1
Friction
A particle P of mass 0.8 kg is placed on a rough horizontal table. The coefficient of friction between
P and the table is -. A force of magnitude 5 N, acting upwards at an angle ! above the horizontal,
where tan ! = 34 , is applied to P. The particle is on the point of sliding on the table.
(i) Find the value of -.
[4]
(ii) The magnitude of the force acting on P is increased to 10 N, with the direction of the force
remaining the same. Find the acceleration of P.
[3]
2
february/march/2016/42/4
Two particles A and B, of masses 0.8 kg and 0.2 kg respectively, are connected by a light inextensible
string. Particle A is placed on a horizontal surface. The string passes over a small smooth pulley P
fixed at the edge of the surface, and B hangs freely. The horizontal section of the string, AP, is of
length 2.5 m. The particles are released from rest with both sections of the string taut.
(i) Given that the surface is smooth, find the time taken for A to reach the pulley.
[5]
(ii) Given instead that the surface is rough and the coefficient of friction between A and the surface
is 0.1, find the speed of A immediately before it reaches the pulley.
[5]
february/march/2016/42/6
3
10 kg
5 kg
!
Two particles of masses 5 kg and 10 kg are connected by a light inextensible string that passes over
a fixed smooth pulley. The 5 kg particle is on a rough fixed slope which is at an angle of ! to the
horizontal, where tan ! = 34 . The 10 kg particle hangs below the pulley (see diagram). The coefficient
of friction between the slope and the 5 kg particle is 12 . The particles are released from rest. Find the
[7]
acceleration of the particles and the tension in the string.
4
may/june/2016/41/5
A particle of mass 30 kg is on a plane inclined at an angle of 20Å to the horizontal. Starting from rest,
the particle is pulled up the plane by a force of magnitude 200 N acting parallel to a line of greatest
slope.
(i) Given that the plane is smooth, find
(a) the acceleration of the particle,
[2]
(b) the change in kinetic energy after the particle has moved 12 m up the plane.
[2]
(ii) It is given instead that the plane is rough and the coefficient of friction between the particle and
the plane is 0.12.
(a) Find the acceleration of the particle.
[4]
(b) The direction of the force of magnitude 200 N is changed, and the force now acts at an angle
[4]
of 10Å above the line of greatest slope. Find the acceleration of the particle.
may/june/2016/41/7
5
TN
20Å
30Å
A block of mass 2.5 kg is placed on a plane which is inclined at an angle of 30Å to the horizontal. The
block is kept in equilibrium by a light string making an angle of 20Å above a line of greatest slope.
The tension in the string is T N, as shown in the diagram. The coefficient of friction between the block
and plane is 14 . The block is in limiting equilibrium and is about to move up the plane. Find the value
of T .
[7]
6
may/june/2016/42/5
A
P
B
0.5 m
A particle A of mass 1.6 kg rests on a horizontal table and is attached to one end of a light inextensible
string. The string passes over a small smooth pulley P fixed at the edge of the table. The other end of
the string is attached to a particle B of mass 2.4 kg which hangs freely below the pulley. The system
is released from rest with the string taut and with B at a height of 0.5 m above the ground, as shown
in the diagram. In the subsequent motion A does not reach P before B reaches the ground.
(i) Given that the table is smooth, find the time taken by B to reach the ground.
[5]
(ii) Given instead that the table is rough and that the coefficient of friction between A and the table is
3 , find the total distance travelled by A. You may assume that A does not reach the pulley. [7]
8
7
may/june/2016/42/7
A particle of mass 15 kg is stationary on a rough plane inclined at an angle of 20Å to the horizontal.
The coefficient of friction between the particle and the plane is 0.2. A force of magnitude X N acting
parallel to a line of greatest slope of the plane is used to keep the particle in equilibrium. Show that the
least possible value of X is 23.1, correct to 3 significant figures, and find the greatest possible value
of X .
[7]
may/june/2016/43/4
8
A particle of mass 0.1 kg is released from rest on a rough plane inclined at 20Å to the horizontal. It is
given that, 5 seconds after release, the particle has a speed of 2 m s−1.
(i) Find the acceleration of the particle and hence show that the magnitude of the frictional force
acting on the particle is 0.302 N, correct to 3 significant figures.
[3]
(ii) Find the coefficient of friction between the particle and the plane.
[2]
october/november/2016/41/2
9
A particle of mass 2 kg is initially at rest on a rough horizontal plane. A force of magnitude 10 N is
applied to the particle at 15Å above the horizontal. It is given that 10 s after the force is applied, the
particle has a speed of 3.5 m s−1 .
(i) Show that the magnitude of the frictional force is 8.96 N, correct to 3 significant figures.
(ii) Find the coefficient of friction between the particle and the plane.
10
[3]
[3]
october/november/2016/42/1
A particle of mass m kg is resting on a rough plane inclined at 30Å to the horizontal. A force of
magnitude 10 N applied to the particle up a line of greatest slope of the plane is just sufficient to stop
the particle sliding down the plane. When a force of 75 N is applied to the particle up a line of greatest
slope of the plane, the particle is on the point of sliding up the plane. Find m and the coefficient of
friction between the particle and the plane.
[6]
october/november/2016/42/5
11
A box of mass 50 kg is at rest on a plane inclined at 10Å to the horizontal.
(i) Find an inequality for the coefficient of friction between the box and the plane.
[2]
In fact the coefficient of friction between the box and the plane is 0.19.
(ii) A girl pushes the box with a force of 50 N, acting down a line of greatest slope of the plane, for
a distance of 5 m. She then stops pushing. Use an energy method to find the speed of the box
when it has travelled a further 5 m.
[5]
The box then comes to a plane inclined at 20Å below the horizontal. The box moves down a line of
greatest slope of this plane. The coefficient of friction is still 0.19 and the girl is not pushing the box.
(iii) Find the acceleration of the box.
12
[2]
october/november/2016/43/7
PN
0.6 kg
21Å
13
A particle of mass 0.6 kg is placed on a rough plane which is inclined at an angle of 21Å to the
horizontal. The particle is kept in equilibrium by a force of magnitude P N acting parallel to a line of
greatest slope of the plane, as shown in the diagram. The coefficient of friction between the particle
and the plane is 0.3. Show that the least possible value of P is 0.470, correct to 3 significant figures,
and find the greatest possible value of P.
[6]
february/march/2017/42/3
A particle of mass 0.8 kg is projected with a speed of 12 m s−1 up a line of greatest slope of a rough
plane inclined at an angle of 10Å to the horizontal. The coefficient of friction between the particle and
the plane is 0.4.
(i) Find the acceleration of the particle.
[4]
(ii) Find the distance the particle moves up the plane before coming to rest.
[2]
may/june/2017/41/2
14
P
0.8 kg
A
B
1.2 kg
60Å
30Å
As shown in the diagram, a particle A of mass 0.8 kg lies on a plane inclined at an angle of 30Å to the
horizontal and a particle B of mass 1.2 kg lies on a plane inclined at an angle of 60Å to the horizontal.
The particles are connected by a light inextensible string which passes over a small smooth pulley P
fixed at the top of the planes. The parts AP and BP of the string are parallel to lines of greatest slope
of the respective planes. The particles are released from rest with both parts of the string taut.
(i) Given that both planes are smooth, find the acceleration of A and the tension in the string.
[6]
(ii) It is given instead that both planes are rough, with the same coefficient of friction, -, for both
particles. Find the value of - for which the system is in limiting equilibrium.
[6]
may/june/2017/41/7
15
PN
30Å
40Å
A particle of mass 0.12 kg is placed on a plane which is inclined at an angle of 40Å to the horizontal.
The particle is kept in equilibrium by a force of magnitude P N acting up the plane at an angle of 30Å
above a line of greatest slope, as shown in the diagram. The coefficient of friction between the particle
and the plane is 0.32. Find the set of possible values of P.
[8]
16
A
may/june/2017/42/5
P
B
1Å
The diagram shows a fixed block with a horizontal top surface and a surface which is inclined at an
angle of 1Å to the horizontal, where sin 1 = 35 . A particle A of mass 0.3 kg rests on the horizontal
surface and is attached to one end of a light inextensible string. The string passes over a small smooth
pulley P fixed at the edge of the block. The other end of the string is attached to a particle B of mass
1.5 kg which rests on the sloping surface of the block. The system is released from rest with the string
taut.
(i) Given that the block is smooth, find the acceleration of particle A and the tension in the string.
[5]
(ii) It is given instead that the block is rough. The coefficient of friction between A and the block is
- and the coefficient of friction between B and the block is also -. In the first 3 seconds of the
motion, A does not reach P and B does not reach the bottom of the sloping surface. The speed
[9]
of the particles after 3 s is 5 m s−1 . Find the acceleration of particle A and the value of -.
may/june/2017/42/6
17
B
4 kg
A
m kg
0.5 m
30Å
Two particles A and B of masses m kg and 4 kg respectively are connected by a light inextensible
string that passes over a fixed smooth pulley. Particle A is on a rough fixed slope which is at an angle
of 30Å to the horizontal ground. Particle B hangs vertically below the pulley and is 0.5 m above the
ground (see diagram). The coefficient of friction between the slope and particle A is 0.2.
(i) In the case where the system is in equilibrium with particle A on the point of moving directly up
the slope, show that m = 5.94, correct to 3 significant figures.
[6]
(ii) In the case where m = 3, the system is released from rest with the string taut. Find the total
distance travelled by A before coming to instantaneous rest. You may assume that A does not
reach the pulley.
[8]
may/june/2017/43/7
18
A
1Å
B
2.5 N
25Å
Two particles A and B of masses 0.9 kg and 0.4 kg respectively are attached to the ends of a light
inextensible string. The string passes over a fixed smooth pulley which is attached to the top of two
inclined planes. The particles are initially at rest with A on a smooth plane inclined at angle 1Å to the
horizontal and B on a plane inclined at angle 25Å to the horizontal. The string is taut and the particles
can move on lines of greatest slope of the two planes. A force of magnitude 2.5 N is applied to B
acting down the plane (see diagram).
(i) For the case where 1 = 15 and the plane on which B rests is smooth, find the acceleration of B.
[5]
(ii) For a different value of 1, the plane on which B rests is rough with coefficient of friction between
the plane and B of 0.8. The system is in limiting equilibrium with B on the point of moving in
the direction of the 2.5 N force. Find the value of 1.
[5]
19
october/november/2017/41/7
Aparticleofmass0.2kgisrestinginequilibriumonaroughplaneinclinedat20Åtothehorizontal.
(i) Show that the friction force acting on the particle is 0.684 N, correct to 3 significant figures. [1]
The coefficient of friction between the particle and the plane is 0.6. A force of magnitude 0.9 N is
applied to the particle down a line of greatest slope of the plane. The particle accelerates down the
plane.
(ii) Find this acceleration.
[4]
october/november/2017/42/1
P
20
!
Q
Two particles P and Q, each of mass m kg, are attached to the ends of a light inextensible string. The
string passes over a fixed smooth pulley which is attached to the edge of a rough plane. The plane is
7 . Particle P rests on the plane and particle Q
inclined at an angle ! to the horizontal, where tan ! = 24
hangs vertically, as shown in the diagram. The string between P and the pulley is parallel to a line of
greatest slope of the plane. The system is in limiting equilibrium.
(i) Show that the coefficient of friction between P and the plane is 34 .
[5]
A force of magnitude 10 N is applied to P, acting up a line of greatest slope of the plane, and P
accelerates at 2.5 m s−2 .
(ii) Find the value of m.
[5]
october/november/2017/42/6
21
A particle is released from rest and slides down a line of greatest slope of a rough plane which is
inclined at 25Å to the horizontal. The coefficient of friction between the particle and the plane is 0.4.
(i) Find the acceleration of the particle.
[4]
(ii) Find the distance travelled by the particle in the first 3 s after it is released.
[2]
october/november/2017/43/3
22
A particle of mass 12 kg is on a rough plane inclined at an angle of 25Å to the horizontal. A force of
magnitude P N acts on the particle. This force is horizontal and the particle is on the point of moving
up a line of greatest slope of the plane. The coefficient of friction between the particle and the plane
is 0.8. Find the value of P.
[6]
february/march/2018/42/4
P
23
0.8 kg
45Å
A
B
1.2 kg
30Å
The diagram shows a triangular block with sloping faces inclined to the horizontal at 45Å and 30Å.
Particle A of mass 0.8 kg lies on the face inclined at 45Å and particle B of mass 1.2 kg lies on the face
inclined at 30Å. The particles are connected by a light inextensible string which passes over a small
smooth pulley P fixed at the top of the faces. The parts AP and BP of the string are parallel to lines
of greatest slope of the respective faces. The particles are released from rest with both parts of the
string taut. In the subsequent motion neither particle reaches the pulley and neither particle reaches
the bottom of a face.
(i) Given that both faces are smooth, find the speed of A after each particle has travelled a distance
of 0.4 m.
[6]
(ii) It is given instead that both faces are rough. The coefficient of friction between each particle and
a face of the block is -. Find the value of - for which the system is in limiting equilibrium. [6]
may/june/2018/41/7
24
A particle of mass 20 kg is on a rough plane inclined at an angle of 60Å to the horizontal. Equilibrium
is maintained by a force of magnitude P N acting on the particle, in a direction parallel to a line of
greatest slope of the plane. The greatest possible value of P is twice the least possible value of P.
Find the value of the coefficient of friction between the particle and the plane.
[7]
may/june/2018/42/5
25
1.6 kg
A
2.5 m
P
B
2.4 kg
1m
30Å
As shown in the diagram, a particle A of mass 1.6 kg lies on a horizontal plane and a particle B of
mass 2.4 kg lies on a plane inclined at an angle of 30Å to the horizontal. The particles are connected
by a light inextensible string which passes over a small smooth pulley P fixed at the top of the inclined
plane. The distance AP is 2.5 m and the distance of B from the bottom of the inclined plane is 1 m.
There is a barrier at the bottom of the inclined plane preventing any further motion of B. The part BP
of the string is parallel to a line of greatest slope of the inclined plane. The particles are released from
rest with both parts of the string taut.
(i) Given that both planes are smooth, find the acceleration of A and the tension in the string.
[5]
(ii) It is given instead that the horizontal plane is rough and that the coefficient of friction between A
and the horizontal plane is 0.2. The inclined plane is smooth. Find the total distance travelled
by A.
[9]
may/june/2018/42/7
26
A particle of mass 3 kg is on a rough plane inclined at an angle of 20Å to the horizontal. A force of
magnitude P N acting parallel to a line of greatest slope of the plane is used to keep the particle in
equilibrium. The coefficient of friction between the particle and the plane is 0.35. Show that the least
possible value of P is 0.394, correct to 3 significant figures, and find the greatest possible value of P.
[6]
27
may/june/2018/43/5
A particle is projected from a point P with initial speed u m s−1 up a line of greatest slope PQR of a
rough inclined plane. The distances PQ and QR are both equal to 0.8 m. The particle takes 0.6 s to
travel from P to Q and 1 s to travel from Q to R.
(i) Show that the deceleration of the particle is 23 m s−2 and hence find u, giving your answer as an
exact fraction.
[6]
(ii) Given that the plane is inclined at 3Å to the horizontal, find the value of the coefficient of friction
between the particle and the plane.
[4]
28
october/november/2018/41/6
A block of mass 5 kg is being pulled by a rope up a rough plane inclined at 6Å to the horizontal. The
rope is parallel to a line of greatest slope of the plane and the block is moving at constant speed. The
coefficient of friction between the block and the plane is 0.3. Find the tension in the rope.
[4]
october/november/2018/42/2
29
0.4 kg
P
Q
0.7 kg
!
Two particles P and Q, of masses 0.4 kg and 0.7 kg respectively, are attached to the ends of a light
inextensible string. The string passes over a fixed smooth pulley which is attached to the edge of a
rough plane. The coefficient of friction between P and the plane is 0.5. The plane is inclined at an
angle ! to the horizontal, where tan ! = 34 . Particle P lies on the plane and particle Q hangs vertically.
The string between P and the pulley is parallel to a line of greatest slope of the plane (see diagram).
A force of magnitude X N, acting directly down the plane, is applied to P.
(i) Show that the greatest value of X for which P remains stationary is 6.2.
[4]
(ii) Given instead that X = 0.8, find the acceleration of P.
[4]
october/november/2018/42/4
30
50 N
20Å
A block is pushed along a horizontal floor by a force of magnitude 50 N which acts at an angle of
20Å to the horizontal (see diagram). The coefficient of friction between the block and the floor is 0.3.
Given that the speed of the block is constant, find the mass of the block.
[5]
october/november/2018/43/2
31
2.5 N
P
15Å
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]
32
february/march/2019/42/1
A block of mass 3 kg is at rest on a rough plane inclined at 60Å to the horizontal. A force of magnitude
15 N acting up a line of greatest slope of the plane is just sufficient to prevent the block from sliding
down the plane.
(i) Find the coefficient of friction between the block and the plane.
[5]
The force of magnitude 15 N is now replaced by a force of magnitude X N acting up the line of greatest
slope.
(ii) Find the greatest value of X for which the block does not move.
[2]
october/november/2019/41/3
33
A block of mass 3 kg is initially at rest on a rough horizontal plane. A force of magnitude 6 N is
applied to the block at an angle of 1 above the horizontal, where cos 1 = 24
. The force is applied for
25
a period of 5 s, during which time the block moves a distance of 4.5 m.
(i) Find the magnitude of the frictional force on the block.
[4]
(ii) Show that the coefficient of friction between the block and the plane is 0.165, correct to
3 significant figures.
[3]
(iii) When the block has moved a distance of 4.5 m, the force of magnitude 6 N is removed and the
block then decelerates to rest. Find the total time for which the block is in motion.
[4]
34
october/november/2019/42/6
A crate of mass 500 kg is being pulled along rough horizontal ground by a horizontal rope attached to
a winch. The winch produces a constant pulling force of 2500 N and the crate is moving at constant
speed. Find the coefficient of friction between the crate and the ground.
[3]
october/november/2019/43/1
35
A particle P of mass 0.4 kg is on a rough horizontal floor. The coefficient of friction between P and
the floor is -. A force of magnitude 3 N is applied to P upwards at an angle ! above the horizontal,
where tan ! = 34 . The particle is initially at rest and accelerates at 2 m s−2 .
(a) Find the time it takes for P to travel a distance of 1.44 m from its starting point.
[2]
(b) Find -.
[4]
36
ferbruary/march/2020/42/2
30Å
TN
The diagram shows a ring of mass 0.1 kg threaded on a fixed horizontal rod. The rod is rough and the
coefficient of friction between the ring and the rod is 0.8. A force of magnitude T N acts on the ring
in a direction at 30Å to the rod, downwards in the vertical plane containing the rod. Initially the ring
is at rest.
(a) Find the greatest value of T for which the ring remains at rest.
[4]
(b) Find the acceleration of the ring when T = 3.
[3]
37
may/june/2020/41/4
TN
60Å
2.5 kg
20Å
A particle of mass 2.5 kg is held in equilibrium on a rough plane inclined at 20Å to the horizontal by a
force of magnitude T N making an angle of 60Å with a line of greatest slope of the plane (see diagram).
The coefficient of friction between the particle and the plane is 0.3.
Find the greatest and least possible values of T .
[8]
may/june/2020/42/3
38
A
3m kg
2m kg
1
B
0.8 m
Two particles A and B, of masses 3m kg and 2m kg respectively, are attached to the ends of a light
inextensible string. The string passes over a fixed smooth pulley which is attached to the edge of a
plane. The plane is inclined at an angle 1 to the horizontal. A lies on the plane and B hangs vertically,
0.8 m above the floor, which is horizontal. The string between A and the pulley is parallel to a line of
greatest slope of the plane (see diagram). Initially A and B are at rest.
(a) Given that the plane is smooth, find the value of 1 for which A remains at rest.
[3]
It is given instead that the plane is rough, 1 = 30Å and the acceleration of A up the plane is 0.1 m s−2 .
1 3.
(b) Show that the coefficient of friction between A and the plane is 10
[5]
(c) When B reaches the floor it comes to rest.
Find the length of time after B reaches the floor for which A is moving up the plane. [You may
assume that A does not reach the pulley.]
[4]
may/june/2020/43/7
39
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]
october/november/2020/41/7
40
A block of mass 5 kg is placed on a plane inclined at 30Å to the horizontal. The coefficient of friction
between the block and the plane is -.
(a)
40 N
5 kg
30Å
Fig. 6.1
When a force of magnitude 40 N is applied to the block, acting up the plane parallel to a line of
greatest slope, the block begins to slide up the plane (see Fig. 6.1).
Show that - < 15 3.
[4]
(b)
40 N
5 kg
30Å
Fig. 6.2
When a force of magnitude 40 N is applied horizontally, in a vertical plane containing a line of
greatest slope, the block does not move (see Fig. 6.2).
Show that, correct to 3 decimal places, the least possible value of - is 0.152.
41
[4]
october/november/2020/42/6
A string is attached to a block of mass 4 kg which rests in limiting equilibrium on a rough horizontal
table. The string makes an angle of 24Å above the horizontal and the tension in the string is 30 N.
(a) Draw a diagram showing all the forces acting on the block.
[1]
(b) Find the coefficient of friction between the block and the table.
[5]
october/november/2020/43/3
42
2 kg A
B
P
10Å
Q
3 kg
20Å
As shown in the diagram, particles A and B of masses 2 kg and 3 kg respectively are attached to the
ends of a light inextensible string. The string passes over a small fixed smooth pulley which is attached
to the top of two inclined planes. Particle A is on plane P, which is inclined at an angle of 10Å to the
horizontal. Particle B is on plane Q, which is inclined at an angle of 20Å to the horizontal. The string
is taut, and the two parts of the string are parallel to lines of greatest slope of their respective planes.
(a) It is given that plane P is smooth, plane Q is rough, and the particles are in limiting equilibrium.
Find the coefficient of friction between particle B and plane Q.
[5]
(b) It is given instead that both planes are smooth and that the particles are released from rest at the
same horizontal level.
Find the time taken until the difference in the vertical height of the particles is 1 m. [You should
assume that this occurs before A reaches the pulley or B reaches the bottom of plane Q.]
[6]
october/november/2020/43/7
Answers
1
february/march/2016/42/4
(i)
[F = 4]
M1
For resolving forces horizontally
Allow use of α = 36.9o throughout
R + 5sin α = 8
[R = 5]
M1
For resolving forces vertically
4 = 5µ
M1
For using F = µR
µ = 0.8
A1
R + 10sin α = 8
and
F = 0.8 × R
(ii)
2
5cos α = F
4
[R = 2]
B1
For resolving forces vertically to find the
new value of R
and using F = µR
10cos α – F = 0.8a
M1
For resolving horizontally
a = 8 ms–2
A1
[F =1.6]
3
february/march/2016/42/6
[T = 0.8a
for A
2 – T = 0.2a for B
0.2g = (0.2 + 0.8)a system]
(i
For applying Newton’s 2nd law either to
particle A or to particle B or to the system
M1
For applying N2 to a second particle (if
needed) and solving for a
M1
[a = 2]
A1
[2.5 = ½ × 2 × t2]
A complete method for finding t such as
using s = ut + ½at2
M1
t = 1.58 s
5
A1
Allow t =
1
10
2
First Alternative Method for (i)
(i)
[0.2 × g × 2.5 or ½(0.2 + 0.8)v2]
M1
Finding PE loss or KE gain (system)
[0.2 × g × 2.5 = ½(0.2 + 0.8)v2]
M1
Using PE loss = KE gain and find v
[v2 = 10]
A1
[2.5 = ½ (0 + √10)t]
M1
t = 1.58 s
A1
For using s = ½(u + v)t
5
Allow t =
1
10
2
Second Alternative Method for (i)
(i)
M1
[T × 2.5 = ½ (0.8) v2]
M1
2
(ii)
Apply N2 to A and B and solve for T
[T = 0.8a
2 – T = 0.2a
→ T = 1.6 N]
[v = 10]
A1
[2.5 = ½ (0 + √10)t]
M1
t = 1.58 s
A1
N = 8 and F = 0.1 × N = 0.8
B1
T – 0.8 = 0.8a and 2 – T = 0.2a
or 0.2g – 0.8 = (0.2 + 0.8)a
M1
a = 1.2
A1
v2 = 0 + 2 × 1.2 × 2.5
M1
v = √6 = 2.45 ms
–1
A1
Use WD by T = KE gain by A, find v
Using s = ½(u + v)t
5
Allow t =
1
10
2
For applying N2 to both particles or to the
system and solving for a
For using v2 = u2 + 2as
5
First Alternative Method for (ii)
(ii)
N = 8 and F = 0.1 × N = 0.8
B1
[0.2 ×g × 2.5 =
½ (0.8 + 0.2) v2 + 0.8 × 2.5]
v = √6 = 2.45 ms–1
M1
Apply work/energy to the system as
PE loss =
KE gain + WD against resistance
A1
Correct Work/Energy equation
M1
For solving for v
5
A1
Second Alternative Method for (ii)
(ii)
3
N = 8 and F = 0.1 × N = 0.8
B1
T – 0.8 = 0.8a and 2 – T = 0.2a
M1
T = 1.76 N
A1
[T × 2.5 = 0.8 × 2.5 + ½ (0.8) v2]
M1
v = √6 = 2.45 ms–1
A1
Apply Work/Energy equation to A
5
may/june/2016/41/5
R = 5g cos α = 4g
F = 0.5 × 4g = 2g
4
Use N2 for A and B and solve for T
B1
For finding the normal reaction R
acting on the 5 kg particle and
using F = µR
M1
For applying Newton’s second
law to one or both particles or to
the system
T – 2g – 5gsin α = 5a→
T – 5g = 5a
A1
10g – T = 10a
A1
[5g = 15a]
M1
a = g/3 = 3.33 ms–2
A1
T = 10g – 10(g/3)
= 20g/3 = 66.7 N
B1
System equation is
10g –5g sin α –2g = 5g = 15a
For eliminating T and solve for a
[7]
may/june/2016/41/7
(i) (a)
(b)
200 – 30g sin 20 = 30a
M1
a = 3.25 ms–2
A1
[v2 = 2 × 3.2465 × 12 = 77.9]
M1
KE change = 0.5 × 30 × 77.9 = 1170 J
A1
For applying Newton’s second
law with 3 terms parallel to the
plane
[[2]
2
a = 3.2465]
For using v2 = u2 + 2as and
attempting to find KE change
[2]
Alternative method for (i)(b)
[KE = 1168.7 J]
(b)
(ii) (a)
(b)
5
KE change =
200 × 12 – 30g × 12 sin 20
M1
Using KE gain =
WD by DF – PE gain
KE change = 1170 J
A1
N = 30g cos 20
B1
[N = 281.9]
F = 0.12 × 30g cos 20 [= 33.8]
M1
Using F = µNa
200 – 30g sin 20 – 33.8 = 30a
M1
For using Newton’s second law
with 4 terms applied to the
particle
a = 2.12 ms–2
A1
N + 200 sin 10 = 30g cos 20
[N = 247.2]
M1
For resolving forces
perpendicular to the plane. Three
term equation.
F = 0.12 N [= 0.12 × 247.2 = 29.66]
M1
N must be from a 3 term equation
200 cos 10 – 29.66 – 30g sin 20 = 30a
M1
For using Newton’s second law
with 4 terms applied to the
particle
a = 2.16 ms–2
A1
[2]
[4]
[4]
may/june/2016/42/5
For resolving forces perpendicular to
the plane
(3 term equation)
M1
R + T sin 20 = 2.5g cos 30
A1
F = 0.25 × R
B1
May be implied
M1
For resolving forces parallel to the
plane (3 term equation)
T cos 20 = F + 2.5g sin 30
A1
For solving and obtaining T
M1
T = 17.5
A1
7
Alternative scheme
F = 0.25 × R
T cos 50 = F cos 30 + R sin 30
B1
May be implied
M1
For resolving forces horizontally
(3 term equation)
A1
For resolving forces vertically
(4 term equation)
M1
R cos 30 + T sin 50 = F sin 30 + 2.5g
A1
For solving and obtaining T
M1
T = 17.5
A1
7
6
may/june/2016/42/7
(i)
[2.4g– T= 2.4aT = 1.6a
or the system equation
2.4g = (1.6 + 2.4)a]
M1
For applying Newton’s second law to
one of the particles or to the
combined system
M1
For applying Newton’s second law to
a second particle if needed and/or
solving for a
a = 6 ms–2
A1
0.5 = ½ × 6 × t2
M1
t = 0.408 s
A1
For using s = ut +½at2
5
Accept t =√6/6
Alternative for (i)
(i)
(ii)
[PE loss = 2.4 × g × 0.5 = 12
KE gain = ½(1.6 + 2.4)v2 = 2v2]
M1
For attempting to find PE and KE as
B reaches the ground
[12= 2v2]
M1
Using PE loss = KE gain
v2 = 6→v = 2.45 ms–1
A1
[0.5 = ½ × (0 + 2.45) × t]
M1
t = 0.408 s
A1
R = 1.6g = 16andF = 3/8 R = 6
B1
System is
[2.4g – 6 = (1.6 + 2.4)a]
M1
For using Newton’s second law for
both particles or the system
2.4g – T = 2.4aandT – 6 = 1.6a
A1
Both or system equation
[a = 4.5]
M1
For finding a and using
v2 = u2 + 2as to find v as B reaches
the ground
v = 2 × 4.5× 0.5 = 4.5 = 2.12 ms–1
A1
–6 = 1.6a → a = –3.75 ms–2
M1
Using s = ½(u + v)t
5
For finding the deceleration of A and
using v2 = u2 + 2as to find s the total
distance travelled by A
0 = 4.5 + 2 × (–3.75) × (s – 0.5)
s = 1.1 m
A1
Accept t =√6/6
7
First Alternative for (ii)
(ii)
R = 1.6g = 16andF = 3/8 R = 6
PE loss = 2.4 × g × 0.5[= 12]
KE gain = ½ × (1.6 + 2.4) × v2[= 2v2]
B1
M1
For attempting PE loss and KE gain
as B reaches the ground
A1
For both PE and KE correct
M1
For using PE loss =
KE gain + WD against F
12 = 2v2 + 6 × 0.5 → v2 = 4.5 → v =
2.12
A1
Loss of KE = WD against F
M1
For considering the motion of A after
B reaches the ground to find s the
total distance travelled
[½ × 1.6 × 4.5 = 6 × (s – 0.5)]
s = 1.1 m
A1
7
7
8
may/june/2016/43/4
R = 15gcos20o
B1
140.95
F = µR = 0.2 × 15gcos20o
B1
28.19
M1
For resolving parallel to the
plane (F acting up plane)
X + 0.2 × 15gcos20o =
15gsin20o
A1
Least value of X is 23.1
A1
AG
[X= 15gsin20o +
0.2 × 15gcos20o]
M1
For resolving parallel to the
plane (F acting down plane)
Greatest value of X is 79.5
A1
7
october/november/2016/41/2
(i)
2 = 5a → a = 0.4 ms-2
B1
For applying Newton’s 2nd law
to the particle
[0.1g sin20 – F = 0.1 × 0.4]
M1
F = 0.302 N
(ii)
AG
A1
[3]
For attempting to find R and
using µ = F/R
[R = 0.1g cos20 ( = 0.9397)]
M1
µ = 0.3020/0.9397 = 0.321
9
A1
[2]
october/november/2016/42/1
(i)
3.5 = 10a
→
a = 0.35 ms-2
B1
Allow a = 3.5 / 10
M1
For applying Newton’s 2nd law to
the particle
[10cos15 – F = 2 × 0.35]
F = 8.96 N
AG
A1
[3]
Alternative to (i)
s = ½ (0 + 3.5) × 10 = 17.5 m
B1
Distanced moved in 10 secs
M1
Work done by 10 N force
= WD against F + KE gain
[10cos15 × 17.5= F × 17.5 + ½ 2 (3.5)2]
F = 8.96 N
(ii)
AG
A1
[3]
[R = 2 g – 10sin15]
M1
Resolving forces vertically
[µ = 8.96 / (2g – 10sin15)]
M1
Using F = µR
µ = 0.515
A1
[3]
10
october/november/2016/42/5
F = µmgcos30
B1
[10 + F – mgsin30 = 0]
M1
Resolving up, first case
[75 – F – mgsin30 = 0]
M1
Resolving up, second case
M1
Either attempt to solve for m
or
Solve a pair of two 3 term
simultaneous equations
for either m or µ
[85 = 2mgsin30]
or
[10 + µmgcos30 – mgsin30 = 0
75 – µmgcos30 – mgsin30 = 0]
11
m = 8.5 kg or µ = 0.442
A1
µ = 0.442 or m = 8.5 kg
B1
[6]
[2]
µ⩾F÷R
october/november/2016/43/7
(i)
R = 50 g cos 10° and
F = 50 g sin 10°
B1
µ ⩾ 0.176
B1
(ii) PE loss = 50g × dsin10o
WD against friction =
0.19 × 50 g cos10° × d
Allow µ ⩾ tan 10°
B1
d = 5 or d = 10
B1
d =5 or d = 10
M1
For using WD by 50 N force + PE loss –
WD against friction = KE gain
50 × 5 + 50 g × 10 sin 10° – 0.19 ×
50 g cos 10° × 10 = 0.5 × 50v2
A1
Speed is 2.70 ms–1
A1
[5]
SC for candidates using Newton’s Second
law: max 2/5
B1 v = 2.94 ms–1 after 5 m
B1 Speed is 2.70 ms–1
12
(iii) 50 g sin 20o –
0.19 × 50 g cos 20o = 50 a
M1
Acceleration is 1.63 ms−2
A1
For using Newton’s Second Law
[2]
february/march/2017/42/3
R = 0.6g cos 21 [= 5.60]
B1
F = 0.3R = 1.8 cos 21 [= 1.68]
M1 Using F = µR
P + F = 6 sin 21[ = 2.15]
M1 Slipping down
P = 2.15 – 1.68 = 0.470
AG
A1 Least possible value
P – F = 6 sin 21
M1 Slipping up
P = 2.15 + 1.68 = 3.83
A1 Greatest possible value
Total:
6
13
may/june/2017/41/2
(i)
R = 0.8g cos 10 [= 7.88]
B1
F = 0.4 × 8 cos 10 [= 3.15]
M1 Use F = µR
–8 sin 10 – 3.2 cos 10 = 0.8a
M1 Newton 2 along the plane
a = –5.68 ms–2
A1
Total:
(ii)
4
0 = 12 2 – 2 × 5.68 × s
M1 Using v2 = u2 + 2as
s = 144/(2 × 5.68) = 12.7 m
A1
Total:
14
2
may/june/2017/41/7
(i)
M1 Resolve along the plane for either A or for B or for the system
T – 0.8g sin 30 = 0.8a
1.2g sin 60 – T = 1.2a
1.2g sin 60 – 0.8g sin 30 = 2a]
T − 4 = 0.8a
For A
A1
A1 System equation is
6 3 − 4 = 6.4 = 2a
For B
6 3 − T = 10.4 − T = 1.2a
M1 Solve for a or T
A1
a = 3 3 − 2 = 3.20 ms–2
T=
(
)
A1
12
1 + 3 = 6.56 N
5
Total:
(ii)
6
RA = 0.8g cos 30 = 4 3
R = 1.2g cos60 = 6
B1 For either RA or RB
FA = 4 3 µ and FB = 6µ
M1 Either FA or FB used
B
M1 Resolve parallel to the plane for both particles A and B or
system
A1 System equation is
12 sin 60 – 8 sin 30 – 6µ – 4√3 µ = 0
12 sin 60 – 6µ – T = 0
or
T – 8 sin 30 – 4√3 µ = 0
M1 Eliminate T and/or find µ
(
)(
µ = 6 √ 3 − 4 / 6 + 4√ 3
)
A1
= 0.494
Total:
15
6
may/june/2017/42/5
M1 Resolve perpendicular to the plane, three terms
R + P sin 30 = 0.12g cos 40
A1 R does not need to be the subject
F = 0.32R
M1 Use F = µR
[Pmin cos 30 + F = 0.12g sin 40]
M1 About to slip down, 3 terms
[Pmax cos 30 – F = 0.12g sin 40]
M1 About to slip up, 3 terms
[P cos 30 = 0.12g sin 40
±0.32 (0.12g cos 40 – P sin 30)]
OR
[P cos 30 ± 0.32R = 0.12g sin 40
R + P sin 30 = 0.12g cos 40]
Must reach P =… in either method
M1 Substitute for F and solve for P in either case, 4 terms
OR solve a pair of simultaneous equations (each with 3
terms) in R and P for P in one of the cases
Pmax = 1.04 Pmin = 0.676
A1 For either correct
0.676 ⩽ P ⩽ 1.04
A1
Total:
8
16
may/june/2017/42/6
(i)
M1 Apply Newton’s second law to A or to B or to the system
A [T = 0.3a]
B [1.5g sin θ – T = 1.5a]
System [1.5g sin θ = 1.8a]
A1 Any two correct equations
M1 Solve 2 simultaneous equations for a and/or T or use the
system equation.
a = 9/1.8 = 5 ms–2
A1
T = 1.5 N
A1
Total:
(ii)
5
[5 = 3a]
M1 v = u + at used with t = 3, u = 0, v = 5
a = 5/3 = 1.67
A1
RA = 3 RB = 15 cos 36.9 = 12
B1 For either reaction
[FA = 3µ FB = 12µ]
M1 Use F = µR for either term
EITHER:
A [T – FA = 0.3a]
B [15 sin 36.9 – T – FB = 1.5a]
System equation is
[1.5g sin 36.9 – FA – FB = 1.8a]
(M1 Apply Newton’s second law to A or to B or to the system
A2/1/0 A1 Correct equation for A or B
A2 Correct equations for A and B
OR A2 Correct system equation
[9 – 15µ = 3]
M1 Solve for µ from equations with correct number of terms
µ = 0.4 = 2/5
A1)
OR:
s = ½ (5/3) × 32 = 7.5
(B1 Find distance travelled in 3 secs
PE loss =
1.5 × 10 × 7.5 × (3/5) = 67.5
B1
KE gain = ½ (1.8) × 52 = 22.5
B1
[67.5 = 22.5 + 3µ × 7.5 + 12µ × 7.5]
M1 Use Work/Energy equation
µ = 2/5 = 0.4
A1)
Total:
17
9
may/june/2017/43/7
(i)
R = mg cos30
B1 Resolves normally
F = 2m cos 30 [= m√3]
M1 Uses F = µR
T = 4g [= 40]
B1 Particle B
T = mgsin30 + F
M1 Resolves parallel to plane for particle A
40 = 5m + m√3
A1 Equation in m
m=
A1 AG All correct and no errors seen
40
= 5.94
5 +√3
Total:
6
(ii)
EITHER:
[R = 3g cos30]
F = 0.2 × 3g cos30 (3√3 = 5.196)
(B1
4g – T = 4a
or T – 3gsin30 – F = 3a
or
g – 3gsin30 –F =7a
M1 Applies Newton’s Second Law to one of the
particles
or forms system equation in a
(mBg – mAgsin30 – F = (mA + mB)a)
T – 3gsin30 – 3√3 = 3a
or 4 – T = 4a
or
g – 3gsin30 – 3√3 = 7a →a = …
M1 Applies Newton’s Second Law to form second
equation in T and a and solves for a
or solves system equation for a
25 − 3√ 3
7
= 2.83.
A1
a=
v2 = 2 × 2.83 × 0.5
v = 1.68…
B1 FT v as T becomes zero
FT on a
–3gsin30 – 0.2(3gcos30) = 3a
–15 – 3√3 = 3a
→a = …(–5–√3 = –6.73)
M1 Applies Newton’s Second Law and solves for a
0 = 1.682 – 2 × 6.73s
s = …(0.210)
M1 Uses v2 = u2 + 2as and solves for s
Total distance = 0.710 m
A1)
OR:
[R = 3g cos30]
F = 0.2 × 3g cos30
(B1
(3√3 = 5.196)
M1 For 4kg mass, uses
PE loss – WDT = KE gain
M1 For 3kg mass, uses
WDT = KE gain + PE gain + WDFr
4g(0.5) – 0.5T = ½ (4v2) and
0.5T = ½ (3v2) + 3g(0.5sin30) + 3√3(0.5)
A1
v2 = (25 – 3√3)/7 or v = 1.68
B1
½ (3)(1.68)2 = 3g(s sin30) + 3√3s
M1 For 3kg mass, uses
KE loss = PE gain + WDFr
s = …(0.210)
M1 Solves for s
Total distance = 0.710 m
A1)
Total:
18
8
october/november/2017/41/7
7(i)
M1 For applying Newton’s 2nd law to either
particle
(correct number of terms)
T – 0.9 g sin 15 = 0.9a
A1
2.5 + 0.4 g sin 25 – T = 0.4a
A1
1.3a = 1.86…
M1 Solving simultaneously for a
a = 1.43 m s–2
A1
5
7(ii)
F = 0.8 × 0.4g cos 25
B1
2.5 + 0.4 g sin 25 – T – F = 0
M1 For using equilibrium of forces acting on
particle B with 4 terms
T – 0.9 g sin θ = 0
M1 For using equilibrium of forces acting on
particle A with 2 terms
M1 For solving for θ
θ = 8.2º
A1
5
19
october/november/2017/42/1
1(i)
F = 0.2g sin 20 = 0.684 N
B1 AG
1
1(ii)
R = 0.2g cos 20
B1
F = µR [= 0.6 × 0.2g cos 20]
M1 Using F = µR
[0.9 + 0.2g sin 20 – F = 0.2a]
M1 Use of Newton’s 2nd law along the plane
(4 relevant terms)
a = 2.28 ms-2
A1
F = 1.1276…
4
20
october/november/2017/42/6
6(i)
R = mg cos α
(R = 9.6m)
B1 Allow use of α = 16.3º throughout
[T = mg
F = mg sin α + T ]
M1 For resolving forces on P and Q and
eliminating T or for considering the
equilibrium of the system
F = mg sin α + mg
A1 (F = 12.8m)
M1 For use of F = µR
Coefficient of friction = 1⅓ =
4
3
A1 AG so must be from exact working
5
6(ii)
EITHER:
P equation is
10 – mg sin α – F – T = 2.5 m
(*M1 For applying Newton’s 2nd law to
P (5 terms) or Q (3 terms)
Q equation is
T – mg = 2.5m
*M1 For applying Newton’s 2nd law to the
other particle and eliminate T
10 – mg sin α – µmg cos α
– mg = 2m (2.5)
A1 If evaluated then this is
10 – 2.8m – 12.8m – 10m = 5m
DM1 For solving this equation for m as far as
m = Dependent on one or other of the
previous M marks having been scored
m = 0.327
OR:
[10 – mg sin α –F – mg = m(2.5 + 2.5)]
A1)
Allow m =
50
153
(*M1 For applying Newton’s 2nd law to the
system. Allow with 5 terms
*M1 System equation with all 6 terms
10 – mg sin α – µmg cos α
– mg = 2m (2.5)
A1
DM1 For solving this equation for m as far as
m = Dependent on one or other of the
previous M marks having been scored
m = 0.327
A1)
Allow m =
50
153
5
21
october/november/2017/43/3
3(i)
R = mg cos 25
B1
[F = 0.4mg cos 25]
M1 Using F = µR
[mg sin 25 – 0.4mg cos 25 = ma]
M1 Use of Newton’s Second Law
a = 0.601 ms–2
A1
4
3(ii)
[s = ½ × 0.601 × 32]
Distance = 2.70 m
M1 Use of s = ut + ½at2
A1 FT FT 4.5 × a from (i)
2
22
february/march/2018/42/4
R = 12g cos 25 + P sin 25
P cos 25 = F + 12g sin 25]
or
[P = F cos 25 + R sin 25
R cos 25 = F sin 25 + 12g]
M1 Attempt resolving of forces in any
one direction, parallel to,
perpendicular to plane
or
horizontally, vertically
A1 Any one correct equation
A1 Any second correct equation
F = 0.8R
M1 Use of F = µR
Complete method to find P from 2
equations(3 terms each)
M1
P = 242
A1
6
23
may/june/2018/41/7
(i)
M1 Apply Newton 2nd law to either A or to B
or to the system
A T – 0.8 g sin 45 = 0.8a
B 1.2g sin 30 – T = 1.2a
System 1.2 g sin 30 – 0.8 g sin 45 = 2a
A1 One correct equation
A1 A second correct equation
a = 0.171
M1 Solve for a
v2 = 2 × a × 0.4
M1 Use v2 = u2 + 2as with u = 0
v = 0.370 so speed of A is 0.370 ms–1
A1
6
Alternative scheme for Question 7(i)
M1 Attempt KE gain or PE loss
KE gain =
A1 v is the required speed of A
1
1
× 0.8 × v2 +
× 1.2 × v2
2
2
PE loss =
1.2 g × 0.4 sin 30 – 0.8 g × 0.4 sin 45
A1
1
1
× 0.8 × v2 +
× 1.2 × v2 =
2
2
1.2 g × 0.4 sin 30 – 0.8 g × 0.4 sin 45
M1 4 term energy equation
M1 Solving for v
v = 0.370 so speed of A is 0.370 ms–1
(ii)
A1
B1 For either RA or RB
RA = 0.8g cos 45 = 4 2
RB = 1.2g cos30 = 6 3
FA = 4 2 µ and FB = 6 3 µ
M1 Either FA or FB used
A 0.8 g sin 45 + FA = T
B 1.2 g sin 30 – FB = T
or system equation:
12 sin 30 – 8 sin 45 = FA + FB
M1 Resolve parallel to the plane either for
both particles A and B or for the system
equation
Correct equation(s)
A1
M1 Eliminate T and solve for µ
µ=
(6 − 4 2 )
(6
3 + 4√ 2
A1
)
= 0.0214
6
24
may/june/2018/42/5
R = 20g cos 60 [= 100]
B1
F = µ × 20g cos 60 [= 100µ]
M1 Use F = µR
M1 Resolve along plane in either case
(Pmax =) 20g sin 60 + F
A1 One correct equation
(Pmin =) 20g sin 60 – F
A1 Second correct equation
20g sin 60 + F = 2(20g sin 60 – F)
M1 Use of Pmax = 2Pmin to give four term equation in F or µ or P
µ=
3
= 0.577
3
A1
7
lternative solution for final 3 marks if Pmin is taken as acting down the plane
Pmin = F – 20g sin 60
A1
20g sin 60 + F = 2(F – 20g sin 60)
M1
µ = 3 3 = 5.196
A1
25
may/june/2018/42/7
(i)
M1 Attempt Newton’s 2nd law for A or B or for the system
[T = 1.6a, 2.4g sin 30 – T = 2.4a]
System is 2.4g sin 30 = 4a
A1 Two correct equations
M1 Solve for a or T
a=3
A1
T = 4.8
A1
5
(ii)
Friction force on A is
F = 0.2 × 1.6g [= 3.2]
B1 From F = µR
T – F = 1.6a
2.4g sin 30 – T = 2.4a
System is 2.4g sin 30 – F = 4a
M1 Attempt Newton’s 2nd law for both particles or for the system
A1 Correct equations for A and B or correct system equation
M1 Attempt to solve for a
a = 2.2
A1
v = 2 × 2.2 × 1
M1 Attempt to find v or v2 when B reaches the barrier
Subsequent acceleration of A is –2
B1
4.4 = 2 × 2 × s
M1 Attempt to find distance A travels while decelerating to v = 0
Total distance travelled is 2.1 m
A1
2
9
(ii)
Alternative method for Q [Work-Energy applied to A and B]
F = 0.2 × 1.6g [= 3.2]
B1 From F = µR = 0.2 × 1.6g = 3.2
M1 Attempt PE loss as B reaches the barrier
PE loss = 2.4g sin 30 [= 12]
A1
M1 Attempt KE gain for both A and B
KE gain =
[2.4g sin 30 =
2
A1
1
(1.6 + 2.4)v2 [= 2v2]
2
M1 Apply work-energy equation for the motion until B reaches
the barrier (Three relevant terms)
1
× 4 × v2 + 3.2 × 1]
2
[v = 4.4]
KE loss =
[
B1 Find KE loss as A comes to rest after B has stopped
1
× 1.6 × 4.4
2
M1 Apply work-energy equation where d is the extra distance
travelled by A leading to a positive value for d
1
× 1.6 × 4.4 = 3.2d]
2
[d = 1.1]
Total distance = 2.1 m
(ii)
A1 Distance = d + 1
Alternative scheme for first 6 marks of (ii) [Work-energy applied to A]
Friction = 0.2 × 1.6g [= 3.2]
B1
[2.4g sin 30 – T = 2.4a
T – F = 1.6a]
M1 Apply Newton’s 2nd law to A and B and solve for T
T = 6.72
A1
1
[ × 1.6 × v2]
2
M1 Attempt KE for A only
1
× 1.6 × v2 + 3.2 × 1]
[6.72 × 1 =
2
M1 Use work/energy equation for A
A1 Correct KE for A
Alternative scheme for first 6 marks of (ii) [Work-energy applied to B]
Friction = 0.2 × 1.6g [= 3.2]
B1
[2.4g sin 30 – T = 2.4a
T – F = 1.6a]
M1 Apply Newton’s 2nd law to A and B and solve for T
T = 6.72
A1
M1 Find energy loss/gain for B
Allow either term
±(
1
× 2.4 × v2 – 2.4g sin 30)
2
2.4g sin 30 =
1
× 2.4 × v2 + 6.72 × 1
2
A1
M1 Use work/energy equation for B
26
may/june/2018/43/5
R = 3 g cos 20°
B1 Correct normal reaction stated or used
[ F = 0.35 × 3 g cos 20° ]
M1 For use of F = µ R
[ P1 + F = 3 g sin 20° ]
M1 Attempted resolving equation for minimum case
P1 = 0.394 (AG)
A1 Correct given answer from correct work
[ P2 = F + 3 g sin 20° ]
M1 Attempted resolving equation for maximum case
P2 = 20.1 (N)
A1
Total:
27
6
october/november/2018/41/6
(i)
M1
1
For using constant acceleration equations such as s = ut + at 2 or
2
equivalent complete methods to find expressions for PQ or QR or
PR
For PQ
0.8 = 0.6u + 0.18a
A1
For PR
1.6 = 1.6u + 1.28a
A1 or for QR 0.8 = (u + a × 0.6) × 1 + 0.5a
M1 Solving simultaneously two relevant equations in u and a
Deceleration =
u=
A1
A1 A
2 –2
ms
3
B1
23
15
6
(ii)
R = mg cos 3
B1
F = µmg cos 3
M1 For use of F = µR
 2
− mg sin 3 − µ × mg cos3 = m ×  − 
 3
M1 For using Newton’s second law (3 terms)
µ = 0.0144 (0.014350…)
A1
4
28
october/november/2018/42/2
R = 5g cos 6
B1
[F = 0.3 × 5g cos 6]
M1 Use of F = µR
[T = 5g sin 6 + F]
M1 For resolving along the plane
T = 20.1 N (20.14425...)
A1
4
29
october/november/2018/42/4
(i)
T = 0.7g
B1
R = 0.4g × 4/5 [ = 16/5 = 3.2]
B1 Normal reaction on particle P
[X + 0.4g × 3/5 – F – T = 0]
M1 Attempt to resolve forces along the plane
X = 6.2
A1 AG
4
(ii)
0.7g – T = 0.7a]
[T – 0.8 – 0.4g × 3/5 – F = 0.4a]
[0.7g – 0.8 – 0.4g × 3/5 – F = (0.7 + 0.4)a] System
M1 For using Newton’s 2nd law for both particle P and particle Q or the
system equation
A1 Both equations correct or system equation correct
M1 Solve either the system equation or solve two simultaneous equations to
find a
a = 2 m s–2
A1
4
30
october/november/2018/43/2
R = mg + 50sin20
B1
[F = 0.3(mg + 50sin20)]
M1 Use of F = µR
M1 Resolving horizontally
50cos20 – 0.3(mg + 50sin20) = 0
A1ft ft R (R containing term in m)
m = 14.0 kg (13.9514…)
A1
5
31
february/march/2019/42/1
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
32
october/november/2019/41/3
(i)
R = 3 g cos 60
B1
Use F = µR
M1
[3gsin 60 – µ3gcos60 – 15 = 0]
M1 Resolve forces parallel to the plane, 3 terms
A1 Correct equation
µ = 0.732
A1 Allow µ = 3 − 1
5
(ii)
[Maximum force = 3gsin60 + F
= 3 sin60 + µ3gcos60]
M1
X = 37(.0)
A1 Allow X = 15 2 3 −1
(
)
2
33
october/november/2019/42/6
(i)
1
4.5 = 0 + × a × 52
2
M1
a = 0.36
A1
6×
24
− F = 3 × 0.36
25
F = 4.68 N
1
For use of s = ut + at 2 to find a
2
M1 Resolving horizontally. Allow use of θ = 16.3
A1
4
(ii)
R = 3g − 6sin16.3 = 3g − 6 ×
7
25
[ = 28.32]
B1
4.68 = µ × 28.32
M1 Use of F = µR
µ = 0.165 (0.165254…)
A1
AG. Allow µ =
39
236
3
v = 5 × 0.36 [= 1.8]
(iii)
or v =
( 2 × 0.36 × 4.5) [ = 1.8]
3a = –0.165 × 3g
0 = 1.8 – 0.165gt
B1FT For velocity at t = 5 ft on their a from 6(i)
M1 Using Newton’s second law with new frictional force
(t = 1.09)
Total time = 5 + 1.09 = 6.09 s
M1 Using constant acceleration equations which would lead to a
positive value of t
A1
4
34
october/november/2019/43/1
F=µ×500g
B1 Use of F=µR
[2500=µ×500g]
M1 Resolving horizontally
µ=0.5
A1
3
35
ferbruary/march/2020/42/2
(a)
[1.44 = 0 + ½ × 2t2]
M1 For using a complete method which would lead to an equation for
finding a value of t such as s = ut + ½ at2 with u = 0, s = 1.44 and
a=2
t = 1.2 s
A1
2
(b)
R = 0.4g – 3 ×
[3 ×
4
5
3
5
= 0.4g – 3 sin 36.9 [= 2.2]
– F = 3 cos 36.9 – F = 0.4 × 2] [F = 1.6]
B1
M1 Use Newton’s 2nd law, 3 terms, to find F.

3 × 54 − 0.4 × 2 1.6 
=
μ =

0.4g − 3 × 53
2.2 

M1
μ = 0.727
A1
Use of μ =
Allow μ =
F
R
8
11
4
36
may/june/2020/41/4
4(a)
Resolving forces in either direction
M1
R = T sin30 + 0.1g, F = T cos30
A1
T cos30 = 0.8 (T sin30 + 0.1g)
M1
T = 1.72 (1.7166...)
A1
4
4(b)
R = 3sin30 + 0.1g
B1
3 cos30 – 0.8(3sin30 + 0.1g) = 0.1a
M1
a = 5.98 ms–2 (5.9807...)
A1
3
37
may/june/2020/42/3
T sin 60 + R = 25cos 20
B1
Attempt at resolving in any direction
M1
T cos 60 = F + 25 sin 20
A1
T cos 60 + F = 25 sin 20
A1
Use of F = μ R
M1
T cos 60 = 25sin 20 ± 0.3(25cos 20 − T sin 60)
25sin 20 ± 0.3 × 25cos 20
cos 60 ± 0.3sin 60
M1
T = 6.26
A1
T = 20.5
A1
T=
8
38
may/june/2020/43/7
(a)
T – 2mg = 0
B1
3mg sin θ – T = 0
(M1 for resolving forces parallel to the plane and solving for θ)
M1
θ = 41.8 (41.810...)
A1
3
(b)
R = 3mgcos30
B1
Use of F = μR
M1
2mg – T = 0.1 × 2m OR T – 3mg sin30 –μ × 3mg cos30 = 0.1 × 3m
M1
2mg – 0.2m – 3mg sin30 – μ × 3mg cos30 = 0.1 × 3m
M1
μ=
A1
3
10
5
(c)
v2 = 0 + 2 × 0.1 × 0.8
M1
(v = 0.4)
–3mg sin30 – μ × 3mg cos30 = 3ma (a = –6.5)
M1
0 = –0.4 – 6.5t
M1
t = 0.4/6.5 = 0.0615 s
A1
4
39
october/november/2020/41/7
(a)
1
0.2 ×10 × 0.5 = × 0.2 × vB2
2
M1 Attempt PE or KE for motion from A to B
vB2 = 10
A1
M1 Attempt PE loss = KE gain from A to B
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)
(a)
THEN, either this method for the next 5 marks
R = 0.2 × 10 × cos 30 = √3
F=
3
3
× 0.2 ×
× 10 = 1.5
2
2
B1
M1 For using F = µR where R must be a component of 0.2g
PE loss = 0.2 × 10 × 0.5 = 1
WD against F = 1.5 × 1
M1 Attempt to find either PE loss or WD against F from B to C
1
1
0.2 ×10 + 0.2 × 10× 0.5 =1.5 ×1 + 0.2vC2
2
2
M1 Apply work-energy equation for motion from B to C as
KE at B + PE at B = WD against F + KE at C with vB ≠ 0
vc =
5 = 2.24 ms–1
A1
OR, this method for the next 5 marks
R = 0.2 × 10 × cos 30 = √3
F=
3
3
× 0.2 ×
× 10 = 1.5
2
2
(B1)
(M1) For using F = µR where R must be a component of 0.2g
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
(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
KE gain = × 0.2 × vC2
2
M1 Attempt KE gain for motion from A to C
Both PE loss and KE gain correct
A1
R = 0.2 × 10 × cos 30 = √3
B1
F=
3
3
× 0.2 ×
× 10 = 1.5
2
2
M1 For using F = µR where R must be a component of 0.2g
WD against F = 1.5 × 1
M1 Attempt WD against F
1
0.2 × 10 ×1 =1.5×1+ × 0.2× vC2
2
M1 Attempt work-energy equation for motion from A to C
5 = 2.24 ms–1
vc =
A1
8
(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
μ=
2
A1 Any correct exact form such as 2/3√3
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
μ=
2
3
A1 Any correct exact form such as 2/3√3
(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
μ=
2
A1 Any correct exact form such as 2/3√3
3
4
40
october/november/2020/42/6
(a)
R = 5g cos 30 [= 25√3]
B1
40 – 5g sin 30 – F > 0
M1 State that the net force up the plane is positive, 3 terms
F = µ × 5g cos 30
M1 For using F = µR with R as a component of 5g to obtain an
equality/inequality in µ only with 3 terms
µ<
1
3
5
A1 AG
Alternative scheme for question 6(a)
R = 5g cos 30 [= 25√3]
B1
40 – 5g sin 30 – F = 5a
M1 Acceleration a > 0
F = µ × 5g cos 30
[40 – 5g sin 30 – µ × 5g cos 30 = 5a]
M1 For using F = µR with R as a component of 5g to obtain an equality
in µ and a
µ<
1
3
5
A1
AG. From µ =
1
a
3 = cos30 with a > 0
5
g
4
(b)
Attempt to resolve forces parallel to or perpendicular to the inclined
plane, 3 relevant terms in either direction
M1
R = 5g cos 30 + 40 sin 30 [= 20 + 25√3 = 63.3]
A1
F = 40 cos 30 – 5g sin 30 [= 20√3 – 25 = 9.64]
A1
µ ⩾ 0.152
B1 AG. Using F ⩽ µR
Alternative method for question 6(b)
41
Attempt to resolve forces horizontally or vertically with 3 relevant
terms in either direction
M1
40 = R sin 30 + F cos 30 [40 = ½R + √3/2F]
A1
5g = R cos 30 – F sin 30 [5g = √3/2R – ½F]
A1
µ ⩾ 0.152
B1 AG. Solve for R and F and use F ⩽ µR
october/november/2020/43/3
(a)
B1 4 forces, labelled
1
(b)
For resolving horizontally or vertically
M1
30 cos 24 = F
(F = 27.406…)
A1
R + 30 cos 24 = 40 (R = 27.797…)
A1
μ=
30cos 24
40 − 30sin 24
μ = 0.986 (0.9859…)
M1 Using µ = F/R
A1
5
42
october/november/2020/43/7
(a)
T = 2g sin 10] or [3g sin 20 = F + T]
M1 Resolve forces parallel to plane P for particle A or parallel to
plane Q for Particle B
T = 2g sin 10 and 3g sin 20 = F + T
A1
R = 30 cos 20 (= 28.19...)
B1 Resolving forces perpendicular to plane Q for particle B
μ=
3g sin 20 − 2 g sin10
30cos 20
µ = 0.241 (=0.2407…)
M1 Using µ = F/R
A1
5
(b)
3g sin 20 – T = 3a or T – 2g sin 10 = 2a
or System: 3g sin 20 – 2g sin 10 = 5a
a=
( 3g sin 20 − 2 g sin10 )
5
M1 For applying Newton’s second law to either A or to B or to the
system
M1 For applying Newton’s second law to the second particle and/or
solving for a
a = 1.3575…
A1
h1 = x sin 20
h2 = x sin 10
x sin 20 + x sin 10 = 1
B1 Using expressions for height change of each particle after each
moves a distance x along the plane, to obtain equation in x
1
1
= 0 + × 1.3575 × t 2
sin10 + sin 20
2
M1 For using s = ut + ½at2 for either particle with s = x, u = 0 and
using their a (= 1.3575)
t = 1.69
A1
6