Mathematics 9709/42 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
v (m s–1)
V
0 t (s)
4 10 T
–3
The velocity of a particle moving in a straight line at time t seconds 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 from t = 0 to t = T . The graph consists of four straight line segments. The particle accelerates from rest to a speed of V m s -1 over a period of 4 s, and then decelerates at 53 m s -2 to instantaneous rest over a period of 6 s.
The particle then travels back towards O, reaching a maximum speed of 3 m s -1 before coming to rest at time t = T .
(a) Find the value of V. [2]
(b) Given that the total distance travelled by the particle from t = 0 to t = T is 68 m, find the value
of T. [3]
Mark scheme
1(a) M1 5
Use v = 0 , a = ±
3
5 0 −V 5 V −0 5
0 = V − × (10 − 4 ) OR = − OR = .
3 10 − 4 3 4−0 2
V = 10 A1 Make sure that 10 comes from a correct initial equation or
0 −V 5
equivalent, if not then A0, e.g. = leading to
10 − 4 3
V = 10 is M1A0.
2
1(b) 1 B1FT SOI
Distance in first 10 seconds = × 10 × their V 1 1
2 OE, e.g. × 4 × their V + × (10 − 4 ) × their V
2 2
If correct distance is 50 m.
1 1 M1 Use distance (68) is total area under graph.
× (T − 10) × 3 + ×10 × their V = 68
2 2 Their 50 must be from 5 × their V .
OR Allow with -3.
1 1 1
× (T − 10 ) × 3 + × 4 × their V + × (10 − 4 ) × their V = 68 Allow
1 1
× X × 3 + ×10 × their V = 68 and use of
2 2 2 2 2
T = X + 10 .
T = 22 A1 WWW
Be aware that V = 10 from wrong work in (a) can be
awarded maximum of B1M1A0.
Special Case for assumption of isosceles triangle for t = 10 to t = T :
1 B1FT SOI
Distance in first 10 seconds = × 10 × their V 1 1
2 OE, e.g. × 4 × their V + × (10 − 4 ) × their V .
2 2
If correct distance is 50 m.
1 T − 10 1 B1 WWW
2 × 2 × 3 = 2 × ( 68 − 50) → T = 22
Special Case for assumption that particle returns to O:
1 1 98 B1 32.66666.
2 × (T − 10 ) × 3 = 2 × 68 → T = 3 or 32.7 AWRT 32.7.
3
A block of mass 20 kg is held at rest at the top of a plane inclined at 30° to the horizontal. The block is projected with speed 5 m s -1 down a line of greatest slope of the plane. There is a resistance force acting on the block. As the block moves 2 m down the plane from its point of projection, the work done against this resistance force is 50 J.
Find the speed of the block when it has moved 2 m down the plane. [4]
Mark scheme
2 Change in PE = ±20 g sin 30 × 2 [ = ±200 ] B1
1 B1 For either expression.
± × 20v2 = ±10v2
1
× 20 ( v − 5) .
2 2
Do not allow
1 2
± × 20 × 52 [ = ±250]
2
1 1 M1 Attempt at work energy equation; 4 terms; dimensionally
× 20v2 − × 20 × 52 = 20 g sin 30 × 2 − 50 correct; allow sign errors; PE term must include a
2 2
component (allow sin/cos mix).
10v2 − 250 = 200 − 50
1
× 20 ( v − 5) for KE.
2
Do not allow with
2
Speed = 6.32 m s-1 OR 40 m s
-1
A1 OE, e.g. 2 10 .
6.324555….
AWRT 6.32.
Special case for assumption of constant resistance force:
50 B1
20a = 20 g sin 30 − [→ a = 3.75]
2
v 2 = 52 + 2 × 2 × 3.75 v = 6.32 m s-1 OR 40 m s-1 B1 OE, e.g. 2 10 .
6.324555…
AWRT 6.32.
4
A cyclist is riding along a straight horizontal road. The total mass of the cyclist and his bicycle is
90 kg. The power exerted by the cyclist is 250 W. At an instant when the cyclist’s speed is 5 m s -1 , his acceleration is 0.1 m s -2 .
(a) Find the value of the constant resistance to motion acting on the cyclist. [3]
The cyclist comes to the bottom of a hill inclined at 2° to the horizontal.
(b) Given that the power and resistance to motion are unchanged, find the steady speed which the cyclist could maintain when riding up the hill. [2]
Mark scheme
3(a) 250 B1 For use of power = Fv e.g. 5 × PF = 250 .
[Peddling force = PF =]
5
their PF − R = 90 × 0.1 M1 Using their PF ≠ 250 . 3 terms; allow sign errors.
Dimensionally correct.
Resistance = 41 N A1
3
3(b) 250 M1 For attempt at resolving up the hill; 3 terms; allow sign
− their 41 − 90 g sin 2 = 0 errors; Must be a component of weight (NOT mass) but
v
allow sin/cos mix; allow use of their 41.
Dimensionally correct.
250
Oe eg v = .
their 41 + 90 g sin 2
Steady speed = 3.45 m s–1 A1 3.45258…
AWRT 3.45.
2
P Q
45° i°
A B
0.2 kg 0.1 kg
The diagram shows two particles, A and B, of masses 0.2 kg and 0.1 kg respectively. The particles are suspended below a horizontal ceiling by two strings, AP and BQ, attached to fixed points P and Q on the ceiling. The particles are connected by a horizontal string, AB. Angle APQ = 45° and BQP = i° .
Each string is light and inextensible. The particles are in equilibrium.
(a) Find the value of the tension in the string AB. [2]
(b) Find the value of i and the tension in the string BQ. [4]
Mark scheme
4(a) TAB sin 45 = 0.2 g cos 45 *B1 BOD for using sin 45 instead of cos 45 , particularly if
T using wrong components in (b).
OR tan 45 = AB
0.2 g
0.2 g
OR tan 45 =
TAB
OR TAB = TAP cos 45 and TAP sin 45 = 0.2 g
TAB = 2 N DB1 Condone TAB = 0.2 g (with or without working) for full
marks.
WWW.
DO NOT ISW.
ALTERNATIVE for 4(a) using LAMI’s THEOREM:
TAB 0.2 g B1
=
sin135 sin135
TAB = 2 N B1 Condone TAB = 0.2 g (with or without working) for full
marks.
WWW.
DO NOT ISW.
2
4(b) TBQ cosθ − their TAB = 0 *M1 For resolving either horizontally OR vertically;
2 terms; allow sin/cos mix; allow their TAB ;
TBQ sin θ − 0.1g = 0
M0 for any use of TAB = TBQ = TAP .
A1FT For both. FT their T ONLY; T ≠ T .
AB AB BQ
Sight of ( their TAB ) tan θ = 0.1g is M1 only without seeing
an equation for TBQ .
0.1g DM1 Solve for θ or solve for T from equations with the
θ = tan −1 or TBQ = ( 0.1g )2 + ( theirTAB )2 BQ
their TAB correct number of relevant terms. Using their TAB .
θ = 26.6 AND TBQ = 5 or 2.24 A1 θ = 26.56505. TBQ = 2.236067 .
AWRT 26.6 and 2.24.
FIRST ALTERNATIVE for 4(b):
TBQ = (theirTAB ) cosθ + 0.1g sinθ M1 For resolving either parallel to BQ or perpendicular to BQ;
2 terms; allow sign errors on 3 term equation only; allow
( their TAB ) sin θ = 0.1g cosθ sin/cos mix; allow their TAB .
Sight of ( their TAB ) tan θ = 0.1g is M1 only without seeing
an equation for TBQ .
M0 for any use of TAB = TBQ = TAP .
A1FT For both. FT their T ONLY; T ≠ T .
AB AB BQ
0.1g DM1 Solve for θ (or solve for T ) from equations with the
θ = tan −1
BQ
( their TAB ) correct number of relevant terms.
Using their TAB .
4(b) θ = 26.6 AND TBQ = 5 or 2.24 A1 θ = 26.56505. TBQ = 2.236067 .
AWRT 26.6 and 2.24.
SECOND ALTERNATIVE for 4(b) using triangle of forces:
= ( 0.1g ) + ( theirTAB ) M1 Using their TAB ONLY;
2 2 2
TBQ
M0 for any use of TAB = TBQ = TAP .
OR 0.1g = TBQ sin (θ )
OR (theirTAB ) = TBQ cos (θ ) A1FT For any 2 equations. FT their T ONLY; T ≠ T .
AB AB BQ
Sight of ( their TAB ) tan θ = 0.1g is M1 only without seeing
an equation for TBQ .
Solve for TBQ or θ DM1 Solve for θ or solve for T from equations with the
BQ
correct number of relevant terms. Using their TAB .
E.g. TBQ = ( 0.1g )2 + ( theirTAB )2
0.1g
θ = tan −1
their TAB
θ = 26.6 AND TBQ = 5 or 2.24 A1 θ = 26.56505. TBQ = 2.236067 .
AWRT 26.6 and 2.24.
THIRD ALTERNATIVE for 4(b) using LAMI’s THEOREM:
*M1 For any 2 fractions correct;
M0 for any use of TAB = TBQ = TAP .
TBQ 0.1g theirTAB TBQ 0.1g their TAB A1FT For all 3 fractions correct. FT their TAB ONLY.
= = = =
Sight of ( their TAB ) tan θ = 0.1g is M1 only without seeing
OR
sin90 sin (180 − θ ) sin ( 90 + θ ) sin 90 sin θ cosθ
an equation for TBQ .
4(b) 0.1g DM1 Solve for θ.
θ = tan −1
their TAB
θ = 26.6 AND TBQ = 5 or 2.24 A1 θ = 26.56505. TBQ = 2.236067 .
AWRT 26.6 and 2.24.
FOURTH ALTERNATIVE for 4(b) for resolving on the whole system:
TBQ sinθ + (theirTAP ) sin 45 = 0.2g + 0.1g M1 For resolving either horizontally or vertically on the whole
system; using their TAP .
TBQ cosθ = (theirTAP ) cos45
M0 for any use of TAB = TBQ = TAP .
If TAP not found in part (a), must have either
TAP sin 45 = 0.2 g or ( theirTAB ) = TAP cos 45 .
Correct number of terms; allow sign errors on the 4 term
equation ONLY; allow sin/cos mix.
A1FT Both equations correct. May see T sin θ = 1 and
BQ
TBQ cos θ = 2 .
Sight of ( their TAB ) tan θ = 0.1g is M1 only without seeing
an equation for TBQ .
0.1g + 0.2 g − ( their TAP ) sin 45 DM1 Solve for θ or solve for T from equations with the
θ = tan −1
BQ
( their TAP ) cos 45 correct number of relevant terms. Using their TAP .
( 0.1g + 0.2 g − ( their TAP ) sin 45) + ( ( their TAP ) cos 45)
2 2
or TBQ =
θ = 26.6 AND TBQ = 5 or 2.24 A1 θ = 26.56505. TBQ = 2.236067 .
AWRT 26.6 and 2.24.
4
Two particles, P and Q, of masses 2m kg and m kg respectively, are held at rest in the same vertical line.
The heights of P and Q above horizontal ground are 1 m and 2 m respectively. P is projected vertically upwards with speed 2 m s -1 . At the same instant, Q is released from rest.
(a) Find the speed of each particle immediately before they collide. [4]
(b) It is given that immediately after the collision the downward speed of Q is 3.5 m s -1 .
Find the speed of P at the instant that it reaches the ground. [5]
Mark scheme
5(a) 1 2 M1
[ sP up =] ± 2t −
1 2 1
gt OR sQ down = ± gt For use of s = ut + at at least once with a = ± g and
2
2 2 2
u = 0 or u = ±2 .
Seen anywhere.
1 2 1 2 1 M1 Use s + s = ± ( 2 − 1) OR ±1 ONLY with s and s of
2t − gt + gt = 2 − 1 t = P Q P Q
2 2 2 the correct form
1 A1 Must be positive.
vP = 2 − g = −3 m s -1 so speed = 3 m s -1
2
1 A1 Must be positive.
vQ = − g = −5 m s -1 so speed = 5 m s -1 If A0A0, allow SCB1 if both are negative
2
4
5(b)
( )
2m × ( their 3[vP ]) + m × their 5 vQ = 2mv + m × 3.5 *M1 Use of conservation of momentum; 4 non- zero terms;
using their ±3 m s -1 and their ±5 m s -1; allow sign errors
and m missing ONLY. Do not allow with vP = ±2 or with
vQ = 0. Do not allow made up values for 3 and 5.
Allow LHS of our equation to have only one term but only
if getting vP = 0 in (a) from s P = sQ .
v = 3.75 A1 Allow v = −3.75 from correct work.
1 1 1
2 *B1 This may be seen in part (a), but do not award the mark
1 1
At t = , s = ±
p 2 × − g = until stated/used in part (b).
2
2 2 2 4
1 1 1
2
1
OR s p = ± −3 × + g =
2 2 2 4
( −3 ) 2 − 2 2 1
OR s p = ± = ±
2g 4
2
1 1 1 5
OR At t = , sQ = ± g = ±
2 2 2 4
1 1 1
2
1 5
OR At t = , sQ = ± 5 × − g = ±
2 2 2 2 4
52 −02
OR sQ = ± = ± 5
2 g 4
3
so height above ground = 4
5(b) 3 DM1 Use of v 2 = u 2 + 2as using their v and a = ± g; their s is
v 2 = ( their 3.75 ) + 2 × ( ± g ) × ±
2
4 1 5
45 either ± 1 + − or ± 2 − OE.
OR 0 = ( their 3.75) − 2 × ( ± g ) × s s = 4
2
2 4
64 Dependent on the previous M mark and B mark being
3 45 awarded.
AND v = 0 + 2 × ( ± g ) × +
2 2
4 64
v = 5.39 m s
-1
A1 465
Or ; AWRT 5.39; 5.390964…
4
5
A particle, P, travels in a straight line, starting from a point O with velocity 6 m s -1 . The acceleration of
P at time t s after leaving O is a m s -2 , where
1 a = -1.5t 2 for 0 G t G 1,
1 1 a = 1.5t 2 - 3t - 2 for t 2 1.
(a) Find the velocity of P at t = 1. [3]
(b) Given that there is no change in the velocity of P when t = 1, find an expression for the velocity of
P for t 2 1. [3]
(c) Given that the velocity of P is positive for t G 4 , find the total distance travelled between t = 0 and t = 4 . [4]
Mark scheme
6(a) Attempt to integrate a for 0 ≤ t ≤ 1 M1 Increase power by 1 and a change in coefficient in at least
one term (which must be the same term);
3
v = at is M0. Expect ( v = ) − t 2 [ +c ] .
1.5 2
3 3 A1 3
(v =) − t + 6 = −t 2 + 6 Allow for −t 2 + c
and c = 6 seen from CWO.
1.5
Allow unsimplified.
Velocity at t = 1 is 5 ms-1 A1 CWO.
3
6(b) Attempt to integrate a for t > 1 *M1 Increase power by 1 and a change in coefficient in at least
3 1 3 1 one term (which must be the same term);
( v = ) 1.5 t 2 − 3 t 2 [ +c ] = t 2 − 6t 2 [ +c ] v = at is M0.
3 1/ 2
2
Use v = their 5 when t = 1 in attempt to find c DM1 Must get a numerical expression for c ; if no substitution
[5 = 1 − 6 + c ] seen, c must be correct for their expression for v .
Their 5 must not be a made up value.
3 1 A1 OE, but must be a complete (possibly unsimplified)
( v = ) t 2 − 6t 2 + 10 expression.
3
6(c) For attempt at integration of their v for either section *M1 Increase power by 1 and a change in coefficient in at least
one term (which must be the same term); Their v ’s have
to come from integration.
s = vt is M0.
Their v ’s have to come from integration.
1 2
5
2
5 A1 For either correct.
( s1 = ) − t + 6t [ +c ] = − t 2 + 6t [ +c ] Allow unsimplified.
5/ 2 5
5 3 5 3
1 2 6 2 2
( s2 = ) t − t + 10t [ +c ] = t 2 − 4t 2 + 10t [ +c ]
5/ 2 3/ 2 5
For use of limits 0 and 1 for s1 and 1 and 4 for s2 DM1 Using correct limits correctly, in expressions that have
come from integration (where all powers must have
increased by 1 with a change in coefficient for non-linear
terms) of a v that came from integration, and using the
limits correctly would lead to 2 positive values when used
in their expressions.
s1 and s 2 must have the correct number of non-constant
terms and must include a linear term.
Total distance = 20 m A1 2 28
s1 = − × 1 + 6 × 1 − 0 = = 5.6 .
5 5
2 5 3
s2 = × 4 2 − 4 × 4 2 + 10 × 4 −
5
2 5 3
72
× 12 − 4 × 12 + 10 × 1 = = 14.4 .
5 5
If either displacement expression has a constant of
integration and is given incorrectly, then award A0 even if
20 m seen. For reference the constant of integration for s2
is –0.8.
6(c) SC for integration not seen:
28 72 B1
s1 = or s2 =
5 5
Total distance = 20 m B1
4
A
0.2 kg
i
B
0.3 kg
0.25 m
Two particles, A and B, of masses 0.2 kg and 0.3 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small fixed smooth pulley which is attached to the bottom of a rough plane inclined at an angle i to the horizontal where sin i = 0.6 . Particle A lies on the plane, and particle B hangs vertically below the pulley, 0.25 m above horizontal ground. The string between A and the pulley is parallel to a line of greatest slope of the plane (see diagram). The coefficient of friction between A and the plane is 1.125 . Particle A is released from rest.
(a) Find the tension in the string and the magnitude of the acceleration of the particles. [7]
(b) When B reaches the ground, it comes to rest.
Find the total distance that A travels down the plane from when it is released until it comes to rest.
You may assume that A does not reach the pulley. [4]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
7(a) R = 0.2 g × 0.8[ = 1.6] B1 Allow cos37 or better for 0.8.
F = 1.125R [ = 1.8] *M1 Where R is a component of weight.
Must have 1.125 × 0.2g × 0.8 or
1.125 × 0.2g × 0.6 or with using cos37 or sin 37 or better
for 0.8 and 0.6 respectively.
These 2 marks may be embedded in the N2L equation(s).
Use of Newton’s second law for A or B or system *M1 Correct number of terms; allow sign errors; allow sin/cos
mix. Dimensionally correct.
0.3g − T = 0.3a A1 For any 2 correct equations.
T + 0.2 g × 0.6 − F = 0.2a Allow sin 37 or better for 0.6.
0.3g + 0.2 g × 0.6 − F = 0.5a Allow their possibly incorrect F .
–2
a = 4.8 m s A1 Must be positive.
For attempt to solve for T DM1 From equations with the correct number of relevant terms.
If a found first then substituting into an equation with the
correct number of relevant terms and solving.
If resolved equations incorrect and no working seen, then
this mark is implied by the correct T value for their
equations.
Dependent on previous 2 M marks
T = 1.56 N A1
7
7(b) [For B or A] v 2 = 02 + 2 × 4.8 × 0.25 *M1 Use of v 2 = u 2 + 2as with u = 0 and using
2 15 s = 0.25 , their 4.8 , a ≠ ± g . Must be a complete method
v = 2.4 or v = = 1.549
2
5 to get an expression for v or v 2 .
Attempt at Newton’s 2nd Law on A when string becomes slack *M1 3 terms; allow sign errors; allow sin/cos mix; allow their
[0.2 g × 0.6 − 1.125 × 0.2 g × 0.8 = 0.2a ] non-zero F from part (a); Dimensionally correct; must be
using 0.2 for the mass.
Allow θ = 37 or better.
For reference a = −3 (or −2.99 if using θ = 36.9 ).
02 = 2.4 + 2 × ( −3) × s [ s = 0.4] DM1 Using constant acceleration formula(e) using a negative
acceleration to get an expression in s only.
Dependent on previous 2 M marks.
Total distance = 0.25 + 0.4 = 0.65 m A1 AWRT 0.650. Allow 0.651 from use of θ = 36.9 .
ALTERNATIVE for 7(b) using energy:
[For B or A] v 2 = 02 + 2 × 4.8 × 0.25 *M1 Use of v 2 = u 2 + 2as with u = 0 .
2 15 Using s = 0.25 , their 4.8 , a ≠ ± g .
v = 2.4 or v = = 1.549
2
5 Must be a complete method to get an expression for v or
v2 .
For attempt at work energy equation DM1 3 terms; dimensionally correct; allow sin/cos mix in PE
term and work done against Friction term; allow sign
errors; allow their non-zero F from part (a).
7(b) 1 A1 For correct equation in d only; must be using 0.2 for the
× 0.2 × 2.4 + 0.2 g × 0.6 × d − 1.125 × 0.2 g × 0.8 × d = 0 mass.
2
Allow θ = 37 or better.
[ d = 0.4m]
Total distance = 0.25 + 0.4 = 0.65 m A1 AWRT 0.650. Allow 0.651 from use of θ = 36.9 .
4
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