Transcript Part I - Otterbein

How many significant figures are
there in 0.0143 10-4 cm ?
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9
5
4
3
What does the prefix “M” in front
of a unit mean?
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Multiply by 10-6
Multiply by 10-3
Multiply by 103
Multiply by 106
How many professors are there at
Otterbein?
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10
100
1000
10000
Somebody claims the correct equation
to calculate the position of an object
from its (constant) velocity v and the
2
elapsed time t is x(t) = x(t=0) + v t .
Can this be correct?
• No
• Yes
• Not enough information
A red and an orange car make a
race. When do they have the same
velocity?
x
t
A
B
At point A
Between point A and B
At point B
Never
A red and an orange car make a
race. Who wins?
x
t
Red car
Yellow car
Insufficient information
Motion at constant acceleration
• If (and only if) acceleration is constant, we can
calculate x,v,a in any instant by using
a(t) = a(t=0)
v(t) = v(t=0) + a(t=0) t
x(t) = x(t=0) + v(t=0) t + ½ a(t=0) t2
Green represent independent variable, darker hues initial values.
Additionally, we may use
v2(t) = v2(t=0) + 2 a(t=0) [x(t) - x(t=0)]
In the equation below, the following is NOT true…
v2(t) = v2(t=0) + 2 a(t=0) [x(t) -x(t=0)]
• We need to know the position of the object in the
moment at which we want to calculate its velocity
• We need to know the initial velocity of the object to
calculate its velocity at time t
• This equation is only true if the acceleration of the
object is constant.
• The equation is only true for positive velocities,
since we have to take the square root eventually.
v(t)
Example for
derivation
v(t2)
v(t1)
∆v
∆t
t1
t2
• Constant acceleration means linear rising
(or falling) velocity
 velocity at later time is velocity at earlier
time plus slope times time elapsed.
• Slope of velocity curve is acceleration
 v(t2) = v(t1) + a (t2-t1)
t
A car accelerates at constant (nonzero) rate. Which of the following
motion diagrams is NOT correct?
v
a
t
v
t
t
x
t
An dropped object accelerates
downward at 9.8 m/s2. If instead you
throw it downward, its downward
acceleration after release is …
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less than 9.8 m/s2
exactly 9.8 m/s2
more than 9.8 m/s2
Insufficient information
Solving Kinematic Problems
1.
2.
3.
4.
5.
6.
7.
8.
Read problem carefully
Draw diagram
List knowns and unknowns
What physics principles do apply?
Find equations that do apply
Solve algebraically (with variables, not values!)
Calculate numerically
Check results: Numbers reasonable? Units
correct?
Vectors
• “Directions with magnitudes” that can be
shifted around
Addition: Tail-to-Tip or
Parallelogram
Subtraction by adding negative
vector
a-b = a + (-b)
3D Vectors
• 3D vector as a sum
of multiples of the
three unit vectors
i,j,k
• Example: a=1.5 i +
2.5j +3 k
To which quadrant does the
following vector belong?
A = - 4 i + 6.5 j
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1st Quadrant
2nd Quadrant
3rd Quadrant
4th Quadrant
To which quadrant does the
following vector belong?
|B| = 4.5, φB = -45º
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1st Quadrant
2nd Quadrant
3rd Quadrant
4th Quadrant
To which quadrant does the
following vector belong?
|c| = 4.5m/s, φc = 3.10
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1st Quadrant
2nd Quadrant
3rd Quadrant
None of the above
To which quadrant does the
following vector belong?
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1st Quadrant
2nd Quadrant
3rd Quadrant
None of the above
   0.9m 

D  
  6.7m 
To which quadrant does the
following vector belong?
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1st Quadrant
2nd Quadrant
3rd Quadrant
Not enough
information
m


2
m

4
t



s

x 
 3.4m  4.9 m t 2 


2
s


What is the length of the following
vector?
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-1 m/s
1 m/s
5 m/s
25 m2/s2
  3m / s 

v  
  4m / s 
What is the
length of this
vector?
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Upper formula
Middle formula
Bottom formula
None of the above
m


2
m

4
t



s

x 
 3.4m  4.9 m t 2 


2
s



m 
 m
| x | 5.4m   4  4.9 2 t t
s 
 s
2
 
m  
m 
| x |  2m  4 t    3.4m  4.9 2 t 2 
s  
s



m  
m 

| x |   4 t    5.4m  4.9 2 t 2 
s  
s


2
Which is a correct statement
concerning this vector?
|B| = 4.5, φB = -45º
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Its magnitude is negative
Its x component is negative
Its y component is positive
Its x and y component have the same absolute
value
Projectile Motion
By aiming a gun higher (increasing
the angle of the barrel w.r.t. the xaxis), I achieve the following:
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Higher initial position
Larger initial y-component of velocity
Higher initial velocity
None of the above
A gun is accurately aimed at a
gutter
dangerous terrorist hanging from
the gutter of a building. The target
is well within the gun’s range, but
the instant the gun is fired and the v0
bullet moves with a speed v0, the
terrorist lets go and drops to the gun
ground. What happens?
• The bullet hits the terrorist regardless of the value of v0
• The bullet hits the terrorist only if v0 is large enough
• The bullet misses the terrorist
• Not enough information
A battleship simultaneously fires two shells at
enemy ships. If the shells follow the parabolic
paths shown, which ship gets hit first?
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A
B
Both at the same time
Need more information
A
B
Circular Motion
An object at the end of an r=1m
string in circular motion completes
10 revolutions in one second. How
long does each revolution take?
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(2 π r)(1s)
1s/(2π r)
0.1 s
Need more information
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An object at the end of an r=1m
string in circular motion completes
10 revolutions in one second. What
is its speed?
6.3 m/s
3.1 m/s
63 m/s
0.31 m/s
An object at the end of an r=1m
string in circular motion completes
10 revolutions in one second. What
is its frequency?
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10 Hz
1 Hz
1/10 Hz
3.14 Hz
Newton’s first Law
• In the absence of a net external force, a
body either is at rest or moves with constant
velocity.
– Motion at constant velocity (may be zero) is
thus the natural state of objects, not being at
rest. Change of velocity needs to be explained;
why a body is moving steadily does not.
Mass & Weight
• Mass is the property of an object
• Weight is a force, e.g. the force an object of
certain mass may exert on a scale
Mass: On the surface of the Moon a
standard block of lead …
• … has a different mass
• … has a different weight
• Its weight is unchanged, but g has a
different value
• Need more information
Newton’s first law states that objects
remain at rest only when they no net
force acts on them. A book on a table
is subject to the force of gravity pulling
it down. Why doesn’t it move?
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Newton’s first law does not apply (obstacle!)
There must be another force opposing gravity
Table shelters book from force of gravity
Not enough information
Newton’s second Law
• The net external force on a body is equal to the
mass of that body times its acceleration
F = ma.
• Or: the mass of that body times its acceleration is
equal to the net force exerted on it
ma = F
• Or:
a=F/m
• Or:
m=F/a
Newton II: calculate Force from
motion
• The typical situation is the one where a
pattern of Nature, say the motion of a
projectile or a planet is observed:
– x(t), or v(t), or a(t) of object are known, likely
only x(t)
• From this we deduce the force that has to
act on the object to reproduce the motion
observed
Calculate Force from motion: example
• We observe a ball of mass m=1/4kg falls to the
ground, and the position changes proportional to
time squared.
• Careful measurement yields:
xball(t)=[4.9m/s2] t2
• We conclude v=dx/dt=2[4.9m/s2]t
a=dv/dt=2[4.9m/s2]=9.8m/s2
• Hence the force exerted on the ball must be
• F = 9.8/4 kg m/s2 = 2.45 N
– Note that the force does not change, since the
acceleration does not change: a constant force acts on
the ball and accelerates it steadily.
Newton II: calculate motion from
force
• If we know which force is acting on an object of
known mass we can calculate (predict) its motion
• Qualitatively:
– objects subject to a constant force will speed up (slow
down) in that direction
– Objects subject to a force perpendicular to their motion
(velocity!) will not speed up, but change the direction
of their motion [circular motion]
• Quantitatively: do the (vector) algebra!
Newton II: calculate motion from
force
• Say a downward force of 4.9N acts on a block of
1 kg (we call the coordinate y and start at y=0)
• We conclude that the block will be accelerated:
ay = F/m= -4.9N / 1kg = -4.9 m/s2
• We use this constant acceleration to calculate
y = -1.225m/s2 t2 + b t + c , where b &c are
constants, c=0
Q: What physical situation is this?
Newton II: calculate mass from force
and acceleration
• Pretty simple, if you know force and acceleration
we have
m=F/a