Mathematics 9709/41 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A winch operates by means of a force applied by a rope. The winch is used to pull a load of mass
50 kg up a line of greatest slope of a plane inclined at 60Å to the horizontal. The winch pulls the load a distance of 5 m up the plane at constant speed. There is a constant resistance to motion of 100 N.
Find the work done by the winch. [3]
Mark scheme
1 Force exerted by winch = 50 g sin 60 + 100 [ = 433.0 + 100 = 533.0] M1 For resolving forces along the plane
Work done = 5 × ( 50 g sin 60 + 100 ) M1 Use of WD = Force × distance
Work done = 2670 J A1
Alternative method for Question 1
PE increase = 50 g × 5sin 60 M1 Correct dimensions
Work done = 50 g × 5sin 60 + 100 × 5 M1 Apply the work-energy equation, 3 terms
Work done = 2670 J A1
3
A B m kg 0.1 kg
0.9 m
Two particles A and B have masses m kg and 0.1 kg respectively, where m > 0.1. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley and the particles hang vertically below it. Both particles are at a height of 0.9 m above horizontal ground
(see diagram). The system is released from rest, and while both particles are in motion the tension in the string is 1.5 N. Particle B does not reach the pulley.
(a) Find m. [4]
(b) Find the speed at which A reaches the ground. [2]
Mark scheme
2(a) 0.1 kg particle T − 0.1g = 0.1a M1 Apply Newton’s 2nd law to either the 0.1 kg particle, the m kg
m kg particle mg − T = ma particle or to the system, correct number of terms
System mg − 0.1g = ( m + 0.1) a A1 Two correct equations
Solve for m [ a = 5] M1 From 2 equations with the correct number of relevant terms
m = 0.3 A1
4
2(b) v2 = 0 + 2 × 5 × 0.9 M1 Use of v2 = u 2 + 2as with u = 0 , s = 0.9 and their a ≠ ± g
v = 3 m s–1 A1 FT FT on 1.8a
2
Three particles P, Q and R, of masses 0.1 kg, 0.2 kg and 0.5 kg respectively, are at rest in a straight line on a smooth horizontal plane. Particle P is projected towards Q at a speed of 5 m s−1 . After P and Q collide, P rebounds with speed 1 m s−1 .
(a) Find the speed of Q immediately after the collision with P. [3]
Q now collides with R. Immediately after the collision with Q, R begins to move with speed V m s−1 .
(b) Given that there is no subsequent collision between P and Q, find the greatest possible value of V . [3]
Mark scheme
3(a) Use of conservation of momentum, 3 terms M1 Correct dimensions
0.1 × 5 + 0 = 0.1 × ( −1) + 0.2 × ( ±v ) A1
v = 3 m s–1 A1 A0 for v = −3
3
3(b) 0.2 × their 3 + 0 = 0.2 × u + 0.5V M1 Use of conservation of momentum, 3 terms, correct dimensions.
Allow u = 0 used or if Q and R coalesce
u −1 B1 Allow u = −1 . Allow equality for finding greatest value of V.
Condition for no collision with P, may be a statement.
Greatest V = 1.6 A1 FT FT on their 3 from 3(a) if u = −1 used.
3
Two cyclists, Isabella and Maria, are having a race. They both travel along a straight road with constant acceleration, starting from rest at point A.
Isabella accelerates for 5 s at a constant rate a m s−2 . She then travels at the constant speed she has reached for 10 s, before decelerating to rest at a constant rate over a period of 5 s.
Maria accelerates at a constant rate, reaching a speed of 5 m s−1 in a distance of 27.5 m. She then maintains this speed for a period of 10 s, before decelerating to rest at a constant rate over a period of 5 s.
(a) Given that a = 1.1, find which cyclist travels further. [5]
(b) Find the value of a for which the two cyclists travel the same distance. [2]
Mark scheme
4(a) Isabella v = 5 × 1.1 [ = 5.5] B1 Isabella’s constant speed for 10 seconds
Use of s = ut + ½at2 or use of v–t graph to find total distance M1 For either Isabella or Maria, all sections included but allow one
error in use of formulae
1 1 A1 For correct expression for Isabella, accept unsimplified
sI = × 1.1 × 52 + 10 × 5.5 + × 1.1 × 52 [ = 82.5]
2 2
1
or sI = × ( 20 + 10 ) × 5.5 [ = 82.5]
2
1 A1 For correct expression for Maria, accept unsimplified
sM = 27.5 + 5 × 10 + × 5 × 5 [ = 90 ]
2
Distances for Isabella = 82.5 and Maria = 90, so Maria goes B1
further
5
4(b) 1 1 M1 Attempt total distance travelled by Isabella and set up an equation
a × 52 + 10 × 5a + a × 52 = 90
2 2 for a, using their value of sM = 90. All parts included, allow one
1 error.
or × ( 20 + 10 ) × 5a = 90
2
a = 1.2 A1
2
A particle moving in a straight line starts from rest at a point A and comes instantaneously to rest at a point B. The acceleration of the particle at time t s after leaving A is a m s−2 , where
1 a = 6t 2 − 2t.
(a) Find the value of t at point B. [3]
(b) Find the distance travelled from A to the point at which the acceleration of the particle is again zero. [5]
Mark scheme
5(a) 1 M1 For integration. v = at is M0.
v = 6t 2 − 2t dt
3 A1 Allow unsimplified coefficients.
v = 4t 2 − t 2 ( + c )
1 A1
v = 0 leading to t = 0 or t 2 = 4 leading to t = 16
3
5(b) 1 M1 Attempt to solve a = 0, using valid algebra, reaching t = …
6t 2 − 2t = 0
t=9 A1
3 2 M1 For integration of their expression for v which includes a term with
s = 4t 2 − t dt a fractional power. Allow unsimplified coefficients. v = at is M0
8 5
1 3
s = t 2 − t ( +c )
5 3
5
8 2 1 3 A1 For correct integral
s= t − t
5 3
Distance = 145.8 m B1 729
Allow or 146 to 3s.f.
5
5
y
20 N 10 N
30Å !Å x
O 60Å
25 N
Three coplanar forces of magnitudes 10 N, 25 N and 20 N act at a point O in the directions shown in the diagram.
(a) Given that the component of the resultant force in the x-direction is zero, find !, and hence find the magnitude of the resultant force. [4]
(b) Given instead that ! = 45, find the magnitude and direction of the resultant of the three forces.
[5]
Mark scheme
6(a) 20cos30 = 25cos60 + 10cosα M1 For resolving forces horizontally, all relevant terms included
[17.32 = 12.5 + 10cosα , → cos α = 0.4821]
α = 61.2 A1 From α = 61.18
Resultant = 20sin30 + 10sin 61.2 − 25sin 60 M1 For resolving forces vertically, all relevant terms included
[ = 10 + 8.761 − 21.651]
Magnitude of resultant force = 2.89 N A1 A0 for –2.89 N or for ±2.89 N.
Allow 2.89 N downwards
4
6(b) X = 25cos60 + 10cos 45 − 20cos30 M1 For either horizontal or vertical component, correct number of
= 12.5 + 7.07107 − 17.32051 = 2.25056 relevant terms. Allow ± X and/or ±Y
Y = 20sin 30 + 10sin 45 − 25sin 60
= 10 + 7.07107 − 21.65064 = −4.57957 A1 For both correct, allow unsimplified
R = X 2 +Y2 M1 OE. Using a method to find the resultant force, using expressions
for X and Y with at least 5 relevant terms.
Y M1 OE. A method to find the direction, using expressions for X and Y
α = tan −1 with at least 5 relevant terms.
X
Resultant = 5.10 N, A1 For both correct, angle clearly explained.
Direction = 63.8° below positive x-axis May use a diagram with a correct arrow and arc for angle.
Allow angle 296° (measured anticlockwise from +ve x-axis)
5
P
35 kg
2.5 m
30Å
A slide in a playground descends at a constant angle of 30Å for 2.5 m. It then has a horizontal section in the same vertical plane as the sloping section. A child of mass 35 kg, modelled as a particle P, starts from rest at the top of the slide and slides straight down the sloping section. She then continues along the horizontal section until she comes to rest (see diagram). There is no instantaneous change in speed when the child goes from the sloping section to the horizontal section.
The child experiences a resistance force on the horizontal section of the slide, and the work done against the resistance force on the horizontal section of the slide is 250 J per metre.
(a) It is given that the sloping section of the slide is smooth.
(i) Find the speed of the child when she reaches the bottom of the sloping section. [3]
(ii) Find the distance that the child travels along the horizontal section of the slide before she comes to rest. [2]
(b) It is given instead that the sloping section of the slide is rough and that the child comes to rest on the slide 1.05 m after she reaches the horizontal section.
Find the coefficient of friction between the child and the sloping section of the slide. [6]
Additional Page
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Mark scheme
7(a)(i) PE = 35 g × 2.5sin 30 M1
1 M1 Use of conservation of energy, 2 terms, correct dimensions
× 35v 2 = 35 g × 2.5sin 30
2
v = 5 m s–1 A1
Alternative method for Question 7(a)(i)
mg sin 30 = ma leading to a = 5 M1 For applying Newton’s 2nd law down the plane, 2 terms, correct
dimensions
v 2 = 0 + 2 × 5 × 2.5 M1 For using v2 = u2 + 2as, using their a ≠ ± g
v = 5 m s–1 A1
3
7(a)(ii) 1 M1 Use of work-energy from the bottom of the slide until motion stops,
× 35 × 52 = 250d 2 terms, correct dimensions, using their v
2
d = 1.75 m A1
Alternative method for Question 7(a)(ii)
35 g × 2.5sin 30 = 250d M1 Use of work-energy from the start until motion stops, 2 terms,
correct dimensions.
d = 1.75 m A1
Alternative method for Question 7(a)(ii)
50 M1 Newton’s 2nd law on the horizontal section with resistance = 250 N
−250 = 35a leading to a = − = −7.14
7 to find a and use v2 = u 2 + 2as with v = 0 , u = 5 and s = d .
0 = 52 + 2 ( a ) d
d = 1.75 m A1
2
7(b) 1 B1 Either use the correct work energy equation for motion on the
× 35v 2 = 250 × 1.05 v 2 = 15 horizontal section or use the fact that the frictional force on the
2
horizontal section is 250 N in order to set up an equation that would
or lead to finding the speed at the bottom of the slide.
50
−250 = 35a leading to a = −
7
50
0 = v2 + 2 × − × 1.05 v 2 = 15
7
R = 35 g cos30 [ = 303.11] B1
v 2 = 0 + 2 × a × 2.5 = 15 leading to a = 3 M1 For using v2 = u 2 + 2as , with their v 2 to set up an equation that
or would lead to finding a .
PE change = 35 g × 2.5sin 30 [ = 437.5]
35 g sin 30 − F = 35a or [175 − F = 35a ] M1 For using Newton’s 2nd law down the slope with correct
or dimensions.
or
1
35 g × 2.5sin 30 = F × 2.5 + × 35 × 15 [ 437.5 = F × 2.5 + 262.5] For using energy equation, 3 relevant terms with correct
2 dimensions.
F =μ×R M1 For using F = µR, where R is a component of 35 g .
μ = 0.231 A1 2 3
Allow μ = OE
15
7(b) Alternative method for Question 7(b)
R = 35 g cos30 B1
PE change = 35 g × 2.5sin 30 [ = 437.5] B1
WD against friction on the flat = 250 ×1.05 B1 WD = 262.5
35 g × 2.5sin 30 = F × 2.5 + 250 × 1.05 [ 437.5 = F × 2.5 + 262.5] M1 For using energy equation, 3 relevant terms with correct
dimensions.
F =μ×R M1 For using F = µR at any stage, where R is a component of 35 g .
μ = 0.231 A1 2 3
Allow μ = OE
15
6
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