Mathematics 9709/41 May/June 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2023. 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 m kg and 0.3 kg respectively, are at rest on a smooth horizontal plane. P is projected at a speed of 5 m s−1 directly towards Q. After P and Q collide, P moves with a speed of 2 m s−1 in the same direction as it was originally moving.
(a) Find, in terms of m, the speed of Q after the collision. [2]
After this collision, Q moves directly towards a third particle R, of mass 0.6 kg, which is at rest on the plane. Q is brought to rest in the collision with R, and R begins to move with a speed of 1.5 m s−1 .
(b) Find the value of m. [2]
Mark scheme
1(a) m × 5 + 0 = m × 2 + 0.3v M1 Attempt at conservation of momentum; 3 non-zero
terms (with m appearing in two terms); allow sign
errors.
Speed = 10m (m s–1) A1 M1A0 if using g in momentum terms.
v = −10m is A0.
2
1(b) 0.3 × 10m + 0 = 0 + 0.6 × 1.5 [3m = 0.9] M1 Attempt at conservation of momentum between Q and
R (so must be using correct masses of 0.3 and 0.6) to
form a linear equation in m using their answer from
(a); 2 non-zero terms; allow sign errors.
m = 0.3 A1FT 3
FT
their +ve coefficient of m from (a)
Condone including kg in answer.
2
A particle P of mass 0.4 kg is projected vertically upwards from horizontal ground with speed 10 m s−1 .
(a) Find the greatest height above the ground reached by P. [2]
When P reaches the ground again, it bounces vertically upwards. At the first instant that it hits the ground, P loses 7.2 J of energy.
(b) Find the time between the first and second instants at which P hits the ground. [4]
Mark scheme
2(a) 0 = 102 + 2( − g ) s s = ... M1 For use of v 2 = u 2 + 2as with v = 0, u = ±10 and
a = ± g or ±10 and solve for s (or any other complete
SUVAT method). Or using an energy method:
1
0.4 gh = (0.4)(10)2 and solve for h (with two terms
2
using correct given values – condone lack of masses in
conservation of energy equation).
Max. height = 5 (m) A1
2
2(b) 1 B1FT Or loss of PE = ± 0.4 × g × 5 using their maximum
KE before impact = × 0.4 × 102 [= 20]
2 height from (a).
1 *M1 1
× 0.4 × v 2 = 20 − 7.2 [v = 8] M1 for × 0.4v 2 = KE/PE before impact − 7.2 (must
2 2
or 0.4 gh = 20 − 7.2 [ h = 3.2] be correct method of subtracting 7.2 (OE)). Or, for
finding the maximum height after first impact. Need
not solve for v (or h) for this mark.
−8 = 8 + (− g )t t = ... DM1 For use of a complete method to find t. Condone sign
1 errors but a = ± g . If calculating the time to the
or 0 = 8t + × − g × t 2 t = ... maximum height between the first and second impacts,
2
then candidates must double this answer (OE). E.g.,
0 = 8 + (− g )T , t = 2T = ... or
3.2 = 0 − 0.5 × − g × T 2 t = 2T = ... .
Time = 1.6 (s) A1
4
A particle moves in a straight line starting from rest. The displacement s m of the particle from a fixed point O on the line at time t s is given by
5 3 s = t 2 − 15
4 t 2 + 6.
Find the value of s when the particle is again at rest. [4]
Mark scheme
3 For an attempt at differentiation. *M1 Decrease power by 1 and a change in coefficient in at
s
least one term. v = is M0.
t
5
3
45
1 A1 Allow unsimplified, including indices (including a +c
(v =) t 2 − t 2 is A0).
2 8
1
1 DM1 Attempting to find t by equating v to 0 and attempt to
v = 0 t 2 (20t − 25) = 0 t = ... solve a linear equation for t (if correct t = 2.25).
8
Must be of the form t = … .
15 A1 Condone 0.938 .
s= [ = 0.9375]
16
4
v (m s−1 )
0.9
0 t (s)
0 3 9 10 T
The velocity of a particle at time t s after leaving a fixed point O is v m s−1 . The diagram shows a velocity-time graph which models the motion of the particle. The graph consists of 5 straight line segments. The particle accelerates to a speed of 0.9 m s−1 in a period of 3 s, then travels at constant speed for 6 s, and then comes instantaneously to rest 1 s later. The particle then moves back and returns to rest at O at time T s.
(a) Find the distance travelled by the particle in the first 10 s of its motion. [2]
(b) Given that T = 12, find the minimum velocity of the particle. [2]
(c) Given instead that the greatest speed of the particle is 3 m s−1 , find the value of T and hence find the average speed of the particle for the whole of the motion. [4]
Mark scheme
4(a) 1 M1 Completely correct method for finding the total area
Distance = (6 + 10) × 0.9 underneath the velocity-time graph from t = 0 to 10
2
only. Can be done as two triangles and a rectangle e.g.,
1 1
× 3 × 0.9 + (9 − 3) × 0.9 + × 1 × 0.9 (allow a slip in
2 2
one value); need not see all three components added
together.
Distance = 7.2(m) A1
2
4(b) 1 M1 1
(12 − 10)vmin = −7.2 Setting (12 − 10)vmin equal to ± their (a).
2 2
Minimum velocity = −7.2 (m s–1) A1 Must be negative – allow those who solve
1
(12 − 10)vmin = 7.2 and obtain vmin = 7.2 and then
2
change to −7.2 without justification.
SC B1 for assuming the triangle is isosceles.
2
4(c) 1 1 *M1 Correct method for finding T or t. Condone sign errors
× (T − 10) × 3 = 7.2 or × t × 3 = 7.2 but must equate to their answer to (a).
2 2
T = 14.8 A1 OE (e.g. from t + 10 = 4.8 + 10 = 14.8).
14.4 DM1 2(their (a)) 2(their (a))
M1 for or .
14.8 their T 10 + their T
36 A1 OE 0.973 [For reference: 0.97297…].
Average speed = (m s–1 )
37
SC *B1 (for T = 14.8). DM1A1 for assuming the
triangle is isosceles.
4
40 N
FN
60Å 1Å 10 N
50 N
Four coplanar forces act at a point. The magnitudes of the forces are F N, 10 N, 50 N and 40 N. The directions of the forces are as shown in the diagram.
(a) Given that the forces are in equilibrium, find the value of F and the value of 1. [6]
(b) Given instead that F = 10 2 and 1 = 45, find the direction and the exact magnitude the resultant force. [3]
Mark scheme
5(a) Resolving either direction. M1 3 terms; allow sign errors and allow sin/cos mix. Must
be an equation with either = 0 or with an attempt to
balance forces.
Vertical: F sin θ + 40sin 60 − 50 = 0 A1 F sin θ = 50 − 20 3 = 15.358...
Horizontal: F cos θ + 10 − 40cos 60 = 0 A1 [ F cosθ = 10]
θ = tan −1 (5 − 2 3) M1 Attempt to solve for θ ; one missing term in total
θ = tan −1 1.535898... .
F = 15.358...2 + 10 2 M1 Attempt to solve for F: one missing term in total.
θ = 56.9, F = 18.3 A1 Both correct (18.327530…, 56.932462…).
6
5(b) (Y =) ± (10 2 sin 45 + 40sin 60 − 50) [= ± (20 3 − 40)] B1 Allow non-exact values for 2 etc. in correct
expression.
( X =) ± (10 2 cos 45 + 10 − 40cos 60) [= 0] B1 Allow non-exact values for 2 etc. in correct
expression. Could be implied by correct answer.
Resultant force is 40 − 20 3 (N) in the same direction as the 50 (N) force. B1 Allow vertically downwards, south, 180°, negative y-
direction.
Resultant force must be exact and positive (so
20 3 − 40 is B0).
3
B
P
0.2 kg
Q
0.1 kg
60Å 1Å
A C
Two particles P and Q, of masses 0.2 kg and 0.1 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. Particle P lies on a smooth plane AB which is inclined at 60Å to the horizontal. Particle Q lies on a plane BC which is inclined at an angle of 1Å 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 1 = 60, the plane BC is rough and the coefficient of friction between Q and the plane BC is 0.7. The particles are released from rest.
Determine whether the particles move. [4]
(b) It is given instead that the plane BC is smooth.
The particles are released from rest and in the subsequent motion the tension in the string is 3 − 1 N.
Find the magnitude of the acceleration of P as it moves on the plane, and find the value of 1. [4]
Mark scheme
6(a) R = 0.1g cos 60[= 0.5] B1 Correctly resolving perpendicular to the plane for Q.
F = 0.7 × 0.1g cos 60 [= 0.35] M1 Use of F = μ R for Q where R is a component of
weight but not mass; allow sin/cos mix.
For whole system: LHS of Newton’s second law: M1 Complete method to determine the resultant force for
0.2 g sin 60 − 0.1g sin 60 − F [= 0.866... − F ] the whole system. Allow sign errors and sin/cos mix,
but must include all required terms and be
dimensionally correct. If considering either the while
Or separately for P and Q: system or P and Q separately then ignore the RHS of
0.2 g sin 60 − T (= 0.2a ) and T − 0.1g sin 60 − F (= 0.1a), and eliminate T to their Newton’s second law equations.
get 0.2 g sin 60 − 0.1g sin 60 − F (= 0.3a )
3 A1 Correct indication (with no incorrect working) that the
As − 0.35 > 0 the particles do move. resultant force is positive (e.g. 0.8660... − 0.35 > 0 or
2
0.516… (to at least 1 sf) which is positive) together
with a correct conclusion. Candidates may calculate
10 3 − 7
the acceleration which is = 1.72008... and
6
then say that the particles are moving.
4
6(b) Attempt to use Newton’s second law for P: M1 Allow sign errors, sin/cos mix. but must be
0.2 g sin 60 − ( 3 − 1) = 0.2a dimensionally consistent.
a = 5 (m s-2) A1
Newton’s second law for system: 0.2 g sin 60 − 0.1g sin θ = 0.3(5) M1 Attempt Newton’s second law for Q, or for the whole
system. Allow sign errors, sin/cos mix, but must be
or Newton’s second law for Q: ( 3 − 1) − 0.1g sin θ = 0.1(5)
dimensionally consistent.
θ = 13.4 A1 13.41784…
4
A car of mass 1200 kg is travelling along a straight horizontal road. The power of the car’s engine is constant and is equal to 16 kW. There is a constant resistance to motion of magnitude 500 N.
(a) Find the acceleration of the car at an instant when its speed is 20 m s−1 . [3]
(b) Assuming that the power and the resistance forces remain unchanged, find the steady speed at which the car can travel. [2]
The car comes to the bottom of a straight hill of length 316 m, inclined at an angle to the horizontal
1 of sin−1 60 . The power remains constant at 16 kW, but the magnitude of the resistance force is no longer constant and changes such that the work done against the resistance force in ascending the hill is 128 400 J. The time taken to ascend the hill is 15 s.
(c) Given that the car is travelling at a speed of 20 m s−1 at the bottom of the hill, find its speed at the top of the hill. [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) 16000 B1 OE e.g. 16000 = 20 × F
Driving force F = [ = 800]
20
F − 500 = 1200a M1 Use of Newton’s second law; allow sign errors but
must be 3 terms. Allow F or any non-zero value for the
driving force (allow 0.8 from using 16 rather than
16000) but not 16000, 16, 20 or 500 for F.
a = 0.25 (m s–2) A1
3
7(b) 16000 M1 Allow sign errors but must be 2 terms. Condone
− 500 = 0
v 16
− 500 = 0 for M1.
v
v = 32 (m s–1) A1
2
7(c) Work done by engine = 16000 × 15 [ = 240000] B1 WD
Or 16000 = .
15
1 1 B1 ±(600v 2 − 240000)
KE change = ± × 1200 × v 2 − × 1200 × 202
2 2
1 B1 Allow 1200 × g × 316 × sin 0.955 or
PE change = ± 1200 × g × 316 × [ = ±63200]
60 79
1200 × g × 316 × sin 0.95 or 1200 × g × or
15
1200 × g × 5.266... .
Attempt at work-energy equation. M1 Use of work-energy principle with 5 terms;
dimensionally correct. Allow sign errors and sin/cos
mix on PE term
1 1 1 A1 Allow a value in the interval [62870,63600] for the PE
16000 × 15 − 128400 = × 1200 × v 2 − × 1200 × 202 + 1200 g × 316 × term from using non-exact values for the given angle
2 2 60
(but not if from incorrect working).
(240000 − 128400 = 600v 2 − 240000 + 63200)
v = 21.9 (m s–1) A1 21.924111…
6
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