Aqa a-level maths: mechanics 1 revision notes

  • Aqa a-level maths: mechanics 1 revision notes

notes fo mech on its way :) sorry, been rushing the notes for C2 and M1 here. time restraints. but will hopefully be able to neaten stuff up/improve and add more info some time soon. hope it helps nethertheless. haha I should write a textbook/revision guide... oh yeah, for mechanics, for every question, draw a clear but detailed labelled diagram. As you are modelling, people, cars etc. as a particle they are drawn as a box. Modelling means a simplified version of the mechanism, it includes only the important factors. As you get further into mechanics, more factors are added (less and less moddeling assumptions are made) and everything gets more complicated, but a model is improved.

Kinematics in 1D

kinematics is the part of mechanics that deals with the motion of objects

Velocity is a measurement that measures speed and direction. It is measured in ms^-1 (meters per second (m/s) written scientifically.)

Displacement is another measurement that factors distance and direction. measured in m, but with direction written in afterwards too.

The gradient of the line in a displacement-time graphs give velocity. Distance and direction travelled in a set time. (displacement (m)/time (s)=velocity (ms^-1))

If the line is straight, it shows constant velocity.

The average velocity is the time it took to travel that overall displacement if it was at a constant speed. (sort of like average speed)

average velocity is worked out by total displacement/total time (average speed is similar, total distance/total time)

The rate of change of velocity is called acceleration. it is measured in ms^-2 (the scientific way of writing meters per second squared m/s^2)

The gradient of a velocity time graph shows acceleration. (velocity (ms^-1)/speed (s) = acceleration ms^-2)

Acceleration is constant when the gradient of the velocity time graph is a straight line.

A negative acceleration is called deceleration, or retardation.

The area under a velocity-time graph is displacement. (ms^-1/s = m)

There are a standard set of letters that are used to represent the motion of objects moving in a straight line with constant acceleration:

  • s= displacement (m)
  • u= initial velocity (ms^-1) (u is used as it is before v in the alphabet)
  • v= final velocity (ms^-1)
  • a= acceleration (ms^2)
  • t= time (s)

so for constant acceleration in one direction, the equations you need to remember can be derived from the velocity time graph:

  • the gradient of the line is acceleration, a=v-u/t => v=u+at
  • the area under the graph is displacement, s=1/2(u+v)t (sideways trapezium)
  • s=ut+1/2 at^2 (area of the ut rectangle + triangle)
  • s=vt-1/2 at^2 (area of the vt rectangle - triangle)
  • v^2=u^2+2as (substituting equations into each other)

kinematics in 2D

There are two types of quantities, scalar and vector.

Scalar refers to the fact that it only measures magnitude (size) , eg. distance, speed

Vector refers to quantities that measures magnitude and direction, eg. displacement, velocity.

Vectors can be expressed by magnitude and direction (angle from horizontal)

They can also be given in column vectors or in ɨ's and ɉ's (underlined/in bold to show theyre vectors) that are resolved (broken) into two components at right angles. (pythag can be used to find the actual direction)

the top number of the column vector or ɨ represents the distance in the x-direction

the bottom number of the column vector or ɉ the distance in the y-direction.

the resultant displacement is the overall displacement travelled, eg. A→B→C , the resultant displacement would be A→→C, which is found by adding together A→B and B→C (draw it as triangle ABC to see what i mean)

The position vector shows the displacement from a fixed origin. The final position vector is the initial position vector + displacement.

The resultant velocity of a boat travelling on a river which has a current is velocity of boat relative to water+velocity of the water. They like this example, as its a 'real life' situation, and the boat moves in the y-direction while the current goes in the x-direction. You'll need a bit of basic trigonometry for vectors.

Acceleration is a vector quantity with both magnitude and direction for constant acceleration, acceleration=change in velocity/time (a=v-u/t)

The constant acceleration equations can be used in 2-D as well, however, v^2=u^2+2as can't be used, as it involves multiplying vectors together which is not allowed...

the vertical and horizontal components of vectors are dealt with individually and seperately.

[resultant vectors are drawn with a double headed arrow]


Forces are also a vector quantity, they have a size and direction. They are usually represented by a line in the direction of the force, and the length shows the size.

Two forces in different directions can be combined together by moving one of the forces (not changing angle or size) head to tail untill it makes a triangle. the resultant force is the last side of the triangle. (called the trapezium rule, as you can also repeat both lines, and make a trapezium shape)

Forces that balance each other/when there is no resultant force is said to be in equilibrium eg. if there are two forces, they are both equal and opposite. If there are three forces, two of the forces would balance the third force etc.

Sometimes resolving forces (breaking it down to the horizontal (↔) and vertical (↨) components are useful.

If the force F and the angle to the horizontal ϴ is known then the components of the force (↔) is Fcosϴ (the line is closing through the angle; close=cos)

The (↨) would be Fsinϴ

coplanar forces refer to three or more forces that act in the same plane. (horizontal or vertical plane)

resolving the coplanar forces mean that you can work out the resultant force. resolve the (↔) and (↨) seperately, and then use the triangle/trapezium rule and pythag.

A few common forces are weight (gravity), tension, and thrust.

As with most of mechanics, the first thing you do when solving a problem is to draw a diagram. You draw the particle and then all the forces acting on it. (draw them as arrows that come out of the object)

You will also need to use modelling assumptions, the object is modelled as particle (forces act on a single point of the object, the size and shape of the object doesnt matter) strings are light (weightless) and inextensible, friction and air resistance ignored when weak, pulleys are smooth (frictionless) tension is equal throughout the string)

The normal reaction force on an object (R) is the force that counteracts the weight. Normal refers to at an right angle to the surface.

μ refers to the coefficient of friction (roughness of an object), which varies between materials

If an object is on a rough surface, the maximum friction is Fr˅max=μR, which occurs when the object moves or is at limiting equilibrium.

When the object is just about to move (when the surface frictional force has reached max) the object is in limiting equilibrium. Fr=μR

When the object is kept in equilibrium by a friction (Fr) [the book uses F which can be mistaken for force so use Fr] then Fr≤μR This is because friction might not be acting at its maximum yet.

When in equilibrium, but not limiting Fr<μR


the momentum of a moving object is a vector quanitity. mass x velocity.

momentum is always conserved, so if two objects collide on a straight line, the total momentum before is the same as total after.

the principle of convservation of momentum can be expressed m˅1 u˅1 + m˅2 u˅2 = m˅1 v˅1 + m˅2 v˅2(m=mass, u/v=initial/final velocity)

however, sometimes the objects coalesce, therefore, they become like one object. m˅1 u˅1 + m˅2 u˅2 = V (m˅1 + m˅2)

when objects are moving backwards, they have a negative velocity.

the principle of momentum can be applied to vectors in 2D too.

Newtons laws of motion (part 1 and 2)

  1. an object at rest or moving in a straight line with constant velocity will continue like that unless acted upon by cleric.
  2. if a force of F newtons acting on an object of mass m kg there will be an acceleration of a ms^-2. The acceleration will continue as long as the force is continually applied. (F=ma)
  3. If object A exerts a force on B, B exerts an equal and opposite force on A. (action and reaction are equal and opposite)

if an object is moving with constant velocity, acceleration is zero. Therefore resultant force is zero/object is in equilibrium.

as force and acceleration are both vector quantities, you might also need to resolve forces, if you are using vectors.

When there is no air resistance, objects accelerate as they fall because of gravity (g) which is approximately 9.8ms^-2 (g does not mean grams! kg is the Si base unit, dont take g=9.81ms^-2, this isnt physics!)

a mass of m kg has a weight of mg newtons. (weiGht, spelt with g, is measure in N as its a force)

When you have to resolve forces on an inclined plane, the best thing to do is to resolve normal/perpendicular to the plane and parllel to it, using the trapezium rule and Fcosϴ/Fsinϴ (sin slope, sin = down the slope)

tension acts on both objects that are connected through a rope (newtons 3rd law)


projectiles are particles that are dropped, thrown or launched, moving in two dimensions, acted on by gravity only.

for projectiles air resistance, spin etc. are negliable. If they werent negliable, they wouldnt be projectiles as other forces would be acting.

the range of the projectile is the horizontal distance covered.

the time of flight is the time that the projectile is in the air.

maximum height is the highest part of the projectile. As the flight is a parabola and so is symetrical, maximum height is in the middle.

at maximum height, the vertical velocity is 0

(↨) the only force acting on the freely falling object is the weight, therefore the magnitude and direction of its velocity changes.

(↔) velocity is constant as air resistance is not taken into account.

the (↔) is independant of the (↨) so the two forces are resolved separately.

To work out things from a time t after release, change the T in the SUVAT equations, to find when the projectile lands, S=0 etc.

When the particle is realeased from a height h, the height is added on the (↨) after the equation. eg. S=vt-1/2 at^2 +h