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Exam 1 – Review SessionPowerPoint Presentation

Exam 1 – Review Session

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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

PHY231

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

PHY231

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

PHY231

Trigonometry

PHY231

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

PHY231

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 !!

PHY231

- Kinematics – constant acceleration
- Time interval Change in position Change in velocity
- Average velocity Average acceleration
- Important kinematic relations

PHY231

9

2-D - Projectile motion

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

The equations of motion become

PHY231

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 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

PHY231

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

PHY231

yG

Relative motionxG

- 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

- Boat velocity w.r.t. Water
- Similar relations exist for displacement instead of velocity

vBW

vWG

vBG

PHY231

15

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

PHY231

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

PHY231

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

PHY231

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

PHY231

20

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

PHY231

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

Vi=80.8 km/h

truckM

fstatic

- Max friction force
- Newton’s law
- Under constant acc, we can write

-Fbrakes

m

Vi=80.8 km/h

-fstatic

M

PHY231

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

PHY231

- A) 2.3 m/s2
- B) 5.9 m/s2
- C) 7.3 m/s2
- D) 9.8 m/s2

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

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