Mathematics 9709/42 October/November 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A block of mass 15 kg slides down a line of greatest slope of an inclined plane. The top of the plane is at a vertical height of 1.6 m above the level of the bottom of the plane. The speed of the block at the top of the plane is 2 m s−1 and the speed of the block at the bottom of the plane is 4 m s−1 .
Find the work done against the resistance to motion of the block. [4]
Mark scheme
1 1 1 B1 For KE at top or bottom.
± × 15 × 22 [ = ±30] ± × 15 × 42 [ = ±120] Need not be evaluated.
2 2
1
× 15 × ( 4 − 2 ) is B0.
2
2
±15 g × 1.6 [ = ±240 ] B1 For PE change.
Need not be evaluated.
240 + 30 = 120 + W M1 Attempt at work energy equation; 4
relevant terms; dimensionally
correct; allow sign errors.
1
× 15 × ( 4 − 2 ) is M0.
2
2
If W = F times a numerical distance
seen, then M0.
Work done = 150 J A1
1 Alternative method for Q1
1.6 *M1 Attempt to use v 2 = u 2 + 2as with
4 2 = 2 2 + 2a ×
sin θ 1.6 1.6
s= or but not
sin θ cos θ
s = 1.6sinθ or 1.6cosθ or 1.6 .
If θ is given a value, then M0.
Must be using speeds 2 and 4 here.
15g sin θ − R = 15a DM1 3 terms; allow sign errors; allow
sin/cos mix but weight must be
resolved; dimensionally correct.
R = 93.75sinθ A1 R = 93.75cosθ
Must be consistent with their s .
1.6 A1
Work done = 93.75sin θ × = 150 J
sin θ
4
A
30Å
R
2N m kg
40Å
B
The diagram shows a smooth ring R, of mass m kg, threaded on a light inextensible string. A horizontal force of magnitude 2 N acts on R. The ends of the string are attached to fixed points A and B on a vertical wall. The part AR of the string makes an angle of 30Å with the vertical, the part BR makes an angle of 40Å with the vertical and the string is taut. The ring is in equilibrium.
Find the tension in the string and find the value of m. [5]
Mark scheme
2 Attempt to resolve in at least one direction to form an equation. *M1 Correct number of terms; allow
sign errors; allow sin/cos mix;
allow with different T ’s.
T sin30 + T sin40 − 2 = 0 A1 If different T ’s then allow
M1A1A0 max.
T cos30 − T cos 40 − mg = 0 A1 Allow with their T.
Attempt to solve for T or m DM1 From equation(s) with correct
number of relevant terms.
Tension T = 1.75, m = 0.0175 A1 T = 1.7501 m = 0.017497
awrt 1.75 for T www, and awrt
0.0175 for m www.
5
120 N
20Å
10 kg
30Å
A block of mass 10 kg is at rest on a rough plane inclined at an angle of 30Å to the horizontal. A force of 120 N is applied to the block at an angle of 20Å above a line of greatest slope (see diagram). There is a force resisting the motion of the block and 200 J of work is done against this force when the block has moved a distance of 5 m up the plane from rest.
Find the speed of the block when it has moved a distance of 5 m up the plane from rest. [5]
Mark scheme
3 Work done by 120 N force = 120 × 5cos 20 [ = 563.81557.] B1
( PE change = )10 g × 5sin 30 [ = 250] B1 For PE change.
Attempt at work energy equation M1 4 relevant terms; dimensionally
correct; allow sign errors; allow
sin/cos mix in relevant resolved
terms.
1 A1
120 × 5cos 20 − 10 g × 5sin 30 − 200 = × 10 × v 2
2
563.815− 250 − 200 = 5v 2
Speed = 4.77 m s–1 A1 awrt 4.77.
3 Alternative method for Question 3
200 *B1 oe e.g. 5 × RF = 200 .
Resistive force = [ = 40]
5
120 cos 20 − RF − 10 g sin30 = 10a *M1 4 relevant terms; dimensionally
correct; allow sign errors; allow
sin/cos mix; allow with their
resistive force or just RF.
a = 2.276… A1 Allow arwt 2.3 to 2sf from correct
work.
v2 = 0 + 2 × ( 2.276) × 5 DM1 Use of v 2 = u 2 + 2as using u = 0 ,
s = 5 and their positive a which
has come from a resistive force
using work done.
Speed = 4.77 m s–1 A1 awrt 4.77.
5
A particle P of mass 0.2 kg lies at rest on a rough horizontal plane. A horizontal force of 1.2 N is applied to P.
(a) Given that P is in limiting equilibrium, find the coefficient of friction between P and the plane.
[3]
(b) Given instead that the coefficient of friction between P and the plane is 0.3, find the distance travelled by P in the third second of its motion. [4]
Mark scheme
4(a) R = 0.2g B1
1.2 = μ × 0.2g M1 Resolve horizontally and using
F = µR to get an equation in µ; 2
relevant terms.
µ = 0.6 A1 oe
3
4(b) 1.2 − 0.3 × 0.2 g = 0.2a *M1 Resolve horizontally using
Newton’s Second Law; 3 relevant
terms; allow sign errors; R = 0.2 g
only.
a=3 A1 0.6 = 0.2a only seen, allow with
BOD, but if 0.6 as friction being
used as resultant force, this is
M0A0.
1 1 DM1 1
s3 = 0 + × 3 × 32 [ = 13.5] s2 = 0 + × 3 × 22 [= 6] For use of s = ut + at 2 (or a
2 2 2
complete method) to find a distance
at least once with u = 0 and their
positive a and t = 2 or t = 3 .
Distance = 13.5 − 6 = 7.5 m A1 www
4
A particle A of mass 0.5 kg is projected vertically upwards from horizontal ground with speed 25 m s−1 .
(a) Find the speed of A when it reaches a height of 20 m above the ground. [2]
When A reaches a height of 20 m, it collides with a particle B of mass 0.3 kg which is moving downwards in the same vertical line as A with speed 32.5 m s−1 . In the collision between the two particles, B is brought to instantaneous rest.
(b) Show that the velocity of A immediately after the collision is 4.5 m s−1 downwards. [2]
(c) Find the time interval between A and B reaching the ground. You should assume that A does not bounce when it reaches the ground. [4]
Mark scheme
5(a) v2 = 252 + 2 ( − g ) × 20 M1 Use of v 2 = u 2 + 2as with u = 25 ,
s = 20 and a = ± g.
1 1 OR using change in KE = ± change
OR × 0.5 × v 2 = × 0.5 × 252 − 0.5 × g × 20
2 2 in PE.
Speed = 15 m s-1 A1
2
5(b) Taking up as positive direction: 0.5 × 15 + 0.3 × ( −32.5 ) = 0.5v + 0 or M1 For use of conservation of
momentum, 3 non-zero terms,
allow sign errors, using their speed
Taking down as positive direction: 0.5 × ( −15 ) + 0.3 × 32.5 = 0.5v + 0 15 m s-1. Must show how ±2.25 is
obtained.
[Taking up as positive direction: velocity of A = −4.5 m s−1] A1 Any error seen in calculating v is
[Taking down as positive direction: velocity of A = 4.5 m s−1] A0.
Speed = 4.5 m s–1 direction downwards Must explicitly say 4.5 m s–1 and
downwards.
2
5(c) 1 M1 Using constant acceleration
Downwards to be positive, for A 20 = 4.5t A + gt A2 and solve for t A formula(e) to get a correct equation
2
1 in t A and solve for t A .
Upwards to be positive, for A −20 = −4.5t A − gt A2 and solve for t A If using quadratic formula, must be
2
the correct formula. If factorising,
when brackets expanded, 2 terms
correct.
1 M1 Using constant acceleration
For B 20 = 0 + gtB 2 [t = 2] and solve for t B formula(e) to get a correct equation
2
in t B and solve for t B .
t A = 1.6 or t B = 2 A1
Difference = 0.4 s only A1
4
A railway engine of mass 120 000 kg is towing a coach of mass 60 000 kg up a straight track inclined at an angle of ! to the horizontal where sin ! = 0.02. There is a light rigid coupling, parallel to the track, connecting the engine and coach. The driving force produced by the engine is 125 000 N and there are constant resistances to motion of 22 000 N on the engine and 13 000 N on the coach.
(a) Find the acceleration of the engine and find the tension in the coupling. [5]
At an instant when the engine is travelling at 30 m s−1 , it comes to a section of track inclined upwards at an angle " to the horizontal. The power produced by the engine is now 4 500 000 W and, as a result, the engine maintains a constant speed.
(b) Assuming that the resistance forces remain unchanged, find the value of ". [4]
Mark scheme
6(a) Engine: 125000 − 120000 g × 0.02 − 22000 − T = 120000a *M1 Attempt at Newton’s second law at
125000 − 120000 g sin (1.145..) − 22000 − T = 120000a least once; correct number of
relevant terms; allow sign errors;
[125000 − 24000 − 22000 − T = 120000a 79000 − T = 120000a ] allow sin/cos mix; allow g
missing; a value for α or sinα must
Coach: T − 60000 g × 0.02 − 13000 = 60000a be substituted.
T − 60000 g sin (1.145..) − 13000 = 60000 a Allow with α = 1.1 or better
[T − 12000 − 13000 = 60000a T − 25000 = 60000a ] [α = 1.145991998].
A1 Any equations correct.
System:
125000 − 120000 g × 0.02 − 60000 g × 0.02 − 22000 − 13000 = (120000 + 60000 ) a A1 Two equations correct.
125000 − 120000 gsin (1.145..) − 60000 gsin (1.145..) − 22000 − 13000 = (120000 + 60000 ) a
If using separate equations for
125000 − 24000 − 12000 − 22000 − 13000 = (120000 + 60000 ) a 54000 = 180000a engine and coach and different T ’s,
then allow M1A1A0 max.
Solve for T or a DM1 Using equations with the correct
number of relevant terms.
If no working seen, must be
solutions to their equation(s) to be
awarded M1.
Acceleration = 0.3 m s–2 A1 Allow 0.299 from use of α = 1.15.
and Awrt 43000 to 3sf from correct
work.
Tension = 43 000 N
5
6(b) 4500000 B1 P
Driving force, DF = [ = 150000] Use of F = ,
30 v
oe e.g. DF × 30 = 4500000 .
Attempt to resolve parallel to the track once if using system equation, twice if using equations for M1 Correct number of relevant terms;
engine and coach separately allow sign errors; allow sin/cos
mix; allow g missing.
Must be correct number of
equations depending on method.
System: 150000 − 120000gsinβ − 60000gsinβ − 22000 − 13000 = 0 A1 Allow DF or their DF.
or
for Engine: 150000 − 120000 gsinβ − 22000 − T ' = 0
and Coach: T '− 60000 gsinβ − 13000 = 0 Must be using same T ' .
Solve to get β = 3.7° A1 3.663058552
awrt 3.7° www.
4
A particle X travels in a straight line. The velocity of X at time t s after leaving a fixed point O is denoted by v m s−1 , where v = −0.1t3 + 1.8t2 − 6t + 5.6.
The acceleration of X is zero at t = p and t = q, where p < q.
(a) Find the value of p and the value of q. [4]
It is given that the velocity of X is zero at t = 14.
(b) Find the velocities of X at t = p and at t = q, and hence sketch the velocity-time graph for the motion of X for 0 ≤ t ≤ 15. [3]
(c) Find the total distance travelled by X between t = 0 and t = 15. [5]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) Attempt to differentiate v *M1 Decrease power by 1 and a change
in coefficient in at least one term
(which must be the same term);
allow unsimplified; allow p or q
for t .
v
a = is M0.
t
dv A1 May be unsimplified.
a = = 3 × −0.1t 3−1 + 2 × 1.8t 2−1 − 6t1−1 = −0.3t 2 + 3.6t − 6 Allow p or q for t.
dt
dv DM1 Allow p or q for t.
Setting a = = 0 and attempt to solve a 3 term quadratic for t.
dt Must get 2 values or numerical
dv expressions for t from their three
a = dt = 0 3t − 36t + 60 = 0 t − 12t + 20 = 0
2 2
term quadratic.
If using quadratic formula, must be
the correct formula. If factorising,
when brackets expanded, 2 terms
correct.
p = 2, q = 10 A1
4
7(b) Velocities are 0 m s-1 and 25.6 m s-1 B1 SOI
Curve with single minimum turning point followed by single maximum turning point *B1 Ignore placement of graph on axes
Not a cusp for the minimum point
or maximum point.
DB1 All correct in 1st and 4th quadrant.
Must go from convex to concave.
Need to label 2 and 14 on the
t-axis where the curve meets the
t-axis.
Do not need to show exact
velocities at t = 0 or
t = 10 or t = 15.
Ignore graph outside 0 t 15 .
3
7(c) Attempt to integrate v *M1 Increase power by 1 and a change
in coefficient in at least one term
(which must be the same term);
s = vt is M0.
0.1 3+1 1.8 2+1 6 1+1 A1 May be unsimplified.
( s =) − t + t − t + 5.6t ( +c )
4 3 2
= −0.025t + 0.6t − 3t + 5.6t [+c]
4 3 2
Attempt distance from t = 0 to t = 14 [= 176.4] DM1 Correct use of limits 0 and 14 for
their s, i.e. F (14 ) − F ( 0 )
May see limits 0 to 2 and 2 to 14
used but must be
( F (14) − F ( 2)) + ( F ( 2) − F (0)).
Attempt distance from t = 14 to t = 15 [= (-) 8.025] DM1 Correct use of limits 14 and 15 for
their s, i.e. ± ( F (15) − F (14) ).
18
For reference F ( 2 ) = = 3.6 ,
5
882
F (14 ) = = 176.4 and
5
1347
F (15) = = 168.375.
8
7377 A1 www
Total distance =176.4 + 8.025 = 184.425 m = Condone 184 or better.
40
7(c) SC for those who show no integration. Max 3 marks.
14 B1
( −0.1t + 1.8t − 6t + 5.6) dt = 176.4
3 2
0
15 B1
( −0.1t + 1.8t − 6t + 5.6) dt = −8.025
3 2
14
15
( −0.1t + 1.8t − 6t + 5.6 ) dt = 8.025
3 2
OR
14
7377 B1 Condone 184 or better.
Total distance =176.4 + 8.025 = 184.425 m =
40
SC for those who show no integration and don’t consider the 2 areas. Max 1 mark.
15
7377 B1 Condone 184 or better.
0
−0.1t 3 + 1.8t 2 − 6t + 5.6 dt = 184.425 m =
40
5
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