Mathematics 9709/41 October/November 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A particle of mass 1.6 kg is projected with a speed of 20 m s−1 up a line of greatest slope of a smooth plane inclined at ! to the horizontal, where tan ! = 34 .
Use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest. [3]
Mark scheme
1 3 1 B1 For either the correct potential energy or kinetic energy
±1.6 g × x × [=9.6x] or ± ×1.6 × 202 [= 320] term. Need not be evaluated.
5 2
1 3 M1 Attempt at energy equation; 2 relevant terms.
× 1.6 × 202 = 1.6 g × x sin α where sinα = Dimensionally correct but allow sign errors.
2 5
Allow sin/cos mix and sin(36.869…) but sin α (oe) must
3
have been substituted. M0 for 1.6 g × x × .
4
100 A1 Allow 33.3.
x=
3
3
A B
35Å 40Å
2.4 kg
A particle of mass 2.4 kg is held in equilibrium by two light inextensible strings, one of which is attached to point A and the other attached to point B. The strings make angles of 35Å and 40Å with the horizontal (see diagram).
Find the tension in each of the two strings. [5]
Mark scheme
2 Attempt to resolve horizontally or vertically to form an equation. *M1 Correct number of terms; allow sin/cos mix; allow sign
errors – do not award this mark if using T for both (see
SC later).
T1 cos35 = T2 cos 40 A1 Must be different Ts.
T1 sin 35 + T2 sin 40 = 2.4 g A1 If same Ts, then SC B2 only for this equation.
Attempt to solve for either tension. DM1 From equations with correct number of relevant terms.
Must get a value for at least one tension.
cos 40
E.g. T2 × sin 35 + sin 40 = 24
cos35
T1 = 20.4 N and T2 = 19.0 N A1 T1 = 19.033621 T2 = 20.353166
awrt 20.4 for T1 www and 19(.0) for T2.
5
v (m s−1 )
12.6
0 t (s)
0 8 48 62 70
The diagram shows the velocity-time graph for the motion of a bus. The bus starts from rest and accelerates uniformly for 8 seconds until it reaches a speed of 12.6 m s−1 . The bus maintains this speed for 40 seconds. It then decelerates uniformly in two stages. Between 48 and 62 seconds the bus decelerates at a m s−2 and between 62 and 70 seconds it decelerates at 2a m s−2 until coming to rest.
(a) Find the distance covered by the bus in the first 8 seconds. [1]
(b) Find the value of a. [3]
(c) Find the average speed of the bus for the whole journey. [4]
Mark scheme
3(a) Distance = 50.4m B1 252
Allow .
5
1
3(b) v1 = 12.6 − ( 62 − 48 ) a M1 Use of suvat for first section of deceleration.
12.6 ± (62 − 48)a only.
0 = v1 − 2a × ( 70 − 62 ) M1 Use of suvat for second section of deceleration. An
expression for the velocity at 62 seconds must be
±2a × (70 − 62).
a = 0.42 A1 –0.42 scores A0.
3
3(c) Speed at time t = 62 is 6.72 m s-1 B1 This may be seen in part (b) but must be used in part (c)
to get this mark.
s2 = ( 48 − 8 ) × 12.6 [ = 504] B2FT B2 FT for any 2 correct, B1 FT for any 1 correct – follow
through their value of v1 where 0 < v1 < 12.6 but must
3381
s3 = 0.5 × (12.6 + their 6.72 ) × ( 62 − 48 ) = 135.24 or oe have come from the correct equations seen in part (b).
25 Allow correct value of v1 from a = −0.42 where
or their 6.72 × ( 62 − 48 ) + 0.5 × ( 62 − 48 ) × (12.6 − their 6.72 ) v1 = 12.6 + (62 − 48)a and v1 = −2a × (70 − 62).
672
s4 = 0.5 × their 6.72 × ( 70 − 62 ) = 26.88 or oe
25
Average speed =10.236 m s-1 B1 2559 59
Allow 10.2 or better oe e.g. , 10 .
250 250
4
Two particles P and Q, of masses 6 kg and 2 kg respectively, lie at rest 12.5 m apart on a rough horizontal plane. The coefficient of friction between each particle and the plane is 0.4. Particle P is projected towards Q with speed 20 m s−1 .
(a) Show that the speed of P immediately before the collision with Q is 10 3 m s−1 . [3]
In the collision P and Q coalesce to form particle R.
(b) Find the loss of kinetic energy due to the collision. [4]
The coefficient of friction between R and the plane is 0.4.
(c) Find the distance travelled by particle R before coming to rest. [2]
Mark scheme
4(a) −0.4 × 6 g = 6a *B1 Resolve horizontally using Newton’s second law; 2
relevant terms; must be either −0.4 × 6 g = 6a or
0.4 × 6 g = 6a.
v2 = 202 + 2 × ( −4) ×12.5 DM1 Use complete suvat method to get an equation in v or v 2 –
must be using u = 20, s = 12.5 and their a.
v 2 = 300 v = 10 3 A1 AG. Condone correct expression for v or v2 followed by
correct answer.
Alternative method for Question 4(a)
RF = 0.4 × 6 g *B1 Correct application of F = μ R for P.
0.5 × 6 × 202 − 0.5 × 6 × v 2 = 12.5 × (0.4 × 6 g ) DM1 3 relevant terms; dimensionally correct; allow sign errors
only.
v 2 = 300 v = 10 3 A1 AG. Condone correct expression for v or v2 followed by
correct answer.
3
4(b) 6 ×10 3 = ( 6 + 2) v ' M1 For use of conservation of momentum, 3 non-zero terms,
allow sign errors. Use of 20 is M0.
v ' = 7.5 3 A1 12.99038…
1
( )
Initial KE = × 6 × 10 3 [ = 900]
B1 Either initial kinetic energy or final kinetic energy correct.
2
2 Allow unsimplified.
1
( )
Final KE = × 8 × 7.5 3 [ = 675]
2
2
Loss of KE = 225 J A1
4
4(c)
( ) + 2 × ( their − 4 ) × s M1 Use complete suvat method to find distance. This must be
2
0 = their 7.5 3
using their v′ from part (b), so it is dependent on scoring
the first M mark in part (b) and either their a from part
(a), or from ±0.4 × 8g = 8a.
[Distance =] 21.1 m A1 21.1 or better (21.09375).
2
P
B
A
1.6 kg
1.2 kg
50Å 40Å
The diagram shows a particle A, of mass 1.2 kg, which lies on a plane inclined at an angle of 40Å to the horizontal and a particle B, of mass 1.6 kg, which lies on a plane inclined at an angle of 50Å to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley P fixed at the top of the planes. The parts AP and BP of the string are taut and parallel to lines of greatest slope of the respective planes. The two planes are rough, with the same coefficient of friction, -, between the particles and the planes.
Find the value of - for which the system is in limiting equilibrium. [7]
Mark scheme
5 Resolving parallel to the slope at A or B to form an equation. *M1 Correct number of terms; allow sign errors; allow sin/cos
mix.
1.6 g sin 50 − T − FB = 0 A1 If using the same Fs, then M1A1A0B1 max.
T − FA − 1.2 g sin 40 = 0 A1 System equation (must be four different terms):
1.6 g sin 50 − FB − FA − 1.2 g sin 40 = 0 only scores
M1A1A1. Any sign errors scores M1 only.
RA = 1.2 g cos 40 or RB = 1.6 g cos50 *B1 Either correct. Must be explicitly linked to the correct
contact (so could be seen on a diagram), or as part of a
resolving parallel to the slope equation(s) (so must be
combined with μ ).
FA = 1.2 g μ cos 40 or FB = 1.6 g μ cos50 *M1 Use of F = μ R at either A or B. Must be explicitly linked
to the correct contact (could be seen on a diagram) or as
part of a resolving parallel to the slope equation(s). Allow
sin/cos mix error only.
1.6 g sin50 − 1.6 g μ cos50 = 1.2 g sin 40 + 1.2 g μ cos 40 DM1 Eliminating T, FA and FB to form an equation in μ only.
1.6 g sin 50 − 1.2 g sin 40 A1 0.23326119…
μ = 1.2 g cos 40 + 1.6 g cos50 μ = 0.233
7
A car of mass 1300 kg is moving on a straight road.
(a) On a horizontal section of the road, the car has a constant speed of 30 m s−1 and there is a constant force of 650 N resisting the motion.
(i) Calculate, in kW, the power developed by the engine of the car. [2]
(ii) Given that this power is suddenly increased by 9 kW, find the instantaneous acceleration of the car. [3]
(b) On a section of the road inclined at sin−1 0.08 to the horizontal, the resistance to the motion of the car is 1000 + 20v N when the speed of the car is v m s−1 . The car travels downwards along this section of the road at constant speed with the engine working at 11.5 kW.
Find this constant speed. [4]
Mark scheme
6(a)(i) Power = 19.5 kW B2 Or B1 for either 650 × 30 or 19 500.
2
6(a)(ii) their19500 + 9000 = DF × 30 B1FT Oe FT their 19.5 in watts only.
DF − 650 = 1300a M1 Newton’s second law horizontally; 3 relevant terms;
dimensionally correct but allow sign errors; allow with
their driving force or just DF.
3 A1 0.23076923…
a= = 0.231 m s–2
13
Alternative scheme for 6(a)(ii)
9000 = DF × 30 *B1 9000
oe e.g. DF = .
30
DF = 1300a DM1 Resolving horizontally using Newton’s second law; 2
relevant terms; dimensionally correct but allow sign
errors.
3 A1 0.23076923…
a= = 0.231 m s–2
13
3
6(b) 11500 B1 oe e.g. DF × v = 11500.
DF =
v
Attempt at Newton’s second law. M1 4 relevant terms, their DF or just DF; allow sign errors:
allow sin/cos mix; allow g missing.
11500 A1 Correct equation.
+ 1300 × g × 0.08 − (1000 + 20v ) = 0
v
Speed = 25 m s–1 A1
4
A particle moves in a straight line starting from a point O before coming to instantaneous rest at a point X . At time t s after leaving O, the velocity v m s−1 of the particle is given by
v = 7.2t2 0 ≤ t ≤ 2, v = 30.6 − 0.9t 2 ≤ t ≤ 8,
1600 v = 2 + kt 8 ≤ t, t where k is a constant. It is given that there is no instantaneous change in velocity at t = 8.
Find the distance OX . [9]
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 1600 *M1 Use velocity at t = 8 to set up a linear equation in k only.
30.6 − 0.9 × 8 = + 8k Allow a slip in one value or sign only.
82
k = −0.2 A1
1600 DM1 Attempt to find the value of t when the particle comes to
+ ( their k ) × t = 0 t = rest using the correct expression for v, set equal to zero
t2
with their negative value of k. Must find a positive value
for t (for reference, t = 20).
Attempt to integrate v for one of the 3 intervals *M1 Increase power by 1 and a change in coefficient in at least
one term (which must be the same term); s = vt is M0.
7.2 3 A1 May be unsimplified (for reference, limits are from 0 to
s= t ( +c ) 2).
3
0.9 2 A1 May be unsimplified (for reference, limits are from 2 to
s = 30.6t − t ( +c ) 8).
2
1600 −1 k 2 A1FT May be unsimplified (for reference limits are from 8 to
s= t + t ( +c ) 20). Follow through their value of k or just k only.
−1 2
Either 19.2 or 156.6 or ±86.4 B1 One correct distance found. Allow unsimplified e.g.
1
( 216 − 59.4 ) or × (8 − 2) × ( 28.8 + 23.4) etc.
2
Distance = 19.2 + ( 216 − 59.4) + ( −120 − ( −206.4)) = 262.2 m B1 This mark can be awarded if no integration is shown oe.
1311
e.g. . Condone 262 www.
5
9
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Mathematics 9709/41 for October/November 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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