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PHYSICS 231
Lecture 9: Revision
Remco Zegers
Walk-in hour: Thursday 11:30-13:30
Helproom
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The slides about friction are in lecture 8!!
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TRIGONOMETRY
SOH-CAH-TOA:
sin=opposite/hypotenuse
cos=adjacent/hypotenuse
tan=opposite/adjacent
Pythagorean theorem:
c  a b
2
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Units
• Convert all units in your problem to be in the SI system
• When adding/subtracting two quantities check whether
their units are the same.
• If you are unsure about an equation that you want to
use, perform the dimensional analysis and make sure
that each part of the equation that is set
equal/subtracted/added have the same dimensions
• When 2 quantities are multiplied, their units do not
have to be the same. The result will have as unit the
multiplied units of the quantities being multiplied.
• Sin, cos and tan of angles are dimensionless
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Solving a quadratic equation
At2+Bt+c=0
• Use the equation:
 b  b  4ac
t 
2a
2
• CHECK THE ANSWER
• OR just calculate it!
t2+(B/A)t+C/A=0 and use (t+d)2=t2+2dt+d2
[t+(B/2A)]2-B2/4A2+C/A=0
t=-B/2A(B2/4A2-C/A)
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Constant motion
x(t)=x0
x(m)
v
m/s
t
v(t)=0
Constant velocity
x(m) x(t)=x0+v0t
t
v
m/s
v(t)=v0+at
t
a
m/s2
a(t)=0
t
t
v
m/s
v(t)=v0
t
a
m/s2
Constant acceleration
x(t)=x0+v0t+½at2
x(m)
a
m/s2
a(t)=0
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t
t
a(t)=a0
t
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x
v
t
time
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Vector operations in equations
 X a b   X a   X b   X a  X b 

        

 Ya b   Ya   Yb   Ya  Yb 
 X a b   X a   X b   X a  X b 

        

 Ya b   Ya   Yb   Ya  Yb 
y
(xb,yb)
B
(xa+b,ya+b)
A+B
A (xa,ya)
x
Xa=Acos()
Ya=Asin()
length/magnitude of A: (Xa2+Ya2)
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Parabolic motion: decompose x and y directions
vx=v0cos
vy=v0sin-2g=0
V=v0
vx=v0cos
vy=v0sin
vx=v0cos
vy=v0sin-1g
vx=v0cos
vy=v0sin-3g
vx=v0cos
vy=v0sin-4g

vx=v0cos
vy=v0sin-5g
t=0
t=1
t=2
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t=3
t=4
t=5
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Parabolic motion
X(t)=X0+V0cost
Y(t)=Y0+V0sint-1/2gt2
X=X0
Y=Y0

t=0
t=1
t=2
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t=3
t=4
t=5
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2D motion
• When trying to understand the motion of an object in
2D decompose the motion into vertical and horizontal
components.
• Be sure of your coordinate system; is the motion of the
object you want to study relative to another object?
• Write down the equations of motion for each direction
separately.
• If you cannot understand the problem, draw motion
diagrams for each of the directions separately.
• Make sure you understand which quantity is unknown,
and plug in the equation of motions the quantities that
you know (givens). Then solve the equations.
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Newton’s Laws
 First Law: If the net force exerted on an object
is zero the object continues in its original state of
motion; if it was at rest, it remains at rest. If it
was moving with a certain velocity, it will keep on
moving with the same velocity.
 Second Law: The acceleration of an object is
proportional to the net force acting on it, and
inversely proportional to its mass: F=ma
 If two objects interact, the force exerted by the
first object on the second is equal but opposite in
direction to the force exerted by the second
object on the first: F12=-F21
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General strategy for problems with forces
• If not given, make a drawing of the problem.
• Put all the relevant forces in the drawing, object by
object.
• Think about the axis
• Think about the signs
• Decompose the forces in direction parallel to the
motion and perpendicular to it.
• Write down Newton’s first law for forces in the
parallel direction and perpendicular direction.
• Solve for the unknowns.
• Check whether your answer makes sense.
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A snowball is launched horizontally from the top of a building
at v=12.7 m/s. If it lands 34 m away from the bottom, how
high was the building?
horizontal
Vertical
V0=12.7 m/s
h
x
y
vx
t
h?
t
Vy
X(t)=x0+voxt
=0+12.7t
2
Y(t)=y
+v
t-0.5gt
0 0y
=34 (hit the
2
=h-0.5gt
ground)
=0 (hits the ground)
t=34/12.7=2.68 s
so: h=0.5gt2
d=34m
h=35 m
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A red ball is thrown upward with a velocity of 26.8 m/s.
A blue ball is dropped from a 13.3 m high building with
initial downward velocity of 5.00 m/s. At what time will
the balls be at the same height.
Vo=-5.00
blue
Blue ball:
red
13.3
yb(t)=y0+v0bt-0.5gt2=
x v0r
=13.3-5.t-0.5gt2
x
Red ball:
2=
y
(t)=y
+v
t-0.5gt
r
o
or
t
t
h=13.3m v0b
=26.8t-0.5gt2
v
v
V0=26.8
yb(t)=yr(t)
13.3-5t-0.5gt2=26.8t-0.5gt2
13.3=31.8t so t=0.42s
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A car starts at rest and travels for 5.09 s with uniform
acceleration of +1.48 m/s2. The driver then brakes, causing
uniform acceleration of -1.91 m/s2. If the brakes are applied
for 2.96 s how fast is the car going after that?
v
5.09
5.05+
2.96
t
In first period: v(t)=v0+at=0+1.48·5.09=7.53 m/s
In 2nd period: v(t)=v0+at=7.53-1.91·2.96=1.88 m/s
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A 2000 kg sailboat is pushed by the tide of the sea with a
force of 3000 N to the East. Because of the wind in its sail
it feels a force of 6000 N toward to North-West direction.
What is the magnitude and direction of the acceleration?
Horizontal
Due to tide: 3000 N
Due to wind: 6000cos(135)=-4243
Sum:
-1243 N
N
6000N
W
3000N
S
Vertical
0N
6000sin(135)=+4243
4243 N
Magnitude of resulting force:
Fsum=[(-1243)2+(4243)2]=4421 N
Direction: angle=tan-1(4243/-1243)=
1060 (calc: -730, add 1800)
E
F=ma so a=F/m=4421/2000=2.21 m/s2
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T
900
T
1kg
A mass of 1 kg is hanging from a rope as shown in the figure.
If the angle between the 2 supporting wires is 90 degrees,
what is the tension in each rope?
TVerL
horizontal
Vertical
TVerR
0
45
0
left rope Tsin(45)
Tcos(45)
45
right rope -Tsin(45)
Tcos(45)
gravity
0
-1*9.81
Sum:
0
2Tcos(45)-9.81
ThorR
ThorL Object
is stationary, so:
2Tcos(45)-9.81=0 so, T=6.9 N
Fg
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PHY 231