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Name:
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Y12 Revision booklet:
Forces _______________________
Class:
_

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Date:
_

Time: 71 minutes

Marks: 59 marks

Comments:

Page 1 of 20
Q1.
Lizzie is sat securely on a wooden sledge.

The combined mass of Lizzie and the sledge is M kilograms.

The sledge is being pulled forward in a straight line along a horizontal surface by means
of a light inextensible rope, which is attached to the front of the sledge.

This rope stays inclined at an acute angle θ above the horizontal and remains taut as the
sledge moves forward.

The sledge remains in contact with the surface throughout.

The coefficient of friction between the sledge and the surface is µ and there are no other
resistance forces.

Lizzie and the sledge move forward with constant acceleration, a m s−2

The tension in the rope is a constant T Newtons.

(a) Show that

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(7)

(b) It is known that when M = 30, y = 30°, and T = 40, the sledge remains at rest.

Lizzie uses these values with the relationship formed in part (a) to find the value
for µ

Explain why her value for µ may be incorrect.

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(2)
(Total 9 marks)

Q2.
The three forces F1 , F2 and F3 are acting on a particle.

F1 = (25i + 12j) N
F2 = (−7i + 5j) N
F3 = (15i − 28j) N

The unit vectors i and j are horizontal and vertical respectively.

The resultant of these three forces is F newtons.

(a) (i) Find the magnitude of F, giving your answer to three significant figures.

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Page 3 of 20
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(2)

(ii) Find the acute angle that F makes with the horizontal, giving your answer to
the nearest 0.1°

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(2)

(b) The fourth force, F4 , is applied to the particle so that the four forces are in
equilibrium.
Find F4 , giving your answer in terms of i and j.

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(1)
(Total 5 marks)

Q3.
In this question use g = 9.8 m s−2, giving your final answers to an appropriate
degree of accuracy.

The diagram shows a box, of mass 8.0 kg, being pulled by a string so that the box moves
at a constant speed along a rough horizontal wooden board.

The string is at an angle of 40° to the horizontal.

The tension in the string is 50 newtons.

Page 4 of 20
The coefficient of friction between the box and the board is µ

Model the box as a particle.

(a) Show that µ = 0.83

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(4)

(b) One end of the board is lifted up so that the board is now inclined at an angle of 5°
to the horizontal.

The box is pulled up the inclined board.

The string remains at an angle of 40° to the board.

The tension in the string is increased so that the box accelerates up the board at
3 m s−2

Page 5 of 20
 (i) Draw a diagram to show the forces acting on the box as it moves.
(1)

(ii) Find the tension in the string as the box accelerates up the slope at 3 m s−2.

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(7)
(Total 12 marks)

Q4.
A box, of mass 3 kg, is placed on a rough slope inclined at an angle of 40° to the
horizontal. It is released from rest and slides down the slope.

(a) Draw a diagram to show the forces acting on the box.

Page 6 of 20
 (1)

(b) Find the magnitude of the normal reaction force acting on the box.

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(2)

(c) The coefficient of friction between the box and the slope is 0.2. Find the magnitude
of the friction force acting on the box.

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(2)

(d) Find the acceleration of the box.

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(3)

(e) State an assumption that you have made about the forces acting on the box.

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Page 7 of 20
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(1)
(Total 9 marks)

Q5.
In this question use g = 9.8 m s–2
A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the
horizontal. The bottom of the ramp is at the point O.

A crate of mass 20 kg is at rest at the point A on the ramp.

The crate is pulled up the ramp using a rope attached to the crate.

Once in motion, the rope remains taut and parallel to the line of greatest slope of the
ramp.

(a) The tension in the rope is 230N

The crate accelerates up the ramp at 1.2 m s–2

Find the coefficient of friction between the crate and the ramp.

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Page 8 of 20
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(7)

(b) (i) The crate takes 3.8 seconds to reach the top of the ramp.

Find the distance OA.

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(3)

(ii) Other than air resistance, state one assumption you have made about the
crate in answering part (b)(i).

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(1)
(Total 11 marks)

Q6.
A particle is in equilibrium under the action of four horizontal forces of magnitudes
5 newtons, 8 newtons, P newtons and Q newtons, as shown in the diagram.

Page 9 of 20
 (a) Show that P = 9.

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(3)

(b) Find the value of Q.

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(2)
(Total 5 marks)

Q7.
Three forces act in a vertical plane on an object of mass 250 kg, as shown in the diagram.

Page 10 of 20
The two forces P newtons and Q newtons each act at 80° to the horizontal. The object
accelerates horizontally at a m s–2 under the action of these forces.

(a) Show that

P = 125
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(5)

(b) Find the value of a for which Q is zero.

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Page 11 of 20
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(3)
(Total 8 marks)

Page 12 of 20
Mark schemes

Q1.

Marking Instructions AO Marks Typical Solution

(a) Resolves vertically to form AO3.1b M1

a three term equation
Condone sign error or
sin/cos error

Obtains fully correct AO1.1b A1

equation for resolving
vertically

Uses Newton’s second law AO3.1b M1 R = T sin θ = Mg

horizontally to form a three
term equation T cos θ − F = Ma
Condone sign error or
consistent cos/sin error F = µR

Obtains fully correct AO1.1b A1 T cos θ − µR = Ma

equation for resolving
horizontally T cos θ − µ(Mg − T sin θ)
= Ma
Uses F = µR to replace F AO3.3 B1
with µR in their horizontal T(cos θ + µ sin θ) = Ma +
equation µMg
Eliminates R to form a AO1.1a M1
single equation

Completes rigorous AO2.1 R1

argument to find required
expression.

Must see T as a factor

before division e.g. T(cos
θ + µ sin θ)
AG

(b) Explains that the AO2.4 B1 The sledge is at rest so

relationship may not be the relationship may not
valid because the sledge be valid as friction may not
is at rest be acting at its limiting
value
Identifies that friction may AO3.5b B1
not be at its limiting value
Accept reference to F ≤
µR
Sledge may not be on the
point of slipping

Total 9 marks

Page 13 of 20
Q2.

Marking Instructions AO Marks Typical Solution

(a) Sums the forces given AO1.1b B1

(i) correctly

Uses Pythagoras to find AO1.1b B1

the magnitude of the
vector and obtains correct
magnitude (given to 3 sig
figs)

(ii) Uses trig expression with AO1.1a M1

appropriate values

Obtains correct angle AO1.1b A1

(given to nearest 0.1°)

= 18.4° (3 sf)
OR

= 18.4° (3 sf)
OR

= 18.4° (3 sf)

(b) States negative of ‘their’ AO2.2a B1F F4 = −33i + 11j

part (a)(i)

Total 5 marks

Q3.

Marking Instructions AO Marks Typical Solution

(a) Resolves horizontally and AO3.4 M1 Resolving vertically

vertically to obtain R = 8 × 9.8 − 50sin40°
expressions for F and R Resolving horizontally
(allow consistent mixing of

Page 14 of 20
 sin and cos) F = 50cos 40°
Obtains correct AO1.1b A1
expressions for R and F

States friction model F = AO1.2 B1 F = µR

µR with ‘their’ values for F 50cos40° = µ(8 × 9.8 −
and R 50sin40°)

Completes a rigorous AO2.1 R1 µ = 0.8279660445 = 0.83

argument that results with (2sf)
correct µ AG
AG
Only award if they have a
completely correct
solution, which is clear,
easy to follow and
contains no slips

(b) Draws correct diagram AO3.3 B1

(i) with exactly four forces
showing arrow heads and
labels
Can use Mg or 8g or 78.4
for W

(ii) Resolves perpendicular to AO3.1b M1 R = 8 × 9.8cos 5° − Tsin 40°

the plane resulting in a = 78.1 − Tsin 40°
three term equation
containing; R, 8gcos 5°
and Tsin 40° (or Tcos 50°)
OR
Resolves horizontally and
vertically to obtain
equations of motion in
horizontal and vertical
directions

Obtains correct expression AO1.1b A1 Tcos 40° − 8 × 9.8sin 5° − F

for R =8×3
OR
Obtains correct horizontal Tcos 40° − 8 × 9.8sin5°
and vertical equations − 0.827… × (8 × 9.8cos 5° −
Tsin 40°) = 24
Forms a four term AO3.1b M1
equation of motion parallel
to the plane with correct T = 73.55939193 = 74 (2
terms
(allow sign errors) sf)
OR

Page 15 of 20
 Eliminates R to solve for T

Obtains correct equation AO1.1b A1

of motion
OR
Obtains correct
expression(s) without R
Uses the friction model in AO3.1b dM1
the four term equation of
motion where R is in the
form a − bT and a and b
are positive constants
OR
Uses the friction model in
the horizontal and vertical
equations
Dependent on previous
M1

Solves for T AO1.1b A1

Obtains correct value of T AO3.2a A1F

with 2 sf accuracy.
FT incorrect value found
for T provided both M1
marks and dM1 mark have
been awarded

Total 12 marks

Q4.
(a)

Diagram with exactly three forces showing arrow heads and

labelled. If components are also shown they must use a
different style e.g. dashed lines then they can be ignored.
Friction must be up the slope.

B1
1

(b) (R =)3 × 9.8 cos 40° = 22.5 N

M1: Resolving perpendicular to the slope. Can use sin40°
or cos50° for method mark, with g or 9.8.
A1: Correct normal reaction. Accept
AWRT 22.5 (Note use of 9.81 still gives 22.5 N.)

Page 16 of 20
 M1A1
2

(c) (F =)0.2R = 4.50 N

M1: Use of F = μR.
A1F: Correct friction. Accept 4.5 N or AWRT 4.50.
(Accept 4.51 N from the use of 9.81.)
Follow through incorrect normal reaction from part (b).

M1A1F
2

(d) 3a = 3 × 9.8sin 40° – 4.504

M1: Three term equation of motion with correct terms,
with 3a, either component of weight and their answer to
part (c) for F.
A1F: Equation of motion with correct terms and signs.

M1A1F

a = 4.80 m s–2
A1F: Correct acceleration. Accept 4.8 or AWRT 4.80.
(Note that using 9.81 still gives 4.80 m s–2).
Follow through friction from part (c).

A1F
3

(e) No air resistance force acting

or
No other forces acting on the box.
or
They (forces in the diagram) are the only forces that act.
OR
No turning effect (due to forces).
or
Forces are concurrent. OE
B1: Correct assumption.
Ignore irrelevant comments

B1
1
[9]

Q5.

Marking Instructions AO Marks Typical Solution

(a) Forms a three or four term equation 3.3 M1

of motion using F=ma parallel to
slope
Condone sign errors and swapped

Page 17 of 20
 trig terms

Resolves parallel to the slope to 1.1b B1 Using Newton’s Second Law

obtain m g sin25 or better PI parallel to slope
T – m g sin25 – F = ma

Resolves perpendicular to the slope 1.1b B1 Weight parallel to slope

to obtain R = m g cos25 or better PI = 196 sin 25 = 82.833…

Uses F= μR 3.3 M1 Resolve perpendicular to slope

R = 196 cos 25 = 177.63 …

Obtains single correct equation with 1.1b A1 F = 196μcos25

μ
Condone inclusion of g in place of
9.8

Obtains correct value of μ to at least 1.1b A1 230 – 196 sin 25 – 196 μ cos 25
one significant figure = 20 × 1.2

μ = 0.69

Obtains μ = 0.69 3.2a A1

CAO

(b) Uses 1.1a M1

(i)

Obtains 8.66 or better 1.1b A1

Obtains 10 – their 8.66 1.1b A1F OA = 1.3 m

Accept answer given to at least 1 dp

(ii) States one valid assumption 3.5b E1 The crate is a particle

May also be:
Crate treated as a point mass
The mass of the crate is constant

Total 11 marks

Q6.
(a) P = 5 + 8cos60°
Both relevant forces, component of 8N attempted

M1
All correct

A1

P=9

Page 18 of 20
 CAO

A1
3

(b) Q = 8cos30°
Component of 8N attempted

M1

Q = 6.93 or
AWRT 6.93

A1
2
[5]

Q7.
(a) P cos80° – Q cos80° = 250a
M1: Horizontal equation of motion in the form
P cos80° ± Q cos80° = 250a or
P sin80° ± Q sin80° = 250a
A1: Correct horizontal equation.
M1A1

P sin80° + Q sin80° = 250g

B1: Correct vertical equation.
Note: the above marks could be awarded for a correct
vector equation.
B1

AG

dM1: Solving for P with an attempt to eliminate Q.

dM1

A1: Correct result from correct working.

Must see an expression for 2P or 2P sin80°cos80°
A1
5

(b) P cos80° = 250a

M1: Using Q = 0 into correct original equation(s) or
resolving without Q.
M1

Page 19 of 20
P sin80° = 250g

dM1: Eliminating P
dM1

A1: Correct a.
Note: use of P = ± Q scores M0dM0A0
Note: use of P = 0 can lead to ± 1.73 but scores
M0dM0A0 unless fully justified by a symmetry argument.
A1
3
[8]

Page 20 of 20