1 Resolving forces perpendicular to the plane or parallel
R = 0.3g cos θ + 4 sin θ = 3 × + 4× [=4]
25 25 to the plane. Allow use of θ = 16.3°
24 7
F = 4 cos θ – 0.3g sin θ = 4 × − 3× [=3] A1 Two correct equations
25 25
3=µ×4 M1 For use of F = µR
3 A1 AG Must be from correct and exact working, not using
µ=
4 16.3
4
7(b) 3 24 54 B1
F = µ× 0.3g cos θ = ×3× = 25 = 2.16
4 25
3 24 7 M1 Use of Newton’s second law
4− × 0.3g × – 0.3g × = 0.3a
4 25 25
10 A1
a= m s–2
3
3
7(c) 1 10 10 B1 FT Distance s1 in 3s and v after 3s; FT a from (b)
s1 = × × 32 = 15 and v = × 3 = 10
2 3 3
–0.3g × sin θ – µ × 0.3g cos θ = 0.3a leading to a = –10 M1 Apply Newton’s 2nd law after 4 N removed, find a and
0 = 102 + 2 × (-10) × s2 use v2 = u2 + 2as to find extra distance s2
[s2 = 5 leading to total distance = s1 + s2 = 15 + 5 =] 20 m A1
Alternative method for Question 7(c)
10 2 B1 FT WD = Fs and s = ½ at2 for 4 N force; FT a from (b)
Work done = 4 × 0.5 × × 3 [= 60 J]
3
60 = µ × 0.3g cos θ × d + 0.3g × d sin θ M1 WD by 4 N force = WD against F + PE gain
d = 20 m A1
3
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