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Exam 1 – Review Session. Clarification: When I write w.r.t , it means “with respect to”. Overview. Wednesday September 30 th 9:10-10 am BPS 1410 Exam 1 includes Introduction Motion in 1D Motion in 2D Laws of Motion Multiple choice questions only. Overview. Introduction

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Exam 1 review session

Exam 1 – Review Session

Clarification:

When I write w.r.t , it means “with respect to”

PHY231


Overview
Overview

  • Wednesday September 30th 9:10-10 am BPS 1410

  • Exam 1 includes

    • Introduction

    • Motion in 1D

    • Motion in 2D

    • Laws of Motion

  • Multiple choice questions only

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Overview1
Overview

  • Introduction

  • Kinematics

    • Motion of an object under constant acceleration

    • 1-D and 2-D motion

  • Laws of motion

    • Newton’s concept of forces, F=ma

    • Apply kinematics to determine object’s motion

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Si units
SI - Units

  • SI - base units

    Length metre m

    Mass kilogram kg

    Time second s

    Electric current ampere A

    Thermodynamic temp. degree Kelvin °K

    Luminous intensity candela cd

    Amount of substance mole mol

  • All other units are derived from these

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Trigonometry
Trigonometry

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Vector components
Vector components

A vector can also be parameterized using its magnitude (I.e. length) and its direction (I.e. angle w.r.t. the coordinate system)

Consider

The magnitude of V and angle w.r.t. the x-axis are

One also has

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Quadratic equation
Quadratic equation

  • Reminder for equations of the type

  • (Real) solutions are given by

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2 equations 2 unknowns
2 Equations – 2 unknowns

Isolate 1 unknown and plug into 2nd equation

2 equations – 2 unknowns

Intermediate calculation

Done, verify that it works in initial system of equations !!

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2 d projectile motion
2-D - Projectile motion

The initial conditions can be broken down into its x- and y-components

The equations of motion become

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Projectile motion at what d x is the shell hitting hill2
Projectile motionAt what Dx is the shell hitting hill2?

Vi =19 m/s

48º

1

h1=20m

2

h2=13m

y

Vi =19 m/s

vyi =vi*sin48º

x

48º

vxi =vi*cos48º


Area under v t is d x
Area under v(t) is Dx

The area under the graph of v(t) is the displacement Dx !!

v(m/s)

2

1

0

t(s)

0

2

4

5

  • Example

    • An object has vi=1 m/s

    • Accel. with a=0.5 m/s2 for 2s

    • Accel. with a=0.0 m/s2 for 2s

    • Decel. with a=-2.0 m/s2 for 1s

  • What is the total displacement of the objects?

  • Answer : Dx = 8m

1m

4m

2m

1m

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Area under a t is d v

The area under the graph of a(t) is the velocity change Dv !!

Area under a(t) is Dv

  • Example

    • Accel. with a=1 m/s2 for 2s

    • Accel. with a=0 m/s2 for 2s

    • Decel. with a=-2 m/s2 for 1s

  • What is the total change of velocity?

  • Answer : Dv = 0 m/s

a(m/s2)

1

2m/s

5

4

0

0

2

t(s)

-2m/s

-2

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Relative motion

yG

Relative motion

xG

  • A boat is moving on a river

    • Boat velocity w.r.t. Water

      • vBW

    • Water velocity w.r.t. Ground

      • vWG

    • Boat velocity w.r.t. Ground

      • vBG = vBW + vWG

  • Similar relations exist for displacement instead of velocity

vBW

vWG

vBG

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yG

xG

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Newton s laws
Newton’s Laws

  • First Law : If F = 0 then a = 0 and v =constant

  • Second Law : Fnet = ma

  • Third Law : FAB = - FBA

  • First Law: To change the velocity of an object, you must apply a force on it

  • Second Law: Describes the relation between the force and the acceleration

  • Third Law: Action-reaction, A applies force F on B necessarily means B applies –F on A

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Mass

  • A measure of the resistance of an object to changes in its motion

  • The larger the mass, the less it accelerates under the action of a given force

  • SI unit: kg

  • Scalar quantity

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Tension in a rope
Tension in a rope

  • Ignore any frictional effects of the rope

  • Ignore the mass of the rope

  • The magnitude of the force exerted along the rope is called the tension

  • The tension is the same at all points in the rope (magnitude of the tension vector T)

  • The Tension follows the rope. Draw it at the junction of the object and the rope, pointing AWAY form the object

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Sliding box on an inclined frictionless table
Sliding box on an inclined frictionless table

  • What is the box overall acceleration if the angle is equal to 30º ?

  • Forces acting on the box

    • Gravity Fg

    • Normal force n

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Cart fan
Cart + fan

  • A cart with a fan mounting on it (1.5 kg total) is on a frictionless inclined surface. The fan can produce a force of Ffan= 5.0 N. What is the angle the inclined surface should have with respect to the horizontal direction so that no net force is acting on the cart (i.e. fan cancels out gravity). g=9.8 m/s2.

  • A) 10º

  • B) 20º

  • C) 30º

  • D) 45º

y

x

Ffan

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y

x

Ffan

-Fgsinq

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Friction forces
Friction forces

  • Force parallel to the surfaces in contact

  • Static friction

    • No motion between Surfaces in contact

    • Max static friction force

    • n is the normal (perpendicular) force between the surfaces in contact. It is the reaction force of the surface being pushed on

  • Kinetic friction

    • Motion between Surfaces in contact

    • Friction force is constant


Truck

m

Vi=80.8 km/h

truck

M

fstatic

  • Max friction force

  • Newton’s law

  • Under constant acc, we can write

-Fbrakes

m

Vi=80.8 km/h

-fstatic

M

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Friction
friction

  • A mass m1=10 kg is on a table and pulled by a mass m2=20 kg through a rope and a pulley.

  • m1 is sliding to the right and the coefficient of kinetic friction between m1 and the table is 0.20.

  • What is the magnitude of the acceleration of either mass? g=9.81 m/s2.

y

  • A) 2.3 m/s2

  • B) 5.9 m/s2

  • C) 7.3 m/s2

  • D) 9.8 m/s2

x

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Gravitational force
Gravitational force

  • Attractive force between every particles in the universe

    • Proportional to the product of the masses

    • Inversely proportional to the square of the distance


Gravity on earth
Gravity on earth

At the surface of the earth

  • ME ~ 6.0 1024 kg

  • RE ~ 6.4 106 m

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