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PHYSICS 231
Lecture 9: Revision
Remco Zegers
Walk-in hour: Thursday 11:30-13:30
Helproom
PHY 231
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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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2
2
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
PHY 231
t
t
a(t)=a0
t
5
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
PHY 231
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
PHY 231
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?
V0=12.7 m/s
h?
d=34m
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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
h=13.3m
V0=26.8
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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?
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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?
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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?
ThorR
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