A Level Mechanics 1 for AQA Revision Cards

• v = u + at
• s = ut + ½at²
• s = ½(u + v)t
• v² = u² + 2as
• s = vt - ½at²

s = displacement (m)

u = initial velocity (ms-1)

v = final velocity (ms-1)

a = acceleration (ms-2)

t = time (seconds)

• These are not given in the exam.
• Questions will usually give 3 variables, your task is to find a missing fourth.

Displacement-Time:

• The gradient = velocity.
• On a distance-time graph the gradient = speed.

Velocity-Time:

• The gradient = acceleration.
• The area under the graph = distance.

Acceleration-Time:

• Area = velocity.

A particle projected with speed u at an angle α to the horizontal has two components of initial velocity - one horizontal (x-axis) and one vertical (y-axis). They are called the x and y components.

The vertical and horizontal motions should be considered separately. The time is the same for both, so can be used to connect the two components.

The vertical motion:

• Has constant acceleration under gravity, so a = 9.81ms-2.
• SUVAT can be used to calculate the required variable.

The horizontal motion:

• Air resistance is ignored, so acceleration = 0.
• distance = speed x time can often be used instead of SUVAT.

Projection at an angle:

• Resolve the initial velocity into horizontal and vertical components.
• Use the vertical component to work out how long it's in the air and / or how high it goes.
• Use the horizontal component to work out how far it goes whilst it's in the air.

• Vectors have magnitude and direction. Scalars have only magnitude.

• Vectors can be added by drawing the arrows nose to tail. The single vector that is formed is called the resultant vector.

Vectors can be described using the i + j unit vectors.

• They each have a magnitude of 1 and a direction.
• i is in the direction of the x-axis (horizontal, or east).
• j is in the direction of the y-axis (vertical, or north).

• A vector can be used to describe the position of a point in relation to the origin, O.

• Weight (W): Weight = mass x gravity. It always acts downwards.
• The Normal Reaction (R or N): The reaction from a surface. It is always at 90º to the surface.
• Tension (T): Force in a taut rope, wire or string.
• Friction (F): Due to the roughness between a body and a surface. Always acts against motion, or likely motion.
• Thrust: Force in a rod.

• Particle: The body is a point so its dimensions don't matter.
• Light: The body has no mass.
• Static: Not moving.
• Rough: The surface will oppose motion with friction / drag.
• Beam or rod: A long particle.
• Uniform: The mass is evenly spread out throughout the body.
• Non-Uniform: The mass is unevenly spread out.
• Rigid: The body does not bend.
• Thin: the body has no thickness.
• Equilibrium: Nothing's moving.
• Plane: A flat surface.
• Inextensible: The body can't be stretched.
• Smooth: The surface doesn't have friction / drag opposing motion.

• This is when you make assumptions, called 'modelling', to make a real-life situation simpler.
• A model can be improved by making more (or fewer) assumptions. Solve the problem using the initial assumptions, compare it to real life, evaluate the model and then use that information to change the assumptions.

• Forces have an i and j component (horizontal and vertical). Trigonometry is required to find these.
• When finding the resultant, the forces should be positioned 'nose to tail'. Pythagoras is used to find the magnitude of the force in N, and trigonometry finds the angle above the horizontal.

• When resolving more than two forces, resolve parallel to the x-axis and parallel to the y-axis, to get two components i and j.
• Resolving (parallel to x) 5 N + 10cos60^oN - 8cos30^oN = 4.86 N (3 s.f.)
• Resolving (parallel to y) - 30 + 10sin60^oN + 8sincos30^oN = - 15.2 N (3 s.f.)
• Turning this into a vector, we get: 4.86i - 15.2j
• Finally, it's just a case of finding the magnitude, 15.9 N, and direction, 72.3^o by finding the resultant of the i + j vector.

• Forces acting on a particle in equilibrium add up to zero force.

• Resolving ---> - 8cos30 + 5 + pcosΘ = 0, so pcosΘ = 1.9282
• Resolving 8sin30 + psinΘ = 0, so psinΘ = 4

• If we divide the second equation by the first, we get:
• tanΘ = 2.07447, so Θ = 64.3^o (3 s.f.)
• Now, if we sub Θ into psinΘ = 4, we can find p:
• 0.9008p = 4 so, p = 4.44 N (3 s.f.)
• Sometimes the sine rule or cosine rule is needed to solve these questions.

Three forces need to be considered:

• Normal force (N) exerted on the body due to the force of gravity (mgcosθ).
• The force due to gravity (mg) acting vertically downwards.
• The frictional force (f) acting parallel to the plane.

• The gravitational force (mg) can be decomposed into two vectors, one perpendicular to the plane and one parallel to the plane.
• Since there is no movement perpendicular to the plane, the component of the gravitational force in this direction (mgcosθ) is equal and opposite to the normal reaction force (N).
• If the component of the gravitational force parallel to the plane (mgsinθ) is greater than the static frictional force (f_s) then the body will slide down the plane with acceleration (g sin θ − f_k/m).
• When θ is 0, sin θ is also 0 so the body doesn't move.

Limiting Equilibrium:

• The frictional force between two objects is not constant, but increases until it reaches a maximum value. When the frictional force is at its maximum, the body in question will either be moving or will be on the point of moving.
• When the friction is at its maximum possible value, it is said to be limiting. If friction is limiting but the body is stationary, the body is said to be in limiting equilibrium.

The Coefficient of Friction:

• Represents the friction between two surfaces.
• The maximum frictional force is equal to the coefficient of friction x the normal reaction force.
• F = µR
• This frictional force 'F' will act parallel to the plane in a direction that opposes movement.

• Resolving up the plane: F - 5gsin30 = 0
• Resolving perpendicular to the plane: R = 5gcos30
• In limiting equilibrium, so F = mR
• 5gsin30 = m5gcos30
• m = sin30/cos30 = 0.577 (3sf)

Newton's First Law

• A body will stay at rest or maintain a constant velocity unless an external force acts to change that motion.
• This means that in order for the acceleration of a body to change, there must be a net force applied to the body.
• If we are told that a body is not accelerating (i.e. moving at a constant velocity), the resultant force in any direction is 0.

Newton's Second Law

• The rate of change in momentum of the body is directly proportional to the net force applied.
• F = ma.
• The overall resultant force 'F' is equal to the mass multiplied by the acceleration. 'F' and 'a' are in the same direction.
• A consequence of Newton's second law is the relation of weight, gravity and mass: W = mg.

Newton's Third Law

• Every action has an equal and opposite reaction.
• For two bodies in contact with each other, the force each applies to the other is equal in magnitude but opposite in direction.

A body of mass 5kg lies on a smooth horizontal table. It is connected by a light inextensible string, which passes over a smooth pulley at the edge of the table, to another body of mass 3kg which is hanging freely. The system is released from rest. Find the tension in the string and the acceleration.

• Since the string is inextensible and the two bodies are connected, they will both accelerate at the same rate. If the body hanging freely accelerates at a certain rate, the string will pull the other body so that it accelerates at the same rate.

• The tension in the string on both sides of the pulley will also be the same. This is because the pulley is smooth.

• Remember, W = mg. The mass of the first body is 5 kg and so its weight is 5g N. The mass of the second body is 3 kg and so its weight is 3g N.

• Remember to mark in the normal reaction force, a consequence of Newton's Third Law.

Using Newton's Second Law on body A (horizontally):

(F = ma)
T = 5a     (1)

Using Newton's Second Law on body B (vertically):

3g - T = 3a    (2)

Solving (1) and (2) simultaneously:
3g - 5a = 3a
8a = 3g
a = 3g/8
T = 15g/8

Therefore the acceleration is 3.68 ms^-2 (3sf) and the tension is 18.4 N (3sf)

Finding the normal:

• The box is not moving in the y-direction, so N must equal the y-component of gravity.
• N = mg cos 30 = 86.6N

Finding the acceleration:

• The force pulling the box in the positive x-direction has a magnitude of mg sin 30.
• Using Newton's Second law, F = ma, acceleration can be found.
• ma = mg sin 30
• a = g sin 30
• a = 5ms-2

• Particles connected together have the same speeds and accelerations.

Frictionless Inclined Planes with Pulleys:

• Assume that mass M > m sin θ, so M will pull m up the plane.
• The particles are connected, so will have the same velocities, tensions, and accelerations.

Finding the acceleration:

• For mass M: µMg - T = Ma (from F = ma)
• For mass m: T - mg = ma (from F = ma)
• mg sin θ - Mg = (m + N)a

• Some problems involve pulleys. In Mechanics 1 you can always assume that the tension in a string will be the same either side of a smooth pulley.
• The heavier weight pulls on the lighter weight causing them both to accelerate in one direction with a common acceleration.

Calculation of acceleration:

Finding the Tension:

• Have the same speeds and accelerations.
• The tension in the string will be the same each side of a smooth pulley.
• Use F = ma in the direction each particle moves.
• F ≤ µR on rough planes.

• Momentum = Mass x Velocity
• Impulse = change in momentum = force x time
• Momentum is always conserved, so the total momentum in a particular direction before collision = the total momentum in a particular direction after collision.

A ball of mass 3kg is moving to the right with velocity 3ms-1 and a ball of mass 1kg is moving to the left with velocity 2ms-1. After the collision, the ball of mass 1kg moves away with velocity 2ms-1.

• The principle of conservation of momentum can be used to determine the velocity of the other ball after collision.
• Initial momentum = 3x3 - 2x1 = 7N.
• Final momentum  = 2x1 - 3x
• 7 = 2 - 3x
• x = -5/3ms-1, 5/3ms-1 to the right.

Particles that join together after impact are said to 'coalesce'. They can be thought of as one object.

Two particles of mass 40g and Mkg move towards each other with speeds of 6ms-1 and 3ms-1 respectively . The particles coalesce after impact and move with a speed of 2ms-1 in the same direction as that of the 40g particle's initial velocity. Find M.

• (0.04 x 6) + (M x - 3) = [(M + 0.04) x 2]
• 0.24 - 3M = 2M + 0.08
• 5M = 0.16
• M = 0.032kg

Momentum of particles moving on a plane:

• The same method is used, just with two components of velocities.
• They will usually be given as column vectors.