Mathematics 9709/43 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2024. 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 0.2 kg and 0.5 kg respectively are at rest on a smooth horizontal plane.
Particle P is projected with a speed 6 m s -1 directly towards Q. After P and Q collide, P moves with a speed of 1 m s -1 .
Find the two possible speeds of Q after the collision. [3]
Mark scheme
1 0.2 × 6 + 0 = 0.2 ×1 + 0.5v or 0.2 × 6 + 0 = 0.2 × −1 + 0.5v M1 For attempt at use of conservation of momentum in at least one
case. Must have three non-zero terms. Allow sign errors.
Must have correct masses with relevant velocities.
Their v may be in opposite direction.
Speed = 2 m s−1 A1 Do not allow negative.
14 4 A1 OE
Speed = 2.8[0] m s−1 or m s−1 or 2 m s−1 Do not allow negative.
5 5
3
30°
XN
0.2 kg
A particle of mass 0.2 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point on a vertical wall. The particle is held in equilibrium by a force of magnitude X N, perpendicular to the string, with the string taut and making an angle of 30° with the wall (see diagram).
Find the tension in the string and the value of X. [3]
Mark scheme
2 X cos30 − T sin 30 = 0 M1 Must have correct number of relevant terms (forces must have
components as required). Allow sin/cos mix. Allow sign errors.
X sin 30 + T cos30 − 0.2 g = 0 M1 Must have correct number of relevant terms (forces must have
components as required). Allow sin/cos mix but must be
consistent with their other equation. Allow sign errors.
X = 1, Tension = 1.73 N [1.7320..] or 3 N A1 For both.
Alternative Method for Question 2: Resolving in directions of X and T or triangle of forces
X − 0.2 gcos60 = 0 (M1) Must have correct number of relevant terms (forces must have
components as required). Allow sin/cos mix. Allow sign errors.
T − 0.2 g sin 60 = 0 (M1) Must have correct number of relevant terms (forces must have
components as required). Allow sin/cos mix but must be
consistent with their other equation. Allow sign errors.
X = 1, Tension = 1.73 N [1.7320..] or 3 N (A1) For both.
Alternative Method for Question 2: Using Lami’s theorem
0.2 g X T (M1M1) First M1 for any two fractions.
= = Second M1 for all three fractions or another pair of fractions.
sin 90 sin150 sin120
X T
Allow and for M1 marks.
sin120 sin150
X = 1, Tension = 1.73 N [1.7320..] or 3 N (A1) For both.
3
A car travels along a straight road with constant acceleration a m s -2 , where a 2 0 . The car passes through points A, B and C in that order. The speed of the car at A is u m s -1 in the direction AB. The distance BC is twice the distance AB. The car takes 8 seconds to travel from A to B and 10 seconds to travel from B to C.
(a) Find u in terms of a. [4]
(b) Find the speed of the car at C in terms of a. [2]
Mark scheme
3(a) 1 B1B1 1 u+v
AB :[ s =] u × 8 + a × 82 For use of s = ut + at 2 or s = t.
2 2 2
u + u + 8a B1 for any one correct expression, B2 for two correct
[ = 8u + 32a ] or ×8 expressions.
2
1
BC : [ 2s =] ( u + 8a ) × 10 + a × 102 [ = 10u + 130a ]
2
u + 8a + u + 8a + 10a
or × 10
2
1
AC : [3s =] u × 18 + a × 182
2
u + u + 18a
[ = 18u + 162a ] or × 18
2
Attempt to solve simultaneously M1 To obtain an equation in u and a only.
1
( u + 8a ) × 10 + a × 102 = 2 u × 8 + a × 82
1 1
Must have come from correct expressions but allow × instead
2 2 3
1 2 3
10u + 130a = 2 ( 8u + 32a )
of ×3 or × instead of ×2 or × instead of × .
2 3 2
1 1
OR u × 18 + a × 182 = 3 u × 8 + a × 82
2 2 1 1
Note: M0 for u × 10 + a × 102 = 2 u × 8 + a × 82 leading to
18u + 162a = 3( 8u + 32a )
2 2
1 3 1 7
u × 18 + a × 182 = ( u + 8a ) × 10 + a × 102 u = − a.
2 2 2 3
Note: M0 for distance AC = 2 AB leading to u = −49a.
3
18u + 162a = 2 (10u + 130a )
u = 11a A1
3(a) Alternative Method for Question 3(a): Using v 2 = u 2 + 2as
( u + 8a ) − u 2 (B1B1) B1 for any one correct expression, B2 for two correct
2
[ s =] expressions.
2a
or [ 2 s = ]
( u + 18a )2 − ( u + 8a )2
2a
or [3s = ]
( u + 18a ) − u 2
2
2a
( u + 18a )2 − u 2 ( u + 8a ) 2 − u 2 (M1) To obtain an equation in u and a only.
= 3
2a 2a
( u + 18a ) − ( u + 8a )
2 2
( u + 8a ) 2 − u 2
or = 2
2a 2a
( u + 18a ) − u 3 ( u + 18a ) − ( u + 8a )
2 2 2 2
or =
2a 2 2a
u = 11a (A1)
4
3(b) v = 11a + a ×18 M1 For use of v = u + at or other complete suvat method
Using their u in terms of a, e.g.
v 2 = (11a ) + 2a (18 × 11a + 162a ) = 841a 2 ,
2
v= (11a )2 + 2a (18 ×11a + 162a ) = 841a 2 .
Speed = 29a A1FT FT their expression for v so their u + 18a.
7 47
Note: If answer to part (a) is u = − a, then speed = a.
3 3
2
A particle travels in a straight line. The velocity of the particle at time t s after leaving a point O is v m s -1 , where
v = kt 2 - 4t + 3.
The distance travelled by the particle in the first 2 s of its motion is 6 m. You may assume that v 2 0 in the first 2 s of its motion.
(a) Find the value of k. [4]
(b) Find the value of the minimum velocity of the particle. You do not need to show that this velocity is a minimum. [3]
Mark scheme
4(a) For attempt at integration M1* The power of t must increase by 1 with a change of coefficient in
the same term.
Use of s = vt scores M0.
1 4 1 A1 Allow unsimplified.
kt 2+1 − t1+1 + 3t = kt 3 − 2t 2 + 3t [ +c ]
2 +1 2 3
1 DM1 Use of limits 0 and 2 with 6 to form an equation in k only
k × 23 − 2 × 22 + 3 × 2 [ −0] = 6 (without c but allow with +c − c ).
3
k =3 A1
4
4(b) 2 × 3t − 4 M1 For attempt at differentiation. Must have expression of the form
−b 4 at + b with a ≠ 3, unless their k = 32 . Allow 2kt − 4.
Or at min value t = =
2a 2 × 3
2 A1FT OE
[ 2 × 3t − 4 = 0 ] t = 2
3 FT their k t = . Allow without working.
their k
2
2 5 A1 OE
2
v = 3 × − 4 × + 3 = m s-1 Allow 1.67 or better for v.
3 3 3
Alternative Method for Question 4(b): Using completing the square
Attempt at completing the square (M1) 2
2
2
2
Must have t − OE, or t − .
3 their k
2 4
2 (A1FT) 2 4
2
3 t − − + 3 FT their k k t − − + 3.
3 3 k k
5 (A1) OE
v= m s-1 Allow 1.67 or better.
3
3
A van of mass 4500 kg is towing a trailer of mass 750 kg down a straight hill inclined at an angle of i to the horizontal where sin i = 0.05 . The van and the trailer are connected by a light rigid tow-bar which is parallel to the road. There are constant resistance forces of 2500 N on the van and 300 N on the trailer.
(a) It is given that the tension in the tow-bar is 450 N.
Find the acceleration of the trailer and the driving force of the van’s engine. [4]
On another occasion, the van and trailer ascend a straight hill inclined at an angle of a to the horizontal where sin a = 0.09 . The driving force of the van’s engine is now 9100 N, and the speed of the van at the bottom of the hill is 20 m s -1 . The resistances to motion are unchanged.
(b) (i) Find the acceleration of the van and the tension in the tow-bar. [5]
(ii) Find the speed of the van when it has travelled a distance of 375 m up the hill. [2]
Mark scheme
5(a) Use of Newton’s second law for van or trailer or system M1* Must have correct number of relevant terms. Allow sign errors.
Note: Trailer has 4 terms Allow sin/cos mix. Allow g missing.
Van has 5 terms Masses must be correct for their equation(s).
System has 7 terms (or 5 if counting van and trailer as one body) Forces must have components (or not) as required.
Must have either 0.05 or sin 2.86 or sin 2.9, not just sin θ .
Trailer: 450 + 750 g × 0.05 − 300 = 750a A1 For any two correct equations.
[525 = 750a ]
Van: D + 4500 g × 0.05 − 2500 − 450 = 4500a
[ D − 700 = 4500a ]
System:
D + 4500 g × 0.05 + 750 g × 0.05 − 2500 − 300 = ( 4500 + 750 ) a
[ D − 175 = 5250a ]
For attempt to solve for a or D DM1 Must get to ‘a =’ or ‘D =’. Must have correct number of relevant
terms in the equation(s) which they are using to find a or D. g
must be present. Allow sign errors. Allow sin/cos mix. If no
working shown to solve their equations, then their answers
should be correct for their equations.
a = 0.7 m s-2 and D = 3850 N A1
4
5(b)(i) Use of Newton’s second law for van or trailer or system M1* Must have correct number of relevant terms Allow sign errors.
Note: Trailer has 4 terms Allow sin/cos mix. Allow g missing.
Van has 5 terms Masses must be appropriate for their equation(s).
System has 7 terms (or 5 if counting van and trailer as one Forces must have components (or not) as required.
body) Must have either 0.09 or sin 5.16 or sin 5.2 not just sin θ .
Trailer: T − 300 − 750 g × 0.09 = 750a A1A1 A1 for one correct equation, second A1 for another correct
[T − 975 = 750a ] equation.
If using Van and Trailer equations, must be using the same T for
both to get the second A1.
Van: 9100 − 2500 − 4500 g × 0.09 − T = 4500a
[ 2550 − T = 4500a ]
System: 9100 − 2500 − 300 − ( 4500 + 750 ) g × 0.09 = ( 4500 + 750 ) a
[1575 = 5250a ]
For attempt to solve for a or T DM1 Must get to ‘a =’ or ‘T =’. Must have correct number of relevant
terms in the equation(s) which they are using to find a or T. g
must be present. Allow sign errors. Allow sin/cos mix.
If no working shown to solve their equations, then their answers
should be correct for their equations.
T = 1200 N and a = 0.3 m s-2 A1
5
5(b)(ii) v 2 = 20 2 + 2 × their 0.3 × 375 M1 For use of v 2 = 20 2 + 2a × 375 or other complete method to find
50
v 2 or v. For info time taken t = .
3
v = 25 m s-1 A1FT FT their value of a, i.e. v = 400 + 750 × their a .
Provided it does not lead to root of negative value.
Alternative Method for Question 5(b)(ii): Using energy
System: (M1) Must include all appropriate terms. Allow sign errors.
g must be present.
1
2
( )
× ( 4500 + 750 ) v 2 − 202 + ( 4500 + 750 ) g × 375 × 0.09 = ( 9100 − 25 Allow their value of T in place of 1200.
or Van:
1
2
( )
× 4500 v 2 − 202 + 4500 g × 375 × 0.09 = ( 9100 − 2500 − their 1200 ) ×
OR Trailer:
1
2
( )
× 750 v 2 − 202 + 750 g × 375 × 0.09 = ( their 1200 − 300 ) × 375
v = 25 m s-1 (A1FT) FT their value of T if using Van or Trailer.
2
A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and her bicycle is
80 kg. There is a constant resistance force of magnitude 32 N to the cyclist’s motion. At an instant when she is travelling at 7 m s -1 , her acceleration is 0.1 m s -2 .
(a) Find the power output of the cyclist. [3]
(b) Find the steady speed that the cyclist can maintain if her power output and the resistance force are both unchanged. [2]
1
The cyclist later descends a straight hill of length 32.2 m, inclined at an angle of sin -1 a20 k to the horizontal. Her power output is now 120 W, and the resistance force now has variable magnitude such that the work done against this force in descending the hill is 1128 J. The time taken to descend the hill is 4 s.
(c) Given that the speed of the cyclist at the top of the hill is 7.5 m s -1 , find her speed at the bottom of the hill. [6]
Mark scheme
6(a) P B1 For P = DF × 7 OE seen at any point in working.
DF =
7 Allow any force term or simply DF,
e.g. 32,80 × 0.1, 32 − 80 × 0.1 , 80 ×10 etc.
D − 32 = 80 × 0.1 M1 For use of Newton’s second law. Must have correct number of
terms. Allow sign errors.
[Power = ] 280 W A1
3
6(b) 280 M1 Attempt at equilibrium equation (a = 0) with their power.
[At steady speed driving force =] 32 =
v
35 A1FT OE
Steady speed = 8.75 m s-1 or m s-1 their 280
4 FT their power from part (a) .
32
2
6(c) 120 × 4 [ 480] B1 Work done by cyclist.
1 B1 For at least one KE term.
× 80v 2 40v 2
2
1
or × 80 × 7.52 [ 2250]
2
1 B1 Change in PE.
80 g × × 32.2 [1288]
20
[80 × 10 × 1.61]
Attempt at work-energy equation M1 Attempt at work energy equation with five relevant terms (four
relevant terms plus work done against resistance); dimensionally
correct. Allow sign errors. Allow sin/cos mix.
A1 For correct equation.
1
120 × 4 + 80 g ×
20
1
(
× 32.2 − 1128 = × 80 v 2 − 7.52
2
)
480 + 1288 −1128 = 40v 2 − 2250
Speed = 8.5[0] m s-1 or 172 A1 OE
Use of constant acceleration scores M0 and cannot score B marks
if the method leading to their answer only uses constant
acceleration.
6
A D
30° 30°
B C
The diagram shows a track ABCD which lies in a vertical plane. The section AB is a straight line
inclined at an angle of 30° to the horizontal and is smooth. The section BC is a horizontal straight line and is rough. The section CD is a straight line inclined at an angle of 30° to the horizontal and is rough.
The lengths AB, BC and CD are each 2 m.
A particle is released from rest at A. The coefficient of friction between the particle and both BC and CD is n. There is no change in the speed of the particle when it passes through either of the points B or C.
(a) It is given that n = 0.1 .
Find the distance which the particle has moved up the section CD when its speed is 1 m s -1 . [5]
(b) It is given instead that with a different value of n the particle travels 1 m up the track from C before it comes instantaneously to rest.
Find the value of n and the speed of the particle at the instant that it passes C for the second time.
[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) For CD R = mg cos30 B1 May be seen in later working without.
Use of F = 0.1R for either BC or CD M1 Note: The first two marks are often gained in the work-energy
3 equation.
FBC = 0.1mg [ = m ] OR FCD = 0.1mg cos 30 = m
2
3 A1 For sum of work done by friction and the change in PE.
0.1mg cos30d + mgd sin 30 + 5 md Note: Allow terms on different sides of a work energy equation
2 as long as they have different signs.
1 M1 Attempt at work energy equation with five relevant terms
mg × 2sin 30 = 0.1mg × 2 + 0.1mg cos30d + mgd sin 30 + m × 12 (dimensionally correct). Allow sign errors. Allow sin/cos errors
2
but must be consistent.
1
10m = 2m + m cos30d + 5md + 2 m Note: Initial PE = mg.
d = 1.28 m or
(
15 10 − 3 ) [1.27854…] A1 ISW if go on to find total distance = 2 + 2 + 1.28 having already
found 1.28.
97
7(a) Alternative Method for Question 7(a): Using Newton’s second law and equations of motion
For CD R = mg cos30 (B1) May be seen in later working without m.
If not seen in working check diagram but must be a reaction
force, not a downward component of the weight.
Use of F = 0.1R for either BC or CD FBC = 0.1mg [ = m ] (M1)
3
or FCD = 0.1mg cos 30 = m
2
For aCD −mg sin 30 − 0.1mg cos30 = ma (A1) For correct equation for a or ma in section CD
Note: Allow if acceleration in the opposite sense and both signs
3 positive.
a = − g sin 30 − 0.1g cos30 = −5.866 = − 5 +
2
For aAB mg sin 30 = ma ⟹ a = 5 ⟹ vB 2 = 0 + 2 × 5 × 2 [ = 20] (M1) Attempt to find d. Allow sign errors in Newton’s second law.
Allow sin/cos errors but must be consistent. Should include a
For aBC −0.1mg = ma a = −1 so vC 2 = 20 − 2 × 1 × 2 [ = 16]
valid attempt at vC 2 to get M1.
3 Must get to final line of working.
12 = 16 − 2 × ( g sin 30 + 0.1g cos30)d 1 = 16 − 2 5 + d
2 Note: this mark can be earned even if A0 above.
Must have 2 term acceleration though could have sign error.
d = 1.28 m or
(
15 10 − 3 ) [1.27854…] (A1) ISW if go on to find total distance = 2 + 2 + 1.28 having already
found 1.28.
97
7(a) Alternative Method for the last 2 marks: Using an energy method for the third phase
For aAB: mg sin 30 = ma ⟹ a = 5 ⟹ vB 2 = 0 + 2 × 5 × 2 [ = 20] (M1) Attempt at work energy equation for the third phase with four
relevant terms (dimensionally correct). Allow sign errors. Allow
For aBC: −0.1mg = ma a = −1 so vC 2 = 20 − 2 × 1 × 2 [ = 16] sin/cos errors but must be consistent. Must get to final line.
1
2
( )
m 12 − 42 = −mgd sin 30 − 0.1mg cos30 × d
d = 1.28 m or
(
15 10 − 3 ) [1.27854…] ignore units (A1) ISW if go on to find total distance = 2 + 2 + 1.28 having already
found 1.28.
97
5
7(b) mg × 2sin 30 = 2 μ mg + 1 × μ mg cos30 + mg × 1sin 30 M1 Attempt at work energy equation with four relevant terms
10m = 20mμ + 10m cos30μ + 5m OR 10m = 20mμ + m5 3μ + 5m (dimensionally correct). Allow sign errors. Allow sin/cos errors
but must be consistent.
4− 3 A1
μ = 0.174 or [0.174457]
13
mg × 1sin 30 − 1 × μ mg cos30 M1 For difference between the change in PE and the work done by
5m − 5 3mμ friction.
Note: Allow terms on different sides of a work energy equation
as long as both have the same sign.
Allow sin/cos errors but must be consistent. Using μ , their μ or
the correct value of μ to at least 2 sf.
Must be as part of an attempt to find speed, not μ , although this
could be the first step.
1 1 A1
mg × 1sin 30 − 1 × μ mg cos30 = mv 2 5m − 5 3mμ = mv 2
2 2
Speed = 2.64 m s-1 [2.64164…]
7(b) Alternative Method for Question 7(b): Newton’s second law and equations of motion
For aAB mg sin 30 = ma ⟹ a = 5 ⟹ vB 2 = 0 + 2 × 5 × 2 [ = 20] (M1)
For attempt at equation for µ. Allow sign errors. Allow sin/cos
For aBC − μ mg = ma a = − μ g so vC 2 = 20 − 2μ g × 2 errors but must be consistent.
3
For aCD −mg sin 30 − μ mg cos30 = ma a = −5.866= − 5 + Must get to fourth line for M1.
2
0 = 20 − 2 μ g × 2 − 2 ( g sin 30 + μ g cos30 ) × 1
20 − 40μ − 10 − 10 3μ = 0
4− 3 (A1)
μ = 0.174 or [0.174457...]
13
a = g sin 30 − μ g cos30 = 5 − 5 3μ (M1) For correct equation for a or ma in section CD down plane
(weight component – friction). Allow sin/cos errors but must be
consistent. Using μ , their μ or the correct value of μ to at least
2sf.
Must be as part of an attempt to find speed, not μ , although this
could be the first step.
v 2 = 0 + 2( g sin 30 − μ g cos 30) × 1 ⇒ Speed = 2.64 m s-1 (A1)
[2.64164…]
7(b) Alternative method for last 2 marks of Question 7(b): Using energy at the start – total work done against friction
1 (M1) For PEA − total work done against friction
mg × 2sin 30 − ( μ mg × 2 + μ mg cos30 × 2 ) = mv 2 [= KEC].
2
Allow sin/cos errors but must be consistent. Using μ , their μ or
the correct value of μ to at least 2sf.
1 2 (A1)
10m − (20mμ + 20m cos30μ = 2 mv
1 2
10m − (20mμ + 10m 3μ = 2 mv
Speed = 2.64 m s-1 [2.64164…]
4
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