Cambridge International AS & A Level
CANDIDATE
NAME
CENTRE
NUMBER
CANDIDATE
NUMBER
*7350285007*
MATHEMATICS
9709/42
Paper 4 Mechanics
February/March 2023
1 hour 15 minutes
You must answer on the question paper.
You will need: List of formulae (MF19)
INSTRUCTIONS
! Answer all questions.
! Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs.
! Write your name, centre number and candidate number in the boxes at the top of the page.
! Write your answer to each question in the space provided.
! Do not use an erasable pen or correction fluid.
! Do not write on any bar codes.
! If additional space is needed, you should use the lined page at the end of this booklet; the question
number or numbers must be clearly shown.
! You should use a calculator where appropriate.
! You must show all necessary working clearly; no marks will be given for unsupported answers from a
calculator.
! Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
! Where a numerical value for the acceleration due to gravity (g) is needed, use 10 m s–2.
INFORMATION
! The total mark for this paper is 50.
! The number of marks for each question or part question is shown in brackets [ ].
This document has 16 pages. Any blank pages are indicated.
JC23 03_9709_42/RP
© UCLES 2023
[Turn over
2
1
A crate of mass 200 kg is being pulled at constant speed along horizontal ground by a horizontal
rope attached to a winch. The winch is working at a constant rate of 4.5 kW and there is a constant
resistance to the motion of the crate of magnitude 600 N.
(a) Find the time that it takes for the crate to move a distance of 15 m.
[2]
........................................................................................................................................................
↳
R ma
........................................................................................................................................................
=
-
........................................................................................................................................................
4.5x103
650 200(0)
........................................................................................................................................................
->
=
-
........................................................................................................................................................
v 7.5
........................................................................................................................................................
=>
=
15
[S ut]
7.5
........................................................................................................................................................
=
=
=
........................................................................................................................................................
..
t2 S
........................................................................................................................................................
=
........................................................................................................................................................
The rope breaks after the crate has moved 15 m.
(b) Find the time taken, after the rope breaks, for the crate to come to rest.
[3]
........................................................................................................................................................
KE los-Work
done
Resistance
........................................................................................................................................................
4.5 min
Omi
against
⑪
⑰
........................................................................................................................................................
..
I
x
600xx
[x200x(-5202
........................................................................................................................................................
->V
=
.ix 9.375m
........................................................................................................................................................
=
........................................................................................................................................................
S
(E)t
=
........................................................................................................................................................
........................................................................................................................................................
9.375 7-5 + 0
xt
........................................................................................................................................................
=
2
........................................................................................................................................................
t
2-55
........................................................................................................................................................
..
© UCLES 2023
=
9709/42/F/M/23
3
2
A particle P is projected vertically upwards from horizontal ground with speed 15 m s−1 .
(a) Find the speed of P when it is 10 m above the ground.
[2]
........................................................................................................................................................
.....
=
v U2+ 2as
........................................................................................................................................................
IOM
=152 + 2x (- 10) x10
........................................................................................................................................................
15,4, ↓
⑰
.iV 5mil
........................................................................................................................................................
u
=
=
/
1
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
At the same instant that P is projected, a second particle Q is dropped from a height of 18 m above
the ground in the same vertical line as P.
(b) Find the height above the ground at which the two particles collide.
[3]
........................................................................................................................................................
Let, they
0
meet I'm above
u
........................................................................................................................................................
t
X
p
=
and 'see later
........................................................................................................................................................
ground
18-U
V
........................................................................................................................................................
FP
↓
I
........................................................................................................................................................
A
S=ut+
I
at
........................................................................................................................................................
-Y
t0
:(18- x) 0+ (-10)t2
........................................................................................................................................................
=
=
7
:18 -x 5t2
........................................................................................................................................................
=
18-5t2 15t-5t2
.ix 18- 5t ........................................................................................................................................................
=
=
-
-
it 1.25
S
........................................................................................................................................................
=
..n 18
2
10.8m
x 15t 2 ( 10)t2
........................................................................................................................................................
=
+
=
-
5x12
=
:in 15t 5+2
N
........................................................................................................................................................
=
© UCLES 2023
-
-
-
-
9709/42/F/M/23
[Turn over
4
3
A particle moves in a straight line starting from rest from a point O. The acceleration of the particle
1
at time t s after leaving O is a m s−2 , where a = 4t 2 .
(a) Find the speed of the particle when t = 9.
[2]
........................................................................................................................................................
fadt=f 4t dt
+
4x
v
e
........................................................................................................................................................
=
=
........................................................................................................................................................
..c 0
at t 0, V 0,
........................................................................................................................................................
=
=
=
........................................................................................................................................................
8+
v 5 9 72 mil
at t 9,
........................................................................................................................................................
..V
=
........................................................................................................................................................
x
=
=
=
........................................................................................................................................................
........................................................................................................................................................
(b) Find the time after leaving O at which the speed (in metres per second) and the distance travelled
(in metres) are numerically equal.
[3]
........................................................................................................................................................
s
=frdt=/8*dt 8x+e bt+e
=
=
........................................................................................................................................................
t 0, S 0 ..e 0
at
........................................................................................................................................................
=
=
=
........................................................................................................................................................
.
4+512
S
........................................................................................................................................................
=
........................................................................................................................................................
..
45k_
........................................................................................................................................................
........................................................................................................................................................
I
=>
2.5
........................................................................................................................................................
=
........................................................................................................................................................
:t
=
215
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
© UCLES 2023
9709/42/F/M/23
5
4
A toy railway locomotive of mass 0.8 kg is towing a truck of mass 0.4 kg on a straight horizontal
track at a constant speed of 2 m s−1 . There is a constant resistance force of magnitude 0.2 N on the
locomotive, but no resistance force on the truck. There is a light rigid horizontal coupling connecting
the locomotive and the truck.
0-2
(a) State the tension in the coupling.
·Fig
I
T
>
-
↳
->
I
[1]
0.8kg
T0
........................................................................................................................................................
=
........................................................................................................................................................
(b) Find the power produced by the locomotive’s engine.
1
[1]
0.2 0.8(0)
........................................................................................................................................................
=
-
........................................................................................................................................................
:. P
0.4W
........................................................................................................................................................
=
The power produced by the locomotive’s engine is now changed to 1.2 W.
(c) Find the magnitude of the tension in the coupling at the instant that the locomotive begins to
accelerate.
[5]
........................................................................................................................................................
0-2
-
I
T 0.40 S
I
Truck:
........................................................................................................................................................
0.8kg
0.4Kg
Locomotive:
T
=
Tr.
12
>
I
->
T-0.2=0.82-N
........................................................................................................................................................
-
........................................................................................................................................................
I N
+
=
........................................................................................................................................................
1
0.2
1.2a
........................................................................................................................................................
-
=
I
...a
........................................................................................................................................................
=
........................................................................................................................................................
0.4(5) 5a0.135
::T
........................................................................................................................................................
=
=
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
© UCLES 2023
9709/42/F/M/23
[Turn over
6
5
C
Tsin45
-
500
500 N
T
100 kg
↑Tz
11
45!
T,
+
450
Ae
-Tz
↑attr ns-serie
5
D
~
45!
B
A
The diagram shows a block D of mass 100 kg supported by two sloping struts AD and BD, each
attached at an angle of 45# to fixed points A and B respectively on a horizontal floor. The block is also
held in place by a vertical rope CD attached to a fixed point C on a horizontal ceiling. The tension in
the rope CD is 500 N and the block rests in equilibrium.
(a) Find the magnitude of the force in each of the struts AD and BD.
[3]
........................................................................................................................................................
4:500+ 2T, Ain95"100g
->:Te095:
[ean 550
........................................................................................................................................................
T
........................................................................................................................................................
Tz
=
T 354N(3St)
=
.
........................................................................................................................................................
........................................................................................................................................................
T,
354N
Am:
........................................................................................................................................................
=
T 354N
=
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
© UCLES 2023
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7
A horizontal force of magnitude F N is applied to the block in a direction parallel to AB.
(b) Find the value of F for which the magnitude of the force in the strut AD is zero.
[3]
........................................................................................................................................................
........................................................................................................................................................
~
450
>F
IT,
........................................................................................................................................................
500
Au
T=O
I
T ........................................................................................................................................................
5
i
-
........................................................................................................................................................
F
-45.
----
450
T,con
........................................................................................................................................................
~
100G
........................................................................................................................................................
4:I
Ain 45%+ 580
100g
........................................................................................................................................................
=
T
........................................................................................................................................................
500/2
..
=
F
T,e0450
........................................................................................................................................................
->:
=
=
500x
........................................................................................................................................................
........................................................................................................................................................
F 500N
=
.
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
© UCLES 2023
9709/42/F/M/23
[Turn over
8
6
TN
20!
B
30!
FN
A block B, of mass 2 kg, lies on a rough inclined plane sloping at 30# to the horizontal. A light rope,
inclined at an angle of 20# above a line of greatest slope, is attached to B. The tension in the rope
is T N. There is a friction force of F N acting on B (see diagram). The coefficient of friction between
B and the plane is $.
(a) It is given that F = 5 and that the acceleration of B up the plane is 1.2 m s−2 .
[3]
20
Tsin
................................................................................................................................................
NRt to
Teazo
(i) Find the value of T .
Ta
................................................................................................................................................
20:5-29sin302 (1.2)
in38 -
................................................................................................................................................
29
~29ed30
:. =
18-5N
FA L
................................................................................................................................................
30................................................................................................................................................
----
. . .
V
................................................................................................................................................
29
................................................................................................................................................
(ii) Find the value of $.
[3]
................................................................................................................................................
F ................................................................................................................................................
↑;
M.R
R+Tsin200=29e030-
=
................................................................................................................................................
Mx(2908300
5
=
-> R
................................................................................................................................................
-
18.5sin28)
29em30"-185sin28
=
................................................................................................................................................
..
M
0.455
................................................................................................................................................
=
................................................................................................................................................
................................................................................................................................................
© UCLES 2023
9709/42/F/M/23
9
(b) It is given instead that $ = 0.8 and T = 15.
Determine whether B will move up the plane.
[3]
Tsin20
........................................................................................................................................................
15
>T=
aRt
Frau-MR
B
........................................................................................................................................................
28.
e-
128°
-
........................................................................................................................................................
(29e30-15sin
200
........................................................................................................................................................
0.8x
=>
Fan
9.75
........................................................................................................................................................
=
........................................................................................................................................................
Egeor300
........................................................................................................................................................
Resultantforce
plane C
to move it
the
without
........................................................................................................................................................
up
friction)
........................................................................................................................................................
15901200-29
sin50
........................................................................................................................................................
-
........................................................................................................................................................
4.09
=
........................................................................................................................................................
Resultant
(Frau
As
........................................................................................................................................................
........................................................................................................................................................
place
........................................................................................................................................................
So I will not move up the
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
© UCLES 2023
9709/42/F/M/23
[Turn over
10
7
O
A
E
1.8 m
F
"!
B
7.0 m
C
The diagram shows a smooth track which lies in a vertical plane. The section AB is a quarter circle
of radius 1.8 m with centre O. The section BC is a horizontal straight line of length 7.0 m and OB is
perpendicular to BC. The section CFE is a straight line inclined at an angle of %# above the horizontal.
A particle P of mass 0.5 kg is released from rest at A. Particle P collides with a particle Q of mass
0.1 kg which is at rest at B. Immediately after the collision, the speed of P is 4 m s−1 in the direction
BC. You should assume that P is moving horizontally when it collides with Q.
(a) Show that the speed of Q immediately after the collision is 10 m s−1 .
[4]
Smil
........................................................................................................................................................
On
->
als
P
I
........................................................................................................................................................
->
- >
4mi
se
........................................................................................................................................................
-M
GPElon-KEgain
0.5x6 + 0 0.5x4 0.1xx
........................................................................................................................................................
059x108
=
x0.5(202)
+
:x 10 mil
........................................................................................................................................................
=
=
=
........................................................................................................................................................
...
6
........................................................................................................................................................
=
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
© UCLES 2023
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11
When Q reaches C, it collides with a particle R of mass 0.4 kg which is at rest at C. The two particles
coalesce. The combined particle comes instantaneously to rest at F. You should assume that there
is no instantaneous change in speed as the combined particle leaves C, nor when it passes through C
again as it returns down the slope.
(b) Given that the distance CF is 0.4 m, find the value of %.
[4]
tom
CLM
Om
........................................................................................................................................................
-
⑨
10X0-1
........................................................................................................................................................
R
0.5xy
=
0.1kgy 0.4kg
2
........................................................................................................................................................
-
:.y
=
⑲
........................................................................................................................................................
0.5kg
........................................................................................................................................................
........................................................................................................................................................
Omit
↑
........................................................................................................................................................
E
①
zam
KE los=
0.4m
........................................................................................................................................................
GPEgain
>
i
F
Ot ------- >
........................................................................................................................................................
e
x0.5x(2202) 0.5940.4 sinc
........................................................................................................................................................
=
........................................................................................................................................................
!.0 300
=
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
[Question 7 continues on the next page.]
© UCLES 2023
9709/42/F/M/23
[Turn over
12
(c) Find the distance from B at which P collides with the combined particle.
[5]
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
10m
->
........................................................................................................................................................
⑨
⑨
........................................................................................................................................................
Im
C
B
........................................................................................................................................................
time
taken
by
s to
from
move
Bloe
........................................................................................................................................................
.
........................................................................................................................................................
7 10t
=
t 0.7s
=
=
........................................................................................................................................................
........................................................................................................................................................
DF
........................................................................................................................................................
za
0.
........................................................................................................................................................
2
........................................................................................................................................................
time
taken
by
combined (OR)
to move
[toF
........................................................................................................................................................
from
........................................................................................................................................................
S
()xt
=
........................................................................................................................................................
...0.4= t
........................................................................................................................................................
:Time taken
DR to return to C=2x0.420.8s
........................................................................................................................................................
by
........................................................................................................................................................
:Distance
during
P
these times
travelled
........................................................................................................................................................
by
t 0.7 + 0.8=
1.55
........................................................................................................................................................
=
S nt
........................................................................................................................................................
=
4x1,5
........................................................................................................................................................
=
.S Gm
........................................................................................................................................................
=
........................................................................................................................................................
© UCLES 2023
9709/42/F/M/23
13
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.
........................................................................................................................................................................
his I amt
........................................................................................................................................................................
........................................................................................................................................................................
6m-5K 1m ->
B*
C
........................................................................................................................................................................
Let,
they
ku -
at
collide
am
........................................................................................................................................................................
from
28
tree
latel
........................................................................................................................................................................
........................................................................................................................................................................
Sp
Sq
+
1
5 5m
:x 2 x
........................................................................................................................................................................
=
=
=
4t + 2t 1
........................................................................................................................................................................
=
I
it
........................................................................................................................................................................
=
........................................................................................................................................................................
:Distance
the
at
which
........................................................................................................................................................................
of
point
they
collide
B 6 + (1
........................................................................................................................................................................
from
=
-
5)
........................................................................................................................................................................
6 he
=>
........................................................................................................................................................................
........................................................................................................................................................................
........................................................................................................................................................................
........................................................................................................................................................................
........................................................................................................................................................................
........................................................................................................................................................................
........................................................................................................................................................................
........................................................................................................................................................................
........................................................................................................................................................................
........................................................................................................................................................................
© UCLES 2023
9709/42/F/M/23