Physics 1201Q Review. 1.7 The Components of a Vector Example A displacement vector has a magnitude of 175 m and points at an angle of 50.0 degrees relative

Physics 1201Q Review

1.7 The Components of a Vector

Example

A displacement vector has a magnitude of 175 m and points atan angle of 50.0 degrees relative to the x axis. Find the x and ycomponents of this vector.

rysin

m 1340.50sinm 175sin ry

rxcos

m 1120.50cosm 175cos rx

yxr ˆm 134ˆm 112

A hiker walks 100m north, 130m northeast, and 120m south. Find the displacement vector and the angle measured from the positive x axis.

N

W E

S

D θ

3.2 Equations of Kinematics in Two Dimensions

tavv xoxx tvvx xox 21

xavv xoxx 222 221 tatvx xox

3.2 Equations of Kinematics in Two Dimensions

tavv yoyy

221 tatvy yoy

tvvy yoy 21

yavv yoyy 222

2.4 Equations of Kinematics for Constant Acceleration

Example 6 Catapulting a Jet

Find its displacement.

sm0ov

??x

2sm31a

sm62v

2.4 Equations of Kinematics for Constant Acceleration

m 62

sm312

sm0sm62

2 2

2222

a

vvx o

2.6 Freely Falling Bodies

Example 12 How High Does it Go?

The referee tosses the coin upwith an initial speed of 5.00m/s.In the absence if air resistance,how high does the coin go aboveits point of release?

2.6 Freely Falling Bodies

y a v vo t

? -9.80 m/s2 0 m/s +5.00 m/s

ayvv o 222 a

vvy o

2

22

m 28.1

sm80.92

sm00.5sm0

2 2

2222

a

vvy o

Free Fall Example• What is the

maximum height the ball reaches?

• How long does it take to reach the maximum height?

• How long is the ball in the air total?

• What is the velocity of the ball just before it hits the ground?

3.3 Projectile Motion

Example 3 A Falling Care Package

The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time requiredfor the care package to hit the ground.


y ay vy voy t-1050 m -9.80 m/s2 0 m/s ?


y ay vy voy t-1050 m -9.80 m/s2 0 m/s ?

221 tatvy yoy 2

21 tay y

s 6.14

sm9.80

m 1050222

ya

yt


Example 4 The Velocity of the Care Package

What are the magnitude and direction of the final velocity ofthe care package?


y ay vy voy t-1050 m -9.80 m/s2 ? 0 m/s 14.6 s


y ay vy voy t-1050 m -9.80 m/s2 ? 0 m/s 14.6 s

sm143

s 6.14sm80.90 2

tavv yoyy

Projectile Motion ExampleA canon is fired with a muzzle velocity of 1000 m/s at an angle of 30°. The projectile fired from the canon lands in the water 40 m below the canon. a) What is the range of the projectile.b) What id the velocity of the projectile in x and y when it landsc) What is the landing angle of the projectile

Knownvo = 1000m/sθ = 30°Δy = -40m

UnknownX, vx, vy, θ, vf

smsmv

smsmv

tvtatvx

oy

ox

oxxox

/50030sin)/1000(

/86630cos)/1000(

221

30°voy

vox

vo

Projectile Motion Example

Use y info to find t

221 tatvy yoy

-40m = (500m/s)t + ½(-9.8m/s2)t2

4.9t2 – 500t – 40 = 0 a b c

Solve for t using quadratic equation

t = -(-500) ± (-500)2 – 4(4.9)(-40) 2(-40)

Take the positive root:t = 102.1s

Projectile Motion Example

mssmx

tvx ox

6.884181.102)/866(

smsmv

msmsmv

yavv

yavv

y

y

yoyy

yoyy

/8.500/8.500

)40)(/8.9(2)/500(

2

2

22

2

22

Calculate x:

Calculate vy

Projectile Motion ExampleFind landing angle and velocity

vx = 866m/s

vf

θ

vy = -500.8m/s

vf = (866m/s)2 + (-500.8m/s)2 = 1000.4m/s

tan θ = -500.8m/s = -0.578 866m/s

θ = tan-1(-0.578)

θ = -30.04°

Inclined Plane Problem

For m1 (taking up the plane as positive)ΣFx = T - W1sin30 = m1a

For m2 (taking down as positive)ΣFy = W2 – T = m2a

Take the direction of motion as positive and use Newton’s Second Law to write equations

M1 = 8kgM2 = 22kg

W = mg\W1 = (8kg)(9.8m/s2) = 78.4N W2 = (22kg)(9.8m/s2)= 215.6N

Inclined Plane ProblemT - W1sin30 = m1a eq. 1

W2 – T = m2a eq. 2

W2 - W1sin30 = m1a + m2a

W2 - W1sin30 = a m1 + m2

215.6N – (78.4N)(sin 30) = 5.88m/s2

8kg + 22kg

M1 = 8kgM2 = 22kg

W = mg\W1 = (8kg)(9.8m/s2) = 78.4N W2 = (22kg)(9.8m/s2)= 215.6N

Solve for T using eq. 2

215.6N – T = (22kg)(5.88m/s2)

T = 86.2N

4.9 Static and Kinetic Frictional Forces

The sled comes to a halt because the kinetic frictional forceopposes its motion and causes the sled to slow down.

4.9 Static and Kinetic Frictional Forces

Suppose the coefficient of kinetic friction is 0.05 and the total mass is 40kg. a) What is the kinetic frictional force?b) How far does the child slide before coming to a stop?

N20sm80.9kg4005.0 2

mgFf kNkk

Child on a sled continued

How far does the child slide before coming to a stop?

Calculate the deceleration of the child:

ΣFx = max

ax = ΣFx = -20N = -0.5m/s2 m 40kg

v = v0 + 2aΔx

Δx = v - v0 0m/s – 4m/s= = 4m

2a 2(-0.5m/s2)

A block having a mass of 5kg is pulled with a force of 20N acting at 30° above the horizontal. The coefficient of friction between the block and the table is 0.2. Find the acceleration of the block.

5.4 Banked Curves

On an unbanked curve, the static frictional forceprovides the centripetal force.

A car rounds a curve having a 100m radius Travelling at 20m/s. What is the minimum Coefficient of friction between the tires and the road required?

Fc = f =Fn

mv2 = mg r

\ = v2 = (20m/s)2

gr (9.8m/s2)(100m)

= 0.41

Fn

f

W = mg

5.5 Satellites in Circular Orbits

r

vm

r

mMG E

2

2

r

GMv E

Fg = Fc


Example 9: Orbital Speed of the Hubble Space Telescope

Determine the speed of the Hubble Space Telescope orbitingat a height of 598 km above the earth’s surface.

hmi16900 sm1056.7

m10598m1038.6

kg1098.5kgmN1067.6

3

36

242211

v


T

r

r

GMv E 2

EGM

rT

232 = 42r3

GM


Global Positioning System

hours 24T

EGM

rT

232 = 42r3

GM

r = 3T2GMe

42

T = (24 hours)(3600s/hour) = 86400s

r = (86400s)2(6.67 x 10-11 Nm2/kg2)(5.98 x 1024kg) 423

r = 42250474m = distance from center of the earth to GPSr = Re + h h = r – Re = 42250474m – 6380000m = 35870474m = 22,300 mi

6.2 The Work-Energy Theorem and Kinetic Energy

Example 4 Deep Space 1

The mass of the space probe is 474-kg and its initial velocityis 275 m/s. If the 56.0-mN force acts on the probe through adisplacement of 2.42×109m, what is its final speed?


2212

f21W omvmv

sF cosW


2

212

f2192- sm275kg 474kg 474m1042.20cosN105.60 v

2212

f21cosF omvmvs

sm805fv

6.5 The Conservation of Mechanical Energy

Conceptual Example 9 The Favorite Swimming Hole

The person starts from rest, with the ropeheld in the horizontal position,swings downward, and then letsgo of the rope. Three forces act on him: his weight, thetension in the rope, and theforce of air resistance.

Calculate the person’sfinal speed if he starts from A height of 8m?

What is the swimmer’s velocity at its lowest point if the rope is 8m long?

of EE

2212

21

ooff mvmghmvmgh

vf = 2gho = (2)(9.8m/s2)(8m)

= 12.5m/s

Example: A marble having a mass of 0.15 kg rolls along the path shown below. A) Calculate the Potential Energy of the marble at A (v=0)B) Calculate the velocity of the marble at B.C) Calculate the velocity of the marble at C

A

CB

10m

3m

Example 7:Two marbles collide in an elastic head-on collision. The first marble has a mass m1 = 0.25kg and a velocity of 5m/s. The second has mass m2 = 0.8kg and is initially at rest. Find the velocities of the marbles after the collision.

ffii vmvmvmvm 22112211

v1i – v2i = v2f – v1f

5m/s – 0 = v2f – v1f

(0.25kg)(5m/s) + 0 = (0.25)v1f + (0.8kg)v2f

v1f = v2f – 5m/s (substitute below)

1.25kgm/s = (0.25kg)(v2f – 5m/s) + (0.8kg)v2f

2.5kgm/s = (0.25kg)(v2f) + (0.8kg)v2f

v2f = 2.38m/s v1f = 2.38m/s – 5m/s = -2.62m/s

Eq 3

Eq 2

7.3 Collisions in One Dimension

Example 8 A Ballistic Pendulim

The mass of the block of woodis 2.50-kg and the mass of the bullet is 0.0100-kg. The blockswings to a maximum height of0.650 m above the initial position.

Find the initial speed of the bullet.


22112211 ooff vmvmvmvm

Apply conservation of momentum to the collision:

1121 of vmvmm

1

211 m

vmmv fo


Applying conservation of energyto the swinging motion:

221 mvmgh

2212

121 ff vmmghmm

221

ff vgh

m 650.0sm80.922 2 ff ghv


1

211 m

vmmv fo

m 650.0sm80.92 2fv

sm896m 650.0sm80.92kg 0.0100

kg 50.2kg 0100.0 21

ov

8.3 The Equations of Rotational Kinematics


Example 5 Blending with a Blender

The blades are whirling with an angular velocity of +375 rad/s whenthe “puree” button is pushed in.

When the “blend” button is pushed,the blades accelerate and reach agreater angular velocity after the blades have rotated through anangular displacement of +44.0 rad.

The angular acceleration has a constant value of +1740 rad/s2.

Find the final angular velocity of the blades.


θ α ω ωo t

+44.0 rad +1740 rad/s2 ? +375 rad/s

222 o

srad542rad0.44srad17402srad375

2

22

2

o

8.5 Centripetal Acceleration and Tangential Acceleration

Example 7 A Discus Thrower

Starting from rest, the throweraccelerates the discus to a finalangular speed of +15.0 rad/s ina time of 0.270 s before releasing it.During the acceleration, the discusmoves in a circular arc of radius0.810 m.

Find the magnitude of the totalacceleration.

8.5 Centripetal Acceleration and Tangential Acceleration

2

22

sm182

srad0.15m 810.0

rac

2sm0.45

s 0.270

srad0.15m 810.0

t

ω-ωrra o

T

22222 sm187sm0.45sm182 cT aaa

8.6 Rolling Motion

rv

The tangential speed of apoint on the outer edge ofthe tire is equal to the speedof the car over the ground.

ra

8.6 Rolling Motion

Example 8 An Accelerating Car

Starting from rest, the car acceleratesfor 20.0 s with a constant linear acceleration of 0.800 m/s2. The radius of the tires is 0.330 m.

What is the angle through which each wheel has rotated?

8.6 Rolling Motion

221 tto

θ α ω ωo t? -2.42 rad/s2 0 rad/s 20.0 s

22

srad42.2m 0.330

sm800.0

r

a

rad 484s 0.20srad42.2 2221

A bucket weighing 15N is wrapped around a pulley having a moment of inertia of 0.385kgm2. Calculate the angular acceleration of the pulley (R=0.33m), the linear acceleration of the bucket, and the tension in the rope.

Write equation for the rotation of the pulley:

Write equation for the acceleration of the bucket

maFmg

maF

IRF

I

t

t

Example, cont.

N

m

sradkgmF

smsradmRa

srad

mkgm

mkg

N

RI

mR

mgR

ImRmg

R

IF

mRFmgRa

maFmgIRF

t

t

t

t

tt

t

44.1033.0

/95.8385.0

/95.2/95.833.0

/95.8

33.0385.0

33.053.1

15

22

22

22

Example

mghImvE 2212

21

smmsmgH

v

gHv

MgHR

vMRMv

Rv

MRI

mghImvmghImv

f

f

ff

iiifff

/74.37.0

)1)(/8.9(

7.0

5

1

2

1

5

2

/5

2

2

2

2

2212

21

2

2212

212

212

21

ENERGY CONSERVATION

H = 1msolid sphere

Find the velocity of the sphere at the bottom of the ramp.

A 0.9 kg mass vibrates according to the equation:x = 1.2cos3t where x is in meters and t is in seconds.Determine: a) amplitude b) frequency c) total energy d) the kinetic and potential energies at x = 0.3m e) vmax f) amax

JmE

kgsmfkm

k

m

k

Tf

k

mT

kAE

sf

fmA

total

total

929.02.1N/m)29.1(

N/m29.1)9.0())477.0(2()2(2

1

2

11

2

k Find

c.

477.02

3

2

2 b.2.1 a.

22

1

212

22

1

1

222max

max

3.03.0

22

122

13.0

m/s8.10)rad/s3)(m2.1(

m/s6.3)rad/s3)(m2.1( f.

871.0058.0929.0

058.0N/m)(0.3m)29.1( d.

Aa

Av

JJJPEEKE

JkxPE

mtotalm

m

13.2 Conduction

Example 4 Layered insulation

One wall of a house consists of plywood backed by insulation. The thermal conductivities ofthe insulation and plywood are, respectively,0.030 and 0.080 J/(s·m·Co), and thearea of the wall is 35m2.

Find the amount of heat conducted through the wall in one hour.

13.2 Conduction

plywoodinsulation

L

tTkA

L

tTkA

But first we must solve for the interface temperature.

plywoodinsulation QQQ

m 019.0

C0.4CmsJ080.0

m 076.0

C0.25CmsJ030.0 tTAtTA

C8.5 T

)( cw TT

13.2 Conduction

J105.9

m 076.0

s 3600C8.5C0.25m 35CmsJ030.0

5

2

insulation

Q

Example: A balloon having a volume of 1.5 cubic meters is filled with ethyl alcohol and is tethered to the bottom of a swimming pool. Calculate the tension in the cord tethering it to the bottom of the swimming pool.

S Fy = B – T – W = 0

Therefore, T = B – W

T = waterVg – alcoholVg

T = (water – alcohol )Vg

T = (1000kg/m3 – 806kg/m3) (1.5m3)(9.8m/s2)

T = 2851.8 N

B

T W

FBD

Example: Water is contained in a tank. The water level is 5 meters above a hole in the tank where water exits through a hole to the atmosphere as shown below. The diameter of the hole is 0.01 meters, and the diameter of the tank is 3 meters. The tank is open to atmosphere. Calculate the velocity of the water exiting the tank.

eeettt gyvPgyvP

gyvP

22

2

2

1

2

1

2

1

:Therefore

constant

•The pressure at the top of the tank is equal to the pressure at the exit since they are both open to atmosphere. Therefore Pt and Pe cancel out.•Since the tank diameter is so much larger than the exit hole, the velocity of the water level drop can be approximated as zero.•The exit height is taken as zero, therefore ye = 0

The equation above reduces to:et

et

vgy

Therefore

vgy

2

:2

1 2

Where yt=hve= 9.9m/s

12.7 Heat and Temperature Change: Specific Heat Capacity

Example 12 Measuring the Specific Heat Capacity

The calorimeter is made of 0.15 kg of aliminumand contains 0.20 kg of water. Initially, thewater and cup have the same temperatureof 18.0oC. A 0.040 kg mass of unknown material is heated to a temperature of 97.0oC and then added to the water.

After thermal equilibrium is reached, thetemperature of the water, the cup, and the material is 22.0oC. Ignoring the small amountof heat gained by the thermometer, find the specific heat capacity of theunknown material.

12.7 Heat and Temperature Change: Specific Heat Capacity

unknownwaterAl TmcTmcTmc

CkgJ1300

C 0.75kg 040.0

C 0.4kg 20.0CkgJ4186C 0.4kg 15.0CkgJ1000.9 2

unknown

waterAlunknown

Tm

TmcTmcc

hotcold QQ

Q=mcwater∆T = (1.5kg)(4186J/kg°C)(20°C) =125,580JQ=mLfusion = (1.5kg)(3.33x105J/kg) =500,000JQ=mcice∆T = (1.5kg)(2100J/kg°C)(12°C) = 37,800J

663,380J

Graph of Ice to Steam

How much energy does a freezer have to remove from 1.5kg of water at 20°C to make ice at -12°C?

Documents

Physics 1201Q Review. 1.7 The Components of a Vector Example A displacement vector has a magnitude of 175 m and points at an angle of 50.0 degrees relative