Mathematics 9709/42 February/March 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, February/March 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
2.5 N
15Å
P
A small ring P of mass 0.03 kg is threaded on a rough vertical rod. A light inextensible string is attached to the ring and is pulled upwards at an angle of 15Å to the horizontal. The tension in the string is 2.5 N (see diagram). The ring is in limiting equilibrium and on the point of sliding up the rod. Find the coefficient of friction between the ring and the rod. [4]
Mark scheme
1 R = 2.5 cos 15 B1
[F = µ × 2.5 cos 15] M1 Using F = µR
[2.5 sin 15 = 0.03g + F] M1 Resolve forces along the rod
µ = 0.144 A1
4
A particle is projected vertically upwards with speed 30 m s−1 from a point on horizontal ground.
(i) Show that the maximum height above the ground reached by the particle is 45 m. [2]
(ii) Find the time that it takes for the particle to reach a height of 33.75 m above the ground for the first time. Find also the speed of the particle at this time. [4]
Mark scheme
2(i) [0 = 302 +2(–g)s] M1 Using v2 = u2 + 2as with v = 0,
u = 30 and a = –g
For any complete method for finding maximum height s
s = maximum height = 900/20 = 45 m A1 AG
2
2(ii) [33.75 = 30t – ½ gt2] M1 Applying s = ut + ½at2 with s = 33.75, u = 30 and a = –g
[5t2 – 30t + 33.75 = 0 or 4t2 – 24t + 27 = 0] M1 Solve a 3-term quadratic for t
t = 1.5 (reject t = 4.5) A1
v = 30 – 1.5g = 15 B1ft Use v = u + at with u = 30 and
t = 1.5
ft on t value found
Alternative method for question 2(ii)
v2 = 302 – 2g(33.75) = 225 → v = 15 B1 Use v2 = u2 + 2as with u = 30,
a = –g and s = 33.75 to find v
[33.75 = ½ (30 + 15) × t] M1 Use s = ½ (u + v) × t with s = 33.75, u = 30 and v as found.
or [15 = 30 – 10t] or Use v = u – gt with u = 30 and v as found
M1 Solve for t
t = 1.5 A1ft ft on v value found
4
FN
!Å
P
20Å 5N
15 N
25 N
Four coplanar forces of magnitudes F N, 5 N, 25 N and 15 N are acting at a point P in the directions shown in the diagram. Given that the forces are in equilibrium, find the values of F and !. [6]
Mark scheme
3 M1 Attempt to resolve forces horizontally or vertically
F cos α = 15 cos 20 – 5 (= 9.095...) A1
F sin α = 15 sin 20 + 25 (= 30.13....) A1
M1 Use Pythagoras or trigonometry to find F
F= (15cos 20 − 5)2 + (15sin 20 + 25)2
(15sin 20 + 25 ) M1 Use trigonometry to find α
∝ = tan −1
(15cos 20 − 5 )
α = 73.2 and F = 31.5 A1
6
A car of mass 1500 kg is pulling a trailer of mass 300 kg along a straight horizontal road at a constant speed of 20 m s−1 . The system of the car and trailer is modelled as two particles, connected by a light rigid horizontal rod. The power of the car’s engine is 6000 W. There are constant resistances to motion of R N on the car and 80 N on the trailer.
(i) Find the value of R. [2]
The power of the car’s engine is increased to 12 500 W. The resistance forces do not change.
(ii) Find the acceleration of the car and trailer and the tension in the rod at an instant when the speed of the car is 25 m s−1 . [5]
Mark scheme
4(i) Driving force = 6000/20 [= 300 N] B1 Using F = P/v
R = 300 – 80 = 220 B1ft Net force on system = 300 – R – 220 = 0 ft on DF found
2
4(ii) [New driving force DF = 12500/25 = 500 N M1 Any one equation from the following:
Car: DF – T – R = 1500a Apply Newton’s 2nd law to the car
Trailer: T – 80 = 300a Apply Newton’s 2nd law to the trailer
System: DF – 80 – R = 1800a] Apply Newton’s 2nd law to the system of car and trailer.
Two correct equations A1ft Correct DF = 500 must be used. ft on R value found
M1 EITHER solve two dimensionally correct simultaneous
equations in a and T to find a or T OR solve the system
equation to find a
a = 0.111 m s–2 A1 Allow a = 1/9
T = 113 N (= 113.3333...) A1 Allow T = 340/3
5
v (m s−1 )
7
0 t (s)
0 3 5 8 16
V
The velocity of a particle moving in a straight line is v m s−1 at time t seconds after leaving a fixed point O. The diagram shows a velocity-time graph which models the motion of the particle from t = 0 to t = 16. The graph consists of five straight line segments. The acceleration of the particle from t = 0 to t = 3 is 3 m s−2 . The velocity of the particle at t = 5 is 7 m s−1 and it comes to instantaneous rest at t = 8. The particle then comes to rest again at t = 16. The minimum velocity of the particle is V m s−1 .
(i) Find the distance travelled by the particle in the first 8 s of its motion. [3]
(ii) Given that when the particle comes to rest at t = 16 its displacement from O is 32 m, find the value of V . [4]
Mark scheme
5(i) Velocity at t = 3 is 3 × 3 = 9 B1
[½ × 3 × 9 + ½ (9 + 7) × 2 + ½ × 3 × 7] M1 Attempt distance travelled in the first 8 seconds using
Distance = area under graph.
Distance = 40 m A1
3
5(ii) [32 = 40 + area of triangle] M1 Use given displacement to set up equation for area of triangle
or attempt to find distance or displacement from t = 8 to t =
16
Area of triangle or displacement/distance = A1
(–)8
[Distance = ½ × 8 × V = (–)8] M1 Set up an equation for the area of triangle involving V
or use suvat equations to set up an equation involving V
V = –2 A1
4
A particle moves in a straight line. It starts from rest at a fixed point O on the line. Its acceleration at
1 time t s after leaving O is a m s−2 , where a = 0.4t3 − 4.8t 2 .
(i) Show that, in the subsequent motion, the acceleration of the particle when it comes to instantaneous rest is 16 m s−2 . [6]
(ii) Find the displacement of the particle from O at t = 5. [3]
Mark scheme
6(i) 1
M1 Attempt to integrate a
[ ∫ 0.4t 3 – 4.8t 2 dt ]
3 A1
4
v = 0.1t – 3.2 t 2 (+ c)
3 DM1 Attempt to solve v = 0, and reach the form ta/b = k
4
[v = 0 → 0.1t – 3.2 t 2 = 0]
5 M1 Attempt to solve an equation of the form ta/b = k
[t2 = 32]
t=4 A1
a = 16 m s–2 B1
6
6(ii) 3 M1 Attempt to integrate v
[s = ∫ 0.1t 4 – 3.2t 2 dt]
5
5 A1 Correct integration.
Displacement = 0.02t 5 − 1.28t 2
0
Displacement = –9.05 m (–9.05417...) A1
3
R
Q
hm
8m
!
P
The diagram shows the vertical cross-section PQR of a slide. The part PQ is a straight line of length
8 m inclined at angle ! to the horizontal, where sin ! = 0.8. The straight part PQ is tangential to the curved part QR, and R is h m above the level of P. The straight part PQ of the slide is rough and the curved part QR is smooth. A particle of mass 0.25 kg is projected with speed 15 m s−1 from P towards
Q and comes to rest at R. The coefficient of friction between the particle and PQ is 0.5.
(i) Find the work done by the friction force during the motion of the particle from P to Q. [4]
(ii) Hence find the speed of the particle at Q. [4]
(iii) Find the value of h. [3]
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(i) R = 0.25g × 0.6 [= 1.5] B1
[F = 0.5 × 0.25g × 0.6] [F = 0.75] M1 Use F = µR
[WD against friction = F × 8] M1 Using WD = Force × distance moved in direction of force
WD = 6 J A1
4
7(ii) [½ × 0.25 ×152 = M1 Work-energy equation in the form
½ × 0.25 × v2 + 6 + 0.25g × 8 × 0.8] Initial KE = Final KE + WD against F + PE gain
A1ft Correct Work–Energy equation for the motion to Q. ft on
WD
M1 Solving the work-energy equation for v
v = 7 m s–1 A1
Alternative method for question 7(ii)
[–F – 0.25g sin α = 0.25a] M1 Applying Newton’s second law to the particle along the plane
a = –11 m s–2 A1ft ft on friction found in (i)
M1 Finding the speed of the particle at Q by applying v2 = u2 +
2as with u = 15, s = 8 or equivalent complete method
v = 7 m s–1 A1
4
7(iii) [½ × 0.25 × 72 = 0.25 × g × H] M1 KE lost from Q to R = PE gain from Q to R
Or H is the height of R above Q
[½ × m × 72 = m × g × H]
H = 72/2g = 2.45 m A1
Total height h = 6.4 + H = 8.85 A1
Alternative method for question 7(iii)
[½ × 0.25 ×152 = 6 + 0.25g × h] M1 Work-energy from P to R
A1 Correct Work-energy equation from P to R
h = 8.85 A1
3
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Mathematics 9709/42 for February/March 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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