Form IV Physics Workshop

8/10/2019 Form IV Physics Workshop

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Form IV

Physics Workshop

May 2013

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Programme

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Physics

Electric circuits

Optics &

WavesMechanics

Click a link


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Friction

The force that opposes the motion of an

object and acts parallel to the surface the

object is in contact with

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Normal Force

The force exerted by a surface on an object in

contact with it.

The normal force acts perpendicular to the

surface irrespective of whether the plane is

horizontal or inclined

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Forces

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Fg

FN

FappliedFFriction

T

Fg

Fg

FN Fapplied

FFriction

Fx= Fappcos

Fy= Fappsin

Fg

FN

F||= mgsinF|= mgcos


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Free body diagrams

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Fg

FN

FappliedFFriction

T

Fg

Fg

FNFapplied

FFriction Fx= Fappcos

Fy= Fappsin

Fg

FN

F||= mgsinF|= mgcos


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Newtons Laws

Newton I:A body will remain at rest or continuemoving at a constant velocity in a straight line

unless acted upon by an external net force

Newton II:The acceleration of a body is directly

proportional to the net force acting on the bodyand inversely proportional to the mass of the

body (Fnet=ma)

Newton III: If object A exerts a force on object B,then object B will exert an equal but oppositeforce on object A (action-reaction pairs)

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Single object on a flat surface

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Fg

3 kg

FN

FappliedFFriction=10 N

=5 N

Calculate the objects acceleration:Fnet= ma (Newton II)

Fnet= FappliedFfrictionma = FappliedFfriction3a = 105a = 1,67 m.s-2to the right


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Force at an angle

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Calculate the objects acceleration:F

net

= ma (Newton II)

Fnet= Fappliedcos 20oFfriction

ma = Fappliedcos 20oFfriction

3a = 10 cos 20o5 = 9,4 - 5

a = 1,47 m.s-2

to the right

3 kg

Fg

FN Fapplied

FFriction

=10 N

=5 N= 20o


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Hanging from a rope moving down

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T = 22 N

Fg= 19,6N

Calculate the objects acceleration:Fnet= ma (Newton II)

Fnet= Fg -T

ma = Fg -T2a = 19,622 = -1,2 m.s-2

a = 1,2 m.s-2upwards

2 kg

NBif the object is moving upwards, it is accelerating.- if the object is moving downwards, it is decelerating!!


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Two bodies

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3 kg5 kgF app= 50 N

F friction= 6N F friction= 10 N

Free body diagrams

T Ffriction

T

Ffriction

F app= 50 N

Fnet= FappTFfrictionFnet= ma (NII)

ma = FappTFfriction3a = 50T6T = 5063a

Fnet = TFfrictionFnet= ma (NII)

ma = TFfriction5a = T10T = 5a + 10

5063a = 5a + 10 -> a = 4,25 m.s-2to the left


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Two bodies

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Free body diagramsT2

Fg

T1Ffriction= 1 N

Fnet= T1FfrictionFnet= ma (NII)

ma = T1Ffriction3a = T11T1= 3a - 1

Fnet = FgT2Fnet= ma (NII)

ma = FgT21a = 9,8T2T2= 9,8 -1a

But, T1= T2(By NIII)

3a1 = 9,8 - a -> a = 2,7 m.s-2

3 kg

1 kg

F friction= 1N


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Object on a plane

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No frictionit will slide down the planeFnet= mgsinFnet= ma5a = (5)(9,8)(sin 20o)

a = 3,35 m.s-2down the slope

=20o

Fg

FN

F||= mgsinF|= mgcos

5 kg object


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Object on a plane

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Frictionobject is standing still / moving at constant v(down the plane)

Fnet= F up plane - F down plane = 0 (Newton I ->constant velocity)

Fnet= Ffrictionmgsin= 0Ffriction= (5)(9,8)(sin20

o) = 16,76 N up the plane

Fg

FN

F||= mgsinF|= mgcos

5 kg object

Ffriction= ???


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Object on a plane

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Friction + ropeobject is standing still / moving at constant v(down the plane)

Fnet= 0; F up plane = Frope+ Ffriction

F down plane = mgsinFrope+ Ffriction= mg sin10 N + Ffriction= (5)(9,8)(sin 20

o)

Ffriction= 6,76 N up the plane

Fg

FN

F||= mgsinF|= mgcos

5 kg object

Ffriction= ???Frope= 10N


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Newton III

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Current electricityCheat Sheet

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Current electricity - Examples

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In the circuit diagram below, the battery has an emf of 12V and an internal

resistance of 0,8 . The resistance of the ammeter and connecting wires may beignored.

Calculate the:

1. Effective resistance of the circuit

2. Reading on the ammeter

3. Reading on the voltmeter


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Current electricity - Solution

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Geometrical Optics Revision: explain reflection

Revision: State the law of reflection

Define the speed of light as being constant when passing through agiven medium and having a maximum value of c = 3 x 108m.s-1in avacuum.

Define refraction

Define refractive index as n=c/v

Define optical density

Know that the refractive index is related to the optical density.

Explain that refraction occurs because of a change of wave speed indifferent media, while the frequency remains constant

Define normal; angle of incidence; angle of refraction

Sketch ray diagrams to show the path of a light ray throughdifferent media

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Geometrical Optics

State the relationship between the angles of

incidence and refraction and the refractiveindices of the media when light passes from onemedium into another (Snells Law)

Apply Snells Law to problems involving light rayspassing from one medium into another

Draw ray diagrams showing the path of light

when it travels from a medium with higherrefractive index to one of lower refractive indexand vice versa

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Snells Law

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Snells Law

A light ray with an angle of incidence of 35

passes from water to air. Find the angle ofrefraction using Snells Law if nwater= 1,33.

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Snells Law

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Geometrical Optics

Explain the concept of critical angle

List the conditions required for total internal

reflection

Use Snells Law to calculate the critical angleat the surface between a given pair of media

Explain the use of optical fibres in endoscopes

and telecommunications

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Critical angle

The critical angle is the angle of incidence for

light moving from an optically more dense toan optically less dense medium above which

total internal reflection occurs.

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Optically more dense


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Total Internal Reflection

Total internal reflection takes place when light

travels from one medium to another of loweroptical density. If the angle of incidence is

greater than the critical angle for the medium,

the light will be reflected back into themedium. No refraction takes place.

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Finding the Critical Angle

Given that the refractive indices of air and

water are 1,00 and 1,33 respectively, find thecritical angle.

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2D and 3D Wavefronts

Define a wavefront as an imaginary line that connects

waves that are in phase (e.g. all at the crest of theircycle)

State Huygensprinciple. Define diffraction as the ability of a wave to spread out

in wavefronts as they pass through a small aperture or

around a sharp edge. Diffraction demonstrates thewave nature of light.

Apply Huygensprinciple to explain diffractionqualitatively. Light and dark areas can be described in

terms of constructive and destructive interference ofsecondary wavelets

Sketch the diffraction pattern for a single slit

Understand that

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Diffraction

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How is diffraction explained?

Diffraction can be explained through Huygens

Principle Huygens Principle:Every point on a wavefront is a

centre of disturbance that sends out circular

secondary wavelets. Superposition of these

secondary wavelets

determines the position

of the wavefront at a

later time


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Single Slit Diffraction

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Finished

Rememberthis has been a summary2hours is not enough time to study!

You need to work through all your notes;

tutorials and old exam paperstutorials.

Good luck!

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Form IV Physics Workshop

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