A Level Maths Notes: M2 – Equations for Projectiles A projectile is a body that falls freely under gravity ie the only force acting on it is gravity. In fact this is never strictly true, since there is always air resistance for example, but these other forces are small compared to gravity in most circumstances so we ignore them. We can resolve the initial velocity of the ball vertically and horizontally as shown:

(1)

The horizontal velocity is constant since there are no horizontal forces acting – gravity acts vertically. Horizontally therefore

Vertically however, the particle is accelerating downwards. We usually take this acceleration as negative and equal

to Vertically, we can apply all the normal suvat equations:

We get,

(2)

(3)

(4)

becomes which we have already.

We can also derive an equation between and by rearranging (1) to give and substituting this into (3):

(5)

We can find the maximum height using with obtaining

We can find the range by substituting in (5) to get Since

the maximum range is given by when

A Level Maths Notes: M2 – Centers of Gravity of Lamina We can find the distance of the center of gravity from a line of a point by taking moments about the point for each part of the lamina. We equate this to the moment of the lamina as a whole and solve the equation to find the distance of the center of gravity from the axis.

/P> The lamina below has uniform density, so the center of gravity of P is at the center of P and the center of gravity of Q is at the center of Q. Taking moments about AB, we can draw up the table:

Section Distance of center of gravity from AB

Area Distance * Area

P

Q

Lamina

Section Distance of center of gravity

from A’B’ Area Distance * Area

P

Q

Lamina

A Level Maths Notes: M2 – Work Done to Overcome Friction Energy is often lost when things move as friction is usually present and energy must be used to overcome it. This usually means that the speed of a body at the end of a period of motion is less than it would have been, or that energy has to be given to a body to enable it to overcome friction,. Or that the potential energy of a body at the end of a period of motion is less than it would otherwise have been. The general equation to describe the loss of energy as work is done to overcome friction is: Initial Energy=Final Energy+Work Done to Overcome Friction or Initial Energy=Final Energy+Force*Distance Moved in the Direction of the Force Initial and Final Energy may be any of:

Kinetic Energy

Gravitational Potential Energy

Elastic potential energy, also known as the energy stored in an elastic string or spring, which is equal to for a

spring where and is equal to for an elastic string, where

As energy is changed from one form to another, some is always lost to overcome friction, and though often it is a good approximation to ignore it, it is always exactly that: an approximation. Even a spring or string loses energy when it is compressed and stretched. This is illustrated on the hysteresis loop below.

The diagram on the loses no energy in being stretched. The diagram on the right illustrated the work done per unit volume, shown by the red area. This is work done to overcome internal forces and cannot be recovered.

A Level Maths Notes: M2 – Force and Power

One of the basic equations of physics is or If the work is done in

a time then we can divide both sides by to obtain

or where and

We can generalize this equation: If a body is travelling at a velocity against a force then the power needed to

overcome the force is It should be noted that at any instant the body may be accelerating or decelerating, so that v is an instantaneous, not necessarily a constant, speed. If a force is applied greater than the force needed to

overcome the resistive force then the body may accelerate, or the potential energy of the body may increase in

some way. Given a power and a velocity we may find the driving force also using Example A lorry of mass 1500 kg moves along a straight horizontal road. The resistance to the motion of the lorry has magnitude 750 N and the lorry’s engine is working at a rate of 36 kW. a) Find the acceleration of the lorry when its speed is 20 m/s?.

The lorry comes to a hill inclined at an angle to the horizontal, where The magnitude of the resistance to motion from non-gravitational forces remains 750 N. The lorry moves up the hill at a constant speed of 20 m/s?. b) Find the rate at which the lorry’s engine is now working.

a)

The resultant force=1800-750=1050N

Now use for the lorry – note that here is the resultant force and not the 750N force in the question.

b) The lorry does work to overcome the force of friction and to increase the gravitational potenital energy by travelling up the hill.

In 1 second the lorry travels a distance v up the hill so from the diagram above, rises a vertical distance

A Level Maths Notes: M2 – Ladders on Rough Ground or Leaning against a Rough Wall Problems involving ladders usually involve taking moments and resolving vertically and horizontally. There are several points to remember: 1. A reaction at a surface can always be resolved into two components.

a) A Normal Reaction which always acts perpendicular to a surface.

b) A Force of friction which always acts parallel to a surface. The relationship between and is

is limiting friction – it is the maximum possible value if the friction force. In practice we often taken friction as equal to the limiting value. 2. If a surface is smooth, there is no frictional force, only a normal reaction. There can never be a frictional force with no normal reaction. 3. It is often best to take moments about that point at which friction acts, especially if neither the frictional force nor the normal reaction force is known. 4. If the ladder – shown in the diagrams below as a rectangle – is uniform, the centre of gravity is at the centre of the ladder.

The ladder is usually assumed to be rigid, so that it does not bend or sag when a weight is put on it.

A Level Maths Notes: M2 – Centre of Gravity Tables

Description Shape Area

Triangle

Quarter Circle

Semicircle

0

Sector of Circle

0

One Quarter the Circumference of a Circle in the 1

st Quadrant

Half the Circumference of a Cicle Above the x – axis

0

Cone

0

Hemisphere

0

A Level Maths Notes: M2 – Restitution, Momentum, Collisions When two particles collide, of course we know that momentum is conserved. On top of this, particles also obey

Newton’s Law of Restitution: where is the coefficient of restitution. The coefficient is taken to be a fixed constant for collisions between any two bodies. We can then write down two equations from the law of conservation of momentum and Newton’s Law of Restitution and solve them simultaneously to find the

velocities of the two particles after the collision. For example, if :

From the first diagram, the momentum

is

Therefore, from Law of Conservation of Momentum,

From Newton’s Law of Restitution, We solve the simultaneous equations,

(1)

(2) (1)+7*(2) gives

+

Sub this value of into (2) to get

Since the collision is inelastic and kinetic energy is lost. The initial kinetic energy is

The final kinetic energy is

So of energy is lost.

A Level Maths Notes: M2 – Finding the Angle a Hanging Lamina Makes With the Vertical

The center of gravity of a hanging body always hangs vertically below the point of suspension. If we know the distance of the center of gravity from two axes, we can use the rules of trigonometry to find the angle a line drawn in the lamina makes with the vertical. The lamina is hung from G, with the center of gravity O vertically below G. GHF is a right angled triangle and

Hanging the lamina from a different point will mean the lamina makes a different angle with the vertical, but the centre of gravity will always be below the point of suspension.

A Level Maths Notes: M2 – Acceleration, Velocity and Displacement

The relationship between displacement, velocity and acceleration is summarized in the following diagram. Given the displacement, to find the velocity, we differentiate. To find the acceleration, we differentiate twice. Given the acceleration, to find the velocity, we integrate. To find the displacement, we integrate twice. Example

A particle P moves on the - axis. At time t seconds, its acceleration is When its velocity is

When the displacement is 2m. Find expressions in terms of for the velocity and displacement.

When

When Example:

The displacement of a particle is Find the velocity and acceleration in terms of t. When is the

particle at rest for ? What are the displacement and acceleration at this time?

When the particle is at rest,

The particle is at rest when or

The displacement when is

The acceleration when is

A Level Maths Notes: M2 Impulse in Two Dimensions Impulse in a vector. An object may experience an impulse acting at any angle to it’s path, forcing it to change direction by any angle. We should always represent impulse by a vector. This may be strictly unnecessary if the impulse acts along the path of the velocity, but it becomes essential if the impulse does not act along the path of the velocity.

We can use the equation where and are all vectors.

We can use this equation to find any of given the other three. It should also be noted that

1. If in a collision a particle experiences an impulse the other particle experiences an impulse of This is Newton’s 3

rd Law – To every action there is an equal and opposite reaction.

2. In general collisions result in a loss of kinetic energy: - individual kinetic energies may increase, but the overall kinetic energy may not. Only in perfectly elastic collisions is kinetic energy conserved. Example

A ball has mass 0.2 kg. It is moving with velocity when it is struck by a bat. The bat

Exerts an impulse of on the ball. Find a) The velocity of the ball immediately after the impact,. b) The angle through which the ball is deflected as a result of the impact. c) The kinetic energy lost by the ball in the impact.

b)

c) Courtesy A Star Maths And Physics (Your Notes Are Brilliant).

A Level Maths Notes: M2 – The Work Energy Principle The Work Energy Principle is a version of the Law of Conservation of Energy. It takes into account that energy is not conserved, or at least that useful energy is not conserved. In almost any process in practice, work must be done to overcome friction, and this work that is done cannot be recovered. The Work Energy Principle states:

Most of the time we want to minimize the work done since this usually represents wasted energy. Sometimes though, we want to maximize the work done, as when a plane lands and needs to decelerate, so undergoes air braking. Example

In a ski-jump competition, a skier of mass 80 kg moves from rest at a point A on a ski-slope. The skier’s path is an arc AB. The starting point A of the slope is 32.5 m above horizontal ground. The end B of the slope is 8.1 m above the ground. When the skier reaches B, she is travelling at 20 m/s. The distance along the slope from A to B is 60 m. The resistance to motion while she is on the slope is modeled as a force of constant magnitude R newton’s. By using the work-energy principle, find the value of R.

is not in fact a constant force. Because the speed and the direction of motion are changing all the time, the

normal reaction is always changing, hence the frictional force is always changing too.