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# Exam 1 – Review Session - PowerPoint PPT Presentation

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

Clarification:

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

PHY231

• 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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• 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 - 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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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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• Reminder for equations of the type

• (Real) solutions are given by

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

The equations of motion become

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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

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 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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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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• 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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• 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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• 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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• 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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• 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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x

Ffan

-Fgsinq

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• 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

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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• 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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• Attractive force between every particles in the universe

• Proportional to the product of the masses

• Inversely proportional to the square of the distance

At the surface of the earth

• ME ~ 6.0 1024 kg

• RE ~ 6.4 106 m

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