M8 Forces and Newton's laws — revision question pack

6 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section M8. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

M8.1 · Understand the concept of a force; understand and use Newton's first law.

Explanation

A force is an interaction with magnitude and direction; common modelled forces include weight, normal reaction, tension, thrust, compression and resistance.
Draw a force diagram for the chosen body, including only forces acting on that body, then resolve forces along useful perpendicular directions.
Newton's first law says that a particle remains at rest or moves with constant velocity when the resultant force on it is zero.
A common error is to infer that a moving particle must have a forward resultant force; constant non-zero velocity also means zero resultant force.

Worked example

A powered trolley moves in a straight line with constant velocity. Its motor exerts a forward force of $8\,\text{N}$ . Find the resistance force and explain your answer.

1.Constant velocity means zero acceleration and hence, by Newton's first law, zero resultant force.
2.The resistance must therefore balance the motor force, so it is $8\,\text{N}$ backwards.

Answer: Resistance $=8\,\text{N}$ opposite to the motion; Constant velocity means the resultant force is zero

Common mistakes

Don't balance only the horizontal forces and declare equilibrium while a vertical resultant remains.
Don't assume motion implies a resultant force even when the velocity is constant.

Exam tip

For constant velocity, invoke zero acceleration and balance the forces with equal magnitudes in opposite directions.

Tier 1 · Easy

1.
A book rests on a horizontal table. Its weight is $35\,\text{N}$ . State the magnitude and direction of the normal reaction on the book and the resultant force on it.
(2)
(Total for Question 1 is 2 marks)
2.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A probe is moving through space with velocity $(5\mathbf{i}-2\mathbf{j})\,\text{m s}^{-1}$ . No resultant force acts on it. State its velocity $12\,\text{s}$ later.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1.
A book of mass $2.5\,\text{kg}$ rests on a horizontal table. Using $g=9.8\,\text{m s}^{-2}$ , find the weight of the book and the normal force exerted by the table, explaining why the forces have equal magnitudes.
(3)
(Total for Question 1 is 3 marks)
2.
A parachutist of weight $735\,\text{N}$ descends vertically at constant speed. Find the magnitude and direction of the air resistance. Explain why the parachutist can be moving although the resultant force is zero.
(3)
(Total for Question 2 is 3 marks)
3.
At one instant a train is stationary at a signal. A student concludes that the resultant force on the train is zero because its velocity is zero at that instant. Explain why the observation is not sufficient to support this conclusion. State the possible forms of motion for an object on which the resultant force is zero.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
A particle moves with constant velocity under three coplanar forces. Two of the forces are $(6\mathbf{i}-8\mathbf{j})\,\text{N}$ and $(-2\mathbf{i}+5\mathbf{j})\,\text{N}$ . Find the third force, giving also its magnitude.
(4)
(Total for Question 1 is 4 marks)
2.
A maintenance platform moves with constant velocity under two cable forces and a resistance force. One cable pulls with force $40\,\text{N}$ due east. The other pulls with force $30\,\text{N}$ at $60^\circ$ north of east. Take $\mathbf{i}$ and $\mathbf{j}$ to be unit vectors due east and due north respectively. Find the resistance force as a vector in terms of $\mathbf{i}$ and $\mathbf{j}$ , its exact magnitude, and its direction to 3 significant figures.
(5)
(Total for Question 2 is 5 marks)
3.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. Three constant forces $(4\mathbf{i}+7\mathbf{j})\,\text{N}$ , $(-9\mathbf{i}+2\mathbf{j})\,\text{N}$ and $(5\mathbf{i}-9\mathbf{j})\,\text{N}$ act on a particle. Initially its position vector is $(2\mathbf{i}-5\mathbf{j})\,\text{m}$ and its velocity is $(3\mathbf{i}+2\mathbf{j})\,\text{m s}^{-1}$ . Show that the particle moves with constant velocity. Find its position vector at time $t$ and the point at which its path crosses the line $y=7$ .
(5)
(Total for Question 3 is 5 marks)
4.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A platform moves with constant velocity while forces $(7\mathbf{i}+3\mathbf{j})\,\text{N}$ and $(-3\mathbf{i}+5\mathbf{j})\,\text{N}$ act on it. A thruster exerts $P\mathbf{i}\,\text{N}$ , where $P>0$ , and a cable exerts $-T(\tfrac35\mathbf{i}+\tfrac45\mathbf{j})\,\text{N}$ , where $T>0$ . Find $P$ and $T$ , and explain why the platform's non-zero velocity can remain unchanged. Give all numerical answers exactly.
(5)
(Total for Question 4 is 5 marks)
5.
A raft moves with constant velocity. One engine force has magnitude $10\,\text{N}$ and acts at $60^\circ$ north of east. A second engine force has magnitude $P\,\text{N}$ and acts at $30^\circ$ south of east. The water resistance acts due west. Find the exact value of $P$ and the magnitude of the water resistance. Give all numerical answers exactly.
(5)
(Total for Question 5 is 5 marks)

M8.2 · Understand and use Newton's second law for motion in a straight line (forces in two perpendicular directions or simple 2-D vectors); extend to situations where forces need to be resolved (restricted to 2 dimensions).

Explanation

Newton's second law is $\sum\mathbf{F}=m\mathbf{a}$ ; in two dimensions it gives one scalar equation in each resolved direction.
Choose axes parallel and perpendicular to the motion where possible, resolve every force onto those axes, then apply $\sum F=ma$ separately.
For motion along a fixed smooth plane, perpendicular acceleration is zero, so the perpendicular force equation determines the normal reaction while the parallel equation determines acceleration.
A common error is to write $F=ma$ for one force instead of the resultant, or to reverse signs for only some components after choosing a positive direction.

Resolve forces in perpendicular directions, then apply

\sum\mathbf F=m\mathbf a

component by component.

Worked example

A $5\,\text{kg}$ particle is on a smooth plane inclined at $30^\circ$ to the horizontal. A force of $40\,\text{N}$ pulls it up the line of greatest slope. Using $g=9.8\,\text{m s}^{-2}$ , find its acceleration.

1.Resolve up the plane.
2.The component of weight down the plane is $5(9.8)\sin30^\circ=24.5\,\text{N}$ .
3.Hence the resultant up the plane is $40-24.5=15.5\,\text{N}$ .
4.From $15.5=5a$ , $a=3.1\,\text{m s}^{-2}$ up the plane.

Answer: $3.1\,\text{m s}^{-2}$ up the plane

Common mistakes

Don't include the normal reaction in the equation parallel to an inclined plane.
Don't resolve weight using the wrong trigonometric component relative to the inclined plane.

Exam tip

Draw a force diagram, choose axes parallel and perpendicular to the plane, then resolve every force consistently.

Tier 1 · Easy

1.
A constant resultant force of $12\,\text{N}$ acts on a particle of mass $3\,\text{kg}$ . Find its acceleration.
(2)
(Total for Question 1 is 2 marks)
2.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A resultant force $(9\mathbf{i}-12\mathbf{j})\,\text{N}$ acts on a particle of mass $3\,\text{kg}$ . Find its acceleration.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A $4\,\text{kg}$ particle moves on a smooth horizontal surface. A force of $20\,\text{N}$ acts at $30^\circ$ above the horizontal. Using $g=9.8\,\text{m s}^{-2}$ , find the horizontal acceleration and the normal reaction.
(5)
(Total for Question 1 is 5 marks)
2.
A $6\,\text{kg}$ sledge moves on a smooth horizontal surface. It is pulled by a force of $32\,\text{N}$ acting at $25^\circ$ above the horizontal, while a horizontal resistance of $5\,\text{N}$ opposes the motion. Using $g=9.8\,\text{m s}^{-2}$ , find the acceleration and the normal reaction, giving both answers to 3 significant figures.
(5)
(Total for Question 2 is 5 marks)
3.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. Constant forces $(10\mathbf{i}+4\mathbf{j})\,\text{N}$ and $(-2\mathbf{i}+8\mathbf{j})\,\text{N}$ act on a particle. During $4\,\text{s}$ its velocity changes by $(4\mathbf{i}+6\mathbf{j})\,\text{m s}^{-1}$ . Find the mass of the particle.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A $6\,\text{kg}$ particle is pulled up a smooth plane inclined at $20^\circ$ to the horizontal by a force of $50\,\text{N}$ acting at $15^\circ$ above the plane. Using $g=9.8\,\text{m s}^{-2}$ , find the normal reaction and the particle's acceleration up the plane.
(7)
(Total for Question 1 is 7 marks)
2.
A $3\,\text{kg}$ particle lies on a smooth plane inclined at angle $\alpha$ to the horizontal, where $\tan\alpha=\dfrac34$ . A horizontal force of magnitude $P\,\text{N}$ acts on the particle towards the upward side of the plane. The particle accelerates up the plane at $1.5\,\text{m s}^{-2}$ . Using $g=9.8\,\text{m s}^{-2}$ , find $P$ and the normal reaction, giving both answers to 3 significant figures.
(6)
(Total for Question 2 is 6 marks)
3.
A particle moves on a smooth horizontal plane under two forces. One force has magnitude $10\,\text{N}$ and acts due east. The other has magnitude $P\,\text{N}$ and acts at $120^\circ$ anticlockwise from east. The particle's acceleration has magnitude $2\,\text{m s}^{-2}$ and direction $60^\circ$ north of east. Find $P$ and the mass of the particle.
(5)
(Total for Question 3 is 5 marks)
4.
A $5\,\text{kg}$ particle rests on a smooth horizontal plane and is pulled by a force of magnitude $P\,\text{N}$ acting upwards and forwards at an angle $\alpha$ to the horizontal, where $\sin\alpha=3/5$ . Take $g=9.8\,\text{m s}^{-2}$ . While the particle remains in contact with the plane, its horizontal acceleration is $6\,\text{m s}^{-2}$ . Find $P$ and the normal reaction. The force is then increased, its direction unchanged. Find the exact value of $P$ at which contact with the plane is just lost and the horizontal acceleration at that instant. Give all numerical answers exactly.
(7)
(Total for Question 4 is 7 marks)
5.
A $4\,\text{kg}$ particle moves upwards while remaining in contact with a smooth vertical wall. A force of magnitude $P\,\text{N}$ acts upwards and towards the wall at an angle $\alpha$ above the horizontal, where $\tan\alpha=3/4$ . The particle's acceleration is $2\,\text{m s}^{-2}$ upwards. Take $g=9.8\,\text{m s}^{-2}$ . Find the exact value of $P$ and the normal reaction from the wall. If the particle starts from rest, find its speed after $3\,\text{s}$ . Give all numerical answers exactly.
(6)
(Total for Question 5 is 6 marks)

M8.3 · Understand and use weight and motion in a straight line under gravity; gravitational acceleration, g, and its value in S.I. units to varying degrees of accuracy.

Explanation

Weight is the gravitational force $W=mg$ , directed vertically downwards; mass is measured in kilograms and does not depend on location.
The value of $g$ is not universal and depends on location; use the stated value, or the Edexcel default $9.8\,\text{m s}^{-2}$ when none is supplied.
For free motion near Earth's surface in the constant- $g$ model, acceleration is $g$ downwards, but a support or tension changes the resultant acceleration.
A common error is to call mass a force or to assume that the normal reaction always equals weight when the body has vertical acceleration.

Worked example

A particle is released from rest and falls freely through $19.6\,\text{m}$ . Using $g=9.8\,\text{m s}^{-2}$ , find the time taken and its speed after falling this distance.

1.Taking downwards as positive, $s=ut+\tfrac12gt^2$ gives $19.6=0+4.9t^2$ , so $t=2.0\,\text{s}$ .
2.Then $v=u+gt=0+9.8(2)=19.6\,\text{m s}^{-1}$ downwards.

Answer: Time $=2.0\,\text{s}$ ; Speed $=19.6\,\text{m s}^{-1}$

Common mistakes

Don't use $g=9.8$ with distances in centimetres and times in seconds without converting units.
Don't use positive gravitational acceleration with an upward-positive displacement equation, producing inconsistent signs.

Exam tip

Declare the positive direction before substituting gravitational acceleration into a constant-acceleration equation.

Tier 1 · Easy

1.
Find the weight of a $2.5\,\text{kg}$ particle, using $g=9.8\,\text{m s}^{-2}$ .
(2)
(Total for Question 1 is 2 marks)
2.
On a planet where $g=3.7\,\text{m s}^{-2}$ , find the weight of a package of mass $6\,\text{kg}$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A particle is projected vertically upwards with speed $14\,\text{m s}^{-1}$ . Using $g=9.8\,\text{m s}^{-2}$ , find its greatest height above the point of projection and the time taken to return to that point.
(4)
(Total for Question 1 is 4 marks)
2.
A stone is projected vertically downwards at $3\,\text{m s}^{-1}$ from a bridge. It falls $20\,\text{m}$ before reaching the water. Using $g=9.8\,\text{m s}^{-2}$ , find the time taken and the speed on reaching the water, giving both answers to 3 significant figures.
(4)
(Total for Question 2 is 4 marks)
3.
A particle is released from rest and falls freely. Model it as a particle moving with constant acceleration and ignore air resistance. Take $g=9.8\,\text{m s}^{-2}$ . Find the distance it falls during the third second of its motion and its speed after $3\,\text{s}$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A person of mass $70\,\text{kg}$ stands on a scale in a lift. The scale exerts an upward force of $770\,\text{N}$ on the person. Using $g=9.8\,\text{m s}^{-2}$ , find the magnitude and direction of the lift's acceleration.
(4)
(Total for Question 1 is 4 marks)
2.
On a moon, a probe is released from rest and falls $18\,\text{m}$ in $2.4\,\text{s}$ . Model the gravitational acceleration as constant. Find the value of $g$ on the moon and the weight there of equipment of mass $12\,\text{kg}$ . Given that $g=9.8\,\text{m s}^{-2}$ on Earth, find the percentage by which the equipment's weight is lower on the moon. Give the percentage to 3 significant figures.
(6)
(Total for Question 2 is 6 marks)
3.
On a planet, a particle is projected vertically upwards. Model the particle as moving under a constant gravitational acceleration of magnitude $g$ and ignore air resistance. The particle is at the same height after $1\,\text{s}$ and after $3\,\text{s}$ . At $t=3\,\text{s}$ its velocity is $8\,\text{m s}^{-1}$ vertically downwards. Find $g$ , the initial speed and the weight on this planet of an object of mass $4\,\text{kg}$ .
(6)
(Total for Question 3 is 6 marks)
4.
A package is released from a balloon that is moving vertically upwards at $14.7\,\text{m s}^{-1}$ . At the instant of release the package is $19.6\,\text{m}$ above horizontal ground. Model the package as a particle moving freely under gravity, ignore air resistance and take $g=9.8\,\text{m s}^{-2}$ . Find the greatest height of the package above the ground, the time from release until it reaches the ground, and its velocity on impact. Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
Particle $A$ is released from rest at a point $63.7\,\text{m}$ above horizontal ground. One second later, particle $B$ is projected vertically upwards from the ground at $19.6\,\text{m s}^{-1}$ . Model both particles as moving freely under gravity, ignore air resistance and take $g=9.8\,\text{m s}^{-2}$ . Find the time after $A$ is released when the particles meet, their height above the ground, and the velocity of each particle at that instant. Give all numerical answers exactly.
(7)
(Total for Question 5 is 7 marks)

M8.4 · Understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; apply to smooth pulleys and connected particles; resolve forces in 2 dimensions; equilibrium of a particle under coplanar forces.

Explanation

Newton's third-law forces are equal and opposite, act on different bodies and arise from the same interaction; they therefore do not cancel on one body's force diagram.
For a particle in equilibrium under coplanar forces, resolve in two independent directions and set both component resultants equal to zero.
For connected particles, draw separate force diagrams, use a common acceleration magnitude while the string is taut, and use one tension for a light string over a smooth pulley before solving simultaneous $F=ma$ equations.
A common error is to treat weight and normal reaction as a third-law pair; both act on the same body, whereas a third-law pair acts on different bodies.

Two particles connected by a light string over a smooth pulley share one tension magnitude while the string is taut.

Worked example

A particle is in equilibrium under three coplanar forces. One force is $14\,\text{N}$ east and another is $10\,\text{N}$ at $120^\circ$ anticlockwise from east. Find the third force as a vector in east-north components and find its magnitude.

1.The first two forces have resultant $(14+10\cos120^\circ)\mathbf{i}+(10\sin120^\circ)\mathbf{j}=9\mathbf{i}+5\sqrt3\mathbf{j}$ .
2.Equilibrium requires zero resultant, so the third force is $-9\mathbf{i}-5\sqrt3\mathbf{j}$ .
3.Its magnitude is $\sqrt{9^2+(5\sqrt3)^2}=\sqrt{156}=2\sqrt{39}\,\text{N}$ .

Answer: Third force $=(-9\mathbf{i}-5\sqrt3\mathbf{j})\,\text{N}$ ; Magnitude $=2\sqrt{39}\,\text{N}$

Common mistakes

Don't assign different tension magnitudes to the two ends of one light inextensible string over a smooth pulley.
Don't treat a Newton third-law pair as two forces on the same particle or assume connected-particle tensions without a force diagram.

Exam tip

Draw a separate force diagram for each particle and distinguish interaction pairs from forces that can balance.

Tier 1 · Easy

1.
A book pushes down on a table with force $P$ . State the Newton's third-law partner to this force.
(2)
(Total for Question 1 is 2 marks)
2.
A tow rope pulls a sledge forwards with force $180\,\text{N}$ . State the Newton's third-law partner to this force.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Particles of masses $3\,\text{kg}$ and $5\,\text{kg}$ hang at the ends of a light inextensible string passing over a smooth pulley. Using $g=9.8\,\text{m s}^{-2}$ , find the acceleration of the system and the tension in the string.
(5)
(Total for Question 1 is 5 marks)
2.
A particle of weight $60\,\text{N}$ is held in equilibrium by two light strings. One string extends horizontally to the left from the particle. The other slopes upwards to the right at $45^\circ$ to the horizontal. Find the tension in each string.
(4)
(Total for Question 2 is 4 marks)
3.
Blocks $A$ and $B$ , of masses $4\,\text{kg}$ and $6\,\text{kg}$ respectively, are in contact on a smooth horizontal plane. A horizontal force of $30\,\text{N}$ is applied to $A$ towards $B$ . Find the acceleration of the blocks and the force exerted by $A$ on $B$ . State the Newton's third-law partner to this contact force.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A $4\,\text{kg}$ particle on a smooth plane inclined at $30^\circ$ is connected by a light inextensible string over a smooth pulley to a freely hanging $3\,\text{kg}$ particle. Using $g=9.8\,\text{m s}^{-2}$ , find the acceleration of the system and the tension in the string.
(7)
(Total for Question 1 is 7 marks)
2.
Particles $A$ and $B$ , of masses $3\,\text{kg}$ and $2\,\text{kg}$ respectively, lie on two smooth planes inclined at $30^\circ$ and $45^\circ$ to the horizontal. They are connected by a light inextensible string passing over a smooth pulley at the common top of the planes. The string lies along a line of greatest slope of each plane. The particles are released from rest. Using $g=9.8\,\text{m s}^{-2}$ , find the direction of motion, the acceleration and the tension, giving numerical answers to 3 significant figures.
(6)
(Total for Question 2 is 6 marks)
3.
A particle $A$ of mass $5\,\text{kg}$ rests on a smooth horizontal table and is connected to a hanging particle $B$ of mass $3\,\text{kg}$ by a light inextensible string passing over a small smooth pulley at the edge of the table. The string is horizontal between $A$ and the pulley. Starting from rest, the particles move with the same acceleration, the string being taut. Take $g=9.8\,\text{m s}^{-2}$ . Find their acceleration and the tension in the string. Find their speed after $B$ has descended $1.5\,\text{m}$ , giving the speed to 3 significant figures. Use unrounded values in your working.
(6)
(Total for Question 3 is 6 marks)
4.
Particles $A$ , $B$ and $C$ , of masses $2\,\text{kg}$ , $3\,\text{kg}$ and $5\,\text{kg}$ respectively, lie in that order on a smooth horizontal plane. Adjacent particles are connected by separate light inextensible strings. A horizontal force of $50\,\text{N}$ is applied to $C$ away from $B$ , and both strings remain taut. Find the acceleration of the particles and the tension in each string. State the Newton's third-law partner to the force exerted by the string joining $B$ and $C$ on particle $C$ . Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
A particle $B$ of mass $4\,\text{kg}$ lies on a smooth horizontal table. It is connected on its left to a hanging particle $A$ of mass $2\,\text{kg}$ by one light inextensible string over a smooth pulley, and on its right to a hanging particle $C$ of mass $6\,\text{kg}$ by a separate light inextensible string over a second smooth pulley. The horizontal parts of both strings are collinear, and both strings remain taut. The system is initially stationary and is then set free. Use $g=9.8\,\text{m s}^{-2}$ . Find the direction and exact magnitude of the acceleration, and find the tension in each string. Give all numerical answers exactly.
(7)
(Total for Question 5 is 7 marks)

M8.5 · Understand and use addition of forces; resultant forces; dynamics for motion in a plane.

Explanation

Forces add as vectors, so a resultant may be written in component form and then converted to magnitude-direction form.
Resolve each force along two perpendicular axes, add corresponding components, and apply $\sum\mathbf{F}=m\mathbf{a}$ to obtain the acceleration vector.
If the resultant is $X\mathbf{i}+Y\mathbf{j}$ , its magnitude is $\sqrt{X^2+Y^2}$ and its direction must be placed in the correct quadrant from the component signs.
A common error is to add force magnitudes without accounting for their directions, or to quote an inverse-tangent angle in the wrong quadrant.

The resultant force is the vector sum of the component forces.

Worked example

Two forces of magnitudes $12\,\text{N}$ and $10\,\text{N}$ act at an angle of $120^\circ$ to each other. Find the magnitude of their resultant and the angle the resultant makes with the $12\,\text{N}$ force.

1.Take the $12\,\text{N}$ force along the positive horizontal direction.
2.The other force has components $10\cos120^\circ=-5$ and $10\sin120^\circ=5\sqrt3$ .
3.The resultant is $7\mathbf{i}+5\sqrt3\mathbf{j}$ , with magnitude $\sqrt{49+75}=2\sqrt{31}\,\text{N}$ .
4.Its angle is $\tan^{-1}(5\sqrt3/7)=51.1^\circ$ .

Answer: Magnitude $=2\sqrt{31}\,\text{N}$ ; Angle $=51.1^\circ$ towards the $10\,\text{N}$ force

Common mistakes

Don't find a resultant magnitude using Pythagoras even though the force components are not perpendicular.
Don't add force magnitudes arithmetically despite a non-zero angle between them.

Exam tip

Resolve forces into perpendicular components or use the cosine rule, then find the resultant direction from its components.

Tier 1 · Easy

1.
A force is $(6\mathbf{i}-8\mathbf{j})\,\text{N}$ . Find its magnitude and its angle below the positive $\mathbf{i}$ direction.
(3)
(Total for Question 1 is 3 marks)
2.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. Forces $(7\mathbf{i}+2\mathbf{j})\,\text{N}$ and $(-3\mathbf{i}+5\mathbf{j})\,\text{N}$ act on a particle. Find their resultant.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A particle of mass $2\,\text{kg}$ is initially at rest. Constant forces $(8\mathbf{i}-2\mathbf{j})\,\text{N}$ and $(-2\mathbf{i}+10\mathbf{j})\,\text{N}$ act on it. Find the resultant force, its acceleration, and its speed after $2\,\text{s}$ .
(5)
(Total for Question 1 is 5 marks)
2.
A particle of mass $4\,\text{kg}$ is acted on by a force of $18\,\text{N}$ due east and a force of $12\,\text{N}$ at $120^\circ$ anticlockwise from east. Find the magnitude and direction of its acceleration. Give the direction to 3 significant figures.
(5)
(Total for Question 2 is 5 marks)
3.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. One of two forces acting on a $2\,\text{kg}$ particle is $(7\mathbf{i}-4\mathbf{j})\,\text{N}$ . Their resultant is $(10\mathbf{i}+5\mathbf{j})\,\text{N}$ . Find the second force, its exact magnitude and the acceleration of the particle.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A particle of mass $5\,\text{kg}$ has initial velocity $(2\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}$ . Constant forces $(15\mathbf{i}+20\mathbf{j})\,\text{N}$ and $(-5\mathbf{i}+10\mathbf{j})\,\text{N}$ act on it. Find its acceleration, velocity after $3\,\text{s}$ and displacement during those $3\,\text{s}$ .
(7)
(Total for Question 1 is 7 marks)
2.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A particle of mass $3\,\text{kg}$ has initial velocity $(2\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}$ . Constant forces $(9\mathbf{i}-3\mathbf{j})\,\text{N}$ and $(p\mathbf{i}+12\mathbf{j})\,\text{N}$ act on it. After $2\,\text{s}$ its velocity is parallel to $\mathbf{i}+2\mathbf{j}$ and has a positive $\mathbf{i}$ component. Find $p$ , the speed after $2\,\text{s}$ and the displacement during these $2\,\text{s}$ .
(6)
(Total for Question 2 is 6 marks)
3.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A particle of mass $3\,\text{kg}$ has velocity $(2\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}$ at time $t=0$ and velocity $(8\mathbf{i}+7\mathbf{j})\,\text{m s}^{-1}$ at time $t=4\,\text{s}$ . A constant resultant force acts throughout. Find this force, its magnitude and its exact direction relative to the positive $\mathbf{i}$ direction. Find also the displacement of the particle during the $4\,\text{s}$ and the exact distance between its initial and final positions.
(6)
(Total for Question 3 is 6 marks)
4.
Two forces of magnitudes $13\,\text{N}$ and $15\,\text{N}$ act on a particle, and their resultant has magnitude $14\,\text{N}$ . Find the exact cosine of the angle between the two forces. Find also the exact cosine of the acute angle between the resultant and the $13\,\text{N}$ force. If the particle has mass $7\,\text{kg}$ , find the magnitude of its acceleration. Give all numerical answers exactly.
(5)
(Total for Question 4 is 5 marks)
5.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A $3\,\text{kg}$ particle has initial velocity $(2\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}$ . A known constant force $(2\mathbf{i}-1.5\mathbf{j})\,\text{N}$ and an unknown constant force act on it. During the next $4\,\text{s}$ its displacement is $(24\mathbf{i}+8\mathbf{j})\,\text{m}$ . Find the unknown force, its exact magnitude and direction relative to the positive $\mathbf{i}$ direction. Find also the particle's velocity after $4\,\text{s}$ . Give all numerical answers exactly.
(7)
(Total for Question 5 is 7 marks)

M8.6 · Understand and use the F ≤ μR model for friction; coefficient of friction; motion of a body on a rough surface; limiting friction and statics.

Explanation

Friction acts to oppose actual or impending relative motion; in equilibrium its magnitude adjusts within $0\leq F\leq\mu R$ .
Find the normal reaction first, decide the likely direction of motion, and use $F=\mu R$ only when friction is limiting or the particle is moving in this model.
For a range of equilibrium values, write the required friction in terms of the applied force and impose $|F|\leq\mu R$ before solving the resulting inequality.
A common error is to set $F=\mu R$ in every static problem; away from limiting equilibrium, friction may be strictly smaller than $\mu R$ .

On a rough horizontal surface, friction opposes impending or actual motion and satisfies

F\leq\mu R

Worked example

A $6\,\text{kg}$ block is moving on a rough horizontal surface with coefficient of friction $0.25$ . A horizontal force of $20\,\text{N}$ acts in the direction of motion. Using $g=9.8\,\text{m s}^{-2}$ , find its acceleration.

1.Vertically, $R=mg=6(9.8)=58.8\,\text{N}$ .
2.Since the block is moving, the friction model gives $F=\mu R=0.25(58.8)=14.7\,\text{N}$ .
3.The horizontal resultant is $20-14.7=5.3\,\text{N}$ , so $a=5.3/6=0.883\,\text{m s}^{-2}$ .

Answer: $0.883\,\text{m s}^{-2}$ in the direction of the applied force

Common mistakes

Don't reverse the direction of friction so that it assists the impending relative motion.
Don't set friction equal to the limiting value in every situation, even when equilibrium requires a smaller force.

Exam tip

Decide whether friction is limiting; otherwise use the force balance to find its actual value within the inequality.

Tier 1 · Easy

1.
A block is in equilibrium on a rough horizontal surface. The normal reaction is $80\,\text{N}$ and the coefficient of friction is $0.30$ . A horizontal force of $10\,\text{N}$ acts on the block. Find the friction force and the greatest possible friction force.
(3)
(Total for Question 1 is 3 marks)
2.
A particle is in limiting equilibrium on a rough surface. The friction has magnitude $18\,\text{N}$ and the normal reaction is $60\,\text{N}$ . Find the coefficient of friction.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A $4\,\text{kg}$ block rests on a rough horizontal surface with coefficient of friction $0.30$ . A horizontal force $P$ is applied. Using $g=9.8\,\text{m s}^{-2}$ , find the greatest value of $P$ for equilibrium and the friction force when $P=10\,\text{N}$ .
(4)
(Total for Question 1 is 4 marks)
2.
A $6\,\text{kg}$ block is on a rough horizontal surface. A force of $24\,\text{N}$ acts at $45^\circ$ above the horizontal and the block is on the point of moving in the horizontal direction of the force. Using $g=9.8\,\text{m s}^{-2}$ , find the coefficient of friction, giving your answer to 3 significant figures.
(5)
(Total for Question 2 is 5 marks)
3.
The angle $\alpha$ between a rough plane and the horizontal satisfies $\tan\alpha=\dfrac34$ . A $4\,\text{kg}$ block slides down the plane and the coefficient of friction is $\dfrac14$ . Model the block as a particle. Take $g=9.8\,\text{m s}^{-2}$ . Find its acceleration down the plane.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A $10\,\text{kg}$ particle rests on a rough plane inclined at $20^\circ$ to the horizontal. The coefficient of friction is $0.30$ . A force of magnitude $P\,\text{N}$ acts up the plane. Using $g=9.8\,\text{m s}^{-2}$ , find the complete range of values of $P$ for which the particle remains in equilibrium.
(7)
(Total for Question 1 is 7 marks)
2.
A $5\,\text{kg}$ block is in equilibrium on a rough horizontal surface with coefficient of friction $0.25$ . A horizontal force of $10\,\text{N}$ acts to the left. A second force of magnitude $P\,\text{N}$ acts upwards and to the right at an angle $\alpha$ above the horizontal, where $\tan\alpha=\dfrac34$ . Given $P\geq0$ and using $g=9.8\,\text{m s}^{-2}$ , find the complete range of values of $P$ for which equilibrium is possible. Give the upper bound to 3 significant figures.
(6)
(Total for Question 2 is 6 marks)
3.
The angle $\alpha$ between a rough plane and the horizontal satisfies $\tan\alpha=\dfrac34$ . A $5\,\text{kg}$ block moves up the plane while a horizontal force of $48\,\text{N}$ acts towards the upward side. The coefficient of friction is $0.20$ . Model the block as a particle. Take $g=9.8\,\text{m s}^{-2}$ . Find the normal reaction and the acceleration of the block. Given that its speed is initially $4\,\text{m s}^{-1}$ , find its speed and direction of motion $3\,\text{s}$ later.
(6)
(Total for Question 3 is 6 marks)
4.
An $8\,\text{kg}$ block slides on a rough horizontal plane. The only horizontal force is friction, modelled with coefficient $\mu$ . The block has speed $10\,\text{m s}^{-1}$ and comes to rest after travelling $25\,\text{m}$ . Take $g=9.8\,\text{m s}^{-2}$ and assume the friction model applies throughout. Find the exact value of $\mu$ , the friction force and the time taken to stop. Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
A $10\,\text{kg}$ block moves on a rough horizontal plane with acceleration $0.8\,\text{m s}^{-2}$ to the right. The coefficient of friction is $1/4$ . A force of magnitude $P\,\text{N}$ acts downwards and to the right at an angle $\alpha$ below the horizontal, where $\tan\alpha=3/4$ . Model the block as a particle, assume the friction model applies and take $g=9.8\,\text{m s}^{-2}$ . Find the exact values of $P$ , the normal reaction and the friction force. Give all numerical answers exactly.
(6)
(Total for Question 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

M8.1 · Understand the concept of a force; understand and use Newton's first law.

Tier 1 · Easy

Mark scheme for M8.1 Tier 1 · Easy
Question	Scheme	Marks
1	Normal reaction $=35\,\text{N}$ upwards Resultant force $=0\,\text{N}$	2
(2 marks)		2
Notes
The stationary book has zero acceleration, so Newton's first law requires zero resultant force. The upward normal reaction therefore balances the $35\,\text{N}$ weight and has magnitude $35\,\text{N}$ .
2	$(5\mathbf{i}-2\mathbf{j})\,\text{m s}^{-1}$	1
(1 mark)		1
Notes
By Newton's first law, zero resultant force means zero acceleration, so the probe continues with the same constant velocity.

Tier 2 · Standard

Mark scheme for M8.1 Tier 2 · Standard
Question	Scheme	Marks
1	Weight $24.5\,\text{N}$ downward Normal force $24.5\,\text{N}$ upward They have equal magnitudes because the book has zero acceleration, so the resultant force is zero.	3
(3 marks)		3
Notes
The weight is $mg=2.5(9.8)=24.5\,\text{N}$ downward. The book is at rest, so its acceleration and resultant force are zero. The only vertical forces are its weight and the table's normal force, so the normal force is $24.5\,\text{N}$ upward.
2	Air resistance $=735\,\text{N}$ upwards Zero resultant force gives constant velocity, not necessarily zero velocity	3
(3 marks)		3
Notes
Constant speed in a fixed downward direction means constant velocity, so the acceleration and resultant force are zero. Air resistance must therefore balance the $735\,\text{N}$ weight and act upwards. Newton's first law permits motion with any constant velocity when the resultant force is zero.
3	Zero velocity at one instant does not imply zero acceleration, so the train could have a non-zero resultant force A zero resultant force would require the train to remain at rest or to move with constant velocity	3
(3 marks)		3
Notes
An object can be instantaneously at rest while its velocity is changing, so the single observation does not establish that its acceleration or resultant force is zero. By Newton's first law, a zero resultant force makes the velocity constant; the train would remain at rest or continue with an unchanged non-zero velocity.

Tier 3 · Hard

Mark scheme for M8.1 Tier 3 · Hard
Question	Scheme	Marks
1	Third force $=(-4\mathbf{i}+3\mathbf{j})\,\text{N}$ Magnitude $=5\,\text{N}$	4
(4 marks)		4
Notes
Constant velocity requires the vector resultant to be zero. The two known forces add to $4\mathbf{i}-3\mathbf{j}$ , so the third force is its negative, $-4\mathbf{i}+3\mathbf{j}$ . Its magnitude is $\sqrt{(-4)^2+3^2}=5\,\text{N}$ .
2	Resistance $=(-55\mathbf{i}-15\sqrt3\mathbf{j})\,\text{N}$ Magnitude $=10\sqrt{37}\,\text{N}$ Direction $25.3^\circ$ south of west	5
(5 marks)		5
Notes
The cable resultant is $40\mathbf{i}+30(\cos60^\circ\mathbf{i}+\sin60^\circ\mathbf{j})=55\mathbf{i}+15\sqrt3\mathbf{j}$ . Constant velocity requires zero resultant force, so resistance is its negative. Its magnitude is $\sqrt{55^2+(15\sqrt3)^2}=10\sqrt{37}\,\text{N}$ . The acute angle from west is $\tan^{-1}(15\sqrt3/55)=25.3^\circ$ , towards the south.
3	The resultant force is $\mathbf{0}$ , so the velocity remains $(3\mathbf{i}+2\mathbf{j})\,\text{m s}^{-1}$ $\mathbf{r}=(2+3t)\mathbf{i}+(-5+2t)\mathbf{j}\,\text{m}$ The path crosses $y=7$ at $(20,7)$	5
(5 marks)		5
Notes
Adding the forces gives $(4-9+5)\mathbf{i}+(7+2-9)\mathbf{j}=\mathbf{0}$ . Newton's first law therefore gives constant velocity. Hence $\mathbf{r}=(2\mathbf{i}-5\mathbf{j})+t(3\mathbf{i}+2\mathbf{j})$ . On the line $y=7$ , $-5+2t=7$ , so $t=6$ and $x=2+3(6)=20$ .
4	$T=10\,\text{N}$ $P=2\,\text{N}$ The forces have zero resultant, so Newton's first law gives constant velocity	5
(5 marks)		5
Notes
Constant velocity requires zero resultant force. Resolving in the $\mathbf j$ direction gives $3+5-4T/5=0$ , so $T=10$ . Resolving in the $\mathbf i$ direction then gives $7-3+P-3(10)/5=0$ , hence $P=2$ . A zero resultant produces zero acceleration, not necessarily zero velocity, so the existing non-zero velocity remains constant.
5	$P=10\sqrt3\,\text{N}$ Water resistance $=20\,\text{N}$ due west	5
(5 marks)		5
Notes
Constant velocity requires the north and east force components to balance. Resolving north, $10\sin60^\circ=P\sin30^\circ$ , so $5\sqrt3=P/2$ and $P=10\sqrt3\,\text{N}$ . The total eastward component of the engine forces is $10\cos60^\circ+10\sqrt3\cos30^\circ=5+15=20\,\text{N}$ . The resistance must therefore be $20\,\text{N}$ due west.

M8.2 · Understand and use Newton's second law for motion in a straight line (forces in two perpendicular directions or simple 2-D vectors); extend to situations where forces need to be resolved (restricted to 2 dimensions).

Tier 1 · Easy

Mark scheme for M8.2 Tier 1 · Easy
Question	Scheme	Marks
1	$4\,\text{m s}^{-2}$ in the direction of the force	2
(2 marks)		2
Notes
Newton's second law gives $F=ma$ , so $a=F/m=12/3=4\,\text{m s}^{-2}$ , directed with the resultant force.
2	$(3\mathbf{i}-4\mathbf{j})\,\text{m s}^{-2}$	2
(2 marks)		2
Notes
Newton's second law gives $\mathbf{a}=\mathbf{F}/m=(9\mathbf{i}-12\mathbf{j})/3=(3\mathbf{i}-4\mathbf{j})\,\text{m s}^{-2}$ .

Tier 2 · Standard

Mark scheme for M8.2 Tier 2 · Standard
Question	Scheme	Marks
1	Exact acceleration $=\dfrac{5\sqrt3}{2}\,\text{m s}^{-2}$ Acceleration to $3$ significant figures $=4.33\,\text{m s}^{-2}$ Normal reaction $=29.2\,\text{N}$	5
(5 marks)		5
Notes
Horizontally, the resultant force is $20\cos30^\circ=10\sqrt3\,\text{N}$ , so $4a=10\sqrt3$ and $a=5\sqrt3/2\,\text{m s}^{-2}$ . Vertically there is no acceleration, so $R+20\sin30^\circ-4(9.8)=0$ . Hence $R=39.2-10=29.2\,\text{N}$ .
2	Acceleration $=4.00\,\text{m s}^{-2}$ Normal reaction $=45.3\,\text{N}$	5
(5 marks)		5
Notes
Horizontally, $32\cos25^\circ-5=6a$ , so $a=4.00\,\text{m s}^{-2}$ to 3 significant figures. Vertically there is no acceleration, so $R+32\sin25^\circ-6(9.8)=0$ . Hence $R=58.8-32\sin25^\circ=45.3\,\text{N}$ to 3 significant figures.
3	Mass $=8\,\text{kg}$	4
(4 marks)		4
Notes
The resultant force is $(8\mathbf{i}+12\mathbf{j})\,\text{N}$ . The acceleration is the velocity change divided by time, so $\mathbf{a}=(\mathbf{i}+1.5\mathbf{j})\,\text{m s}^{-2}$ . From $\mathbf{F}=m\mathbf{a}$ , either component gives $m=8/1=12/1.5=8\,\text{kg}$ .

Tier 3 · Hard

Mark scheme for M8.2 Tier 3 · Hard
Question	Scheme	Marks
1	Normal reaction $=42.3\,\text{N}$ Acceleration $=4.70\,\text{m s}^{-2}$ up the plane	7
(7 marks)		7
Notes
Perpendicular to the plane, $R+50\sin15^\circ-6(9.8)\cos20^\circ=0$ , so $R=6(9.8)\cos20^\circ-50\sin15^\circ=42.3\,\text{N}$ . Parallel to the plane, $50\cos15^\circ-6(9.8)\sin20^\circ=6a$ . Therefore $a=[50\cos15^\circ-58.8\sin20^\circ]/6=4.70\,\text{m s}^{-2}$ .
2	$P=27.7\,\text{N}$ Normal reaction $=40.1\,\text{N}$	6
(6 marks)		6
Notes
Here $\sin\alpha=3/5$ and $\cos\alpha=4/5$ . Resolving up the plane gives $(4/5)P-3(9.8)(3/5)=3(1.5)$ , so $P=27.675\,\text{N}=27.7\,\text{N}$ to 3 significant figures. Perpendicular to the plane, the horizontal force presses into the plane, so $R=3(9.8)(4/5)+(3/5)P=40.125\,\text{N}=40.1\,\text{N}$ to 3 significant figures.
3	$P=10\,\text{N}$ Mass $=5\,\text{kg}$	5
(5 marks)		5
Notes
The resultant force has components $(10-\tfrac12P)\,\text{N}$ east and $(\tfrac{\sqrt3}{2}P)\,\text{N}$ north. Its direction is $60^\circ$ north of east, so $\tfrac{\sqrt3}{2}P=\sqrt3(10-\tfrac12P)$ . Hence $P=10\,\text{N}$ . The resultant components are then $5$ and $5\sqrt3$ , so its magnitude is $10\,\text{N}$ . From $F=ma$ , $10=2m$ and $m=5\,\text{kg}$ .
4	Initially $P=37.5\,\text{N}$ Initially $R=26.5\,\text{N}$ Contact is lost when $P=\dfrac{245}{3}\,\text{N}$ Horizontal acceleration then $=\dfrac{196}{15}\,\text{m s}^{-2}$	7
(7 marks)		7
Notes
Here $\cos\alpha=4/5$ . Horizontally, $(4/5)P=5(6)$ , so $P=37.5\,\text{N}$ . Vertically, $R+(3/5)P=5(9.8)$ , giving $R=49-22.5=26.5\,\text{N}$ . Contact is just lost when $R=0$ , so $(3/5)P=49$ and $P=245/3\,\text{N}$ . The horizontal acceleration is then $[(4/5)(245/3)]/5=196/15\,\text{m s}^{-2}$ .
5	$P=\dfrac{236}{3}\,\text{N}$ Normal reaction $=\dfrac{944}{15}\,\text{N}$ Speed after $3\,\text{s}$ is $6\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
Since $\tan\alpha=3/4$ , $\sin\alpha=3/5$ and $\cos\alpha=4/5$ . Vertically, $(3/5)P-4(9.8)=4(2)$ , so $(3/5)P=47.2$ and $P=236/3\,\text{N}$ . Perpendicular to the wall there is no acceleration, so the normal reaction balances the horizontal component: $R=(4/5)P=944/15\,\text{N}$ . From rest with constant acceleration $2\,\text{m s}^{-2}$ , the speed after $3\,\text{s}$ is $v=2(3)=6\,\text{m s}^{-1}$ .

M8.3 · Understand and use weight and motion in a straight line under gravity; gravitational acceleration, g, and its value in S.I. units to varying degrees of accuracy.

Tier 1 · Easy

Mark scheme for M8.3 Tier 1 · Easy
Question	Scheme	Marks
1	$24.5\,\text{N}$ vertically downwards	2
(2 marks)		2
Notes
$W=mg=2.5(9.8)=24.5\,\text{N}$ , and weight acts vertically downwards.
2	$22.2\,\text{N}$ vertically downwards	2
(2 marks)		2
Notes
Weight is $W=mg=6(3.7)=22.2\,\text{N}$ and acts vertically downwards.

Tier 2 · Standard

Mark scheme for M8.3 Tier 2 · Standard
Question	Scheme	Marks
1	Greatest height $10\,\text{m}$ Exact return time $=\dfrac{20}{7}\,\text{s}$ Return time to $3$ significant figures $=2.86\,\text{s}$	4
(4 marks)		4
Notes
At the greatest height, $v=0$ . Taking upwards as positive, $v^2=u^2+2as$ gives $0=14^2-2(9.8)h$ , so $h=196/19.6=10\,\text{m}$ . For the return to the starting level, $0=14t-4.9t^2$ . The non-zero solution is $t=14/4.9=20/7\,\text{s}$ .
2	Time $=1.74\,\text{s}$ Speed $=20.0\,\text{m s}^{-1}$	4
(4 marks)		4
Notes
Taking downwards as positive, $20=3t+4.9t^2$ . The positive root is $t=(-3+\sqrt{401})/9.8=1.74\,\text{s}$ to 3 significant figures. Also $v^2=u^2+2gs=3^2+2(9.8)(20)=401$ , so $v=\sqrt{401}=20.0\,\text{m s}^{-1}$ to 3 significant figures.
3	Distance during the third second $=24.5\,\text{m}$ Speed after $3\,\text{s}$ is $29.4\,\text{m s}^{-1}$	4
(4 marks)		4
Notes
The distance fallen by time $t$ is $s=\tfrac12gt^2=4.9t^2$ . The distance during the third second is $s(3)-s(2)=4.9(9-4)=24.5\,\text{m}$ . The speed after $3\,\text{s}$ is $v=gt=9.8(3)=29.4\,\text{m s}^{-1}$ .

Tier 3 · Hard

Mark scheme for M8.3 Tier 3 · Hard
Question	Scheme	Marks
1	$1.2\,\text{m s}^{-2}$ upwards	4
(4 marks)		4
Notes
The person's weight is $70(9.8)=686\,\text{N}$ downwards. Taking upwards as positive, Newton's second law gives $770-686=70a$ . Hence $a=84/70=1.2\,\text{m s}^{-2}$ upwards.
2	Moon's $g=6.25\,\text{m s}^{-2}$ Moon weight $=75\,\text{N}$ Weight is $36.2\%$ lower	6
(6 marks)		6
Notes
From $s=\tfrac12gt^2$ , $18=\tfrac12g(2.4)^2$ , so $g=36/5.76=6.25\,\text{m s}^{-2}$ . The moon weight is $12(6.25)=75\,\text{N}$ ; the Earth weight is $12(9.8)=117.6\,\text{N}$ . The percentage reduction is $(117.6-75)/117.6\times100=36.2\%$ to 3 significant figures.
3	$g=8\,\text{m s}^{-2}$ Initial speed $=16\,\text{m s}^{-1}$ Weight $=32\,\text{N}$	6
(6 marks)		6
Notes
Taking upwards as positive and writing the initial speed as $u$ , the height above the projection point is $s=ut-\tfrac12gt^2$ . Equality of the heights at $t=1$ and $t=3$ gives $u-\tfrac12g=3u-\tfrac92g$ , so $u=2g$ . Also $v=u-gt$ , and at $t=3$ the velocity is $-8\,\text{m s}^{-1}$ . Thus $2g-3g=-8$ , giving $g=8\,\text{m s}^{-2}$ and $u=16\,\text{m s}^{-1}$ . The weight is $mg=4(8)=32\,\text{N}$ .
4	Greatest height $=30.625\,\text{m}$ Time to ground $=4\,\text{s}$ Impact velocity $=24.5\,\text{m s}^{-1}$ vertically downwards	6
(6 marks)		6
Notes
Taking upwards as positive, at greatest height $0=14.7^2-2(9.8)h$ , so the additional height is $11.025\,\text{m}$ and the greatest height above ground is $19.6+11.025=30.625\,\text{m}$ . Ground level satisfies $0=19.6+14.7t-4.9t^2$ , or $0=4+3t-t^2$ , whose positive root is $t=4$ . The impact velocity is $14.7-9.8(4)=-24.5\,\text{m s}^{-1}$ , namely $24.5\,\text{m s}^{-1}$ downwards.
5	Meeting time $=3\,\text{s}$ after $A$ is released Meeting height $=19.6\,\text{m}$ $A$ has velocity $29.4\,\text{m s}^{-1}$ downwards $B$ is instantaneously at rest	7
(7 marks)		7
Notes
Let $t$ be the time after $A$ is released and take upwards as positive. Then $y_A=63.7-4.9t^2$ . For $t\geq1$ , $y_B=19.6(t-1)-4.9(t-1)^2$ . Equating and cancelling the common $-4.9t^2$ terms gives $63.7=29.4t-24.5$ , so $t=3$ . The height is $63.7-4.9(9)=19.6\,\text{m}$ . The velocities are $v_A=-9.8(3)=-29.4\,\text{m s}^{-1}$ and $v_B=19.6-9.8(3-1)=0$ .

M8.4 · Understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; apply to smooth pulleys and connected particles; resolve forces in 2 dimensions; equilibrium of a particle under coplanar forces.

Tier 1 · Easy

Mark scheme for M8.4 Tier 1 · Easy
Question	Scheme	Marks
1	The table pushes upwards on the book with force $P$	2
(2 marks)		2
Notes
The partner must be the same interaction with the bodies reversed. It is the force of the table on the book, equal in magnitude to $P$ and opposite in direction.
2	The sledge pulls the tow rope backwards with force $180\,\text{N}$	2
(2 marks)		2
Notes
The partner force belongs to the same rope-sledge interaction, has equal magnitude and opposite direction, and acts on the other body. It is therefore the $180\,\text{N}$ backward force of the sledge on the rope.

Tier 2 · Standard

Mark scheme for M8.4 Tier 2 · Standard
Question	Scheme	Marks
1	Acceleration $2.45\,\text{m s}^{-2}$ , with the $5\,\text{kg}$ particle moving down Tension $36.75\,\text{N}$ , or $36.8\,\text{N}$ to 3 significant figures	5
(5 marks)		5
Notes
Treating both particles as one system, the driving force is $(5-3)g=19.6\,\text{N}$ and the total mass is $8\,\text{kg}$ , so $a=19.6/8=2.45\,\text{m s}^{-2}$ . For the $3\,\text{kg}$ particle moving upward, $T-3g=3a$ , so $T=3(9.8+2.45)=36.75\,\text{N}$ .
2	Tension in the sloping string $=60\sqrt2\,\text{N}$ Tension in the horizontal string $=60\,\text{N}$	4
(4 marks)		4
Notes
Let the sloping tension be $T$ and the horizontal tension be $H$ . Vertical equilibrium gives $T\sin45^\circ=60$ , so $T=60\sqrt2\,\text{N}$ . Horizontal equilibrium then gives $H=T\cos45^\circ=60\,\text{N}$ .
3	Acceleration $=3\,\text{m s}^{-2}$ Force exerted by $A$ on $B=18\,\text{N}$ in the direction of motion The partner is the $18\,\text{N}$ force exerted by $B$ on $A$ in the opposite direction	5
(5 marks)		5
Notes
Treating both blocks as one system, $30=(4+6)a$ , so $a=3\,\text{m s}^{-2}$ . For block $B$ , the only horizontal force is the contact force from $A$ , so it is $6(3)=18\,\text{N}$ in the direction of motion. By Newton's third law, $B$ exerts an $18\,\text{N}$ force on $A$ in the opposite direction.

Tier 3 · Hard

Mark scheme for M8.4 Tier 3 · Hard
Question	Scheme	Marks
1	Acceleration $=1.4\,\text{m s}^{-2}$ , with the $3\,\text{kg}$ particle moving down Tension $=25.2\,\text{N}$	7
(7 marks)		7
Notes
The downslope weight component of the $4\,\text{kg}$ particle is $4(9.8)\sin30^\circ=19.6\,\text{N}$ , while the hanging weight is $29.4\,\text{N}$ , so the $3\,\text{kg}$ particle moves down. For the system, $29.4-19.6=7a$ , giving $a=1.4\,\text{m s}^{-2}$ . For the hanging particle, $29.4-T=3(1.4)$ , so $T=25.2\,\text{N}$ .
2	$A$ moves down the $30^\circ$ plane and $B$ moves up the $45^\circ$ plane Acceleration $=0.168\,\text{m s}^{-2}$ Tension $=14.2\,\text{N}$	6
(6 marks)		6
Notes
The downslope weight components are $3g\sin30^\circ=14.7\,\text{N}$ for $A$ and $2g\sin45^\circ=9.8\sqrt2\,\text{N}$ for $B$ . Since $14.7>9.8\sqrt2$ , $A$ moves down its plane. For the system, $14.7-9.8\sqrt2=5a$ , so $a=0.168\,\text{m s}^{-2}$ to 3 significant figures. For $A$ , $14.7-T=3a$ , giving $T=14.2\,\text{N}$ to 3 significant figures.
3	Acceleration $=3.675\,\text{m s}^{-2}$ Tension $=18.375\,\text{N}$ Speed $=3.32\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
Let the acceleration be $a$ and the tension be $T$ . For $A$ , $T=5a$ . For $B$ , taking downwards as positive, $3g-T=3a$ . Hence $29.4=8a$ , so $a=3.675\,\text{m s}^{-2}$ and $T=18.375\,\text{N}$ . Using $v^2=u^2+2as$ with $u=0$ and $s=1.5$ , $v=\sqrt{2(3.675)(1.5)}=3.320\ldots\,\text{m s}^{-1}$ , which is $3.32\,\text{m s}^{-1}$ to 3 significant figures.
4	Acceleration $=5\,\text{m s}^{-2}$ Tension between $A$ and $B=10\,\text{N}$ Tension between $B$ and $C=25\,\text{N}$ The partner is the equal and opposite force exerted by $C$ on that string	6
(6 marks)		6
Notes
For the complete system, $50=(2+3+5)a$ , so $a=5\,\text{m s}^{-2}$ . For $A$ , the first tension is the only horizontal force, so $T_1=2(5)=10\,\text{N}$ . The second tension accelerates $A$ and $B$ together, so $T_2=(2+3)(5)=25\,\text{N}$ . The third-law partner to the string's force on $C$ is the force of $C$ on the same string, equal in magnitude and opposite in direction.
5	$C$ moves down, $B$ moves right and $A$ moves up Acceleration $=\dfrac{49}{15}\,\text{m s}^{-2}$ Left-string tension $=\dfrac{392}{15}\,\text{N}$ Right-string tension $=\dfrac{196}{5}\,\text{N}$	7
(7 marks)		7
Notes
The greater hanging weight is that of $C$ , so take $C$ down, $B$ right and $A$ up as positive. For the whole system the driving force is $(6-2)g=39.2\,\text{N}$ and the total mass is $12\,\text{kg}$ , giving $a=39.2/12=49/15\,\text{m s}^{-2}$ . For $A$ , $T_L-2g=2a$ , so $T_L=19.6+98/15=392/15\,\text{N}$ . For $C$ , $6g-T_R=6a$ , so $T_R=58.8-294/15=196/5\,\text{N}$ .

M8.5 · Understand and use addition of forces; resultant forces; dynamics for motion in a plane.

Tier 1 · Easy

Mark scheme for M8.5 Tier 1 · Easy
Question	Scheme	Marks
1	Magnitude $=10\,\text{N}$ Angle $=53.1^\circ$ below the positive $\mathbf{i}$ direction	3
(3 marks)		3
Notes
The magnitude is $\sqrt{6^2+(-8)^2}=10\,\text{N}$ . Since the components place the force in the fourth quadrant, the angle below the positive $\mathbf{i}$ direction is $\tan^{-1}(8/6)=53.1^\circ$ .
2	$(4\mathbf{i}+7\mathbf{j})\,\text{N}$	2
(2 marks)		2
Notes
Add corresponding components: $(7-3)\mathbf{i}+(2+5)\mathbf{j}=4\mathbf{i}+7\mathbf{j}$ .

Tier 2 · Standard

Mark scheme for M8.5 Tier 2 · Standard
Question	Scheme	Marks
1	Resultant force $=(6\mathbf{i}+8\mathbf{j})\,\text{N}$ Acceleration $=(3\mathbf{i}+4\mathbf{j})\,\text{m s}^{-2}$ Speed after $2\,\text{s}$ is $10\,\text{m s}^{-1}$	5
(5 marks)		5
Notes
Add the two forces component by component: $\mathbf F=(8-2)\mathbf i+(-2+10)\mathbf j=6\mathbf i+8\mathbf j$ . Newton's second law gives $\mathbf a=\mathbf F/m=3\mathbf i+4\mathbf j$ . From rest, after $2$ seconds $\mathbf v=2\mathbf a=6\mathbf i+8\mathbf j$ , whose magnitude is $\sqrt{6^2+8^2}=10\,\text{m s}^{-1}$ .
2	Acceleration magnitude $=\dfrac{3\sqrt7}{2}\,\text{m s}^{-2}$ Direction $=40.9^\circ$ north of east	5
(5 marks)		5
Notes
The resultant is $(18+12\cos120^\circ)\mathbf{i}+12\sin120^\circ\mathbf{j}=12\mathbf{i}+6\sqrt3\mathbf{j}$ . Its magnitude is $\sqrt{12^2+(6\sqrt3)^2}=6\sqrt7\,\text{N}$ , so the acceleration magnitude is $6\sqrt7/4=3\sqrt7/2\,\text{m s}^{-2}$ . Its direction is $\tan^{-1}(6\sqrt3/12)=40.9^\circ$ north of east.
3	Second force $=(3\mathbf{i}+9\mathbf{j})\,\text{N}$ Its magnitude is $3\sqrt{10}\,\text{N}$ Acceleration $=(5\mathbf{i}+2.5\mathbf{j})\,\text{m s}^{-2}$	5
(5 marks)		5
Notes
The second force is the resultant minus the first force: $(10\mathbf{i}+5\mathbf{j})-(7\mathbf{i}-4\mathbf{j})=3\mathbf{i}+9\mathbf{j}$ . Its magnitude is $\sqrt{3^2+9^2}=3\sqrt{10}\,\text{N}$ . Newton's second law gives $\mathbf{a}=\mathbf{F}/m=(10\mathbf{i}+5\mathbf{j})/2=5\mathbf{i}+2.5\mathbf{j}\,\text{m s}^{-2}$ .

Tier 3 · Hard

Mark scheme for M8.5 Tier 3 · Hard
Question	Scheme	Marks
1	Acceleration $=(2\mathbf{i}+6\mathbf{j})\,\text{m s}^{-2}$ Velocity $=(8\mathbf{i}+17\mathbf{j})\,\text{m s}^{-1}$ Displacement $=(15\mathbf{i}+24\mathbf{j})\,\text{m}$	7
(7 marks)		7
Notes
The resultant force is $10\mathbf{i}+30\mathbf{j}$ , so $\mathbf{a}=\mathbf{F}/m=2\mathbf{i}+6\mathbf{j}$ . Then $\mathbf{v}=\mathbf{u}+\mathbf{a}t=(2\mathbf{i}-\mathbf{j})+3(2\mathbf{i}+6\mathbf{j})=8\mathbf{i}+17\mathbf{j}$ . Finally $\mathbf{s}=\mathbf{u}t+\tfrac12\mathbf{a}t^2=(6\mathbf{i}-3\mathbf{j})+(9\mathbf{i}+27\mathbf{j})=15\mathbf{i}+24\mathbf{j}$ .
2	$p=-\dfrac{33}{4}$ Speed $=\dfrac{5\sqrt5}{2}\,\text{m s}^{-1}$ Displacement $=(\dfrac92\mathbf{i}+4\mathbf{j})\,\text{m}$	6
(6 marks)		6
Notes
The acceleration is $[(9+p)/3]\mathbf{i}+3\mathbf{j}$ . After $2\,\text{s}$ , the velocity is $[2+2(9+p)/3]\mathbf{i}+5\mathbf{j}$ . Parallel to $\mathbf{i}+2\mathbf{j}$ with positive components requires the $\mathbf{j}$ component to be twice the $\mathbf{i}$ component, so $5=2[2+2(9+p)/3]$ , giving $p=-33/4$ . Thus $\mathbf{a}=\tfrac14\mathbf{i}+3\mathbf{j}$ and $\mathbf{v}=\tfrac52\mathbf{i}+5\mathbf{j}$ , whose magnitude is $5\sqrt5/2$ . Finally $\mathbf{s}=2\mathbf{u}+\tfrac12\mathbf{a}(2^2)=(4,-2)+(1/2,6)=(9/2,4)$ metres.
3	Resultant force $=(4.5\mathbf{i}+6\mathbf{j})\,\text{N}$ Magnitude $=7.5\,\text{N}$ Direction $=\tan^{-1}(4/3)$ above the positive $\mathbf{i}$ direction Displacement $=(20\mathbf{i}+12\mathbf{j})\,\text{m}$ Distance between the positions $=4\sqrt{34}\,\text{m}$	6
(6 marks)		6
Notes
The acceleration is the change in velocity divided by time: $\mathbf{a}=[(8-2)\mathbf{i}+(7-(-1))\mathbf{j}]/4=1.5\mathbf{i}+2\mathbf{j}$ . Hence $\mathbf{F}=3\mathbf{a}=4.5\mathbf{i}+6\mathbf{j}\,\text{N}$ . Its magnitude is $\sqrt{4.5^2+6^2}=7.5\,\text{N}$ and its direction is $\tan^{-1}(6/4.5)=\tan^{-1}(4/3)$ above $\mathbf{i}$ . Constant acceleration gives displacement $\tfrac12(\mathbf{u}+\mathbf{v})t=\tfrac12(10\mathbf{i}+6\mathbf{j})(4)=20\mathbf{i}+12\mathbf{j}\,\text{m}$ , whose magnitude is $\sqrt{20^2+12^2}=4\sqrt{34}\,\text{m}$ .
4	Cosine of angle between forces $=-\dfrac{33}{65}$ Cosine of angle between resultant and $13\,\text{N}$ force $=\dfrac5{13}$ Acceleration magnitude $=2\,\text{m s}^{-2}$	5
(5 marks)		5
Notes
If $\theta$ is the angle between the forces, $14^2=13^2+15^2+2(13)(15)\cos\theta$ , so $\cos\theta=(196-169-225)/390=-33/65$ . In the force triangle, if $\phi$ is the angle between the resultant and the $13\,\text{N}$ force, the side opposite $\phi$ has length $15$ , so $15^2=13^2+14^2-2(13)(14)\cos\phi$ . Hence $\cos\phi=5/13$ . Newton's second law gives acceleration magnitude $14/7=2\,\text{m s}^{-2}$ .
5	Unknown force $=(4\mathbf{i}+6\mathbf{j})\,\text{N}$ Magnitude $=2\sqrt{13}\,\text{N}$ Direction $=\tan^{-1}(3/2)$ above the positive $\mathbf{i}$ direction Velocity after $4\,\text{s}$ is $(10\mathbf{i}+5\mathbf{j})\,\text{m s}^{-1}$	7
(7 marks)		7
Notes
Using $\mathbf{s}=\mathbf{u}t+\tfrac12\mathbf{a}t^2$ , $(24,8)=4(2,-1)+8\mathbf a$ , so $8\mathbf a=(16,12)$ and $\mathbf a=(2,3/2)\,\text{m s}^{-2}$ . The resultant force is $3\mathbf a=(6,9/2)\,\text{N}$ . Subtracting the known force $(2,-3/2)$ gives the unknown force $(4,6)\,\text{N}$ . Its magnitude is $2\sqrt{13}$ and its direction is $\tan^{-1}(6/4)=\tan^{-1}(3/2)$ above $\mathbf i$ . Finally $\mathbf v=\mathbf u+4\mathbf a=(10,5)\,\text{m s}^{-1}$ .

M8.6 · Understand and use the F ≤ μR model for friction; coefficient of friction; motion of a body on a rough surface; limiting friction and statics.

Tier 1 · Easy

Mark scheme for M8.6 Tier 1 · Easy
Question	Scheme	Marks
1	Friction $=10\,\text{N}$ opposite to the applied force Greatest possible friction $=24\,\text{N}$	3
(3 marks)		3
Notes
Equilibrium requires friction to balance the $10\,\text{N}$ applied force, so $F=10\,\text{N}$ . The limiting value is $\mu R=0.30(80)=24\,\text{N}$ ; the actual friction is smaller because the block is not in limiting equilibrium.
2	$0.30$	2
(2 marks)		2
Notes
At limiting equilibrium $F=\mu R$ , so $18=60\mu$ and $\mu=0.30$ .

Tier 2 · Standard

Mark scheme for M8.6 Tier 2 · Standard
Question	Scheme	Marks
1	Greatest $P=11.76\,\text{N}$ , or $11.8\,\text{N}$ to 3 significant figures When $P=10\,\text{N}$ , friction is $10\,\text{N}$ opposite to $P$ .	4
(4 marks)		4
Notes
The normal reaction is $R=4g=39.2\,\text{N}$ , so the limiting friction is $\mu R=0.30(39.2)=11.76\,\text{N}$ . This is the greatest horizontal force that static friction can balance. When $P=10\,\text{N}<11.76\,\text{N}$ , friction adjusts to $10\,\text{N}$ ; it is not automatically equal to $\mu R$ .
2	$\mu=0.406$	5
(5 marks)		5
Notes
Vertical equilibrium gives $R+24\sin45^\circ=6(9.8)$ , so $R=58.8-12\sqrt2\,\text{N}$ . At limiting equilibrium the friction is $\mu R$ and horizontal equilibrium gives $\mu R=24\cos45^\circ=12\sqrt2$ . Hence $\mu=12\sqrt2/(58.8-12\sqrt2)=0.406$ to 3 significant figures.
3	Acceleration $=3.92\,\text{m s}^{-2}$ down the plane	4
(4 marks)		4
Notes
Since $\tan\alpha=3/4$ , $\sin\alpha=3/5$ and $\cos\alpha=4/5$ . The normal reaction is $R=4g(4/5)$ , so the friction opposing the downward motion is $\mu R=(1/4)4g(4/5)=4g/5\,\text{N}$ . Resolving down the plane, $4a=4g(3/5)-4g/5=8g/5$ , hence $a=2g/5=3.92\,\text{m s}^{-2}$ .

Tier 3 · Hard

Mark scheme for M8.6 Tier 3 · Hard
Question	Scheme	Marks
1	$5.89\leq P\leq61.1$	7
(7 marks)		7
Notes
The normal reaction is $R=98\cos20^\circ$ , so limiting friction is $\mu R=29.4\cos20^\circ$ . The component of weight down the plane is $98\sin20^\circ$ . Equilibrium requires friction of magnitude $\|98\sin20^\circ-P\|$ , so $\|98\sin20^\circ-P\|\leq29.4\cos20^\circ$ . Hence $98\sin20^\circ-29.4\cos20^\circ\leq P\leq98\sin20^\circ+29.4\cos20^\circ$ , giving $5.89\leq P\leq61.1$ .
2	$0\leq P\leq\dfrac{445}{19}\,\text{N}$ , or $0\leq P\leq23.4\,\text{N}$ to 3 significant figures	6
(6 marks)		6
Notes
Since $\cos\alpha=4/5$ and $\sin\alpha=3/5$ , vertical equilibrium gives $R=49-3P/5$ . The friction required has magnitude $\|4P/5-10\|$ . Equilibrium is possible when $\|4P/5-10\|\leq(1/4)(49-3P/5)$ . The lower inequality gives $P\geq-45/13$ , which is replaced by the stated $P\geq0$ . The upper inequality gives $4P/5-10\leq49/4-3P/20$ , so $19P/20\leq89/4$ and $P\leq445/19=23.4\,\text{N}$ to 3 significant figures. This range also keeps $R>0$ .
3	Normal reaction $=68\,\text{N}$ Acceleration $=0.92\,\text{m s}^{-2}$ down the plane After $3\,\text{s}$ the block moves up the plane at $1.24\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
Here $\sin\alpha=3/5$ and $\cos\alpha=4/5$ . Perpendicular to the plane, the weight contributes $5g(4/5)=39.2\,\text{N}$ into the plane and the horizontal force contributes $48(3/5)=28.8\,\text{N}$ into the plane, so $R=68\,\text{N}$ . Since the block moves up the plane, friction $0.20R=13.6\,\text{N}$ acts down the plane. Taking up the plane as positive, the resultant is $48(4/5)-5g(3/5)-13.6=38.4-29.4-13.6=-4.6\,\text{N}$ . Thus $a=-4.6/5=-0.92\,\text{m s}^{-2}$ . After $3\,\text{s}$ , $v=4-0.92(3)=1.24\,\text{m s}^{-1}$ , which is still up the plane.
4	$\mu=\dfrac{10}{49}$ Friction force $=16\,\text{N}$ opposite to the motion Stopping time $=5\,\text{s}$	6
(6 marks)		6
Notes
Taking the direction of motion as positive, $0=10^2+2a(25)$ gives $a=-2\,\text{m s}^{-2}$ . The friction magnitude is therefore $m\|a\|=8(2)=16\,\text{N}$ . Vertically, $R=8(9.8)=78.4\,\text{N}$ , so $\mu=F/R=16/78.4=10/49$ . Finally $0=10-2t$ , giving $t=5\,\text{s}$ .
5	$P=50\,\text{N}$ Normal reaction $=128\,\text{N}$ Friction force $=32\,\text{N}$ to the left	6
(6 marks)		6
Notes
Here $\cos\alpha=4/5$ and $\sin\alpha=3/5$ . Vertical equilibrium gives $R=98+3P/5$ , so friction is $R/4$ . Resolving horizontally to the right, $4P/5-(98+3P/5)/4=10(0.8)$ . Thus $13P/20-24.5=8$ , so $13P/20=32.5$ and $P=50\,\text{N}$ . Hence $R=98+30=128\,\text{N}$ and friction is $128/4=32\,\text{N}$ to the left.