Mathematics 9709/41 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.
A car starts from rest and accelerates at 2 m s -2 for 10 s. It then travels at a constant speed for 30 s. The car then uniformly decelerates to rest over a period of 20 s.
(a) Sketch a velocity-time graph for the motion of the car. [2]
(b) Find the total distance travelled by the car. [2]
Mark scheme
1(a) Trapezium on the t-axis starting at (0, 0) B1
Fully correct with correct labels B1 With height 20, intersecting the t-axis at 60, horizontal line segment
between 10 and 40.
Axis need not be labelled with t and v but if they are they must be correct.
If not labelled, then must assume t is on the horizontal axis.
2
1(b) 1 M1 Must be considering the area of a trapezium. Allow a single slip in one
Area = ( 30 + 60 ) × 20 term only.
2
or Must be adding all terms together if considering two triangles and a
1 1 rectangle. Using their 20 from (a). If no value for the height shown on the
Area = × 10 × 20 + 30 × 20 + × 20 × 20 graph in (a), then it must be correct.
2 2
= 900 m A1FT FT their 20 ( = 45 × their 20 ) but A0 FT if using 2.
2
y
20 N
60°
P x
FN
Two forces of magnitudes 20 N and F N act at a point P in the directions shown in the diagram.
(a) Given that the resultant force has no component in the y-direction, calculate the value of F. [2]
(b) Given instead that F = 10 , find the magnitude and direction of the resultant force. [5]
Mark scheme
2(a) F = 20sin 60 M1 Attempt to resolve in y-direction; 2 terms; must be 20 cos 60 or 20 sin 60
and must be linked to F (can be implied by the correct answer seen only).
= 17.3 N A1 AWRT 17.3 (17.320508…) or 10 3.
2
2(b) For resolving in any direction *M1 Correct number of terms; allow sin/cos mix; allow sign errors.
(Horizontal component = X = R cosθ ) = ± ( 20cos 60 ) [ ±10 ] A1 For both correct.
(Vertical component = Y = R sin θ ) = ± ( 20sin 60 − 10 )
[= ±7.3205]
DM1 OE – correct number of terms.
Magnitude = ( 20sin 60 − 10 )2 + ( 20cos 60 )2
[=12.393136…]
20sin 60 − 10 DM1 OE (e.g. reciprocal) - correct number of terms.
Angle = tan −1 [=36.206023…]
20cos 60
Magnitude = 12.4 N A1 OE for direction e.g. 36.2 anticlockwise from (positive) x-direction,
and Direction = 36.2° above (positive) x-axis
36.2 above the horizontal.
Possibly seen on a diagram.
(Radians: 0.63191… to 3sf or better)
5
A train of mass 180 000 kg ascends a straight hill of length 1.5 km, inclined at an angle of 1.5° to the horizontal. As it ascends the hill, the total work done to overcome the resistance to motion is 12 000 kJ and the speed of the train decreases from 45 m s -1 to 40 m s -1 .
Find the work done by the engine of the train as it ascends the hill, giving your answer in kJ. [4]
Mark scheme
3 PE gained = 180000 g × 1500sin1.5 [= 70677760.4] B1 180000 g × 39.2654
1 B1 For initial KE or final KE (for reference: difference in KE is 38 250 000).
KEInitial = × 180000 × 452 [ = 182 250 000]
2
1
KEFinal = × 180000 × 40 2 [= 144 000 000]
2
For work energy equation: M1 Correct number of terms; dimensionally correct; allow sign errors and
minor slip(s) in values; allow sin/cos mix on PE term.
WD = 180000 g × 1500sin1.5 + 12000 000 Work done = ( 70677760.4 + 12000000 − 38250000 )
(
− 12 × 180000 × 452 − 12 × 180000 × 402 ) J = ( 70677.7 + 12000 − 38250 ) kJ
= 44400 kJ [44427.7604…] A1 Must be in kJ.
Alternative Method for Question 3: Newton’s second law and equations of motion
a = −0.142 [= − 0.141666…] (B1) Correct acceleration from 402 = 452 + 2a (1500).
Allow AWRT −0.14 or exact − 120
17
.
1500( D − 12 1500
000 000
− 180000 g sin1.5) = 180000 × − 120
17
× 1500 (M1) M1 for applying Newton’s second law parallel to the hill.
Correct number of terms, allow sin/cos mix on weight component,
dimensionally correct and multiplying both sides by 1500.
Allow their a or a for acceleration (and minor slip(s) in values).
(B1) Correct weight component multiplied by 1500.
WD (= 1500 D ) = 44400 kJ [44427.7604…] (A1) Must be in kJ.
4
A car of mass 1700 kg is pulling a trailer of mass 300 kg along a straight horizontal road. The car and trailer are connected by a light inextensible cable which is parallel to the road. There are constant resistances to motion of 400 N on the car and 150 N on the trailer. The power of the car’s engine is
14 000 W.
Find the acceleration of the car and the tension in the cable when the speed is 20 m s -1 . [6]
Mark scheme
4 14000 B1 OE, e.g. 20 × DF = 14000.
Driving force = [ = 700]
20
Attempt at Newton’s second law on car or trailer or system *M1 Correct number of relevant terms (e.g. correct masses); allow sign errors.
700 − 400 − T = 1700a A2 A1 for one correct and A2 for any two correct.
T − 150 = 300a May have DF (or their DF) for 700 but A0 if using 14000 as DF without
700 − 400 − 150 = (1700 + 300 ) a first stating DF.
Must have same T if using first two equations for A2.
Solving for a or T DM1 From equations with the correct number of relevant terms.
3 For both correct answers.
Acceleration = = 0.075 m s-2
40
Tension = 172.5 N A1 Condone 173.
6
A straight slope of length 60 m is inclined at an angle of 12° to the horizontal. A bobsled starts at the top of the slope with a speed of 5 m s -1 . The bobsled slides directly down the slope.
(a) It is given that there is no resistance to the bobsled’s motion.
Find its speed when it reaches the bottom of the slope. [3]
(b) It is given instead that the coefficient of friction between the bobsled and the slope is 0.03 .
Find the time that it takes for the bobsled to reach the bottom of the slope. [5]
Mark scheme
5(a) a = 2.08 [ 2.07911] B1 From mg sin12 = ma.
Allow exact (e.g. a = g sin12).
v 2 = 52 + 2a × 60 M1 For use of v 2 = u 2 + 2as with u = 5 and s = 60.
Allow sign errors but a must be either g sin12 or g cos12 only.
Speed = 16.6 m s−1 [16.567861…] A1 AWRT 16.6
Alternative Method for Q5(a)
For attempt at work energy equation (M1) 3 terms, dimensionally correct.
Allow sign errors; allow sin/cos mix on PE term – condone m missing
from all terms. Must be a weight component.
1 2 1 (A1) Correct equation.
mv = m × 52 + mg × 60sin12 (for reference: 60 sin12 = 12.4747…)
2 2
Speed = 16.6 m s−1 [16.567861…] (A1) AWRT 16.6
3
5(b) R = mg cos12 B1 Resolving correctly perpendicular to the plane.
mg sin12 − F = ma *M1 Use of Newton’s second law, correct number of terms; allow sign errors;
allow sin/cos mix (must be a weight component).
For use of F = 0.03R to get equation in a (and m) only DM1 Where R is a component of weight only (dimensionally correct but allow
sin/cos mix).
a = 1.79 [1.78567.]
1 DM1 Dependent on previous two M marks.
60 = 5t + at 2 and solve for t
2 1
For use of s = ut + at 2 with s = 60, u = 5 and their a or other complete
2
method to find positive value(s) of t.
Time = 5.86 s [5.86260…] A1 AWRT 5.86 (from using a = 1.79 or better).
AWRT 5.85 (from using a = 1.8).
AWRT 5.87 (from correct working).
5(b) Alternative method for Question 5(b): Using energy
R = mg cos12 (B1) Resolving correctly perpendicular to the plane.
1
mv 2 − 12 m × 52 = 60 × mg sin12 − 60 × F (*M1) Use of work-energy principle, correctly number of relevant terms; allow
2
sign errors; allow sin/cos mix on PE term (must be a weight component).
For use of F = 0.03R to get equation in v (and m) only (DM1) Where R is a component of weight only (dimensionally correct but allow
sin/cos mix)
v = 15.5 [15.46870.] .
1 (DM1) Dependent on previous two M marks.
60 = ( 5 + v ) t and solve for t 1
2 Use of s = ( u + v ) t with s = 60, u = 5 and their v or other complete
2
method to find positive value(s) of t.
Time = 5.86 s [5.86260…] (A1)
5
A particle moves in a straight line, starting from a point O. The velocity of the particle at time t s after
1 leaving O is v m s -1 . It is given that v = kt 2 - 2t - 8 , where k is a positive constant. The maximum velocity of the particle is 4.5 m s -1 .
(a) Show that k = 10 . [5]
(b) (i) Verify that v = 0 when t = 1 and t = 16 . [1]
(ii) Find the distance travelled by the particle in the first 16 s. [5]
Mark scheme
6(a) For an attempt at differentiation *M1 Decrease power by 1 and a change in coefficient in at least one term
1 1
d −
(which must be the same term). Therefore, must see (kt 2 ) = ct 2 with
dt
c ≠ k.
1 −1
1
1 −
1 A1 May be unsimplified.
a = kt 2 − 2t1−1 + 0 = kt 2 − 2
2 2
1 −
1
k2 1 DM1 Equate a to 0 and attempt to solve for t or k.
a = 0 kt 2 − 2 = 0 t = or k = 4t 2 1
2 16 Must be of the correct form, e.g. t = λ k 2 or k = δ t 2 (possibly implied by
forming a correct equation in k).
1 M1 Or complete method to find k or an equation involving k,
k 2 2 k2 1
( v = ) k − 2 − 8 = 4.5
1 1
e.g. 4t 2 × t 2 − 2t − 8 = 4.5 t = 6.25 therefore k = 4(6.25) 2 .
16 16
Dependent on both previous M marks.
k2 A1 AG - Allow verification.
= 12.5 k = 10 Any errors seen is A0.
8
Alternative method for last two marks of Question 6(a): Completing the square method
1 (M1) Dependent on both previous M marks.
(Let x = t2 )
k k k k2
2
v = −2 x 2 − x + 4 = −2 x 2 − x + + −8
2 2 16 8
k2 (A1) AG
so − 8 = 12.5
8
so k = 10
5
6(b)(i) 1 1 B1 AG
10 × 12 − 2 × 1 − 8 = 0 and 10 × 16 2 − 2 × 16 − 8 = 0 Any errors seen is B0.
Or equate v = 0 and solve for t to get t = 1 and t = 16.
1
6(b)(ii) For integration (do not penalise missing c) *M1 Increase power by 1 and a change in coefficient in at least one term
(which must be the same term).
10
3
2t 2 20
3 A1 Allow unsimplified; allow with k not substituted.
( s = ) 3 t 2 − − 8t [ +c ] = t 2 − t 2 − 8t [ +c ]
2
2 3
3 3
k 2t 2 2k 2 2
or ( s = ) 3 t 2 − − 8t [ +c ] = t − t − 8t [ +c ]
2
2 3
20 3 2 DM1 Attempt at correct use of limits 0 and 1 or 1 and 16 – allow a single slip.
7
12 − 1 − 8 × 1 − 0 = − and
3 3
20 3 20 3 2
2 − 16 2 − 8 × 16 −
16 12 − 1 − 8 × 1 =
3 3
128 7
3 − − 3 = 45
For both set of limits applied correctly DM1 Allow at most one slip in both.
142 A1 Allow 47.3 or better (47.3333…)
m
3
SC if no integration shown:
7 142
B1 for ± or 45, B1 for (so 2 marks max).
3 3
5
A particle P of mass 0.2 kg is projected vertically upwards from horizontal ground with speed 25 m s -1 .
(a) Show that the speed of P when it reaches 20 m above the ground is 15 m s -1 . [2]
When P reaches 20 m above the ground it collides with a second particle Q of mass 0.1 kg which is moving downwards at 20 m s -1 . P is brought to instantaneous rest in the collision.
(b) Find the velocity of Q immediately after the collision. [2]
When P reaches the ground it rebounds back directly upwards with half of the speed that it had immediately before hitting the ground.
(c) Find the height above the ground at which P and Q next collide. [6]
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) v 2 = 252 − 2 g × 20 M1 For use of v 2 = u 2 + 2as or equivalent to get an equation in v only with
u = 25, s = 20 and a = ± g.
⇒ speed = 15 m s–1 A1 AG
Allow verification – at least one intermediate step from equation of
motion to given result. Any errors seen is A0.
Alternative method for Question 7(a):
1 1 (M1) Attempt at energy with m = 0.2, h = 20, u = 25; correct number of terms,
0.2 g × 20 = × 0.2 × 252 − × 0.2 × v 2
2 2 allow sign errors.
⇒ speed = 15 m s-1 (A1) AG
Allow verification - at least one intermediate step from conservation of
energy to given result. Any errors seen is A0.
2
7(b) 0.2 ×15 − 0.1× 20 = 0 + 0.1v M1 OE
or 0.2 ×15 − 0.1× 20 = 0 − 0.1V Attempt at conservation of momentum; 3 non-zero terms; allow sign
errors – use of 25 is M0.
v = 10 m s-1 upwards A1 Must have direction (possibly seen on diagram).
If using mgv for momentum, then M1 A0 max.
2
7(c) Speed of P at impact is 20 m s–1 B1 From v 2 = 0 + 2 × g × 20 (OE) - possibly implied by speed of P after
impact being stated at 10.
Time to when P reaches ground = 2 s B1 From 20 = 0 + 1 × g × t 2 (OE).
2
sP = 10t + 12 × − g × t 2 M1 Distance travelled by P after impact with the ground.
Must be using their 10 (speed of P after impact with the ground) and
a = ± g.
sQ = 10t + 12 × g × t 2 M1 Distance travelled by Q after P’s impact with the ground.
Must be using their ±10 (the speed/vel. of Q after their 2 s
(= '10'+ (− g ) × 2 where ‘10’ is the value from (b)) and a = ± g.
sP + sQ = 20 t = 1 A1 Time after P hits the ground to next collision.
Height = 5 m A1 CWO
7(c) Alternative Method for last 4 marks:
sQ = 10t + 12 × − g × t 2 (M1) Expression for the displacement of Q after first impact of P and Q.
Must be using their 10 (from (b)) and a = ± g.
sP = 10 × (t − 2) + 12 × − g × (t − 2) 2 (M1) Expression for the displacement of P (for values of t 2 ) measured from
point of first collision between P and Q.
Must be using their 2 (time for P to reach ground), their 10 (speed of P
after impact with the ground), a = ± g.
sQ + 20 = sP t = 3 (A1) Correct time between collisions of P and Q.
Height = 5 m (A1) CWO
6
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