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Physics Review
Fall Semester
D.
Newtonian Mechanics (60 %)
A.
Description of motion
1. displacement, d = x – xo (m)
2. time interval t = t – to (s)
3. velocity, vav = d/t (m/s)
4. acceleration, aav = (v – vo)/t (m/s2)
a. instantaneous velocity (vt or v)
b. also includes changing direction
c. vav = ½(vo + vt) and if vo = 0, then v = 2vav
d. constant acceleration formulas
1. 5 variables; d, vo, v, a and t
2. data summary chart
d
vo
vt
a
v
E.
t
3. formula selection based on missing data
no v
no a
no d
no t
d = vot + ½at2 d = ½(vt + vo)t v = vo + at v2 = vo2 + 2ad
g. falling objects: acceleration due to gravity, g, at
sea level is about 9.80 m/s2 (can use 10 m/s2)
B. Graphing one dimensional motion
1. position, velocity and acceleration are graphed as
dependent variables (y-axis) with respect to the
independent variable, time (x-axis)
2. three graphs, x vs. t, v vs. t and a vs. t
a. constant velocity graphs
x
v
a
t
b.
t
t
constant acceleration graphs (a  0)
x
v
a
t
t
t
graphs are related
1. graphing slope vs. t generates graph to the
right
2. graphing area vs. t generates graph to the
left
Addition of vectors—component method
1. x-component and y-component for each vector are
calculated (R = magnitude;  = direction—measured
counterclockwise from +x)
2. Rx = Rcos
3. Ry = Rsin 
y
Rx = Ax + Bx
Bx= BcosB
c.
C.
By = BsinB
B
Ry = Ay + By
R
A

 Ax = AcosA
Ay = AsinA
x
4.
5.
6.
7.
Ax + Bx = Rx
Ay + By = Ry
R = (Rx2 + Ry2)½
tan = Ry/Rx   = tan-1(Ry/Rx)
add 180o to  when Rx is negative
Relative motion
example: a boat is traveling at certain velocity with respect
to the water (vboat), but the water has a velocity with
respect to the earth (vwater)  the boat velocity with respect
to the earth is vector sum: v = vboat + vwater
vboat
vwater
Projectile motion (horizontal and vertical dimensions)
1. data chart
d
vo
vt
a
t
direction
dy
vyo
vy
vertical
-10 m/s2
t
dx
vx
horizontal
a. solve for unknown in the y direction with
no vy
no dy
no t
dy = vyot + ½at2
vy = vyo + at
vy2 = vyo2 + 2ady
b. solve unknown in the x-direction with vx = dx/t
2. helpful shortcuts when a ball is kicked at ground level
across a horizontal field
a. vy = -vyo when the ball hits the ground
b. vy = 0 when the ball reaches its highest point
c. it takes half the time to reach its highest point
F. Uniform circular motion (north-south and east-west
dimensions)
1. occurs when there is a constant acceleration at right
angles to the velocity
2. constant perimeter (tangential) speed: vt = 2r/T
a. distance = circumference of the circle: 2r
b. time = time for one revolution: T (period)
3. constant inward (centripetal) acceleration: ac = v2/r
G. Newton’s laws of motion
1. multiple forces often act on an object
a. if vector sum of forces is zero, Fnet = 0, object is
stationary or velocity is constant
b. if vector sum of forces is not zero, Fnet = ma
1. Fnet  and v : increase v
2. Fnet  and v : decrease v
3. Fnet  and v : turn in a circle
2. vector quantity, measured in Newtons: 1 N = 1 kg•m/s
3. paired forces
a. forces always come in pairs (action and reaction),
which are always equal and in the opposite
direction (Newton's third law)
b. third law forces vs. first law forces
1. third law forces act on different objects  one
or both objects can accelerate
2. first law forces act on the same object 
there is no acceleration
H. Types of forces
1. push and pulls (Fp), pulls can be tensions (Ft)
2. spring force: Fs = kx
3. weight (Fg) is the force of attraction between the object
and the earth—gravity, Fg = mg
4. normal force (Fn) is the force that the surface exerts
on an object
a. perpendicular away from the surface
b. not calculated in isolation, but is determined by
other perpendicular forces so that Fnet- = 0
5. friction (Ff) is parallel to surface and opposes motion
a. when stationary: Ff is part of F|| = 0, but cannot
exceed Ff  sFn
b. when moving: Ff kFn
I.
Forces that cause linear acceleration
1. define system in motion
a. draw a free-body diagram of all forces
b. resolve forces into and || components
Fn
Fp

L.
Fp-
Ff
Fp-||
Fg

Fn
Fp

Ff
Fg
Fg
Fg-||


assign positive directions
two equations
1. Fnet- = 0
2. Fnet-|| = ma (m is all moving mass)
use kinematics to solve for vt, d, or t
solve for internal forces (tension)
c.
d.
2.
3.
Ff
J.
K.
m1
Ft-1
Ft-2
m2
F
a. solve for a: F – Ff = (m1 + m2)a
b. isolate m2 and solve for Ft-2: F – Ft-2 = m2a
4. vertical acceleration: Fp – Fg = ma
Forces that cause circular and rotational acceleration, Fr
1. torque,  = rFr
2. Newton’s laws for rotation
a. first law: rotating/orbiting object remains
rotating/orbiting unless acted by a torque
(moment of inertia, I = mr2)
b. second law: Fr = ma
3. rolling: Frolling = (1 + )ma
Forces that keep an object in circular motion
1. centripetal force needed to keep object in uniform
circular motion, Fc = mac = mv2/r
2. horizontal loop problem (mass on the end of a string)
a. Ft-x = Fc, Ft-y = Fg
b. Ft = (Fc2 + Fg2)½
c. tan = Fc/Fg ( is measured from vertical)
3. vertical loop problem (mass on the end of a string)
a. top: Fnet = Fc = Fg – Ft  Ft = Fc – Fg
b. bottom: Fnet = Fc = Ft – Fg  Ft = Fc + Fg
c. if on a roller coaster: Fn = Ft
4. turning on a road
a. when the road is horizontal: Fc = Ff = smg
b. roads are banked in order to reduce the amount
of friction needed to keep the car on the road
5. orbital problems
a. force of gravity supplies centripetal force needed
to keep an object in orbit: Fg = Fc
b. universal gravity, Fg = GMm/r2
G = 6.67 x 10-11 N•m2/kg2
c. related formulas
1. g = GMplanet/rplanet2
2. orbital velocity, v = (GM/r)½
Equilibrium forces
1. static vs. dynamic equilibrium
a. static: object is stationary
b. dynamic: object is in constant motion (inertia)
2. forces act through the center of mass
a. calculating center of mass (cm)
1. uniform: rcm = geometric center
2. complex: rcm = (m1r1 + m2r2 + ...)/(m1 + m2 + ...)
b. resolve forces into || and  components
1. F|| = 0
2. F = 0
c. special case short cut
1. works with 3 forces, 2 perpendicular
2. vector sum of forces forms right triangle
3. use trigonometry to solve for unknown
3. forces act away from the center of mass
a. off center forces generate torques   = 0
b. solve problems
1. draw free body diagram
2. select pivot point that eliminates an unknown
3. use:  = 0, F|| = 0, F = 0
4. solve for unknown
M. Work
1. W = F||d
2. measured in Joules (J = N•m)
3. no work is done when F is perpendicular to d
4. work is a scalar quantity: work can be positive or
negative, but has no direction
5. work equals area under graph
N. Power: P = W/t = Fvaverage measured in Watts (W = J/s)
O. Mechanical energy
1. work done on or by an object changes its amount of
mechanical energy (work-energy theorem)
2. object in motion has kinetic energy, K = ½mv2
a. rotation: Kr = ½mv2
b. rolling: Krolling = ½(1 +  )mv2
3. object elevated above the Earth’s surface has
gravitational potential energy, Ug = mgh
(for orbiting system: Ug = -GMm/r
4. stretched/compressed spring Us = ½kx2
P. Conservation of energy
1. work done by an "external" force (push or pull) adds
energy to an object
2. work done by Wg or Ws transforms U  K
3. all mechanical energy ultimately becomes random
kinetic energy of molecules (temperature rises)
4. solving problems
a. determine initial energy of the object, Eo
1. if elevated h distance: Ug = mgh
2. if moving v velocity: K = ½mv2
3. if spring compressed x distance: Us = ½kx2
b. determine energy added/subtracted due to a push
or pull: Wp = ±F||d
c. determine energy removed from the object by
friction: Wf = Ffd = (mgcosd
1. d is the distance traveled
2.  is the angle of incline (0o for horizontal)
d. determine resulting energy, E' = Eo ± Wp – Wf
e. determine d, h, x or v
1. if E' is 0, then 0 = Eo – Wf  Eo = mgcosd'
2. if E' is Ug: E' = mgh'
3. If E' is Us: E' = ½kx'2
4. if E' is K: E' = ½mv'2
f. general equation
K + U ± Wp – Wf = K' + U'
½mv2 + mgh + ½kx2 ± Fpd – Ffd = ½mv'2 + mgh' + ½kx'2
Q. Linear momentum, force and kinetic energy
1. linear momentum: p = mv (measured in kg•m/s)
2. vector quantity (must be added vectorily)
3. impulse: J = Ft = mv = p
4. K  p: K = p2/2m
R. Conservation of linear momentum
1. object explodes into two pieces mA and mB
Mtotalvcm = (mA + mB)vcm = mAvA’ + mBvB’
2. inelastic collision: mAvA + mBvB = (mA + mB)v’
3. elastic collision: mAvA + mBvB = mAvA’ + mBvB’ and
vA + vA’ = vB + vB’
4. collisions in two dimensions
mAvAx + mBvBx = (mA + mB)vx' or mAvAx’ + mBvBx’
mAvAy + mBvBy = (mA + mB)vy' or mAvAy’ + mBvBy’
S. Ballistics problem
1. bullet (mass = m) imbeds in a block (mass = M)
2. conservation of linear momentum: mvbullet = (M + m)v'
3. block swings or slides
a. swing: K = Ug  ½(M + m)v'2 = (M + m)gh
b. slide: K = Wf  ½(M + m)v'2 = (M + m)gd
T. Conservation of angular momentum, L
1. angular momentum, L = rmv, is constant as long as
no external rotating force, , acts on the system
2. orbiting planet r1v1 = r2v2
U. Simple Harmonic Motion (SHM)
1. mass (m) on a stretched spring, when released, will
oscillate about a midpoint = SHM
2. formulas at midpoint, 0, and extremes, A
midpoint
extreme
displacement
0
±A
vA = 0
velocity
vo = 2A/T = A(k/m)½
ao = 0
aA = vo2/A = A(k/m)
acceleration
0
UA = ½kA2
potential energy
K = ½mv2
0
kinetic energy
V. Simple Pendulum: T = 2(L/g)½
(period only depends on the length and g)
Practice Problems
1. What is the average velocity for a 100 m dash in 10.5 s?
v = d/t = 100 m/10.5 s = 9.5 m/s
2.
What is a car's acceleration that goes from 20 m/s to 0 in 5 s?
a = v/t = -20 m/s/5 s = -4
3.
4.
5.
6.
Bob, at a stop sign, sees Jane drive by at a constant 20 m/s.
Bob accelerates at 3.2 m/s2 to catch up to Jane.
a. How fast is Bob traveling when he catches up to Jane?
vav = ½(vo + vt)
20 m/s = ½(0 + v) vt = 40 m/s
b. How long does it take Bob to catch up to Jane?
vt = vo + at
40 m/s = 0 + (3.2 m/s2)t t = 12.5 s
c. How far has Bob traveled when he catches up to Jane?
d = 20t = (20 m/s)(12.5 s) = 250 m
8.
A ball is thrown upward with vo = 25 m/s.
a. What is the direction of acceleration?
up
down
(1) on the way up
(2) at the highest point
(3) on the way down
b. How high does the ball rise?
vt2 = vo2 + 2ad
02 = (25 m/s)2 + 2(-10.0 m/s2)d  d = 31.25 m
c. How much time does it take to reach the highest point?
vt = vo + at
0 = 25 m/s + (-10 m/s2)t  t = 2.5 s
d. What is the ball's velocity when it returns to its original
position?
(-)25 m/s
e. How much time is the ball in the air?
vt = vo + at
-25 m/s = 25 m/s + (-10 m/s2)t  t = 5 s
9. A cart moves on a straight horizontal track. The graph of
velocity, vx, versus time for the cart is given below.
vt (m/s)
4
2
d = vt = (30 m/s)(0.75 s) = 22.5 m
b. How far does the car travel during deceleration?
vt2 = vo2 + 2ad
02 = (30 m/s)2 + 2(-5.0 m/s2)d  d = 90 m
c. What is the total distance traveled?
t (s)
0
-2
0
1
3
5
7
9
11
On the grid below, graph the acceleration of the cart
as a function of time. (slope of the v vs. t graph)
a (m/s2)
a.
m/s2
A plane can accelerate at 2.5 m/s2. How long is the take
off if the plane must reach a final velocity of 45 m/s?
vt2 = vo2 + 2ad
(45 m/s)2 = 0 + 2(2.5 m/s2)d  d = 405 m
How long does it take for a rock to fall 45 m?
d = vot + ½at2
-45 m = 0 + ½(-10 m/s2)t2 t = 3 s
What is the final velocity of the space shuttle if it
accelerates at 7.5 m/s2 for 3 minutes?
vt = vo + at
vt = 0 + (7.5 m/s2)(180 s) = 1350 m/s
Calculate the minimum braking distance for a car traveling
at 30 m/s, which can decelerate at -5.0 m/s2 and whose
driver has a reaction time of 0.75 s.
a. How far does the car travel during the reaction time?
90 m + 22.5 m = 112.5 m
7.
1
0
t (s)
-1
0
1
3
5
7
9
11
On the grid below, graph the displacement of the cart
as a function of time. (area under the v vs. t graph)
d (m)
b.
6
4
2
t (s)
-0
-2
0
1
3
5
7
9
11
10. Determine which graph (1-4) is described below.
1
2
3
4
c.
t
t
t
t
(assume the initial position and velocity are zero)
1 2
3 4
d vs. t for a stationary object?
v vs. t for a stationary object?
d vs. t for an object in constant motion?
v vs. t for an object in constant motion?
a vs. t for an object in constant motion?
d vs. t for an accelerating object?
v vs. t for an accelerating object?
a vs. t for an accelerating object?
11. Determine the resultant for each situation below.
Displacement Direction (angle)
14 m
90o
6 m north + 8 m north
d.
6 m north + 8 m south
2m
-90o
10 m
143o
6 m north + 8 m west
o
12. A student jogs 100 m at 30 north of east. What are the x
and y components of the displacement vector?
x = (100 m)cos30 = 86.6 m, y = (100 m)sin30 = 50 m
13. Consider the following vectors, A, B and C.
a. Calculate the x-component and y-component of
vectors A, B and C and then determine the xcomponent and y-component of the resultant.
Vector Magnitude Angle x-component y-component
10cos30
10sin30
A
10 cm
30o
= 8.66
= 5.00
4cos145
4sin145
o
B
4 cm
145
= -3.28
= 2.29
5cos-60
5sin-60
C
5 cm
-60o
= 2.50
= -4.33
7.88
2.96
Resultant
b. Calculate the magnitude and angle of the resultant
vector.
Vector
Magnitude
Angle
R
(7.882 + 2.962)½ = 8.42
tan = dy/dx = -100 m/270 m   = -20o
What is the vertical velocity, vy?
v = vo + at = 0 + (-10 m/s2)(4.5 s) = -45 m/s
e.
With what speed do the supplies land?
v = (vx2 + vy2)½ = (602 + 452)½ = 75 m/s
17. A soccer ball is kicked at an angle of 45o above the
horizontal at 20 m/s.
a. What is vxo?
vxo = vocos = (20 m/s)cos45 = 14 m/s
b.
What is vyo?
vyo = vosin = (20 m/s)sin45 = 14 m/s
c. What is the total time in the air?
vy = vyo + at
-14 m/s = 14 m/s + (-10 m/s2)t  t = 2.8 s
d. What is the total distance the ball travels horizontally?
dx = vxot = (14 m/s)(2.8 s) = 39 m
e. What is the maximum height?
vy2 = vyo2 + 2ady
02 = (14 m/s)2 + 2(-10 m/s2)dx  dy = 9.8 m
f. Graph the following.
x
y
t
vx
v = vboat + vwater = 4.0 m/s + -3.0 m/s = 1 m/s east ( = 0o)
b. if the boat is headed due north?
v = (vboat2 + vwater2)½ = (42 + 32)½ = 5 m/s
 = tan-1(4/-3) = -53o + 180o = 127o
15. A ball is thrown horizontally at 15 m/s from a bridge that is
45 m above the water.
a. How long does it take for the ball to hit the water?
d = vot + ½at2
45 m = 0 + ½(-10 m/s2)t2  t = 3 s
b. How far horizontally does the ball travel?
d = vt = (15 m/s)(3 s) = 45 m
16. A rescue plane, traveling horizontally at 60 m/s, wants to
drop supplies to climbers on a rocky ridge 100 m below.
a. How long does it take for the supplies to fall 100 m?
d = vot + ½at2
100 m = 0 + ½(-10 m/s2)t2  t = 4.5 s
b. How far in advance must the supplies be dropped?
t
vy
t
ax
t
ay
t
tan-1(2.96/7.88) = 20.6o
14. The boat’s speedometer reads 4.0 m/s. The boat is
traveling in a river with a current of 3.0 m/s due west. How
fast is the boat traveling relative to the land?
a. if the boat is headed due east?
d = vt = (60 m/s)(4.5 s) = 270 m
What is the line of sight angle from the plane to ridge?
t
18. A student throws a ball, which lands on a cliff that is 45 m
away and 15 m above the student. The ball is in the air for
3 s. Determine the following for the ball's initial velocity
a. horizontal component
vx = dx/t = 45 m/3.0 s = 15 m/s
b. vertical component
dy = vyot + ½at2
15 m = vyo(3 s) + ½(-10 m/s2)(3 s)2  vyo = 20 m/s
c. magnitude
v = (vx2 + vy2)½ = (152 + 202)½ = 25 m/s
d.
direction
tan = vy/vx = 20 m/s/15 m/s = 53o
19. A car is traveling west on the north side of a circular
track (radius = 100 m) takes 31.4 s for one lap.
a. What is the direction of centripetal acceleration? south
b. What is the car's velocity?
v = 2r/T = 2(100 m)/31.4 s = 20 m/s
c.
What is the centripetal acceleration?
ac = v2/r = (20 m/s)2/(100 m) = 4 m/s2
d.
Which direction will the car skid if the section of track
on the north side was icy? west
20. The moon is 3.84 x 108 m away from earth and completes
one circular orbit in 2.33 x 106 s.
a. What is the moon’s orbital velocity?
v = 2r/T = 2(3.84 x 108 m)/2.33 x 106 s = 1040 m/s
b.
What is the centripetal acceleration of the moon
toward the earth?
ac = v2/r = (1040 m/s)2/3.84 x 108 m = 0.00282 m/s2
21. A 10-kg block is pulled vertically upward by a force of 150 N.
a. Draw all the forces acting on the block.
Ft
d.
normal force (Fn).
Fn = Fg- = 87 N
e.
force of friction, Ff. (k = 0.25)
Ff = Fn = 0.25(87 N) = 22 N
f.
overall || force acting on the block.
F|| = Fg-|| – Ff = 50 N – 22 N = 28 N
g.
block's acceleration.
F|| = ma  28 N = (10 kg)a  a = 2.8 m/s2
24. A 3-kg mass is suspended over a pulley which is
connected to a 1-kg mass
a. Draw all forces acting on the 3-kg and 1-kg weights.
Determine the
Fg
b. overall force acting on the block.
F|| = Ft – Fg = 150 N – (10 kg)(10 m/s2) = 50 N
c.
block's acceleration.
F|| = ma  50 N = (10 kg)a  a = 5
d.
m/s2
block's vertical velocity after 2.0 s of motion.
v = vo + at = 0 + (5 m/s2)(2.0 s) = 10 m/s upward
22. A 0.50-kg hockey puck traveling at 10 m/s on a horizontal
surface slows to a stop in 50 m.
a. Draw all the forces acting on the hockey puck
Fn
Ff
Determine the
Fg
b. puck's acceleration.
v2 = vo2 + 2ad
(0 m/s)2 = (10 m/s)2 + 2a(50 m)  a = -1 m/s2
c. force of friction.
Ff = ma = (0.50 kg)(-1 m/s2) = -0.50 N
d.
normal force.
Ft-1
Ft-3
1 kg
3 kg
Fg-1
Fg-3
Determine the
b. overall force acting on the system.
F|| = Fg-3 – Ft-3 + Ft-1 – Fg-1
F|| = (m3 – m1)g = (3 kg – 1 kg)(10 m/s2) = 20 N
c. acceleration.
F|| = ma
20 N = (4 kg)a  a = 5 m/s2
d. tension in the cable above the 3-kg mass.
F|| = Fg-3 – Ft-3 = m3a
30 N – Ft-3 = (3 kg)(5 m/s2)  Ft-3 = 15 N
e. tension in the cable above the 1-kg mass.
F|| = Ft-1 – Fg-1 = m1a
Ft-1 – 10 N = (1 kg)(5 m/s2)  Ft-1 = 15 N
f. time it takes to travel 10 m starting from rest.
d = vot + ½at2
10 m = 0 + ½(5 m/s2)t2  t = 2 s
25. A 15-kg block initially at rest on a table (k = 0.20) is attached
to a 5-kg mass, which hangs over the table's edge.
a. Draw all forces acting on the block and mass.
Fn
Fn = Fg = mg = (0.50 kg)(10 m/s2) = 5 N
e.
Ff
coefficient of kinetic friction.
Ff = Fn  0.50 N = (5 N)   = 0.10
23. A 10-kg block slides down a 30o incline.
a. Draw and Label all forces acting on the block. Break
up the force of gravity into its components.
Ff
Fn
Ft-15
Fg-15
Ft-5
Fg-5
Determine the
b. force of friction on the block.
Ff = Fn = m15g = (0.20)(15 kg)(10 m/s2) = 30 N
 Fg
Fg
Fg-||

Determine the
b. parallel component of the force of gravity, Fg-||.
Fg-|| = Fgsin = (10 kg)(10 m/s2)sin30 = 50 N
c.
perpendicular component, Fg-
Fg- = Fgcos = (10 kg)(10 m/s2)cos30 = 87 N
c.
overall || force on the system.
F|| = Fg-5 – Ff = (5 kg)(10 m/s2) – 30 N = 20 N
d. acceleration of the system.
F|| = (m15 + m5)a
20 N = (15 kg + 5 kg)a  a = 1 m/s2
e. tension in the rope.
F|| = Fg-5 – Ft = m5a
(5 kg)(10 m/s2) – FT = (5 kg)(1 m/s2)  Ft = 45 N
f. velocity of the mass 2 seconds after is starts moving.
v = vo + at
v = 0 + (1 m/s2)(2 s) = 2 m/s
26. A student wants to see what happens to a 1-kg piece of
clay when it slips off of a potter's wheel ( = ½). He places
it on the edge of a 0.5-m, 100-kg potter's wheel. He then
exerts a tangential force of 200 N along the edge of the
potter's wheel. The coefficient of friction, s, is 0.80.
a. What is the maximum force of friction between the
weight and potter's wheel?
31.
Ff = Fn = (0.80)(10 N) = 8 N
b. What is the maximum velocity before the weight slips?
Ff = Fc = mv2/r
8 N = (1 kg)v2/(0.5 m)  v = 2 m/s
c. How much can he accelerate the potter's wheel?
F = ma
200 N = ½(100 kg)a  a = 4 m/s2
d. How much time will it take the student to reach the
maximum velocity before the weight slips?
v = vo + at
2 m/s = 0 + (4 m/s2)t  t = 0.5 s
27. a. How much would a 800-N person weigh on the
following planets?
(1) A planet with the same radius as earth but twice
the mass?
Fg = GMplanetm/rplanet2  2 x Mplanet  2 x Fg
Fg = 2(800 N) = 1600 N
(2) A planet with the same mass but twice the earth's
radius?
Fg = GMplanetm/rplanet2  2 x rplanet  ¼ x Fg
Fg = ¼ (800 N) = 200 N
b. How would the following affect the force of gravity
between the earth and moon?
(1) The earth's mass is doubled
32.
33.
34.
35.
36.
Fg = GMplanetm/rorbit2  2 x Mplanet  2 x Fg
(2) The earth-moon distance was doubled.
Fg = GMplanetm/rorbit2  2 x rorbit  ¼ x Fg
c.
What is the acceleration due to gravity (g) on Mars?
(m = 6.4 x 1023 kg, r = 3.4 x 106 m)
g = GM = (6.67 x 10-11 N•m2/kg2)(6.4 x 1023 kg) = 3.7 m/s2
r2
(3.4 x 106 m)2
28. A 3-kg mass is connected to a 2-kg mass by a 2-m rod.
3 kg
2 kg
37.
Where is the center of mass measured from the left end?
rcm = (3 kg)(0 m) + (2 kg)(2 m)/(3 kg + 2 kg) = 0.8 m
53o
29. A 40-kg box is anchored to the ceiling
and wall by cords as shown to the right.
a. Draw a triangle showing the vector
sum of the three forces.
Ft-w
Ft-c
40 kg
Fg
38.
b. Calculate the tension in the ceiling cord.
sin53 = Fg/Ft-c
sin53 = 400 N/Ft-c  Ft-c = 500 N
c. Calculate the tension in the wall cord.
tan53 = Fg/Ft-w
tan53 = (400 N)/Ft-w  Ft-w = 300 N
30. A 3300-kg trailer is attached to a stationary truck.
39.
Determine the
a. normal force on the trailer tires at A.
A = cm
(8 m)FA = (5.5 m)(33000 N)  FA = 23,000 N
b. normal force on the support B.
FA + FB = Fg
23,000 N + FB = 33,000 N  FB = 10,000 N
What is the area under a force vs. displacement graph?
Work
State whether the form of energy increases (+), decreases
K
Ug or Us
(–) or doesn't change (0).
+
+
A book is accelerated upward.
0
+
A spring is stretched.
0
0
The earth orbits the sun.
+
–
A ball is dropped from a building.
Barry hits a 0.1-kg baseball straight up into the air with a
1000-N force. The ball remains on the bat for 0.10 m. What
is the maximum height reached by the ball?
F||d = mgh
(1000 N)(0.10 m) = (0.10 kg)(10 m/s2)h  h = 100 m
How much power does it take to lift 100 kg a distance of 4 m
in 5 s?
P = W/t = mgh/t
P = (100 kg)(10 m/s2)(4 m)/(5 s)  P = 800 W
An 800-kg car maintains a constant speed of 25 m/s against
a combined friction and air resistance force of 200 N. How
much power is needed to cruise at 25 m/s?
Pf = Ffv
Pf = (200 N)(25 m/s)  Pf = 5,000 W
A 1-kg block is on a level table ( = 0.2) and held against a
spring (k = 500 N/m), which is compressed 0.1 m. The block
is released and propelled horizontally across the table.
a. Determine the potential energy of the spring.
Us = ½kx2
Us = ½(500 N/m)(0.1 m)2 = 2.5 J
b. Determine the distance that the block travels until it
comes to rest.
Us = Wf = Ffd = mgd
2.5 J = (0.2)(1 kg)(10 m/s2)d  d = 1.25 m
A 10-kg box is initially at the top of a 5-m long ramp set at
37o. The box slides down to the bottom of the ramp. The
force of friction is 35 N. Determine the
a. loss in potential energy during the slide.
Ug = mgh = mgLsin
Ug = (10 kg)(10 m/s2)(5 m)sin37 = 300 J
b. work done by friction during the slide.
Wf = Ffd
Wf = (35 N)(5 m) = 175 J
c. velocity of the box at the bottom of the ramp.
Ug – Wf = K = ½mv2
300 J – 175 J = ½(10 kg)v2  v = 5 m/s
A 1-kg pendulum bob initially hangs from a 2-m string. It is
swung away from vertical until it is reaches an increase in
height of 0.25 m.
a. What is the velocity of the pendulum bob when it
returns to the vertical position?
Ug = Kmgh = ½mv2
v = (2gh)½ = [(2)(10 m/s2)(0.25 m)]½ = 2.2 m/s
b. What is the tension in the string when the bob is at the
bottom of its swing?
Ft = Fc + Fg = mv2/r + mg
Ft = (1 kg)(2.2 m/s)2/(2 m) + (1 kg)(10 m/s2) = 12.5 N
A 1-kg, disk ( = ½) is placed on a 5-m ramp where the top
is 3 m above the base of the ramp. The disk is placed at the
top and rolls down to the base of the ramp.
a. What is the disk's velocity when it reaches the base?
43. A 0.010-kg bullet traveling at 1,000 m/s penetrates a 2-kg
block of wood.
mgh = ½(1 + ½)mv2
a. How far would the block of wood travel along a rough
(10 m/s2)(3.0 m) = ¾v2 v = 6.3 m/s
surface ( = 0.25) before stopping?
b. How much time does it take for the disk to travel the 5 m?
mAvA + mBvB = (mA + mB)v'
d = ½(v + vo)t
(0.010 kg)(1,000 m/s) + 0 = (2.01 kg)v' v' = 5 m/s
5 m = ½(6.3 m/s + 0)t t = 1.6 s
K = W = F||d = Ffd  ½mv2 = mgd
c. Predict how the following alterations would change the
½(5 m/s)2 = (0.25)(10 m/s2)d  d = 5 m
disk's velocity at and time to reach the base of ramp?
b. How high would the block rise if it were suspended
Adjustment
Final Velocity
Time
from a long string?
no change no change
0.5-kg disk is used
K = Ug  ½mv2 = mgh
increase
decrease
1.0-kg ball ( = 2/5) is used
½(5 m/s)2 = (10 m/s2)h  h = 1.25 m
A steeper ramp is used (h = 3 m) no change decrease
44.
A satellite has a velocity of 3500 m/s when it is 9 x 106 m
40. A string attached to a 1-kg box resting on a table ( = 0.25)
from the earth's center. How fast is it traveling when it is
passes over a pulley ( = ½, m = 2 kg, r = 0.1 m) and
1.5 x 107 m from the earth's center?
attaches to a 0.5-kg mass hanging over the edge of the
r1v1 = r2v2
table. The system moves 1 m.
(9 x 106 m)(3500 m/s) = (1.5 x 107 m)v2  v2 = 2100 m/s
45. What is the angular momentum of mars?
Determine
1m
a. kinetic energy.
m (kg)
rmars (m)
rorbit (m)
Trotation (s)
Torbit (s)
6.4 x 1023 3.4 x 106 2.3 x 1011 8.8 x 104
5.9 x 107
K = Ug – Wfr = mgh – mgd
K = (.5 kg)(10 m/s2)(1 m) – (.25)(1 kg)(10 m/s2)(1 m) = 2.5 J
a. about its axis of rotation ( = 2/5)
b. velocity.
L = rmv = r(2/5)m(2r/T) = 4mr2/5T
K = K1 + Kpulley + K0.5 = ½m1v2 + ½m2v2 + ½m0.5v2
L = 4(6.4E23 kg)(3.4E6 m)2/(5)(8.8E4 s) =2.1E32 kg•m2/s
2.5 J = ½(1 kg + 1 kg + 0.5 kg)v2  v = 1.4 m/s
b. in its orbit around the sun
41. A 0.06-kg tennis ball at rest is hit by a racket, which makes
L = rmv = r(1)m(2r/T) = 2mr2/T
contact with the ball for 0.025 s, after which, the ball's
L = 2(6.4E23 kg)(2.3E11 m)2/(5.9E7 s) = 3.6E39 kg•m2/s
velocity is 55 m/s.
46. A student (m = 60 kg) runs at 8 m/s tangentially toward a
a. Determine the change in the ball's momentum.
stationary merry-go-round ( = ½, m = 120 kg, r = 1.6 m),
p = mv = (0.06 kg)(55 m/s)
jumps on the merry-go-round and sets it rotating.
p = 3.3 kg•m/s
a. What is the velocity of the student after she jumps on
b. Determine the average force exerted by the racket.
to the merry-go-round?
Ft = mv
rsmsvs = r(sms +  mmm)v'
F(0.025 s) = -3.3 kg•m/s  F = 132 N
(1)(60 kg)(8 m/s) = [(1)(60 kg) + (½)(120 kg)]v'  v' = 4 m/s
42. Consider two masses, mA = 1 kg and mB = 2 kg, which can
b. What is the minimum coefficient of friction needed to
move on a frictionless surface.
keep the student on the merry-go-round?
a. A spring (k = 500 N/m) compressed 0.2 m is placed
Ff = Fc  mg = mv2/r
between the two stationary masses and then released.
(10 m/s2) = (4 m/s)2/(1.6 m)  = 1
What are the velocities of the two masses?
47. Ball A (1 kg) and ball B (1 kg) hang
(mA + mB)vcm = mAvA' + mBvB'
from 0.8-m strings. Ball A is raised
0 = (1)vA' + (2)vB'  vA' = -2vB'
to a horizontal position and swings
Us = KA + KB  ½kx2 = ½mAvA'2 + ½mBvB'2
toward Ball B and collides with it.
(500 N/m)(0.2 m)2 = (1 kg)(-2vB')2 + (2 kg)vB'2
a. What is the velocity of ball A before the collision?
vB' = 1.8 m/s and vA' = -2vB' = -2(1.8 m/s) = -3.6 m/s
Ug = K  mAgh = ½mAvs2
b. Mass A, moving east with a speed of 2 m/s, has a
(10 m/s2)(0.8 m) = ½vs2 vs = 4 m/s
head-on inelastic collision with mass B, moving west
Assume that the collision is inelastic.
with a speed of 4 m/s. What is the resulting velocity of
b. What is the velocity of ball B after the collision?
the combined masses?
r(mAvA + mBvB) = r(mA + mb)v'
mAvA + mBvB = (mA + mB)v'
(1 kg)(4 m/s) + 0 = (1 kg +1 kg)v'  v' = 2 m/s
(1 kg)(2 m/s) + (2 kg)(-4 m/s) = (1 kg + 2 kg)v'  v' = -2 m/s
c. What is the maximum change in height reached by
c. Mass A, moving east with a speed of 3 m/s, has a
ball B after the collision?
head-on elastic collision with mass B initially at rest.
Ug = K  (mA + mB)gh = ½(mA + mB)v'2
What are the resulting velocities?
(10 m/s2)h = ½(2 m/s)2  h = 0.2 m
vA + vA' = vB + vB'  3 + vA' = 0 + vB'  vB' = vA' + 3
d. What is the maximum angle that ball B makes with the
mAvA + mBvB = mAvA' + mBvB'
vertical?
(1 kg)(3 m/s) = (1 kg)(vA') + (2 kg)(vA' + 3)
vA' = -1 m/s and vB' = vA' + 3 = -1 m/s + 3 = 2 m/s
cos = (L – h)/L = 0.6/0.8   = 41o
d. Mass A, moving 37o north of east at 5 m/s collides and
e. What is the maximum height reached after the
sticks to mass B moving south at 3 m/s. What is the
collision if both balls were raised to the horizontal
resulting speed and direction of the combined
position in the opposite direction and released?
masses?
r

(m
AvA + mBvB) = r (mA + mb)v'
'
px: mAvAcosA + mBvBcosB = (mA + mA)vx
(1 kg)(4 m/s) + (1 kg)(- 4 m/s) = (1 kg +1 kg)v'
(1)(5cos37) + (2)(3cos-90) = (1 + 2)vx'  vx' = 1.33 m/s
v' = 0 m/s  h = 0 m
py: mAvAsinA + mBvBsinB = (mA + mA)vy'
Assume the collision is elastic and ball A is raised to the
(1)(5sin37) + (2)(3sin-90) = (1 + 2)vy'  vy' = - 1 m/s
horizontal position initially.
2
2
½
2
2
½
v' = (vx' + vy' ) = [(1.33 m/s) + (-1 m/s) ] = 1.7 m/s
o
tan = vy'/vx' = -1 m/s/1.33 m/s = -0.75   = -37
f. What is ball B's velocity after the collision?
vA + vA' = vB + vB'
4 m/s + vA' = 0 + vB'  vA' = vB' – 4
r(msvs + mbvb) = r(msvs' + mbvb')
(1 kg)(4 m/s) + 0 = (1 kg)(vB' – 4) + (1 kg)vB'  vb' = 4 m/s
g. Ideally will the system ever stop? Explain your answer.
No, conservation of energy will keep the balls going
back and forth (Newton's Cradle).
48. A 4-kg bob is suspended from a light spring (k = 10 N/m).
The mass is pulled downward 5 cm and then released.
The system is set into simple harmonic motion.
a. What is the period of vibration? 4 s
b. What is the frequency of vibration?
f = 1/T
f = ¼ s = 0.25 s-1
c. What is the amplitude in meters?
5 cm x 1 m/100 cm = 0.05 m
d.
Determine the time for each of the following.
A
a
v
Us
3
1
2
1, 3
+ max
zero
0, 2, 4
0, 2, 4
1, 3
0, 2, 4
K
0, 2, 4
1, 3
1
3
0, 4
- max
e. What is the maximum acceleration?
a = A(k/m)
a = (0.05 m)(10 N/m/4 kg) = 0.125 m/s2
f. What is the maximum velocity?
v = A(k/m)½
v = (0.05 m)(10 N/m/4 kg)½ = 0.08 m/s
g. What is the maximum kinetic energy?
K = ½mv2
K = ½(4 kg)(0.08 m/s)2 = 0.013 J
h. What is the maximum potential energy?
Us = ½kA2
Us = ½(10 N/m)(0.05 m)2 = 0.013 J
49. A pendulum is made with a 1-kg bob on the end of a string.
a. What is the length of the string if the pendulum has a
period of 2 s?
Tp = 2(L/g)½  Tp2 = 42L/g
(2 s)2 = 42L/(10 m/s2)  L = 1 m
b. The 1-kg bob is replaced with a 2-kg bob. What is the
period of oscillation for the new pendulum? 2 s
c. The pendulum is transported to Mars, where g = 4 m/s2.
What is the period of oscillation on Mars?
Tp = 2(L/g)½
Tp = 2(1.0 m/4 m/s2)½ = 3.1 s
Waves (25 %)
A.
B.
Wave motion
1. characteristics of a periodic wave
a. amplitude, A: distance from midpoint
b. wavelength, : distance between crests
c. frequency, f: the number of waves per s (Hz or s-1)
d. period, T: time per wave (s), T = 1/f
e. velocity, v: speed of the waveform, vw = /T = f
2. types of waves
a. transverse wave: disturbance  wave 
b. longitudinal wave: disturbance  wave 
Interference
1. amplitudes combine when two waves meet
(superposition principle)
a. crest + crest = constructive interference
b. crest + trough = destructive interference
2. similar frequency produce beats: fbeat = |fB – fA|
3. same frequency = standing wave
a. constructive interference = “antinode”
b. destructive interference = “node”
c. vibrating string (transverse wave)
1. velocity: vw = [FT/(m/L)]½
2. harmonics (first, second, etc.)
a. number of harmonic = number of loops
b. n = 2L/n and fn = nf1
4. resonance: vibrating a structure at its natural
frequency will cause amplitude magnification
C. The Doppler Effect
1. when source and/or observer are moving with respect
to each other, the perceived frequency changes
2. frequency heard, f', compared to frequency generated, f
a. f’ = f(vw ± vo)/(v ± vs)
b. f' > f when approaching (+ vo and/or – vs)
c. f' < f when receding (– vo and/or + vs)
D. Reflection off of a surface
1. law of reflection: incident = reflection
2. phase shift when reflecting surface has greater n
E. Refraction within a medium
1. light slows when entering a transparent medium
a. vn = c/n = 3 x 108/n
b. v = f and fn = fo  n = o/n
2. rays that enter at an angle from normal
a. bend toward normal if nincidence < nrefraction
b. bend away from normal if nincidence > nrefraction
3. Snell's law: n1sin 1 = n2sin 2
4. n increases with frequency (violet > red) = dispersion
5. total reflection
a. only when nincidence > nrefraction
b. critical angle, c, sin c = (nlow/nhigh)
F. Lenses and Mirrors
1. parabolic mirror: focus = ½ radius of curvature, f = ½r
2. shape vs. function
shape
convex
concave
lens
converging
diverging
mirror
diverging
converging
3. Lens/mirror equation: 1/do + 1/±di = 1/±f
a. +f for converging lens or mirror,
–f for diverging lens or mirror
b. do = object distance, use + for single optic
c. di = image distance
1. +, image is real—visible on screen
2. –, image is virtual—visible through lens/mirror
d. ho = object height, use + for upright object
e. hi = image height
1. +, image is upright
2. –, image is upside down (inverted)
f. magnification, M = hi/ho = -di/do
1. > |1|, image is larger than object
2. < |1|, image is smaller than object
3. generalizations
a. converging: |M| increases as do  f
b. for diverging: M increases as do  0
g. small or partially covered lens reduces amount of
light (brightness of image) but not size or nature
G. Interference
1. 2 or more openings (slits)
a. light intensity decreases as m increases
b. angular deflection () from center to band
1. tan  = x/L
2. x = distance from center to band
3. L = distance between slits and screen
c.
angular deflection vs. order
1. constructive: sinc = m/d
2. destructive: sind = (m + ½)/d
3. d = distance between slits
d. multiple slits (grating)
1. more pronounced effect
2. d = grating space = length/lines
2. one opening: width of zero order maxima, W  2L/d
3. partial reflection
minimum thickness of a film, T (film = o/n)
nfilm is middle value
nfilm is extreme value
Reflection
Bright
T = ½film
T = ¼film
Dark
T = ¼film
T = ½film
Practice Problems
50. A 20 s-1 standing wave is generated on a 0.1 kg string.
54. An object 2 cm high is placed on the radius of curvature of
a concave mirror of focal length 1 cm.
a. Mathematically determine the following.
image distance, di
magnification, M
image height, hi
1/do + 1/di = 1/f
M = -di/do
M = hi/ho
1/2 + 1/di = 1/1
M = -2/2 = -1
-1 = hi/2
di = 2 cm
hi = -2 cm
b. Circle the correct characteristic of the image.
(1) Image is (real/virtual).
(2) Image is (upright/inverted).
(3) Image is (larger/smaller/same size) as the object.
55. 700-nm laser shines on two slits, which are 2.5 mm apart,
and forms an interference pattern on a screen 10 m away.
a. What is the angular deflection of the first bright fringe?
sin = m/d = (1)(700 x 10-9 m)/(0.0025 m)
  = 0.016o
b. What is the distance, in mm, from the central
maximum to the first order bright fringe?
tan = x/L
4
x = Ltan = (10 m)tan(.016o) = 2.8 x 10-3 m (2.8 mm)
a. Which harmonic is diagramed? Second
56. The hydrogen spectrum is analyzed using a 500-lines/mm
b. What are the wavelengths and frequencies of the first
diffraction grating, which is 1.00 m away from the light
harmonic?
source. An aqua line appears 25.4 cm from the light
n = 2L/n  1 = 2(4 m)/1 = 8 m
source and a red line appears 35.7 cm from the light
-1
-1
fn = nf1  f1 = 20 s /2 = 10 s
source. Determine the
c. What are the wavelengths and frequencies of the
a. wavelength of aqua light.
fourth harmonic?
tan = (25.4 x 10-2 m)/(1.00 m)   = 14.3o
n = 2L/n  4 = 2(4 m)/4 = 2 m
sin = m/d   = dsin/m = (10-3/500)sin(14.3)/1 = 4.94E-7
fn = nf1  f4 = (4)(10 s-1) = 40 s-1
b. wavelength of red light.
d. What is the wave velocity?
tan = (35.7 x 10-2 m)/(1.00 m)   = 19.6o
vw =  f
sin = m/d   = dsin/m = (10-3/500)sin(19.6)/1 = 6.71E-7
vw = (4 m)(20 s-1) = 80 m/s
57. 600-nm-laser beam is incident on a single slit of width 0.200
d. What is the tension in the string?
mm. How wide is the central maximum on a screen 1.00 m
vw = [FT/(m/L)]½  vw2 = FtL/m
away?
(80 m/s)2 = Ft(4 m)/(0.1 kg)  Ft = 160 N
W  2L/D = 2(600 x 10-9 m)(1.00 m)/(0.200 x 10-3 m)
51. A police car emits a 450 Hz sound while traveling 33 m/s
W  6 x 10-3 m (6 mm)
toward a stationary observer. Approximately what frequency 58. Determine the thickness for each of the following.
does the observer hear if the speed of sound is 330 m/s?
a. A soap bubble that appears yellow.
f/f  v/vw  f/450  33/330  f = 45 Hz
( = 600 nm, n = 1.35)
f' = 450 Hz + 45 Hz = 495 Hz
(nextreme bright T = film/4)
52. Light strikes a flat piece of glass (n = 1.50) at an incident
T = /4n = 600 nm/(4)(1.35) = 111 nm
angle of 45o. Some of the light is reflected and some light
b. A coating (n = 1.22) on glass (n = 1.50) that will
passes out of the glass on the opposite side.
eliminate 600-nm light.
o
a. What is the angle of reflection? 45
(nmiddle dark  T = film/4)
b. What is the angle of refraction inside the glass?
T = /4n = 600 nm/(4)(1.22) = 123 nm
nrsinr = nisini
Atomic and Nuclear Physics (15 %)
o
o
1.50sinR = 1.00sin45  R = 28 
A.
Wave/Particle
Duality
c. With what angle does light exit the glass on the
1. wave nature of electro-magnetic (EM) radiation
opposite side?
a. travel at the speed of light, c = 3 x 108 m/s
nrsin R = nisini
b. polarize
1.50sin28 = 1.00sini  i = 45o
c. Doppler shift (red = receding, blue = approaching)
d. What is the critical angle if the light originated in the
2. particle nature of EM radiation
glass surrounded by air?
a. hot objects glow a color (frequency)  T  f
b. Ephoton = hf (h = 6.63 x 10-34 J•s)
sinc = (nlow/nhigh) = 1/1.50 c = 42o
c. Ephoton = hf = mc2 and pphoton = mc = h/ = E/c
53. An object 1 cm high is placed 3 cm from a convex lens of
3. De Broglie wavelength: particle = h/p = h/mv
focal length 4 cm.
B.
Quantum Theory
a. Mathematically determine the following.
1. electrons energy levels: En = -B/n2 (BH = 13.6 eV)
image distance, di
magnification, M
image height, hi
2. photon emissions: Ephoton = En-high – En-low
1/do + 1/di = 1/f
M = -di/do
M = hi/ho
3. photoelectric effect Kelectron = Ephoton - 
1/3 + 1/di = 1/4
M = -(-12)/3 = 4
4 = hi/1
a. Ephoton = 1240 eV•nm/nm
di = -12 cm
hi = 4 cm
b. Circle the correct characteristic of the image.
b.  called "work function" in eV
c. Kelectron = ½mv2 = p2/2m
(1) Image is (real/virtual).
1. convert eV to joules (1 eV = 1.60 x 10-19 J)
(2) Image is (upright/inverted).
2. melectron = 9.11 x 10-31 kg
(3) Image is (larger/smaller/same size) as the object.
C.
Structure and properties of the nucleus
1. 2 types of particles--nucleons
a. positively charged proton and neutral neutron
b. rest mass chart (u = 1.66 x 10-27 kg)
2. nuclide (nucleus without orbiting electrons)
a. number of protons = atomic number (Z)
b. number of protons + neutrons = mass number (A)
c. nuclide symbol: AZX (X is atomic symbol)
1. X corresponds to Z number
2. A varies = isotopes
D. Binding energy and nuclear forces
1. takes energy to break a stable nucleus into its parts 
energy is added to the nucleus
a. nuclide + BE = protons + neutrons
b. BE equals binding energy
c. Einstein established that matter and energy are
interchangeable: E = mc2
1. energy, E (J)
2. mass, m (kg)
3. c = 3 x 108 m/s (speed of light)
d. mnuclide + mBE = mp + mn  mBE = mp + mn – mnuclide
2. nuclear processes
a. mproducts – mreactants = m (E = mc2)
1. m > 0 for non-spontaneous process
2. m < 0 for spontaneous process
b. fusion combines small nuclides
c. fission splits large nuclides
E. Naturally occurring radioactivity
symbol penetration
example
4  = 4 He
226 Ra  222 Rn + 4 
alpha
low
2
2
88
86
2
0  = 0 e
14 C  14 N + 0 
beta
medium
-1
-1
6
7
-1
0 
0 - + 0  +  
gamma
high
0
-1
1
F. Artificial induced nuclear reaction (transmutation)
1. mass and charge are conserved
2. 10n + 147N  146C + 11p
G. Rate of decay and half-life
half-life, t½: time it takes to reduce radioactivity by half
(1  ½  ¼  1/8  1/16 ...)
Practice Problems
59. Yellow light has a wavelength of 600 nm. Determine the
a. frequency.
f = c/ = 3 x 108 m/s/600 x 10-9 m
f = 5.0 x 1014 s-1
b. energy.
E = hf = (6.63 x 10-34 J•s)(5.0 x 1014 s-1)
E = 3.3 x 10-19 J
c. relativistic mass.
m = E/c2 = 3.3 x 10-19 J/(3 x 108 m/s)2
m = 3.7 x 10-36 kg
d. momentum.
p = mc = (3.7 x 10-27 kg)(3 x 108 m/s)
p = 1.1 x 10-27 kg•m/s
e. the de Broglie wavelength of an electron with the
same momentum as the yellow light.
 = h/p = 6.63 x 10-34/(1.1 x 10-27 kg•m/s)
 = 6.0 x 10-7 m
60. Consider an electron transition from the third energy level
to the second energy level on a hydrogen atom.
a. What are E3 and E2?
En = -B/n2 = -13.6 eV/n2
E3 = -13.6 eV/32 = -1.5 eV
E2 = -13.6 eV/22 = -3.4 eV
b. What is E of the emission line?
E = En-high – En-low
E = -1.5 – (-3.4) = 1.9 eV
c. What is the wavelength of the photon?
Ephoton = 1240 eV•nm/nm
nm = 1240 eV•nm/1.9 eV = 653 nm
d. What is the frequency of the photon?
f = c/ = 3 x 108 m/s/653 x 10-9 m
f = 4.6 x 1014 s-1
61. A barium surface ( = 2.48 eV) is illuminated by 400-nm
light.
a. What is the maximum kinetic energy of the
photoelectrons?
K = Ephoton – = 1240 eV•nm/nm – 
K = 1240 eV•nm /600 nm – 2.48 eV = 0.62 eV
b. What is the threshold wavelength for photoemission
from the surface of barium?
 = 1240 eV•nm/nm
nm = 1240 eV•nm/2.48 eV = 500 nm
62. Complete the following chart for the isotope, Rn-226.
Number of Number of Number of
Z value
A value
nucleons
protons
neutrons
226
86
140
86
226
63. Determine the binding energy (in joules) of an average
nucleon in 2H (2.013553 u) by completing the following
sequence of calculations
m m = 1mp + mn – mH-2
(u) m = 1.007276 + 1.008665 – 2.013553 = 0.002388 u
m = 0.002388 u x 1.66 x 10-27 kg/1 u
m
(kg) m = 3.96 x 10-30 kg
BE = mc2 = (3.96 x 10-30 kg)(3 x 108 m/s)2
BE
BE = 3.57 x 10-13 J
BE/A BE/A = 3.57 x 10-13 J/2 = 1.78 x 10-13 J/nucleon
64. Fill in the missing symbols.
45
20Ca
 4521Sc + 
n + 23592 U  14156Ba + 9236Kr + 3 n
65. Stationary Uranium-232 (231.98667 u) undergoes an alpha
decay (4.001504 u) to produce Thorium-228 (227.97936).
a. Write a nuclear equation for this radioactive decay.
232
92U
 22890Th + 42He
b. Calculate the change in mass in kg for this reaction.
227.97936 + 4.001504 – 231.98667 = -0.005806 u
-0.005806 u x 1.66 x 10-27 kg/1 u = -9.64 x 10-30 kg
c. Calculate the energy is joules released per decay.
E = mc2
E = (9.64 x 10-30 kg)(3 x 108 m/s)2 = 8.67 x 10-13 J
d. If all the released energy becomes kinetic energy,
what is K/KRa?
mRavRa = mv (228)(4) = (4)(228)
K = ½mv2 = (4)(228)2 = 228 = 57
KRa ½mRavRa2 (228)(4)2
4
e. What is the alpha particle's velocity if 98 % of the
energy is absorbed by the particle?
E = ½mv2
(.98)(8.67 x 10-13 J) = ½(4 x 1.66 x 10-27 kg)v2 
v = 1.6 x 107 m/s
f. What is the De Broglie wavelength of the alpha particle?
 = h/mv
 = 6.63 x 10-34 J•s/(4 x 1.66 x 10-27 kg)(1.6 x 107 m/s)
 = 6.24 x 10-15 m
66. How long does it take for a sample of 32P (t1/2 = 14 days) to
lose 3/4 of its activity?
The loss of 3/4 activity means that 1/4 remains
It takes 2 half-lives to reach 1/8 (1  1/2  1/4 )
2 x 14 = 28 days