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PHYSICS REVIEW FOR CFA 3
Displacement – A change of position in particular direction. A distance in a given direction. Vector. Unit: meter (m)
Average Velocity =
total displacement
⃗ avg =
v
total time
Average Speed = average speed =
x
∆t
total distance traveled
vavg =
total time
d
t
(Instantaneous) Velocity – Value of velocity at a particular time.
(Instantaneous) Speed – Value of speed at a particular time.
Acceleration =
change in velocity
(vector) (m/s2)
time taken for that change
Acceleration can cause:
1. change in speed (speeding up: v and a in the same direction;
slowing down: v and a in the opposite direction)
2. changing direction
3. both
Motion with constant velocity (equal displacements in the equal amounts of time)
x = vt
magnitude of velocity = speed
v avg = v
Motion with constant acceleration a
v = u + at
vavg =
u+v
x =
2
u+v
2
t
x = ut +
𝑎
2
t2
v2 = u2 + 2ax
Free Fall formulas – Formulas are the ones for uniform accelerated motion with a = g
v = u + gt
vavg =
u+v
2
y =
u+v
t
2
y = ut +
g 2
t
2
v2 = u2 + 2gy
g = 9.8 m/s2, downward ≈ 10 m/s2.
Remember that in the coordinate system in which upward is chosen to be positive, g is negative and vice versa.
If air resistance is not mention it is assumed that we ignore air resistance.
When the object reaches maximum height, the velocity of the object is 0 m/s, but acceleration is still g = 9.8 m/s2
downward. Velocity changes, but g does NOT!!!
Terminal speed – When air resistance is taken into account object in free fall will not accelerate forever. The speed of the
object will increase until the object reaches a maximum, constant speed
Graphs for:
motion with constant velocity
positive direction is away from
the initial position
motion with constant acceleration
positive direction is away from
the initial position
free fall (up and down)
positive direction is upward
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PROBLEMS:
1. A car starts from rest and uniformly accelerates to a final speed of 20.0 m/s in a time of 15.0 s.
How far does the car travel during this time?
2. A car starts from rest and accelerates at 0.80 m/s2 for 10 s. It then continues at constant velocity. Twenty seconds after it began to
move, what is the cars velocity and how far has it traveled from the beginning? the car has:
3. A racing car traveling with constant acceleration increases its speed from 10 m/s to 30 m/s over a distance of 80 m? How long does
this take?
4. A brick is dropped from rest from a height of 4.9 m. How long does it take for the brick to reach the ground?
5. What maximum height will be reached by a stone thrown straight up with an initial speed of 35 m/s?
(This one you can solve with conservation of energy law)
6. How long does it take for a stone dropped off a 175-m high building to land on the ground ?
7. A kid dropped the ball on a slide. It starts rolling with constant acceleration. It rolls 2 m in first second. How far will it roll after 3
seconds?
8. The graph represents tile relationship between distance and time
for an object that is moving along a straight line.
Between what times did the object have a non-zero acceleration?
A. 0 s only
B. 0 s to 5 s
C. 5 s to 8 s
B. 0 s to 8 s
E. the object was not accelerating at any time.
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9. The motion of a circus clown on a unicycle moving in a straight line is
shown in the graph.
a. What would be the acceleration of the clown at 5 s?
b. What would be the acceleration of the clown at 1 s?
c. What would be the acceleration of the clown at 3 s?
d. What would be the acceleration of the clown at 10 s?
e. How far from the starting point is clown after 6 s?
f. After 12 seconds, how far is the clown from her original starting point?
1. 150 m
2. velocity 8.0 m/s and has traveled 120 m. 3. 4.0 s
4. 1.0 s
8. C
9. a. 2.0 m/s2
b. 3.0 m/s2
c. 0.0 m/s2
d. – 2.5 m/s2
5. 62 m
e. 34 m
6. 6 s 7. a=4m/s2 , x=18 m
f. 74 m,
Projectile motion
HORIZONTAL MOTION
VERTICAL MOTION
ux = u cos 0
uy = u sin 0
vx = u x
vy = uy + gt
x = ux t
y = uy t +
g 2
t
2
v2y =u2y + 2gy
y=
uy  v y
2
t
1. Zac accidentally falls out of a helicopter that is traveling horizontally at 60m/s. He plunges into the water below
3 seconds later. Assuming to air resistance, what is the horizontal distance he travels while falling? (180m)
(Where Zac is going to fall depends ONLY on horizontal velocity and time. Time depends ONLY on how high is helicopter)
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2. The Essex county sheriff was catching speeding cars with laser gun at the end of a small bridge on a snowy New England
night. A person in a car driving at 28 m/s (63 miles/hour) saw him, turned his steering wheel, and landed in a snow pile
20.00 m below the level of the road (very dangerous). How long did it take him to land and how far from the bridge did
he land.
(Where the car is going to land depends ONLY on horizontal velocity and on how high is the bridge)
(t=2s, x=56m)
3. A boy on the tower in the figure below throws a ball a distance of 60 m, as shown. At what speed, in m/s, is the ball
thrown?
(30 m/s - it takes 2 s to fall 20 m)
4. During the Canadian Amateur Open Tennis Match, Eli serves the ball with a horizontal velocity of 23.6 m/s 2.37 m
above the court surface. By how much does the ball clear the net which is 12.0 m away and 0.900 m high? [ 20.0 cm]
5. Emanuel Zacchini, the famous human cannonball of the Ringling Bros. and Barnum & Bailey Circus, was fired out of a
cannon with a speed of 24.0 m/s at an angle of 40.0° to the horizontal. If he landed in a net at the same height from which
he was fired, how long was Zacchini in the air? g = 10 m/s2, down
ux = u sin  = (24m/s) sin 400 = 15.4 m
t = 3.08 s
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FORCES
Inertia is resistance an object has to a change of velocity
Mass is numerical measure of the inertia of a body / is a measure of the amount of matter in the object unit: kg
• doesn’t depend on the location of the object . Object of mass of 1 kg here on earth would have the mass
of 1 kg on the moon, even though it would weigh only one-sixth as much.
Weight is the gravitational force acting on an object .
• W = mg
Net force, Fnet, is the vector sum of all forces acting on an object
unit: Newton (N)
Free Body Diagram/ Force diagram is a sketch of a body and all forces acting on it.
Newton’s first law: An object continues in motion with constant speed in a straight line (constant velocity)
or at rest unless acted upon by a net external force.
 If net force is zero, acceleration is zero, velocity is constant (or zero).
The object is in equilibrium. Any force acting on it is balanced.
Newton’s second law: If a net force is acting on an object of mass m, object will acquire acceleration proportional to
the net force and inversely proportional to the mass of the object. Direction of acceleration is direction of the net force.
𝑎 =
𝐹𝑛𝑒𝑡
𝑚
Newton’s third law:
Whenever object A exerts a force on object B, object B exerts an equal in magnitude but
opposite in direction force on object A
FA - force object A exerts on object B
We are talking about forces
acting on two different bodies.
FB - force object B exerts on object A
Tension T is a force that the end of the rope exerts on whatever is attached to it.
Direction of tension is along the rope.
Normal force Fn is the force which is preventing an object from falling through the surface of another body .
That’s why normal force is always perpendicular (normal) to the surfaces in contact.
Friction force Ffr is the force that opposes slipping (relative motion ) between two surfaces in contact;
it acts parallel to surface in direction opposed to slipping.
 Friction depends on type and roughness of surfaces and normal force.
Ffr = μ Fn
μ is called coefficient of friction
• μ has no units
• it is a measure of surface-to-surface roughness
• depends on characteristics of both surfaces
• different values for static and kinetic coefficient of friction (tables)
• kinetic μ is smaller than static μ. You probably noticed that once you moved
something from rest it becomes easier to push around.
PROBLEMS:
1. Howard, the soda jerk at Bea’s diner, slides a 0.60-kg root beer from the end of the counter to a
thirsty customer. A force of friction of 1.2 N brings the drink to a stop right in front of the customer.
a) What is the acceleration of root beer?
b) What is the coefficient of kinetic friction between the glass and the counter?
Vertical direction:
Horizontal direction:
Vertical acceleration = 0
Net force = friction force: Fnet = Ffr =1.2 N
Vertical net force = 0
Fnet = ma
Fn = mg = 6.0 N
1.2 = 0.60 a
a = 2.0 m/s2
Ffr = 𝜇 Fn
𝜇 = Ffr / 𝐹𝑛 = 1.2/6.0 = 0.20 (no units)
2. A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2 along a horizontal surface.
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a. How large is the frictional force?
b. What is the coefficient of friction?
m = 5.0 kg
a = 6.0 m/s2
F = 40.0 N
Vertical direction:
a = 0, so Fnet = 0
Fn = mg = 50 N → Ffr = μ Fn = 50 μ
horizontal direction: a = 6.0m/s2
F = ma
net
F – F = ma
fr
40.0 – F = 30
fr
F = 10 N
fr
Ffr = μ Fn
𝜇 = Ffr /Fn = 10/50 = 0.2
3. A cute bear, m = 60 kg, is sliding down an iced incline 300. The ice can support up to 550 N.
Will bear fall through the ice? If the coefficient of the friction is 0.115, what is the acceleration of
the bear?
m = 60 kg
Ffr = 300 N
θ = 300
g = 10 m/s2
vertical direction :
F sin θ + F = mg
Horizontal direction:
F cos θ – F = ma
n
fr
50 + F = 800
n
86.6 – 40 = 80 a
F = 750 N
n
4. A 1000-kg elevator is rising and its speed is increasing at 3 m/s2. The tension in the elevator cable is:
a = 3 m/s2 ↑
Fnet = ma
T – 10000 = (1000)(3)
T = 13000 N
5. A system of two cables supports a 150-N ball as shown.
a) What is the tension in the right-hand cable?
b) What is the tension in the horizontal cable?
y:
x:
T1 sin 300 = 150
T2 = T1 cos 300
T1 = 300 N
T2 = 260 N
6. A 70 N block and an 35-N block are connected by a string as shown.
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a. If the pulley is massless and the surface is frictionless, find the magnitude of the acceleration of the 70-N block, and tension in the
string.
b. There is friction between 70 N block and the table with µ=0.05. Find the magnitude of the acceleration of the 70-N block, and
tension in the string.
If the initial position of the block of 35 N is 2m above the ground, how long does it take for it to reach the ground. What is the speed at
the ground?
Newton’s Law of Universal Gravitation: Force between masses m1 and m2 that are at distance r from each other
attract each other with the force
F=G
m1 m 2
r2
PROBLEMS:
1. Find the change in the force of gravity between two objects when both masses are doubled and the distance between
them is also doubled.
2. Two point masses m and M are separated by a distance d. If the distance between the masses is increased to 3d, how
does the gravitational force between them change?
3. Find the change in the force of gravity between two objects when both masses are doubled and the distance between
them is also doubled.
4. If the moon orbited Earth at a distance twice what it is now, would the moon be moving faster, slower, or with the
same speed as today? Explain. What would happen to the length of the month?
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5. Two satellites are orbiting the Earth. Mass of the second one is 4 times greater than the first one (space station). The
distance of the second one is 2 times greater than the first one. What is the ratio of the gravitational force on the second
one to the gravitational force to the first one?
1. F = Gm1m2/r2
Fnew = G(2m1)(2m2 )/(2r)2 = G 4m1m2/4r2 = F the force is unchanged
2. The force will be one-ninth as great.
3. F = Gm1m2/r2 Fnew = G(2m1)(2m2 )/(2r)2 = G 4m1m2/4r2 = F the force is unchanged
4. If the moon orbited Earth at twice its present distance, the gravitational attraction between Earth and moon would be
one fourth as much. Because of this reduced attraction, the speed of the moon required to maintain a circular orbit would
be less (its present speed with less gravity would overshoot a circle, making an elliptical orbit). Because it would be
moving more slowly and also because it would have more distance to cover, the length of the month would increase.
𝑀𝑒𝑎𝑟𝑡ℎ 4m1
(2r)2
𝑀
m1
G 𝑒𝑎𝑟𝑡ℎ
r2
G
5.
=1
MOMENTUM and IMPULSE
Momentum, p is mass times velocity:
p=mv
Impulse F∆t will produce change in momentum Δp:
vector!
unit: (p) = kg m/s
F∆t = ∆p
Δp = mv - mu
o
Example: You want to throw a ball (m=0.5 kg) over a tree. You hit it at 60 so it leaves your hand at the speed
of 10 m/s. Unfortunately that was not enough. Your ball is now stuck in the tree. It was just at its maximum
height. What impulse did you impart on the ball. What impulse did the tree exert on the ball?
you: impulse = change in momentum = 5 kg m/s.
tree: at the top speed is equal to the horizontal component of the velocity = 10 cos 600 = 5 m/s
impulse = change in momentum = 2.5 kg m/s
Law of conservation of momentum:
In collision
pafter = pbefore
m1v1 + m2v2 = m1u1 + m2u2
WORK – ENERGY - (measured in Joules)
 Work done by a constant force F exerted on an object through distance d is:
W = Fd cos Ѳ
 Work done by a varying force F graphically
The area under a Force - distance graph equals
the work done by that force
(Fd = F cos Ѳ)
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 Work done by force F is
positive
negative
• when the force and direction of
motion are generally in the
same directions (cos θ = +)
• when the force and direction of
motion are generally in the
opposite directions (cos θ = – )
• the force helps the motion
• force opposes the motion
maximum work
minimum work
W = – Fd
W = Fd
(the work done by friction force is always negative)

Work done by force F is
• the force is exerted but
no motion is involved:
d = 0, W = 0
zero if:
• the force is perpendicular to the direction of motion (cos 900 = 0)
for example
work done by normal
force can be zero
normal force is
perpendicular to v
W=0
 Work done by
normal force is
parallel to v
W = Fd = mgh
the work done by centripetal force is zero
Wnet = 0 → Wnet = ∆KE
→ no change in KE
no change in speed;
centripetal force can not
change the speed, only
direction
for example, gravitational
force on the moon does not
change speed of the moon
applied force changes potential energy (when net force is zero, so there is no
acceleration).
What work should be done in raising an object of mass 6 kg to the top of the incline?
W=mgh = 180 J
What (minimal) force should be applied to push it along the incline to the top:
F = mg sin θ = 60 (3/5) = 36 N
 Work done
by net force changes kinetic energy (net force gives acceleration, therefore can change speed).
KE = ½ mv2

Kinetic energy

Work – Kinetic energy relationship: work done by net force changes kinetic energy
W = ∆KE = KEf – KEi = ½ mv2 – ½ mu2
W = Fd cos Ѳ
When the net force and direction of motion are in generally the same directions, work is positive and KE is increasing.
When the net force and direction of motion are in generally opposite directions, work is negative and KE is decreasing.
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 Conservation of energy law
For the system that has only mechanical energy (ME = PE + KE)
and there is no frictional force acting on it, so no mechanical energy
is converted into heat, mechanical energy is conserved
ME1 = ME2 = ME3 = ME4
mgh1 + ½ mv12 = mgh2 + ½ mv22 = • • • • • •
 Conservation of energy law with friction included
Friction converts part of kinetic energy of the object into heat energy. We say that the frictional force has dissipated energy.
This energy equals to the work done by the friction and it doesn’t belong to the object alone but is shared with environment.
ME1 – Ffr d = ME2
(Wfr = – Ffr d)
P=
 Power is the work done in unit time or energy converted in unit time 𝑷 =
𝑾
𝒕
W
t
or P =
or 𝑷 =
E
t
𝑬
𝒕
measures how fast work is done or how quickly energy is converted. Power is a scalar quantity.
Units: 1 W(Watt) = 1 J/ 1s

There is another way to calculate power P = F v
Example: Firework explodes into three pieces of equal mass. They all move in three different directions each with the
speed v.
What work was done on firework?
W = ∆KE = 3(½ mv2)
In addition remember that momentum must be conserved !!!!!
PROBLEMS:
1.An engineer is asked to design a playground slide such that the speed a child reaches at the bottom does not exceed 6.0 m/s.
Determine the maximum height that the slide can be.
2.A care package is dropped from rest from a helicopter hovering 25 m above the ground. What is the speed of the package just before
it reaches the ground? Neglect air resistance
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3. An escalator is used to move 20 people (60 kg each) per minute from the first floor of a department store to the second floor, 5 m
above. The power required is approximately:
4.What power is needed to lift a 49-kg person a vertical distance of 5.0 m in 20.0 s?
5. A bullet with a mass of 5.00 x 10–3 kg is loaded into a gun. The loaded gun has a mass of 0.52 kg. The bullet is fired, causing the empty
gun to recoil at a speed of 2.1 m/s. What is the speed of the bullet?
6. Two carts with masses of 1.5 kg and 0.7 kg, respectively, are held together by a compressed spring. When released, the 1.5 kg cart
moves to the left with a velocity of 7 m/s. What is the velocity of the 0.7 kg cart? (Disregard the mass of the spring.)
7. A 1000-kg car traveling east at 20 m/s collides with a 1500-kg car traveling west at 10 m/s. The cars stick together after the collision.
What is their common velocity after the collision?
8. A 75-kg man is riding in a 30-kg cart at 2.0 m/s. He jumps off in such a way as to land on the ground with no horizontal velocity. The
resulting change in speed of the cart is:
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1. mg h = ½ m v2
g h = ½ v2
10h = 18
2. mg h = ½ m v2
g h = ½ v2
250 = ½ v2
3. P = W/t = mgh/t = 20(60)(10)(5)/(60) = 1000 W
4. P = W/t = mgh/t = (49)(9.80)(5.0)/(20.0) = 120 W
5.
220 m/s
6. 15 m/s to the right
v = 1.8 m
v = 22 m/s
7. 2 m/s, east
8. 5.0 m/s
1. What forces are acting on pendulum? In which direction is the net force? Where is the pendulum accelerating the
most? The least? Where is it moving the fastest? Stopped?
Discuss how energy conservation applies to a pendulum. Where is potential energy the most? The least? Where is
kinetic energy the most? The least?
Solution: two forces: tension and gravitational force (weight).
Net force is tangential to the path. The farther pendulum is
from equilibrium position the greater net force is. Greatest
force at the ends, smallest (=0) at bottom position.
Net energy is never created or destroyed. It can change from
one form to another form. A swinging pendulum has the most
gravitational potential energy at the top of its swing. At that
point it has no kinetic energy. At the bottom of its swing its
potential energy is at a minimum, or zero relative to that
lowermost point, and its kinetic energy is at a maximum.
Halfway down, it has half kinetic and half potential energy.
Everywhere along the swing the sum of the kinetic and
potential energies is the same. When air resistance and
friction are taken into account, energy is transferred from the pendulum to the surroundings in the form of heat.
Time for one period doesn’t depend on mass!!!! This is like free fall. Time needed for an object to fall down from certain
height doesn’t depend on the mass.
REVIEW CIRCUITS
 Resistance of a wire when the temperature is kept constant is:
𝑅=𝜌
𝐿
𝐴
 OHM’S LAW: Current through resistor is proportional to potential difference
across the resistor and inversely proportional to resistance
of that resistor.
𝐼 =
𝑉
𝑅
𝐼(𝐴)
𝑉(𝑉)
𝑅(𝛺)
• length, L
• cross-sectional area, A
• material/resistivity, ρ
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 Electric power, P, is the rate at which energy is supplied to or used by a device in which electric energy is
converted into another form such as mechanical energy, thermal energy, or light.
Power dissipated in a resistor:
P=IV
P=
𝑉2
𝑅
= 𝐼2 𝑅
Power of the source = ε I
Electric energy is: 𝐸 = 𝑃 𝑡
𝑠𝑜
𝐸 (𝐽𝑜𝑢𝑙𝑒𝑠) = 𝑃(𝑊𝑎𝑡𝑡𝑠) × 𝑡(𝑠)
𝐸 (𝑘𝑊ℎ) = 𝑃(𝑘𝑊) × 𝑡(ℎ)
 Electromotive force, 𝜺, is the voltage generated by battery (how much energy per unit charge is available for
the circuit including internal resistance)
Resistors in Series
• connected in such a way that all components have the same current through them.
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3
𝑉
𝐼=𝑅
𝑒𝑞
Resistors in Parallel
• Electric devices connected in parallel are connected to the same two points of an electric circuit, so all components
have the same potential difference across them.
• The current flowing into the point of splitting is equal to the sum of the currents flowing out at that point:
𝐼 = 𝐼1 + 𝐼2 + 𝐼3
𝑎𝑛𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑟𝑜𝑝 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎𝑙𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟𝑠: 𝐼1 𝑅1 = 𝐼2 𝑅2 = 𝐼3 𝑅3
The greater resistance, the smaller current.
1
𝑅𝑒𝑞
=
1
𝑅1
1
1
+𝑅 +𝑅
𝐼=
2
3
𝑉
𝑅𝑒𝑞
A device that transforms mechanical energy into electrical energy is called a generator.
A device that transforms electrical energy into mechanical energy is called an electric motor.
A transformer is a device that transforms/change voltage.
REVIEW PROBLEMS:
1. A circuit is wired with a power supply, a resistor and an ammeter (for measuring current). The ammeter reads a current
of 24 mA (milliAmps). Determine the new current if the voltage of the power supply was ...
a. increased by a factor of 3 and the resistance was held constant.
b. held constant and the resistance was increased by a factor of 2.
c. increased by a factor of 3 and the resistance was decreased by a factor of 2.
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d. decreased by a factor of 2 and the resistance was increased by a factor of 2.
2. A 541-Watt toaster is connected to a 120-V household outlet. What is the resistance (in ohms) of the toaster?
3. Consider two appliances that operate at the same voltage. Appliance A has a higher power rating then appliance B.
a. How does the resistance on A compare with that of B. Is it (1) larger, (2) smaller, or (3) the same?
4. A computer system includes a color monitor with a power requirement of 200 W, whereas a countertop
broiler/toaster oven is rated at 1500 W. What is the resistance of each if both are designed to run at 120 V?
5. Calculate the resistance of an aluminum (ρ = 2.8x10-8 Ωm) wire that is 2.0 m long and of circular cross section
with a diameter of 1.5 mm.
6. The resistance of a certain wire is 10 ohms. What would the resistance of the same wire be if it were twice as long? If it were twice as
thick?
7. A refrigerator operates on average for 10.0 h a day. If the power rating of the refrigerator is 700 W, how much electrical energy does
the refrigerator use in 1 day?
8. An electrical appliance is rated as 2.5 kW, 240 V.
(a) Determine the current needed for it to operate.
(b) Calculate the energy it would consume in 2.0 hours.
9. What is the equivalent resistance for the resistors in the figure ?
10. Find equivalent resistance for the circuit shown.
11. A 100 Ω, 120 Ω, and 150 Ω resistor are connected to a 9-V battery in the circuit shown. Which of the three resistors
dissipates the most power?
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12. What is the current in the battery of the circuit shown below?
13.
a. Simplify the above circuit so that it consists of one equivalent resistor and the
battery.
b. What is the total current through this circuit?
c. Find the current through each resistor. Find the voltage across each resistor.
d. Find the power of the battery. Find the power dissipated in each resistor.
1. a. Inew = 72 mA
b. Inew = 12 mA c. Inew = 144 mA d. Inew = 6 mA
2. 26.6 Ω
3. smaller
4. 72 W, 9.60 W
5. 32 mΩ
6. 20 2.5 , because twice the diameter gives four times the cross sectional area and one-fourth the resistance
7. E = Pt = (700 W)(36000 s ) = 25.2 x 106 J
or
E = Pt = (0.7 kW)(10.0 h) = 7 kWhours
8. (a) I = P/V = 2500W/240V = 10.4 = 1.0 x 101 A (b) energy = Pt = VIt = (240V)(10.4A)(7200s) = 1.8 x 107 J
𝑅 𝑅
9. 5.2 Ω
10. 𝑅𝑒𝑞 = 1 2 + 𝑅3 + 𝑟
11. 100 Ω
𝑅1 +𝑅2
12. Req = 5 +l5 =20 I = V/R = 0.5 A
13. only the first step :
I = V/Req = 0.5 A
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REVIEW MAGNETISM
The direction of a magnetic field line is defined as the direction in which the north
pole of a compass points when it is placed in the magnetic field.
Outside the magnet, the field lines emerge from the magnet at its north pole and
enter the magnet at its south pole.
Inside the magnet, there are no isolated poles on which field lines can start or stop,
so magnetic field lines always travel inside the magnet from the south pole to the
north pole to form closed loops.
Magnetic field is measured in Tesla
1 T(Tesla) =
N∙s
C∙m
1. An electric charge experiences a magnetic force when moving in a magnetic field.
Magnetic force acting on a charge q
Magnetic force on a wire carrying current I
in a magnetic Field B: F = qvB sinin a magnetic field B: F = I LB sin
 

q = charge [C]
v = velocity [m/s]
B = magnetic field [Tesla T]
 = angle between v and B
I = current [A]
L = length [m]
B = magnetic field [T]
 = angle between I and B
R-H-R 1: The direction of the magnetic force on a charge/current is given by the right-hand rule 1:
Outstretch fingers in the direction of v (or current I).
Curl fingers as if rotating vector v (I ) into vector B.
Magnetic force on a positive charge (or I) is in
the direction of the thumb.
Magnetic force on a negative charge points in opposite direction.
2. A moving charge produces a magnetic field.
R-H-R 2: The direction of the magnetic field produced by electric current is given by the right-hand rule 2:
If a wire is grasped in the right hand with the thumb in the direction of current flow, the fingers will curl in the direction of
the magnetic field.
= the permeability of free space 4×10-7 T·m/A
𝜇0 𝐼
Magnetic field B around a wire with current I B =
I = current [A]
2𝜋 𝑟
r = distance from the center of the conductor
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Two parallel current carrying wires
attract each other
Two antiparallel current carrying wires
repel each other
Magnetic field of a solenoid with steady current is:
1. uniform from south to north pole inside the solenoid.
(constant in magnitude and direction )
2. not uniform from north to south pole outside of the solenoid.
REVIEW PROBLEMS:
1. The magnetic field of a bar magnet is shown in the figure.
Is the magnet’s north pole at A or B?
2. The direction of the force on a current-carrying wire in an external magnetic field is
a. perpendicular to the current only.
b. perpendicular to the magnetic field only.
c. perpendicular to the current and to the magnetic field.
d. parallel to the current and to the magnetic field.
3. If a proton is released at the equator and falls toward Earth under the influence of gravity, the magnetic force on the
proton will be toward the
a. north.
c. east.
b. south.
d. west.
4. What is the path of an electron moving perpendicular to a uniform magnetic field?
a. a straight line
c. an ellipse
b. a circle
d. a parabola
5. What is the path of an electron moving parallel to a uniform magnetic field?
a. straight line
c. ellipse
b. circle
d. parabola
6. Find the direction of the force on an proton moving
through the magnetic field shown.
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7. Find the direction of the force on an electron moving through the magnetic field.
8. A negative charge is moving through a magnetic field. The direction of motion and the direction
of the force acting on it at one moment are shown. Find the direction of the magnetic field.
9. The direction of the force on a current-carrying wire in an external magnetic field is
a. perpendicular to the current only.
b. perpendicular to the magnetic field only.
c. perpendicular to the current and to the magnetic field.
d. parallel to the current and to the magnetic field.
10. If a proton is released at the equator and falls toward Earth under the influence of gravity, the magnetic force on the
proton will be toward the
a. north.
c. east.
b. south.
d. west.
11. What is the path of an electron moving perpendicular to a uniform magnetic field?
a. a straight line
c. an ellipse
b. a circle
d. a parabola
12. What is the path of an electron moving parallel to a uniform magnetic field?
a. straight line
c. ellipse
b. circle
d. parabola
13. An electron that moves with a speed of 3.0 x 104 m/s perpendicular to a uniform magnetic field of 0.40 T experiences
a force of what magnitude? (e = 1.60 x 10–19 C)
14. An electron moves north at a velocity of 4.5 x 104 m/s and has a force of 7.2 x 10–18 N exerted on it. If the magnetic
field points upward, what is the magnitude of the magnetic field?
15. A 2.0 m wire segment carrying a current of 0.60 A oriented perpendicular to a uniform magnetic field of 0.50 T
experiences a force of what magnitude?
16. A 2.0 m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.50 T experiences a
force of what magnitude?
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17. A current-carrying wire 0.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A
and there is a resultant force of 3.0 N on the wire due to the interaction of the current and field, what is the magnetic
field strength?
18. A current in a long, straight wire produces a magnetic field. These magnetic field lines
a. go out from the wire to infinity.
c. form circles that pass through the wire.
b. come in from infinity to the wire.
d. form circles that go around the wire.
19. Two long parallel straight wires carry equal currents in opposite directions. At a point midway between the wires, the
magnetic field they produce is:
A) zero
B) non-zero and along a line connecting the wires
C) non-zero and parallel to the wires
D) non-zero and perpendicular to the plane of the two wires
E) none of the above
20. Magnetic field lines inside the solenoid shown are:
A) clockwise circles as one looks down the axis from the top of the page
B) counterclockwise circles as one looks down the axis from the top of the page
C) toward the top of the page
D) toward the bottom of the page
E) in no direction since B = 0
21. A solenoid is in an upright position on a table. A counterclockwise current of electrons causes the solenoid to have
a(n) ____ magnetic pole at its bottom end. If a compass is placed at the top of the solenoid, the north pole of the compass
would be
a.
north; attracted c.
north; repelled
b.
south; attracted d.
south; repelled
22. A solenoid is in an upright position on a table. A clockwise current of electrons causes the solenoid to have a(n) ____
magnetic pole at its bottom end. If a compass is placed at the top of the solenoid, the north pole of the compass would be
a.
north; attracted c.
north; repelled
b.
south; attracted d.
south; repelled
23. A long straight wire carries current as shown. Two electrons move with velocities that are parallel and perpendicular to
the current. Find the direction of the magnetic force experienced by each electron.
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1. A 2. c
3. c 4. b. 5. a.
6. up, toward the top of the page
7. down, toward the bottom of the page
8. to the left
9. c.
10. c
11. b
12. a
13. 1.9 x 1015 N 14. 1.0 mT
18. d 19. D 20. C 21. d. 22. a
23. A. south
B. west
15. 0.60 N 16. 0.0 N
17. 0.60 T
REVIEW – SHM and WAVES
Periodic Motion - A motion that repeats itself over and over.
Period, T is the time required for one cycle of a periodic motion.
((T) = seconds/ cycle = s)
Frequency, f is the number of oscillations per second.
((f) = cycles/second = 1/s = 1 Hz)
1
1
f 
T 
T
f
Displacement: directed distance (angle) of the oscillating object from equilibrium position.
Amplitude: the magnitude of maximum displacement of an object from its equilibrium position
Oscillations: Periodic motion caused by restoring force
Restoring force:. a force that is directed toward the equilibrium position, so it is always opposite to the displacement. The restoring
force changes continually as the displacement changes leading to oscillations. It is maximum at maximum displacement and zero at
equilibrium position.
If displacement (angle in the case of pendulum and extension in the case of spring) increases, restoring force increases, and vice versa.
SHM:
Whenever the force acting on a particle is linearly proportional to the displacement and directed toward equilibrium, the
particle undergoes simple harmonic motion.
Remember that a = F/m, so acceleration is also linearly proportional to the displacement and is directed
towards the equilibrium position,
 Properties of SHM
• at x = ± xmax
PE is max
v = 0 so KE = 0
F is max, so a is max
• at x = 0,
(equilibrium position)
PE = 0
v is max so KE is max
F = 0 so a = 0
When are pendulum and spring oscillations SHM ?
Pendulum:
Spring: m is the mass attached to the spring
SHM for amplitude less then 120
sin  ≈  , so F = – mg 
that is restoring force bringing
pendulum to equilibrium position
L is the length of the string
g is gravitational acceleration
T  2
SHM for x not too long compared to L
F = – kx
k is the spring constant
Spring force F – “restoring force”: the spring exerts its
force on the block in the direction opposite the
displacement x, acting to return it to its normal length
(toward the equilibrium position) ( the “-“ sign)
L
g
T  2
m
k
Period is CONSTANT for a given pendulum or spring and does NOT depend on amplitude.
Damping is decreasing of amplitude of SHM due to the action of forces of friction on the oscillator.
• In real system there is always friction at the support and often air resistance.
That results in the loss of energy of the object that oscillates.
Although amplitude decreases the period/frequency remains constant.
• oscillating systems are often classified by the degree of damping.
Light damping: the decay in amplitude is relatively slow and the oscillator will make
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quite a few oscillations before finally coming to rest. There is a slow energy loss.
Heavily damped oscillations: the amplitude of the heavily damped oscillations decay very
rapidly and the oscillator quickly comes to rest. There is a rapid energy loss.
Critical damping: Critical damping occurs if the resistive force is so big that the system
returns to its equilibrium position without passing through it. The mass comes to rest at its
equilibrium position without oscillating..
Natural frequency: is a frequency an object vibrates when displaced from its equilibrium position and the left on its own (most solids
are elastic – they vibrate when given an impulse)
Forced Oscillations: Oscillations of an object produced by vibrating force acting on it.
Resonance: The increase in amplitude of oscillation of a system exposed to a periodic driving force with a frequency equal to the
natural frequency of the system.
Even if an object is at rest, when exposed to the driving force with the frequency equal to the natural frequency of that object, the
amplitude of the vibrating object can become very large. The resonance can result in a quite dramatic increase in amplitude that
sometimes can be very unfortunate.
1.
a.
b.
In simple harmonic motion, the restoring force must change the same way as:
amplitude c.
velocity e.
distance squared
frequency
d.
distance
2. Tripling the displacement from equilibrium of an object in simple harmonic motion will change the object’s maximum acceleration by what
factor?
a. one-third
b. 1
c. 3
d. 9
3. What is the period of a 4.12 m long pendulum?
a. 2.01 s
c. 4.07 s
3.11
s
b.
d. 9.69 s
4.By what factor should the length of a simple pendulum be changed if the period of vibration were to be tripled?
a. 3 c.
9
b. 6 d.
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5. On the planet Xenos, an astronaut observes that a 1.00 m long pendulum has a period of 1.50 s. What is the free-fall acceleration on
Xenos?
6. An object attached to one end of a spring makes 20 vibrations in 10 seconds. Its frequency is:
7. A clown is rocking on a rocking chair in the dark. His glowing red nose moves back and forth a distance of 0.42 m exactly 30 times a
minute, in a simple harmonic motion.
a. What is the amplitude of this motion?
b. What is the period of this motion?
c. What is the frequency of this motion?
8. A block attached to a spring oscillates in simple harmonic motion along the x axis. The limits of its motion are x = 10 cm and x = 50 cm
and it goes from one of these extremes to the other in 0.25 s. Its amplitude and frequency are:
1. d
2. 3
7. A = d/2 = 0.21 m
3. c
4. c
5. 17.5
6.
f = 30/60 Hz = 0.5 Hz
T = 1/f = 2s
f = 2 Hz
f = 0.5 Hz
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8. A = 20 cm
f = 2 Hz
WAVES
Wave Pulse – A single non-repeated disturbance in a medium.
Wave – A disturbance that repeats regularly in space and time, traveling through a medium by which energy is transferred from one
particle of the medium to another without causing any permanent displacement of the medium itself.
Waves transfer energy without transporting matter because each part of the medium oscillates around its equilibrium position (moves
away from its normal position and then returns) (energy of a wave is proportional to amplitude squared: E  A2; at the same time E 
freq.)
A medium – is a material through which a wave passes.
Two different types of waves regarding medium:
1. Mechanical waves – a disturbance that travels through a medium. The particles of the medium oscillate around their equilibrium
positions.
2. Electromagnetic waves – do not need medium. The electric and magnetic
field oscillate
(change magnitude and direction) even when travelling in a medium.
Waves may also be classified according to direction of the oscillations of the particles relative to the direction of energy
propagation. Two most common are:
Transverse waves – The vibration of the particles of the medium is transverse (or perpendicular) to the direction of the
propagation of the wave (energy).
(vibrating strings of musical instruments, earthquake, water ripples, EM waves, rope, spring,..)
Longitudinal waves – The vibration of the particles of the medium is along direction of the propagation
of the wave (energy).
Compressions – regions of higher density and pressure (particles are closer than normal)
Rarefactions – regions of lower density and pressure (particles are further apart than normal)
Compressions correspond to crests, and rarefactions correspond to troughs.
(sound waves in any medium, shock waves in an earthquake, compression wave along a spring.
A displacement/position graph shows the displacement of the different
sections of a medium.
A snapshot of a wave in an instant of time.
Transverse wave: displacement vs. x,
Amplitude, A – of a wave is the largest distance
Longitudinal wave: density vs. x
from the normal position that the medium is displaced.
or pressure vs. x
Wavelength, λ – of a wave is the distance from one point to the next
corresponding point (e.g. from crest to crest, or from compression to
compression)
A displacement/time graph shows the displacement of one point of a
medium as time elapses.
Period, T – of a wave is the time in seconds that it takes one wavelength to
pass by. In this time the medium will complete one oscillation.
Frequency, f – of a wave is the number of wavelengths that pass each second.
Transverse wave: displacement vs. t,
Longitudinal wave: density vs. t
or pressure vs. t
Frequency and period are reciprocals. T = 1/f ; f= 1/T.
The speed of a mechanical wave depends on the medium ONLY, and the type
of the
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wave !!!!!!! How? It always depends on two properties of the medium—stiffness and inertia. A stiff medium is one that is difficult to
stretch or compress. Stiffness depends on the intermolecular forces in a medium. When stiffness is greater, velocity is greater since
each segment of medium is in tighter contact with its neighbor and transfer of energy is easier. Inertia of the medium depends on the
density of the medium. It is more difficult to change the velocity of larger masses than smaller ones because they have more inertia.
In general, the speed of transverse waves in solids are about 0.6 times the speed of longitudinal waves in solids.
In earthquake first the longitudinal wave reach, then more devastating transverse wave.
Transverse waves cannot travel in gases like air.
Sound: air – 343 m/s (200C, 1 atm), helium – 1005 m/s (200C, 1 atm), water – 1500 m/s,
bone - 3000 m/s; steel rod – 5000 m/s glass – 4500 m/s
For transverse wave in the string the speed of the wave depends on the tension in string and the mass of the string per metre
(density).
v
T
m/ L
Violin string: A-string – 288 m/s, G-string – 128 m/s
Wave Equation – Golden rule for waves –
Waves in medium can have different frequencies and different wavelength, but product of wavelength and frequency for all waves in a
given medium is constant and equal to the speed of wave in that medium:
v

T
 f
Wave fronts propagating from a point source
 Spherical wave – The center of the circle is the source of the oscillations. If there is 3-D medium the wave will spread out in all
directions. And if the medium is uniform these waves are approximately spherical.
 Ray shows direction of wave/ direction of energy propagation
 Wavefront is the set of crest points
 Plane waves: far away from the source circular wavefronts can be approximated with straight parallel lines / planes in 3-D. These are
known as plane waves.
 EM waves striking the earth are plane waves
Sound
 Sound is mechanical, longitudinal wave. Can be spread in gases, liquids and solids. The wave consists of compressed regions
alternating with rarefied regions.
 The maximum (minimum) pressure during normal conversation is 3  10 5 % higher (lower) than normal pressure. Ear can detect
such small changes.
 Just like a speed of a wave on a string, the speed of sound is determined by the properties of the medium through which it
propagates. In air, under normal atmospheric pressure and temperature, the speed of sound is approximately 343 m/s.
 Frequency of the sound determines the pitch of the sound. The pitch is perceived frequency of the sound.
 Humans’ audible range is 20 Hz – 20 kHz
 Infrasonic sound – frequencies below 20 Hz
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 Ultrasonic sound – frequencies above 20 kHz
 Infrasound is used in the nature for communication: elephants (~ 15Hz) couple of kilometers, whales – as sound travels faster in
water (v ~ 1500 m/s) than in air, the call can be heard over distances of thousands kilometers.
 Sources of infrasonic waves include earthquakes, thunder, volcanoes, and waves produced by vibrating heavy machinery.
 Ultrasound is used for echolocation: dolphins, bats, sonar, sonograms ....
 Sound needs a medium – won’t travel in a vacuum since nothing to compress and expand
 Frequency is determined by the source of oscilations, so when guitar string plays a note, the air (or water in the case of underwater
concert) vibrate at the that frequency. As the speed is different in string and air, wavelengths are too.
The Doppler Effect
is an apparent (observed) change in frequency and wavelength of a wave occurring
source and observer are in motion relative to each other, with the perceived frequency
when the source and observer approach each other and decreasing when they move
when the
increasing
apart.
Doppler effect is the characteristic of EM waves too.
Astronomy: the velocities of distant galaxies can be determined from the Doppler shift.
As the star moves toward us the observed frequency increases, we say it shifts toward the blue spectrum.
That’s why we call it a blue shift.
As the star moves away from us the observed frequency decreases, we say it shifts toward the red spectrum.
That’s why we call it a red shift.
Most distant galaxies are observed to be red-shifted in the color of their light, which indicates that they are moving away from the
Earth. Some galaxies, however, are moving toward us, and their light shows a blue shift.
Edwin Hubble discovered the red shift in the 1920's.
His discovery led to him formulating the Big Bang Theory of the Universe's origin.
Reflection and refraction of waves
When a wave is incident upon a boundary (barrier) between two media
reflected, some is absorbed and some of it is transmitted.
How much of each?
That depends on the media.
Reflection of waves
All waves can be reflected.
The end of the rope if fixed – reflected pulse returns inverted.
Free end – reflected pulse is not inverted.
Law of reflection
The angle at which the wave is reflected from the surface is equal to angle of incidence.
• the angles are measured from the normal to the ray.
• all waves, including light, sound, water obey the law of
reflection at the boundary between two different media.
Refraction
some of it is
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When a wave passes from one medium to another, its velocity changes. The change in speed results in a change in direction of
propagation of the refracted wave.
If a wave passes from faster medium into slower, the wave will be refracted toward the normal. Angle of refraction will be smaller
than angle of incidence. If a wave passes from slower medium into faster, the wave will be refracted away the normal. Angle of
refraction will be greater than angle of incidence.
medium 1: air
medium 2: water
frequency is determined by the source so it doesn’t change. Only
wavelength changes. Wavelength of the same wave is smaller in
the medium with smaller speed.
f=
v1
v
= 2
λ1
λ2
A mathematical law which will tell us exactly HOW MUCH the direction has changed is called SNELL'S LAW.
Law of refraction – Snell’s law
v1
λf
λ
= 1 = 1
v2
λ2f
λ2
For a given pair of media, the ratio
sin θ1
v
= 1
sin θ2
v2
is constant for the given frequency.
The Snell’s law is of course valid for all types of waves.
The speed of light inside matter
 The speed of light in vacuum is: c = 3 x 108 m/s
 In any other medium such as water or glass, light travels at a lower speed.
 INDEX OF REFRACTION, n, of the medium is the ratio of the speed of light in a vacuum, c, and the speed of light, v, in that medium:
no units
As c is greater than v for all media, n will always be > 1.
n=
c
v
greater n – smaller speed of light.
As the speed of light in air is almost equal to c, nair ≈ 1
 for light only, Snell’s law can be expressed as:
n1 sin 1 = n2 sin 2
n1 – index of refraction of the medium 1
n2 – index of refraction of the medium 2
 Refraction of light
The refracted ray is refracted more in the medium with greater n, slower speed of light
Diffraction
When waves pass through a small opening, or pass the edge of a barrier, they always spread out to some extent into the region that is
not directly in the path of the waves. - into the region of the geometrical shadow
The spreading of a wave into a region behind an obstruction is called diffraction.
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Diffraction effects depend on λ of the waves and are most obvious when the object or aperture causing the diffraction is similar in
size to λ of the waves.
remember: big λ, big diffraction effects
AM radio waves have longer wavelengths then FM. So, they diffract around obstacles (buildings, mountains) much better then FM
radio waves.
Ultrasound (f > 20 kHz, λ < 1.7 cm) is used for echolocation: dolphins, bats, sonar.
Low frequency sound has longer wavelength, so they will be diffracted, so not being able to detect the prey. That’s why infrasound is
used for communication. High frequency sound has smaller wavelength, so it will be reflected back from the object. and detected.
Interference - Superposition
 two waves can be at the same place at the same time! and when they meet they generally interfere, superimpose
 After two waves overlap they carry on with exactly the same properties as before, as if nothing had happened.
Principle of superposition
When two or more waves overlap, the resultant displacement at any point and at any instant is the sum of the
displacements of the individual waves at that point.
constructive interference – increased amplitude,
increased energy (E  A2 ) – increased intensity – brighter light or loud sound at point the waves are in phase
destructive interference – decreased amplitude, decreased energy – decreased intensity – no light or no sound
the waves are out of phase
At point P, d1 distance from left source, and d2 distance from right source anything can happen
path difference = d1 – d2
 Constructive interference at point P will happen if the path difference is:
d1 - d2
n = 0, ±1, ± 2,
 Destructive interference at point P will happen if the path difference is:
d1 - d2
n = 0, ±1, ± 2,
 Partially destructive interference.
Standing waves are the result of the interference of two identical waves with the same frequency and the same
amplitude traveling in opposite direction.
A node is a point where the standing wave has minimal amplitude
A antinode is a point where the standing wave has maximal amplitude
The frequencies at which standing waves are produced are called natural frequencies or resonant frequencies
of the string or pipe or...
the lowest freq. standing wave is called FUNDAMENTAL or the FIRST HARMONICS
The higher freq. standing waves are called HARMONICS (second, third...) or OVERTONES
only standing wave that has wavelength λn = 2L/n can be formed on the string of length L.
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Beats are a periodic variation in loudness (amplitude) – throbbing - due to interference of two tones of slightly
different frequency. f = |f1 – f2|
1.What is the wavelength of a wave with a speed of 12 m/s and a period of 0.25 s?
a. 0.25 m
b. 1.5 m
c. 3.0 m
d. 24 m
e. 48 m
2. A sinusoidal transverse wave is traveling on a string. Any point on the string:
a. moves in the same direction as the wave
b. moves in simple harmonic motion with a different frequency than that of the wave
c. moves in simple harmonic motion with the same frequency as the wave
d. moves in uniform circular motion with a different speed than the wave
e. moves in uniform circular motion with the same speed as the wave
3.The tension in a string with a linear density of 0.0010 kg/m is 0.40 N. A 100 Hz sinusoidal wave on this string has a wavelength of:
a.
b.
c.
d.
e.
0.05 cm
2.0 cm
5.0 cm
20 cm
100 cm
4.A long string is constructed by joining the ends of 2 shorter strings. The tension in the strings is the same but string I has 4 times the linear
density of string II. When a sinusoidal wave passes from string I to string II:
a. the frequency decreases by a factor of 4
b. the frequency decreases by a factor of 2
c. the wave speed decreases by a factor of 4
d. the wave speed decreases by a factor of 2
e. the wave speed increases by a factor of 2
5. Two identical but separate strings, with the same tension, carry sinusoidal waves with the same frequency. Wave A has an amplitude
that is twice that of wave B and transmits energy at a rate that is __________ that of wave B.
a. half
b. twice
c. one-fourth
d. four times
e.
eight times
6. A certain string on a piano is tuned to produce middle C (f = 261.63 Hz) by carefully adjusting the tension in the string. For a fixed
wavelength, what is the frequency when this tension is doubled?
a. 130.08 Hz
b. 185.00 Hz
c. 370.00 Hz
d. 446.63 Hz
e. 523.26 Hz
7.
Which one of the following statements concerning the index of refraction for a given material is true?
A)
It may be less than 1.
B)
It may be measured in nanometers.
C)
It does not depend on the frequency of the incident light.
D)
For a given frequency, it is inversely proportional to the wavelength of light in vacuum.
E)
For a given frequency, it is inversely proportional to the wavelength of light in the material.
8.
The bending of light as it moves from one medium to another with differing indices of refraction is due to a change in what
property of the light?
A)
amplitude B) period C) frequency D) speed
E) color
9.
When certain light rays pass from a vacuum into a block of an unknown material, the measured index of refraction of the
material is 3.50. What is the speed of light inside the block?
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10.Blue light with a wavelength of 425 nm passes from a vacuum into a glass lens; and the index of refraction is found to be
1.65. The glass lens is replaced with a plastic lens. The index of refraction for the plastic lens is 1.54. In which one of the
two lenses does the light have the greatest speed and what is that speed?
11.A beam of light passes from air into water. Which is necessarily true?
A)
The frequency is unchanged and the wavelength increases.
B)
The frequency is unchanged and the wavelength decreases.
C)
The wavelength is unchanged and the frequency decreases.
D)
Both the wavelength and frequency increase.
E)
Both the wavelength and frequency decrease.
12.A ray of light passes from air into a block of glass with a refractive index of 1.50 as shown in the figure.
What is the value of the distance D?
13.The figure shows the path of a portion of a ray of light as it passes through three different materials. Note: The figure is
drawn to scale.
What can be concluded concerning the refractive indices of these three
materials?
A) n1 < n2 < n3
C) n1 > n2 > n3
B) n2 < n1 < n3
D) n1 < n3 < n2
E) n3 < n1 < n2
14.Light with a wavelength of 589 nm in a vacuum strikes the surface of an unknown liquid at an angle of 31.2° with respect
to the normal to the surface. If the light travels at a speed of 1.97  108 m/s through the liquid, what is the angle of
refraction?
15.A ray of light propagates in water (n = 1.333) and strikes a sheet of crown glass (n = 1.523). If the angle of refraction in
the glass is 35.2°, determine the angle of incidence.
16.
A.
B.
C.
D.
A type of electro-magnetic wave with a wavelength longer than that of visible light is
gamma ray
X ray
ultra-violet
infra-red.
29
17.
A.
B.
C.
D.
The observed frequency of a source of sound will be greater than the true frequency when the
source moves away from a stationary observer
observer moves towards a stationary source
observer moves away from a stationary source
source and observer both move with the same speed in the same direction.
18.
Waves moving across the surface of water cover a distance of 10·5 cm in 0·5 s. The horizontal distance between a
crest and the nearest trough is 15 mm. The vertical distance from crest to trough is 4 mm. Which line below gives a correct
description of the waves ?
Frequency in Hz
Amplitude in mm
A.
1·4
2
B.
7
2
C.
14
4
D.
7
4
19.
be
A.
B.
C.
D.
In a ripple tank, when waves go from deep water to shallow water their speed is reduced. This means that there will
no change in frequency or wavelength
a decrease in frequency and wavelength
a decrease in wavelength only
a decrease in frequency only
20. The diagram below shows ocean waves incident on a stone barrier protecting boats anchored behind it.
The boats could still be at risk of damage by waves mainly as a result of
A.
refraction.
B.
standing waves.
C.
diffraction.
D.
reflection.
21.
Two identical triangular wave pulses of
amplitude X travel toward each other long a string. At
instant shown on the diagram, point M is midway
between the two pulses.
the
The amplitude of the disturbance in the string as the pulses move through M is
A.
2X
B.
X
C.
X/2
D.
0
22. A guitar string produces 4 beats/s when sounded with a 250 Hz tuning fork and 9 beats per second when sounded with a 255 Hz tuning
fork. What is the vibrational frequency of the string?
a. 240 Hz
b. 246 Hz
c. 254 Hz
d. 259 Hz
e. 263 Hz
23. Which one of the following superpositions will result in beats?
a. the superposition of waves that travel with different speeds
b. the superposition of identical waves that travel in the same direction
c. the superposition of identical waves that travel in opposite directions
d. the superposition of waves that are identical except for slightly different amplitudes.
e. the superposition of waves that are identical except for slightly different frequencies.
24. Two timpani (tunable drums) are played at the same time. One is correctly tuned so that when it is struck, sound is produced that has a
wavelength of 2.20 m. The second produces sound with a wavelength of 2.08 m. If the speed of sound is 343 m/s, what beat frequency
is heard?
a. 7 beats/s
b. 9 beats/s
c. 11 beats/s
d. 13 beats/s
e. 15 beats/s
25.The largest number of beats per second will be heard from which pair of tuning forks?
a. 200 and 201 Hz
c. 534 and 540 Hz
e. 8420 and 8422 Hz
b. 256 and 260 Hz
d. 763 and 774 Hz
26. Which one of the following will result in standing waves?
30
a.
b.
c.
d.
e.
the superposition of waves that travel with different speeds
the superposition of identical waves that travel in the same direction
the superposition of identical waves that travel in opposite directions
the superposition of nearly identical waves of slightly different amplitudes
the superposition of nearly identical waves of slightly different frequencies
27. What is the distance from the fixed end of a guitar string to the nearest antinode?
a. 
b. 2
c. /2
d. /4
e. 3/4
28. A rope of length L is clamped at both ends. Which one of the following is not a possible wavelength for standing waves on this rope?
a. L/2
b. 2L/3
c. L
d. 2L
e. 4L
29. A 4-m long string, clamped at both ends, vibrates at 200 Hz. If the string resonates in six segments, what is the speed of transverse
waves on the string?
a. 100 m/s
b. 133 m/s
c. 267 m/s
d. 328 m/s
e. 400 m/s
30. Four standing wave segments, or loops, are observed on a string fixed at both ends as it vibrates at a frequency of 140 Hz. What is the
fundamental frequency of the string?
a. 23 Hz
b. 28 Hz
c. 35 Hz
d. 47 Hz
1.C 2.D 3.D 4.E
5. D 6. C
7.E
8. D
10. plastic, 1.95  108 m/s
11. B
12. 2.38 cm
13. E 14. 19.9°
15.41.2°
16. D
17. B 18. B.
22. B 23.E
24.B
25.D
26.C
27.D
28.E
The type of reflection is dependent on the size of the surface
irregularities relative to the incident wavelength ( ).
Sharp
image
DIFFUSE
Surface is rough relative to the
incident .
Light is reflected (scattered)
in all directions.
Fuzzy or
no image
All reflections follow
the law of reflection
Smooth (flat) surface
 > irregularities in the surface
9. 8.6  107 m/s
9. 3.76  1014 Hz
19. C
29.C
21. D
20. C
30.C
IMAGE FORMED BY PLANE MIRROR
- bunch of parallel rays encounters an obstacle:
SPECULAR - 'mirror' reflection
Surface particles are small
relative to the .
Light is reflected in a
single direction.
e. 70 Hz
An object is in front of a plane mirror.
The light is spreading in all directions.
Shown is the path of several rays.
This light reflects from the mirror.
The reflected light doesn’t meet (intersect) in the real space, but
extended rays behind mirror in the virtual space do. For the eyes it
seems as if these reflected light rays were coming from
another object back BEHIND the mirror at the intersection of
the extended rays !! We call this virtual space because the
light never really exists back there....it just SEEMS to be
coming from there. We call this apparent source of the light
rays a VIRTUAL IMAGE.
rough surface –
 <_ irregularities
Many natural surfaces act as
Different eyes at different positions; yet - the same image location.
a diffuse reflector to some extent.
Convex - Converging lenses
The lenses used in optical instruments (eyeglasses, cameras, telescopes, ...) are made from
transparent materials that refract light.
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Lenses work by refracting light at a glass-air boundary. Although refraction occurs at the boundary, we will treat all lenses
as bending the rays at the lens axis. In other words we are dealing with thin lenses.
symbol for thin lens
Optical axis – axis of symmetry of a lens
Focal point – Focus – the point to which light rays parallel to
the optical axis converge after passing through a converging
lens.
It is a real image of an object at infinite distance from lens.
Because light can go both ways lens has two focal points.
Focal length – distance between center of lens and focal point.
Three strategic and predictable light rays: Standard rays to help us draw an image formed by a lens
Ray 1 parallel to optical axis is refracted through the
lens through focal point F
Ray 2 coming through the centre of the lens will
continue in the same direction
Ray 3 passing through F in front of the lens is
refracted parallel to the axis



object
between ∞ and 2F
image is:
real
inverted
shrunken/diminished
object
at 2F
object
between 2F and F
image is:
real
inverted
same size
image is:
real
inverted
magnified/enlarged
object
between F and lens
image is:
virtual
upright
magnified/enlarged
All refracted rays coming from one point of an object intersect at one point called image.
32
First you locate image of one point of an object sitting on optical axis; now you know how to draw the
whole object.
Real image is formed when the light passes through the actual image location. Such image can be
caught on the screen.
Virtual image is image which is formed at the position where extended rays cross; you can not catch
it on the screen, but you can see it with your eyes. That is actually only thing you see. The real
refracted light doesn’t intersect, only extended rays in virtual space do. For the eyes (and brain) it
seems as if these refracted rays were coming from virtual image – intersection of the extended rays.
The Thin-Lens Equation and the Magnification Equation
1 + 1 =1
u
v
f
Magnification:
(lateral)
m=
hi
= -v
h
u
f is + for a converging lens
object is real: u is + object is virtual: u is –
image is real: v is + image is virtual: v is –
Power of a lens – measure of the extent of refraction of light:
P  dioptres  =
1
f  m
f greater – less curvature – less bending of light – less P
f smaller – more curvature – more bending of light – more P
Aberrations
In an ideal lens, all light rays from one point of the object would meet at the same point of the image,
forming a clear image. The influences which cause different rays to converge to different points are
called aberrations.
Lenses do not form perfect images, and there is always some degree of
distortion or aberration introduced by the lens which causes the image to
be an imperfect replica of the object. Careful design of the lens system for
a particular application ensures that the aberration is minimized. There are
several different types of aberration which can affect image quality. (Wikipedia)
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Spherical Aberration occurs because spherical surfaces are not the ideal shape with which to make a
lens, but they are by far the simplest shape to which glass can be ground and polished (the least
expensive) and so are often used.
perfect lens
spherical lens
paralel light rays striking the outer edges of a
lens are focused in a slightly different place
than beams close to the axis.
This problem is not limited to parallel light.
Any incident ray which strikes the outer edges
of the lens is subject to this departure from
the expected or proper course for the ideal lens. This manifests itself as a blurring of the image. Lenses
in which closer-to-ideal, non-spherical surfaces are used are called aspheric lenses.
Correction for spherical aberration
this or money
remember that all rays incident on the lens from the object will be
focused, and that the image will be formed even if part of the lens is
covered. The image will be simply dimmer.
Chromatic Aberration
A lens will not focus different colors in exactly the same
place because the focal length depends on refraction and
the index of refraction for blue light (short wavelengths)
is larger than that of red light (long wavelengths). The
amount of chromatic aberration depends on the
dispersion of the glass.
One way to minimize this aberration is to use glasses of
different dispersion in a doublet or other combination
This effect can be reduced by having a combination of a
convex and a concave lens made of glasses having different
refractive indices.
Chromatic aberration can be minimized using additional lenses
In an Achromat, the second lens cancels the dispersion of the first.
Achromats use two different materials, and one has a negative focal length.
34
1.An object is 2 m in front of a plane mirror. Its image is:
a. virtual, inverted, and 2 m behind the mirror
b. virtual, inverted, and 2 m in front of the mirror
c. virtual, erect, and 2 m in front of the mirror
d. real, erect, and 2 m behind the mirror
e. none of the above
2.An Easter Bunny is standing 1.0 m in front of a flat mirror. Standing 4.0 m behind the bunny is Godzilla. How far from the real
Easter Bunny will the image of Godzilla appear?
a. 5.0 m
c. 1.0 m
e. none of the above
b. 4.0 m
d. 6.0 m
3.An object is placed 1 m in front of a plane mirror. An observer stands 3 m behind the object. For what distance must the
observer focus his eyes in order to see the image of the object?
a. 1 m
b. 2 m
c. 3 m
d. 4 m
e. 5 m
4.Suppose that a mirror and three lettered cards are set up as in the figure. If a person’s eye is at point P, which of the lettered
cards will be seen reflected in the mirror? Explain.
1. E
2. D
3. E
4. By the law of reflection, only light from card B reaches her or his eyes.
1.Which one of the following statements is true concerning the focal length of a lens?
a. The focal length is the same for all colors.
b. The focal length is different for different colors because of reflection.
c. The focal length is different for different colors because of dispersion.
d. The focal length is different for different colors because of polarization.
e. The focal length is different for different colors because of spherical aberration.
2. An object is placed at the focal point of a converging lens of focal length f . What is the image distance?
a. f
b. 2 f
c. 1/f
d. 2/f
e. at an infinite
distance
3. An object is placed at the focal point of a thin diverging lens of focal length f . What is the image distance?
a. f
b. 2 f
c. 1/f
d. f /2
e. at an infinite
distance
4. Where must an object be placed in front of a converging lens in order to obtain a virtual image?
a. At the focal point
35
b.
c.
d.
e.
At twice the focal length
Greater than the focal length
Between the focal point and the lens
Between the focal length and twice the focal length
5. An erect object placed outside the focal point of a converging lens will produce an image that is:
a. erect and virtual
c. erect and real
e. impossible to locate
b. inverted and virtual
d. inverted and real
6. An object is 30 cm in front of a converging lens of focal length 10 cm. The image is:
a. real and larger than the object
b. real and the same size than the object
c. real and smaller than the object
d. virtual and the same size than the object
e. virtual and smaller than the object
7. An object is placed 4.0 cm from a thin converging lens with a focal length of 12 cm. Which one of the following statements is
true concerning the image?
a. The image is virtual and 6.0 cm from the lens.
b. The image is virtual and 12 cm from the lens.
c. The image is real and 3.0 cm from the lens.
d. The image is real and 6.0 cm from the lens.
e. The image is real and 12 cm from the lens.
8. A converging lens is used to focus light from a small bulb onto a book. The lens has a focal length of 10 cm and is located
40 cm from the book. Determine the distance from the lens to the light bulb.
a. 8 cm
b. 13 cm
c. 20 cm
d. 33 cm
e. 50 cm
9. When an object is placed 25 cm from a lens, a real image is formed. Which one of the following conclusions is incorrect?
a. The image is upright.
b. The lens is a converging lens.
c. The image may be reduced or enlarged.
d. The image distance can be less than 25 cm.
e. The focal length of the lens is less than 25 cm.
11.A 6.0-cm object is placed 30.0 cm from a lens. The resulting image height has a magnitude of 2.0 cm; and the image is
inverted. What is the focal length of the lens?
a. 7.5 cm
b. 15.0 cm
c. 22.5 cm
d. 30.0 cm
e. 45.0 cm
12. A camera with a lens of focal length 6.0 cm takes a picture of a 1.4-m man standing 11 m away. The height of the image is
about:
a. 0.39 cm
b. 0.77 cm
c. 1.5 cm
d. 3.0 cm
e. 6.0 cm
1.C
2. E
3. D
4. D
5. D
6. C
7. A
8. B
9. A
10. B
11. A
12. B
36