Physics 1201Q Review
Jan 11, 2016
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.
3.3 Projectile Motion
y ay vy voy t-1050 m -9.80 m/s2 0 m/s ?
3.3 Projectile Motion
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
3.3 Projectile Motion
Example 4 The Velocity of the Care Package
What are the magnitude and direction of the final velocity ofthe care package?
3.3 Projectile Motion
y ay vy voy t-1050 m -9.80 m/s2 ? 0 m/s 14.6 s
3.3 Projectile Motion
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
5.5 Satellites in Circular Orbits
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
5.5 Satellites in Circular Orbits
T
r
r
GMv E 2
EGM
rT
232 = 42r3
GM
5.5 Satellites in Circular Orbits
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?
6.2 The Work-Energy Theorem and Kinetic Energy
2212
f21W omvmv
sF cosW
6.2 The Work-Energy Theorem and Kinetic Energy
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.
7.3 Collisions in One Dimension
22112211 ooff vmvmvmvm
Apply conservation of momentum to the collision:
1121 of vmvmm
1
211 m
vmmv fo
7.3 Collisions in One Dimension
Applying conservation of energyto the swinging motion:
221 mvmgh
2212
121 ff vmmghmm
221
ff vgh
m 650.0sm80.922 2 ff ghv
7.3 Collisions in One Dimension
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
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.
8.3 The Equations of Rotational Kinematics
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?