Mathematics 9709/41 October/November 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A particle B of mass 5 kg is at rest on a smooth horizontal table. A particle A of mass 2.5 kg moves on the table with a speed of 6 m s−1 and collides directly with B. In the collision the two particles coalesce.
(a) Find the speed of the combined particle after the collision. [2]
(b) Find the loss of kinetic energy of the system due to the collision. [3]
Mark scheme
1(a) 6 × 2.5 = 2.5v + 5v M1 Apply conservation of momentum, 3 terms implied
v = 2 ms–1 A1
2
1(b) Use KE = ½ mv2 either before or after collision M1 Allow this for either particle
KE(before) = 0.5 × 2.5 × 62 A1 FT Both correct FT on v
KE(after) = 0.5 × 7.5 × 22
Loss of KE = 30 J A1
3
A car of mass 1400 kg is moving along a straight horizontal road against a resistance of magnitude
350 N.
(a) Find, in kW, the rate at which the engine of the car is working when it is travelling at a constant speed of 20 m s−1 . [2]
(b) Find the acceleration of the car when its speed is 20 m s−1 and the engine is working at 15 kW.
[3]
Mark scheme
2(a) P = 350 × 20 M1 Using P = Fv
P = 7 kW A1
2
2(b) 15 000 = DF × 20 [DF = 750] B1 Using P = Fv
DF – 350 = 1400a M1 Use Newton’s 2nd law, 3 terms
2 A1 a = 0.286
a= ms–2
7
3
8N
PN
1Å 30Å
45Å 12 N
10 N
Coplanar forces of magnitudes 8 N, 12 N, 10 N and P N act at a point in the directions shown in the diagram. The system is in equilibrium.
Find P and 1. [6]
Mark scheme
3 Resolve forces either horizontally or vertically M1 Correct number of relevant terms
P cos θ = 12 + 8 cos 30 – 10 cos 45 [= 11.857] A1
P sin θ = 10 sin 45 – 8 sin 30 [= 3.071] A1
P= (11.857 + 3.071 )
2 2 M1 OE. Use of correct method for finding P
3.071 M1 OE. Use of correct method for finding θ
θ = tan −1
11.857
P = 12.2 and θ = 14.5 A1 Both correct
6
A particle P moves in a straight line. It starts from rest at a point O on the line and at time t s after leaving O it has acceleration a m s−2 , where a = 6t − 18.
Find the distance P moves before it comes to instantaneous rest. [6]
Mark scheme
4 [v = 3t2 – 18t (+ C)] *M1 Attempt to integrate a
[s = t3 – 9t2 (+ C)] #M1 Attempt to integrate v
v = 3t2 – 18t A1 Both integrals correct
s = t3 – 9t2
v = 0, 3t2 – 18t = 0 [t = 6] *DM1 Attempt to find t when v = 0
s = 63 – 9 × 62 – [0] #DM1 Substitute limits correctly into s
s = 108 m A1 Answer must be positive
6
0.8 kg 0.2 kg
0.5 m
Two particles of masses 0.8 kg and 0.2 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The system is released from rest with both particles 0.5 m above a horizontal floor (see diagram). In the subsequent motion the 0.2 kg particle does not reach the pulley.
(a) Show that the magnitude of the acceleration of the particles is 6 m s−2 and find the tension in the string. [4]
(b) When the 0.8 kg particle reaches the floor it comes to rest.
Find the greatest height of the 0.2 kg particle above the floor. [3]
Mark scheme
5(a) 0.8g – T = 0.8a, T – 0.2g = 0.2a, M1 Apply Newton’s 2nd law to either particle or to the system
For system: 0.8g – 0.2g = (0.8 + 0.2)a A1 Any 2 correct equations
Attempt to solve for either a or T M1
a = 6 ms–2 and T = 3.2 N A1 AG. Both correct
4
5(b) v2 = 2 × 6 × 0.5 M1 Attempt to find v or v2 as 0.8 kg particle reaches the ground using
a from 5(a)
0 = 6 – 20s M1 Attempt to find the extra height reached by 0.2 kg particle
using v2 from previous M1 mark
Greatest height = 0.5 + 0.5 + 0.3 = 1.3 m A1
3
A car of mass 1500 kg is pulling a trailer of mass 750 kg up a straight hill of length 800 m inclined at an angle of sin−1 0.08 to the horizontal. The resistances to the motion of the car and trailer are 400 N and 200 N respectively. The car and trailer are connected by a light rigid tow-bar. The car and trailer have speed 30 m s−1 at the bottom of the hill and 20 m s−1 at the top of the hill.
(a) Use an energy method to find the constant driving force as the car and trailer travel up the hill.
[5]
After reaching the top of the hill the system consisting of the car and trailer travels along a straight level road. The driving force of the car’s engine is 2400 N and the resistances to motion are unchanged.
(b) Find the acceleration of the system and the tension in the tow-bar. [4]
Mark scheme
6(a) KE (final) = ½ × 1500 × 202 + ½ × 750 × 202 B1 Use KE = ½mv2 for any two of the four elements
KE (initial) = ½ × 1500 × 302 + ½ × 750 × 302
PE gain = 2250 × 10 × 800 × 0.08 B1
WD against friction = 600 × 800 B1
½ × 2250 × 302 + DF × 800 = 600 × 800 M1 Use energy equation.
+ ½ × 2250 × 202 + 2250 × 10 × 800 × 0.08
DF = 1700 N A1 DF = 1696.875 N
5
6(b) 2400 – 600 = 2250a M1 Apply Newton’s second law to the system or to each of the car
or and trailer separately
T – 200 = 750a and 2400 – 400 – T = 1500a
A1 Two correct equations
Attempting to solve for a or for T M1
T = 800 N and a = 0.8 ms–2 A1
4
0.2 kg
A
1m
B 1m
30Å
C
Three points A, B and C lie on a line of greatest slope of a plane inclined at an angle of 30Å to the horizontal, with AB = 1 m and BC = 1 m, as shown in the diagram. A particle of mass 0.2 kg is released from rest at A and slides down the plane. The part of the plane from A to B is smooth. The part of the plane from B to C is rough, with coefficient of friction - between the plane and the particle.
(a) Given that - = 12 3, find the speed of the particle at C.
[8]
(b) Given instead that the particle comes to rest at C, find the exact value of -. [4]
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) 1 M1 Attempt PE or KE for motion from A to B
0.2 ×10 × 0.5 = × 0.2 × vB2
2
M1 Attempt PE loss = KE gain from A to B
vB2 = 10 A1
Alternative method for the first 3 marks
0.2 × 10 × sin 30 = 0.2a, a = 5 (M1) Attempt to find acceleration a for motion from A to B
vB2 = 0 2 + 2 × 5 × 1 (M1) Use v2 = u2 + 2as in attempt to find speed at B
vB2 = 10 (A1)
7(a) THEN, either this method for the next 5 marks
R = 0.2 × 10 × cos 30 = √3 B1
3 3 M1 For using F = µR where R must be a component of 0.2g
F= × 0.2 × × 10 = 1.5
2 2
PE loss = 0.2 × 10 × 0.5 = 1 M1 Attempt to find either PE loss or WD against F from B to C
WD against F = 1.5 × 1
1 1 M1 Apply work-energy equation for motion from B to C as
0.2 ×10 + 0.2 × 10 × 0.5 = 1.5 ×1 + 0.2vC2 KE at B + PE at B = WD against F + KE at C with vB ≠ 0
2 2
vc = 5 = 2.24 ms–1 A1
OR, this method for the next 5 marks
R = 0.2 × 10 × cos 30 = √3 (B1)
3 3 (M1) For using F = µR where R must be a component of 0.2g
F= × 0.2 × × 10 = 1.5
2 2
0.2 × 10 sin 30 – 1.5 = 0.2a a = –2.5 (M1) Attempt to find acceleration a for motion from B to C
vc2 = 10 + 2 × −2.5 × 1 (M1) Use v2 = u2 + 2as in attempt to find vc using vB ≠ 0
vc = 5 = 2.24 ms–1 (A1)
8
7(a) Alternative method for question 7(a)
PE loss = 0.2 × 10 × 2 sin 30 = 2 M1 Attempt PE loss for motion from A to C
1 M1 Attempt KE gain for motion from A to C
KE gain = × 0.2 × vC2
2
Both PE loss and KE gain correct A1
R = 0.2 × 10 × cos 30 = √3 B1
3 3 M1 For using F = µR where R must be a component of 0.2g
F= × 0.2 × × 10 = 1.5
2 2
WD against F = 1.5 × 1 M1 Attempt WD against F
1 M1 Attempt work-energy equation for motion from A to C
0.2 × 10 ×1 = 1.5 ×1 + × 0.2 × vC2
2
vc = 5 = 2.24 ms–1 A1
8
7(b) 0 = 10 + 2a [a = –5] M1 Attempt to find a for motion from B to C, using vB2 = 10 , vC = 0
0.2 × 10 × sin 30 – F = 0.2 × -5 M1 Attempt Newton’s 2nd law for motion from B to C
2=μ 3 M1 Use F = µR where R is a component of 0.2g but R = 0.2g is M0
2 A1 Any correct exact form such as 2/3√3
μ=
3
Alternative method for question 7(b)
PE loss = 0.2 × 10 × 1 sin 30 = 1 M1 Attempt PE loss for motion from B to C
1 + ½ × 0.2 × 10 = F × 1 M1 Work-Energy equation for motion from B to C in the form
PE at B + KE at B = WD against F using vB2 = 10 , vC = 0
F =μ 3 M1 Use F = µR leading to an equation in µ where R is a component of
0.2g
2 A1 Any correct exact form such as 2/3√3
μ=
3
7(b) Alternative method for question 7(b)
PE loss = 0.2 × 10 × 2 sin 30 = 2 M1 Attempt PE loss for motion from A to C
2=F×1 M1 Work-Energy equation for motion from B to C
F =μ 3 M1 Use F = µR leading to an equation in µ where R is a component of
0.2g
2 A1 Any correct exact form such as 2/3√3
μ=
3
4
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Mathematics 9709/41 for October/November 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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