AP Physics Unit 1 Notes 3. Two Dimensional Motion 3.1. · Chapter 3 Jason M. Roberts Kenwood...

AP Physics Unit 1 Notes Chapter 3

Jason M. Roberts Kenwood Academy 1

3. Two Dimensional Motion All two dimensional problems are simply two, one dimensional problems.

3.1. Vectors and Scalars Any given quantity is either a vector or a scalar. While scalars have just size (magnitude) vectors have both size and direction (magnitude and orientation). Because of this, we have to change how we look at the negative.

3.1.1. In the past, the negative would tell you to take away. If you have negative five dollars, you probably owed five dollars. But that is not the case with vectors.

It is all relative to the main object. Two cars are moving at

!

10 ms but in opposite directions. While their speeds

are the same, their vectors are not.

!

v1 =10 ms

v2 = "10 ms

In this case, the first car is the main object and has set the coordinate system.

3.1.2. Every vector is 3-D, but we primarily look at the 2-D vector. y y y

!

A

!

B

!

C x x x Vector A is only in the y-direction, vector B is only in the x-direction, but vector C is in both the y-direction and x-direction.

3.2. Vector Components

A given vector may have some amount in the x and some in the y-axis. When working with vectors, it is important to remember that the various components of each vector only interact with the same components of another vector. 3.2.1. Vector Addition

3.2.1.1. Use the Head Tail method by placing the tail of the second vector on the head of the first vector. (Maintain size and direction!)

Add vector A to vector B

!

A

!

B

!

R

!

A

!

B



!

R = A + B Vector Addition

!

R2 = A2 + B2 Scalar addition 3.2.1.2. Use trig to find the scalar quantities of each vector component.

When

!

" is measured from the x-axis, cosine will give you the x component and sine will give you the x component.

A vector of length 25 has what x and y components for the following angles

!

" = 20o," = 30o," = 37o? Knowns Unknowns Equations

!

"1 = 20o

"2 = 30o

"3 = 37o

!

V1x = ?V1y = ?V2x = ?V2y = ?V3x = ?V3y = ?

!

V cos" =Vx

V sin" =Vy

25cos20o = 23.5 =V1x25sin20o = 8.5 =V1y25cos30o = 21.65 =V2x25sin30o =12.5 =V2y25cos37o =19.97 =V3x25sin37o =15.04 =V3y

A cannon ball is fired with a velocity of

!

32 ms at an angle of

!

29o. What are the horizontal and vertical components of the cannon ball’s velocity?

3.3. Displacement, velocity, and acceleration in 2-D 3.3.1. While position and speed are scalar quantities, displacement and velocity

are vector components. There is no scalar form of acceleration. Because velocity is a vector, if an object changes direction but not speed, the objects velocity changes. This is due to an acceleration.

3.4. Motion in 2-D

Knowns Unknowns Equations

!

v = 32 ms

" = 29o

!

vx = ?vy = ?

!

v cos" = vxv sin" = vy32 m

s cos(29o) = 27.99 m

s # 28 ms

32 ms sin(29

o) =15.51ms #16 ms



3.4.1. A cannon fires a 10kg ball

!

30 ms at an angle of

!

27o with the horizontal. What can you find?


!

m =10kgv0 = 30 m

s

" = 27o

ax = 0 ms2

ay = #9.8 ms2

x0 = 0my0 = 0my = 0mvy,top = 0 m

s

!

v0x = ?v0y = ?ymax = ?x = ?t = ?

!

v0 cos" = v0xv0 sin" = v0yv 2 = vo

2 + 2a#yv = v0 + atx = v0xt

30 ms cos(27

o) = 26.7 ms

30 ms sin(27

o) =13.62 ms

(0 ms )2 = (13.62 m

s )2 + 2($9.8 m

s2)ymax $ 0m

0 m 2

s2=185.5 m 2

s2+ ($19.6 m

s2)ymax

$185.5 m 2

s2= $19.6 m

s2ymax

ymax = 9.5mvy2 = v0

2 + 2a#y

vy2 = (13.62 m

s )2 + 2($9.8 m 2

s )0m

vy2 =185.5 m 2

s2

vy = ±13.62 ms

vy = v0y + ($9.8 m 2

s )t

$13.62 ms =13.62 m

s = ($9.8 m 2

s )t

$27.24 ms = ($9.8 m 2

s )tt = 2.8sx = v0xtx = (26.7 m

s )2.8sx = 74.76m



3.4.2. An object falling off of a 110m building from rest.


!

y = 0my0 =110ma = "9.8 m

s2

v0 = 0 ms

!

v = ?t = ?

!

y =y 0+v0t + 12 at

2

y " y0 = v0t + 12 at

2

0m "110m = 0 ms t + 1

2 ("9.8 ms2 )t

2

"110m = "4.9 ms2 t

2

22.448s2 = t 2

t = 4.738sv = v0 + atv = 0 m

s + ("9.8 ms2 )(4.738s)

v = "46.45 ms

Experiment: Design an experiment that will measure the acceleration of gravity on objects. -What do you suspect will happen? Show Steps -How will you measure your data? Be clean, be neat, be clear -How many trials do you need? -How will you record your data?

3.4.3. An object is thrown up at

!

11ms from the ground and falls back down.




!

v0 =11msa = "9.8 m

s2

y0 = 0my = 0m

!

v = ?t = ?

!

y = y0 + v0t + 12 at

2

y " y0 = v0t + 12 at

2

0m =11ms (t) + 12 ("9.8 m

s2 )t2

0m =11ms (t) + ("4.9 ms2 )t

2

Use quadratic equation or…

!

vtop = 0 ms

upv = v0 + at0 m

s =11ms + ("9.8 ms2)t

"11ms = "9.8 ms2t

t =1.12sy = y0 + v0t + 1

2 at2

y = 0m + (11ms )(1.12s) + 12 ("9.8 m

s2)(1.12s)2

y =12.32m " 6.14my = 6.17mdownv0 = 0 m

s

y0 = 6.17my = 0ma = "9.8 m

s2

y = y0 + v0t + 12 at

2

"6.17,= "4.9 ms2t 2

t 2 =1.259s2

t =1.12s



3.4.4. A cannon fires a ball off of a 100m cliff at 20 meters per second with the horizontal. How far does it travel?


!

v0 = 20 ms

" = 0o

vx = v0 cos"vy = v0 sin"

vx = 20 ms

vy = 0 ms

y0 =100my = 0mx0 = 0mag = #9.8 m

s2

!

x = ?t = ?

!

x = x0 + vtx = 20 m

s t

we need t.

!

y = y0 + v0t + 12 at

2

0m =100m + 12 ("9.8 m

s2)t 2

"100m = "4.9 ms2t 2

20.408s2 = t 2

t = 4.517sx = 20 m

s (4.517s)x = 90.35m

3.4.5. An object launched at an angle of

!

30o off the horizontal at

!

47 mS .

v

!

" 3.4.5.1. Break up the velocity into vertical and horizontal components

!

vx = 47 ms cos30

o

vy = 47 ms sin30

o

vx = 40.7 ms

vy = 23.5 ms

For x direction For y direction

!

x0 = 0mvx = 40.7 m

s

ax = 0 ms2

t = ?

!

y0 = 0mv0y = 23.5 m

s

ag = "9.8 ms2

And we have seen this situation before. (Sect 3.4.3)

!

t = 4.8symax = 28.2mx =195.36m



3.4.6. A cannon atop a 100m cliff fires a cannon ball

!

30o off the horizontal at

!

47 mS . How far does it travel? How high does it get? How long is it in the air?

Much of the work has been done for us already.

!

vx = 40.7 ms

v0y = 23.5 ms

t = 4.8s

x =195.36my =100mytop =128.2m

vy = "23.5 ms

#

$

% %

&

% %

Do some trig and…

!

Vy

!

V

!

" '

!

Vx

!

v = "47 ms

tan# ' =vyvx

# ' = 30o

!

Vx does not change, so this looks just like we threw a ball downward at

!

23.5 ms .

!

yo =100my = 0mvo = "23.5 m

s

a = "9.8 ms2

y = yo + vot + 12 at

2

0m =100m + ("23.5 ms )t + 1

2 ("9.8 ms2)t 2

0m =100m + ("23.5 ms )t " 4.9 m

s2t 2

Use the quadratic equation



!

23.5 ± ("23.5)2 " 4("4.9)(100)2("4.9)

23.5 ± 552.25 +1960"9.8

23.5 ± 2512.25"9.8

23.5 ± 50.12"9.8

73.62"9.8

or"26.62"9.8

"7.512or2.716t = 2.716

Because time can not be negative Now we move to the x-direction

!

x =xo+vtxo=195.36mv = 40.7 m

s

t = 2.716sx =195.36m + 40.7 m

s (2.716s)x =195.36m +110.54mx = 305.9m

For a recap

!

ytop =128.2mx = 305.9mt = 7.4s

AP Physics Unit 1 Notes 3. Two Dimensional Motion 3.1. · Chapter 3 Jason M. Roberts Kenwood...

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Transcript of AP Physics Unit 1 Notes 3. Two Dimensional Motion 3.1. · Chapter 3 Jason M. Roberts Kenwood...