unit 1 - 1.1,1.2,1.3,1.4 - vector, motion, forces

7/24/2019 Unit 1 - 1.1,1.2,1.3,1.4 - Vector, Motion, Forces

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HERIOT WATT UNIVERSITY (MALAYSIA)

Foundation in Science

1

Unit 1 : Mechanics & Properties of Matter

1.1 Vector Analysis

Vectors & Scalars Qty

Difference between distance & displacement

Distance total path length travelled, only the magnitude/ size is concerned.Displacement - direct length from start point to end point, both magnitude ( length) &

direction must be given.

Vectors Scalars

Displacement (m) Distance (m)

Velocity (ms-1) Speed (ms-1)

Acceleration (ms-2) Time (s)

Force (N) Mass (kg)

Momentum (kgms-1) Energy (J)


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2


Distance & Displacement

A

B

3m

4m

1)Determine the distance from point A to point B.

2) Determine the displacement from point A to pint B.(hint: how far it is from A to B? )

a) Using scale diagram b) Using vector diagram1cm : 1m

FiS HWUM Ph sics TWW


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3


Speed & Velocity

avg. speed , =

avg. velocity, =

* need to include the direction* ex: Car travelling at 20 ms-1at N45E

Vector AdditionWhen vectors are being added, both their magnitude & direction need to be considered.Vector can be represented by a line with arrow where

the length of the line represent the magnitude of the vectorthe arrow of the line represent the direction of the vector

Vectors can be added by using(i) scale diagram (ii) Vector diagram

step 1: choose an appropriate scale, ex: 1cm: 1N * similar to scale diagram, but instead ofstep 2: re-arrange the lines so that the arrows show measuring using ruler, the length and

1 start point & 1 end point direction of the resultant vector isstep 3: joint the start point & end point to measure calculated using mathematical method.

the length ( magnitude) & use protactor to

determine the angle ( direction)



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4

1.1 : Vector Preliminaries

Exercise:1) Three forces, A,B and C, with 10N each, acting in the direction as shown below:

Construct a vector diagram for the case below and hence, find the resultant force:a) FA+ FB

45

FA= 10N

FB= 10N FC= 10N



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5


Exercise:1) b) FA+FC

c) FAFB



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Exercise:d) FA- FC

e) FA+ FB+ Fc



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1.1 Vector Preliminaries

Vector Resolution

http://www.frontiernet.net/~imaging/vector_calculator.html

-All vectors can be resolved into 2 mutually components. i.e vxand vy.-By solving the right-angled triangle,

v = vx+ vy

wherevx= v cos ,

vy= v sin

-Inversely, we can find the vector, v from its x- , y- components using Pythagoras theorem where

Magnitude: Direction

v2= vx2+ vy

2v = tan = = tan-1

2

y

2

x vv

vv

vy

v

vx

vy

vx

x

y

v

v

x

y

v

v



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8


Example:1)Two horizontal forces act at the same point on a body. One force of magnitude 10N acts towards the

east. The second force of magnitude 10N acts at north 50 east. Determine the resultant of the twoforces.

40

50

F1= 10N

F2= 10N

N

40

F1= 10N

F2= 10N

R

F1= 10N

F2= 10N

R

C

BA

Resultant Force, R can be calculated by solving the vector diagram using triangle rules:1) Magnitude

from a2= b2+ c22bc cos A AC2= AB2+ BC2(AB)(BC) cos

R2= 102+ 102(10)(10) cos 140

R = 18.8N

2) Angle / Direction of R,

So,= = 20(isosceles triangle)

Resultant force, R = 18.8N in the direction N70E / above the horizontal.

ACBBAC

ABC

40140

2

140180



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9

CHAPTER 1 : Vector Preliminaries

Exercise:

2) Two dogs are pulling a sandal in the direction as shown below. Jack tries to save his sandal fromthe dogs and he pulls the sandal towards him. Construct a vector diagram and calculate how muchforce he needs to pull the sandal so that his sandal is at static equilibrium.

FA

= 20NFB= 20N

FJack= ? N

90



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1.2 Equation of Motion

Acceleration rate of change of velocity

For an object moving in a straight line with constant/uniform acceleration, the graphs ofvelocity Vs time are shown as below:

where v = final velocityu = initial velocitytf= final time takenti = initial timeif tt

uv

t

va

i) a= 0 ms-2 i i ) a= + ve value i i i ) a= -ve value

V / ms-1 V / ms-1V / ms-1

t / s t / s t / s


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Constant velocity & constant acceleration

v / ms-1

t / s

A B

CO

A car travel with velocity changes over time as shownon the graph. What can you conclude about thevelocity and acceleration of the car from the graph ?

path OA

path AB

path BC

velocity = linearly increasingacceleration = constant +ve a

velocity = constantacceleration = 0

velocity = linearly decreasing

acceleration = constant ve a

i) a= 0 ms-2 i i ) a= + ve value i i i ) a= -ve value

a / ms-2 a / ms-2a / ms-2

t / st / s

t / s


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In a linear / straight line motion with constant acceleration , we can analyze the motion via linear motionequation as below:

Derivation of the above equation :

Knowing that and the graph of v vs t is shown as below, derive the above equations.

1) v = u + at2) v

2= u 2+ 2as3) s = ut + at2

where v = final velocity at time tu = initial velocity at t = 0t = time takena = acceleration

s = displacement travelled

t

uva


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Derivation of the equation of motion :

1) v = u + at2) s = ut + at23) v

2= u 2+ 2as

1) v = u + at

From

Rearrange the formula:

v = u + at

2) s = ut + at2

From the graph of v vs t above,Displacement (s ) = v t

= area underthe graph

s = area of A + area of B

=

(v-u) t + ut

=

(at) t + ut

=

at2+ ut

3) v2= u2+ 2as

From v = u + at, square theequation become

(v) 2= (u +at )2

= u2+ 2uat + a2t2

= u2+ 2a ( ut +

at2)

= u2+ 2as

t

uva


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Examples of motion:

1) linear motionEx: A car is accelerated at 6 ms-2 from an initial velocity of 2ms-1for 10s. What is

(a) the final velocity?(b) the distance moved?

u

v

st = 0 t = 10s


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Examples of motion:

2) Vertical motion upwardsEx: An object is thrown vertically upwards at a speed of 40 ms-1. Determine

(a) the time taken by the object to reach maximum height(b) the maximum height reached

(c) the velocity at 2.0s after throwing the object

At max. height, v = 0

Since object only experience

gravitational force while travelling

In the air, , a = - g

vector directed up+ve value

vector directed down-ve value


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u

ux

uy Height = Sy

A

B

C

ux

vvyay= -g


Examples of motion:

3) 2 Dimensional Motion - Projectile motionany object that has been projected/ launched/ thrown/ fired at some ANGLE,into the air, near the

surface of the Earth and move in a parabolic path

At max. height, vy

= 0

Since object only experience

gravitational force while travelling

In the air, , a = - g at any point

vector directed up+ve value

vector directed down-ve value

Range = Sx

Resolve all vectors into vertical &

horizontal component, i.e

uy, ux,

vy, vx ay= -g, ax= 0

sy, sx


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Examples of motion:

3) Projectile motion

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initial point A B(max height) at any point

1)ux= ucos 1)instantaneous velocity, v = ux 1) instantaneous velocity, v

uy= usin (vy= 0) v

=

2) Neglect air resistance, 2) tan =the horizontal velocity, uxisalways constant at all point ( ux= vx)

3) ax= 0 at all point

* Acceleration of the object at any point, ay = - g

22

yx vu

x

y

v

v

Then we can use the kinematic

equation to study the

Projectile motion

1) Sx= uxt

2) vy= uy+ ayt

3) vy2= uy

2+ 2 ays

4) sy= uyt + ayt2



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Example:

1) An object is thrown upwards at a speed of 30ms-1at an angle of projection of 60. Determinea) the time taken by the object to reach maximum heightb) the maximum height reachedc) the ranged) the time taken to reach a height of 30me) the velocity of the object at a height of 30m(Assume g= 9.8ms-2)



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condition When m isconstant

When a isconstant

Whenforce isconstant

When a isconstant

When m isconstant

Tick if true

1.3 Newtons 2ndLaw, energy & power

Newtons 1stLaw object will remain at rest or continue to move with constant speed in straight lineunless it is acted by an external force

Newtons 2ndLaw - maFwhere F = force / N

m = mass of the objecta = acceleration

The SI unit for Force ( F) = kgms-2Common written as Newton ( N) where

1N = resultant force which will cause a mass of 1 kg to accelerate at 1ms -2.

this also implies that

aF

mF

ma

Fm

Fa

where F = force / Nm = mass of the objecta = acceleration


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When there are more forces acting on an object, the resultant/ net force acting on the object need tobe found considering both the magnitude & direction ( Force is a vector) .

Examples of forces acting on an object and its motion :

1) Free Body DiagramEx: When a rocket take off, the thrust on a rocket of mass 8000kg is 200 kN, find the acceleration of

the rocket.

Thrust Force = 200kN

Weight/

Gravitational Force = 200kN


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2) In a lift ( apparent force)Ex: A student of mass 40kg stands inside a stationary lift initially. Determine the apparent weight of the

student if the lift moves(a) downwards with constant speed(b) downwards with a constant acceleration of 0.50 ms-2.(c) upwards with a constant acceleration of 0.50 ms-2


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3) On an inclined plane ( resolution of force)Ex:A wooden block of mass 2 kg is placed on a slope at 30to the horizontal as shown. A frictional force

of 4 N acts up the slope. The block slides down the slope for a distance of 3 m.Determine(i) the acceleration of the block(ii) the speed of the block at the bottom of the slope.


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Energy & Power

Power = rate of transformation of energy from one form to another form

Ex:

t

W

time

doneWork

time

EnergyP

whereP = power ( unit: Watt (W) / Js-1 )

W = work done ( J)

t = time (s)

1) A trolley is released down a slope from a height of 0.3 m. If its speed at the bottom is found to be2 ms-1, find a) the energy difference between the Epat top and Ek at the bottom.

b) the work done by frictionc) the force of friction on the trolley


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2) A student of mass 55kg runs up a flight of 50 steps, which each has 20cm high as shown below. Whatis the power generated by the boy if he takes 22s to climb to the top of the stairs ?

20 cm


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1.4 Momentum & Impulse

Momentum = the product of mass and velocity

Law of conservation of momentumTotal momentum before collision is the same as the total momentum after collision, in aclosed/isolated system ( i.e no external force applied to the system.

Generally, we encounter 2 types of collisions:1) Elastic collision both momentum and kinetic energy are conversed2) Inelastic collisionmomentum is conserved but kinetic energy is NOT.

vmp wherep = momentum ( kgms-1)m = mass of the object ( kg)v = velocity ( ms-1)

finalfinalinitialinitial pppp 2121

22112211 vmvmumum


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Exercise

1) ) A bullet of mass 2.0g strikes a metal block of mass 3.0kg suspended freely on a string. The bulletrebounds in the opposite direction at a speed of 250ms-1and the block starts to move at a speed of0.5ms-1. Determine the initial speed of the bullet.

2) An object of mass 400kg was at stationary initially and exploded . After explosion, it breaks up into2 parts, one having a mass of 50kg moving at a speed of 120ms-1in the opposite direction.Determine the velocity of the second part.

3) An object A moving towards object B on the same straight line and finally collide with object B. Afterthe collision, object B moves at a speed of 3.0ms-1to the right. Determine the velocity of object Aafter collision.

2kg

5ms-1 2ms-13kgA B

Before

collision


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Impulse

28Simulation: http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/Collision/jarapplet.html

Newtons 2ndLaw of Motion states that the acceleration of a body is directly proportional to the forcethat causes it.

For a given force, when mass of the body, m , the acceleration, a

For a given force, when mass of the body, m , the acceleration, aF is vector, a is vector, m is scalar. Force F is always in the

SAME direction as the direction of acceleration vector, a.

Also, from , and

can be re-define as

amF

aF

Masscan be used as a measureof an objects resistance to

acceleration (changes of motion)measure of INERTIA of a body

Ex: when a fat boycollide with a thin boy,the fat boy will standstill but the thin boyfalls (change motion!

amF

t

vva

if

amF

tvvmF

if

t

vmF

tvmvmF

if

http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/Collision/jarapplet.htmlhttp://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/Collision/jarapplet.html


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Impulse

2929

From redefined that

** the force on a body is a measure of the rate at which the quantity of changes.

where is defined as a bodys MOMENTUM, p such that

thus, momentum, p is a vector and its direction is the same as v.also, we can then defined the Newtons 2ndlaw of motion as the rate of change of momentum of the

the body which it acts on.

therefore, unit p can also be represented in sN (sec Newtons)

tvmF

vm

vm

if vmvmp

)()()( 11

msvkgmkgmsp

t

vm

F

t

pF

ptF

F is also known as Impulsive Force inthis Eq. Impulsive force is a force

which exerts on an object for a shortwhile.

Basic ideas used when consideringcar safety design.-> increase to reduce theimpulsive force acted on occupantof car.Ex: seat belt, rubber bumper bars,

Collapsible steering wheel columns

t

t

vmF

Ft = Impulse= change of momentum



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The Law of Conservation of MomentumIn an isolated system ( i.e no external force acts), the momentum of the system remains constant.p1+ p2+ p3+ p4+ .pn= constant if net external force acting on the system is ZERO.

In 2 Body CollisionFor 2 objects moving on the same line and colliding as below:

Before CollisionObject A has mass m1with velocity u1. Object B has mass m2with velocity u2.

After Collision

Object A has mass m1with velocity v1. Object B has mass m2with velocity u2.

Therefore, each body experiences a change in the momentum, and

During impact, each object exerts a force on the other body. (Newtons third law)

111 mumvp

222 mumvp

m2m1m1

u1m1

u2m2 m2

v2

F1

F2 v1

BeforeCollision During

Collision

AfterCollision

21 FF

t

p

t

p

21

21 pp

021

pp

Sum of changes of momentumof the 2 bodies is ZERO.

Simulation: http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/Collision/jarapplet.html

http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/Collision/jarapplet.htmlhttp://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/Collision/jarapplet.html


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On a graph of F vs t, impulse ( Ft) is the area under the graph as shown below:

31

21 FF

t

p

t

p

21

21 pp

021

pp

Sum of changes of

momentum of the 2 bodies isZERO.

Change of momentumexperienced by Object 1 =

change of momentumexperienced by Object 2,but in OPPOSITE direction

Force acting oneach object isequal in sizebut opposite indirection

t is the same forboth since both

object colliding eachother at the sameduration

F / N

t / s

Area under the graph = impulse



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Exercise:1) A tennis ball of mass 0.1 kg hit horizontally on a vertical wall at initial speed of 10ms-1, is rebound in

the opposite direction by the wall, with a speed of 10ms-1. the duration of the impact occur for 50 ms.Calculate the impulsive force exerted on the ball by the wall.

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Extended Readings:

1) Free fall projectile

2) Frictional force

3) Heat & thermodynamics

This will help you to understand more on

Kinematics of Motion

unit 1 - 1.1,1.2,1.3,1.4 - vector, motion, forces

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