AnnouncementsLAST chance to do Math Skills test is Friday (1/29) at 4:30 pm
in HL G59.No class after 5pm Thursday April 1.
Instead of Lab that week, Statistics Quiz.Can be taken during any scheduled lab time except after 5pm
Thursday.
Reminder do Pre-Lab for lab 1 Before Lab.If late to Lab (by more than 10 min) you are considered
absent.Mastering Physics assignment due Monday (1/25), next due
Saturday (1/30).Bring your “Student Workbook” and Homework notebook to
your discussion section.Tuesdays at 4:30 pm in G59, Review the current materials
using MCAT Questions.
Inclined Plane
g
g Sin()
We effectively have a way adjust the value of g.
€
v f2 = v0
2 + 2Δxa = (2.3 ms )2 +
3m
Cos(35o)2 × 9.8 m
s2 Sin(35o) = 46.46
v f = 6.8 ms
Rolling down the a hill
f
l=3m
Ignore change in direction
a=g Sin
x=3m/Cos()
a=g Sin a=0 a=0
f
Rolling up hillf
0
Ignore change in direction
a=g Sin
lxCos()=0.38m
a=g Sin a=0
0
€
v f2 = 0 = v0
2 + 2Δxa = (2.3 ms )2 − 2Δx9.8 m
s2 Sin(35o)
(2.3 ms )2 = 2Δx9.8 m
s2 Sin(35o)
Δx =(2.3 m
s )2
2 × 9.8 ms2 Sin(35o)
= 0.47mf
hxSin()=0.27m
Recall
9m6m
A B
€
yA 0 = 9m
vA 0 = ? ms
g ≡ 9.8 ms2
A
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yB 0 = 6m
vB 0 = 0 ms
g ≡ 9.8 ms2
B
€
yA ( ′ t B ) = 0 = 9m + vA 0 ′ t B − 12 9.8 m
s2 ′ t B2
12 9.8 m
s2 ′ t B2 − 9m = vA 0 ′ t B
vA 0 =12 9.8 m
s2 (1.11s)2 − 9m
1.11s= −2.67 m
s
€
yB ( ′ t B ) = 0 = 6m + 0 ′ t A − 12 9.8 m
s2 ′ t B2
′ t B2 =
2 × 6m
9.8 ms2
=1.22s2
′ t B =1.11s What can we through ball A down such that it will hit the ground at the same time at ball B?
So if I through ball A down (-) with an initial speed of 2.67m/s, ball A will hit the ground at the same time as ball B.
Graphically
Velocities are not the same because vA0≠0.Slope are the same because a=g are the same.
What aboutCan we through ball B such that it will hit the ground at the same time at ball A?
9m6m
A B
€
yA 0 = 9m
vA 0 = 0 ms
g ≡ 9.8 ms2
A
€
yB 0 = 6m
vB 0 = ? ms
g ≡ 9.8 ms2
B
€
yA ( ′ t A ) = 0 = 9m + 0 ′ t A − 12 9.8 m
s2 ′ t A2
′ t A2 =
2 × 9m
9.8 ms2
=1.84s2
′ t A =1.35s
€
yB ( ′ t A ) = 0 = 6m + vB 0 ′ t A − 12 9.8 m
s2 ′ t A2
12 9.8 m
s2 ′ t A2 − 6m = vB 0 ′ t A
vB 0 =12 9.8 m
s2 (1.35s)2 − 6m
1.35s= 2.17 m
s
So if I through ball B up (+) with an initial speed of 2.17m/s, ball B will hit the ground at the same time as ball A.
Graphically
Velocities are not the same because vA0≠0.Slope are the same because a=g are the same.
By throwing the ball upwards we can delay its impact time with the ground.
General Relations
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a(t) = a ⇒ a(t) = a
ν (t) = a( ′ t )d ′ t t0
t
∫ ⇒ ν (t) = ν 0 + at
x(t) = ν ( ′ t )d ′ t t0
t
∫ ⇒ x(t) = x0 + ν 0t + 12 at 2
€
x(t) = x0 + ν 0t + 12 at 2
ν (t) =dx(t)
dt= ν 0 + at
a(t) =dν (t)
dt=
d2x(t)
dt 2= a
€
rx (t) =
r x 0 +
r v 0t + 1
2
r a t 2
r v (t) =
v v 0 +
r a t
€
v 2(t) = vo2 + 2Δ
r x (t) •
r a
Near the earth’s surface Gravity is just a constant acceleration a=g=9.8m/s2 Down (it is a vector).
Scalars, Vectors, Tensors• Scalar (A measurable quantity describable by a
singular number)– Temperature, length, speed, brightness, time,
pressure, volume, area, energy,….
• Vector (A measurable quantity describable by a one or more numbers including a direction of some sort).– Position, velocity, acceleration, force, momentum,
color, a row or column of numbers, electric field, magnetic field, some sets of functions, ….
• Tensor (More complicated mathematical structure).– Matrix, spinners, ….
VectorsVectors are things which obey the algebra of
vectors. (a lot of help)A Vector must have a magnitude and a
direction.
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rC =
r A +
r B =
r B +
r A
r C =
r A −
r B = −
r B +
r A
Vectors Commute
Vector Addition Vector Scalar (dot) product
€
C2 =r C •
r C = (
r A +
r B ) • (
r A +
r B )
=r A •
r A +
r A •
r B +
r B •
r A +
r B •
r B
= A2 + 2r A •
r B + B2
Vectors Distributive
€
C =r C ≡ C2 =
r C •
r C
Magnitude of a Vector
€
rA
€
rB
€
rC
€
rA
€
rB
€
rC
Vector times a scalar
€
rA = a
r B
€
rB
€
rA = 2
r B
€
rB + a Not Defined
€
rA = −
r B
€
rv = (34 m
s ,25o east of north)
Some Vector examples
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rr = (22m along x axis,−12m along y axis)
€
rp ixel = (17row,9column)
More on Vectors scalar product
€
rA •
r B =
r A
r B Cos(θ)
€
rA
€
rB
€
rA •
r B =
r A
r B Cos(θ) = 0
€
rB
€
rA
Vector Scalar (dot) product
€
C2 =r C •
r C = (
r A +
r B ) • (
r A +
r B )
=r A •
r A +
r A •
r B +
r B •
r A +
r B •
r B
= A2 + 2r A •
r B + B2
= A2 + 2r A
r B Cos(θ) + B2
Vectors Distributive
€
If r B =1 then
€
rA •
r B is the length of A along B
€
rA
€
rB
€
rA Cos(θ)
Unit Vectors
A vector whose magnitude =1.
€
ˆ i , ˆ j , ˆ k or ˆ x , ˆ y , ˆ z Orthogonal unit vectors
€
) i •
) i =1
) i •
) j = 0
) i •
) k = 0
) j •
) i = 0
) j •
) j =1
) j •
) k = 0
) k •
) i = 0
) k •
) j = 0
) k •
) k =1
€
) i
€
) j
€
rA = (
r A •
) i )
) i + (
r A •
) j )
) j + (
r A •
) k )
) k
r A = Ax
) i + Ay
) j + Az
) k
Completing the AlgebraVector, Cross,Product
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rC =
r A ×
r B ≠
r B ×
r A
Does not commute
Will better define and use later