9. Cons of p (2-D)

7/29/2019 9. Cons of p (2-D)

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UNIT 1: CONSERVATION LAWS

Lesson 9:

Conservation of Momentum (2-D)

CENTRE HIGH: PHYSICS 30

7/29/2019 9. Cons of p (2-D)


2-D Momentum

Recommended Reading:

Heath pp. 250 - 260

Ladner pp. 49 - 51

7/29/2019 9. Cons of p (2-D)


F2. Total Momentum (2-D)

Before we take on 2-D momentum,we need to review how to add 2-D vectors

7/29/2019 9. Cons of p (2-D)


F2. Total Momentum (2-D)

- there are two methods to add 2-D vectors:1. Tail-to-tip

- use when vectors are perpendicular (90)

2. Components

- general method

- works for all cases

7/29/2019 9. Cons of p (2-D)


Method 1: Tail-to-Tip (90)

How? Place one vector right after the other

i.e. Place the tail of the second vector on the tip

of the first vector

The resultant is the "start to finish" vector

Use Pythagorean formula and Soh Cah Toa to solve

the triangle

7/29/2019 9. Cons of p (2-D)


Animation:

2-D Addition:http://www.phy.ntnu.edu.tw/java/vector/vector.html

Note: In this animation,

- you create the first two vectors

- then, it will show you how to add them tail-to-tip

http://www.phy.ntnu.edu.tw/java/vector/vector.html

http://www.phy.ntnu.edu.tw/java/vector/vector.html

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


Step 1: Place one right after the other ("tail-to-tip")

The resultant is the "start to finish" vector

F

p2 R

40 kgm/s

p1 S

30 kgm/s

Place the angle at the base of the unknown vector.

7/29/2019 9. Cons of p (2-D)


Step 2: Solve the triangle

c

b p2 R 40 kgm/s

Pythag:

p1

c2 = a2 + b2 30 kgm/s

R2 = (30)2 + (40)2 a

R = 50 kgm/s

7/29/2019 9. Cons of p (2-D)


Step 2: Solve the triangle

hyp

opp p2 R 40 kgm/s

Soh Cah Toa:

p1

tan = opp 30 kgm/sadj adj

tan = 40 = tan -1 (1.3333) = 53.1

30

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


N

R

53.1°

In RCS format: W E

= 180 - 53.1 S

= 127

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


Vector Diagram: N

R p2 15.00 kgm/s

W p1 = 9.00 kgm/s E

S

Note:

- ref system always goes at base of vector

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


N

hyp R p2 opp

W p1 E

Soh Cah Toa: adj

tan Angle = 15 S

9

= tan -1 (1.6667) = 59.0 degrees N of E

N from the E axis

7/29/2019 9. Cons of p (2-D)


Method 2: Components (General)

Find the components of each vector V

- find the RCS angle (standard position)

Vx = V cos

Vy = V sin

Add up the x, add up the yRx = x1 + x2 + ...

Ry = y1 + y2 + ...

Add Rx and Ry tail-to-tip

Use Pythag and Soh Cah Toa to solve the triangle

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


Step 1: Find components y

rcs = 270 x

x1 = 20 cos 270 p1

= 0 20 kgm/s

y1 = 20 sin 270

= -20 kgm/s

7/29/2019 9. Cons of p (2-D)


y

p2 = 25 kgm/srcs = 150 150

y2

x2 = 25 cos 150 30

= -21.65 kgm/s x2 x

y2 = 25 sin 150

= 12.5 kgm/s

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


Rx = 21.65 kgm/s

Ry

7.5 kgm/s

R

Soh Cah Toa:

tan = 7.5

21.65

= tan -1 (0.346) = 19 S of W (199 rcs)

7/29/2019 9. Cons of p (2-D)


Animation

Adding by components:http://www.upscale.utoronto.ca/GeneralInterest/Harrison/F

lash/Vectors/VectorAddComponents.html

This animation will show why adding by components

actually works.

http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/Vectors/VectorAddComponents.html




7/29/2019 9. Cons of p (2-D)


Ex. 2 After an explosion, two objects are moving apart:

1 160 kgm/s2

30.0

140 kgm/s

Find the total momentum (magnitude and direction).

7/29/2019 9. Cons of p (2-D)


Step 1: Find x- and y- components

y

p1 = 160 kgm/s

x

RCS Angle: 0 deg

x1 = 160 cos 0 = 160 kgm/s

y1 = 160 sin 0 = 0

7/29/2019 9. Cons of p (2-D)


y

210

x2 30 x

y2

p2 = 140 kgm/s

RCS Angle: 210

x1 = 140 cos 210 = -121.24 kgm/s

y1 = 140 sin 210 = -70 kgm/s

7/29/2019 9. Cons of p (2-D)


Step 2: Add up x, Add up y

Rx = x1 + x2 = 160 - 121.24 = 38.76 kgm/s

Ry = y1 + y2

= 0 - 70 = -70 kgm/s

7/29/2019 9. Cons of p (2-D)


Step 3: Add Rx and Ry (tail-to-tip)

Rx = 38.76 kgm/s

Pythag: R2 = Rx

2 + Ry2 Ry

R = 80.0 kgm/s R 70 kgm/s

Soh Cah Toa:

tan = 70

38.76

= tan

-1

(1.806) = 61.0

S of E

(-61.0 rcs)

7/29/2019 9. Cons of p (2-D)


Practice Problems

http://www.physicsclassroom.com/morehelp/vectaddn/practice.html

This website uses scale diagrams to solve the problems.

See if you can get the same answers using tail-to-tip

or the component method.





7/29/2019 9. Cons of p (2-D)


F3. Conservation of Momentum (2-D)

We are now ready to try conservation of momentumin two dimensions.

This is a challenging process to learn, so pay

close attention as you go through these slides.

7/29/2019 9. Cons of p (2-D)


F3. Conservation of Momentum (2-D)

- there are 2 methods to solve 2-D momentum questions:

1. Tail-to-Tip

- especially good for vectors at right angles (90)

2. Components

- works in all cases

7/29/2019 9. Cons of p (2-D)


Method 1: Cons of p (Tail-to-Tip)

State the law of conservation of momentum

pT = pT'

Sketch the vector addition diagram suggested by

this equation

Solve the triangle

7/29/2019 9. Cons of p (2-D)


To illustrate, consider two objects that collide and stick together.

p'

42 kgm/s

STICK!

80 kgm/s

How would we find the momentum of the combined masses

after the collision (i.e. p' and )

7/29/2019 9. Cons of p (2-D)


Step 1: Use the conservation of momentum equation

to discover how to draw the tail-to-tip diagram:

pT = pT '

p1 + p2 = p'

Add the two momentum The Resultant (R)

vectors before the collision is the momentum vector

(tail-to-tip) after the collision

7/29/2019 9. Cons of p (2-D)


p1 + p2 = p'

Add tail-to-tip

p2

80 kgm/s

p1 = 42 kgm/s

7/29/2019 9. Cons of p (2-D)


p1 + p2 = p'

Resultant (R)("start to finish" vector)

F

p2

80 kgm/s p'

S

p1 = 42 kgm/s

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


Ex. 3 An object is travelling East with a momentum of

20 kgm/s. It then explodes into two pieces (on a two

dimensional plane). After the explosion, one piece hasa momentum of 15 kgm/s due North.

What is the momentum of the second piece

(magnitude and direction)?

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


Based on the law of conservation of momentum,

pT = pT'

p = p1' + p2'

Resultant Add the two vectors

after the explosion

That is, when you add p1' and p2', you must get

the horizontal resultant p

7/29/2019 9. Cons of p (2-D)


p = p1' + p2'

Add tail-to-tip

p1' p2'

15 kgm/s

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


p1' p2'

15 kgm/s

p = 20 kgm/s

Pythag: (resultant)

c2 = a2 + b2

(p2')2 = 152 + 202

p2' = 25 kgm/s

7/29/2019 9. Cons of p (2-D)


p1' p2'

15 kgm/s

p = 20 kgm/s

Soh Cah Toa: (resultant)

tan = 20

15

= tan -1 (1.333) = 53 E of S (-37, 323)

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


p2' 7.0 kgm/s

p1 = 10.0 kgm/s 4.0 kgm/sBoom!

Rest x1'

xT = xT' p1' y1'

x1 + x2 = x1' + x2'

10.0 + 0 = x1' + 4.0

x1' = 6.0 kgm/s

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


Ex. 4 Find: a) x2' and y2'

b) pT before the collision (mag and dir)

6.0 kgm/s 12.0 kgm/s

10.0 kgm/s 4.0 kgm/s

Boom!

x2'

17.0 kgm/s

y2'

9.0 kgm/s

7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


7/29/2019 9. Cons of p (2-D)


b) 6.0 kgm/s

Step 1: Find components

- done already 10.0 kgm/s

17.0 kgm/s

Step 2: Add up x, Add up y 9.0 kgm/s

Rx = x1 + x2 Ry = y1 + y2

= 6.0 + 9.0 = -10.0 + 17.0

= 15.0 kgm/s = 7.0 kgm/s

7/29/2019 9. Cons of p (2-D)


b) Step 3: Add Rx and Ry (tail-to-tip)

Pythag: Ry

R2 = 152 + 72 R 7.0 kgm/s

R = 17 kgm/s

Soh Cah Toa: Rx

15.0 kgm/s

tan = 7.0

15.0

= tan -1 (0.467) = 25 N of W

7/29/2019 9. Cons of p (2-D)


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9. Cons of p (2-D)