1. Physics 121, Sections 9, 10, 11, and 12 Lecture 6
Today’s Topics:
Homework 2: Due Friday Sept. 16 @ 6:00PM
Ch.3: # 2, 11, 18, 20, 25, 32, 36, 46, 50, and 56.
Chapter 4: Motion in 2-D
Review of vectors
Projectile motion
Relative velocity
More examples of FBD’s 1

2. Katzenstein Distinguished Lecture
Prof. Franck Wilczek
Nobel Prize in Physics 2004
From MIT
Title: The Universe is a Strange Place
Where: Room P-36
When: Friday at 4:00PM
Refreshments at 3:00 PM in front of P-36
Come for a great talk 1

3. Unit Vectors: A Unit Vector is a vector having length 1 and no units.
It is used to specify a direction.
Unit vector u points in the direction of U.
Often denoted with a “hat”: u = û U û x y z i j k
Useful examples are the cartesian unit vectors [ i, j, k ]
point in the direction of the x, y and z axes.
R = rxi + ryj + rzk

4. Vector addition using components:
Consider C = A + B.
(a) C = (Ax i + Ay j ) + (Bx i + By j ) = (Ax + Bx )i + (Ay + By )j
(b) C = (Cx i + Cy j )
Comparing components of (a) and (b):
Cx = Ax + Bx
Cy = Ay + By By C B Bx A Ay Ax

5. Lecture 6, ACT 1 Vector Addition
Vector B = {3,0}
Vector C = {1,-4}
What is the resultant vector, D, from adding A+B+C?
(a) {3,-4} (b) {4,-2} (c) {5,-2}

6. Review (1-D): vav A few other useful formulas :
For constant acceleration we found: x a v t
vav
A few other useful formulas :

7. 2-D Kinematics
For 2-D, we simply apply the 1-D equations to each of the component equations.
Which can be combined into the vector equations:

8. 2-D Kinematics So for constant acceleration we get: a = const
v = v0 + a t
r = r0 + v0 t + 1/2 a t2
(where a, v, v0, r, r0, are all vectors)

9. 3-D Kinematics
Most 3-D problems can be reduced to 2-D problems when acceleration is constant;
Choose y axis to be along direction of acceleration.
Choose x axis to be along the “other” direction of motion.
Example: Throwing a baseball (neglecting air resistance).
Acceleration is constant (gravity).
Choose y axis up: ay = -g.
Choose x axis along the ground in the direction of the throw.

10. “x” and “y” components of motion are independent.
A man on a train tosses a ball straight up in the air.
View this from two reference frames:
Reference frame
on the moving train.
y motion: a = -g y
x motion: x = v0t
Reference frame
on the ground.

11. Projectile Motion.
If I set something moving near the earth, it reduces to a 2-D problem we call projectile motion.
Use a coordinate system with x along the ground, y vertical with respect to the ground. (Notice no change in third direction.)
Equations of motion reduce to:
X: Dx = voxt ax = 0
Y: y = yo + voyt – g t2 /2 y positive upwards

12. Lecture 6, ACT 2 2-D Motion Alice Bill C) B) A)
Alice and Bill are playing air hockey on a table with no bumpers at the ends. Alice scores a goal and the puck goes flying off the end of the table. Which diagram best describes the path of the puck ?
Alice
Bill C) B) A)

13. Problem:
Sammy Sosa clobbers a fastball toward center-field. You are checking out your new fancy radar gun which can detect ball velocity, i.e. speed and direction. You measure that the ball comes off the bat with initial velocity is 36.5 m/s at an angle of 30o above horizontal. Since Sammy was hitting a high fastball, you estimate that he contacted the ball about one meter off of the ground. You know the dimensions of Wrigley field and the center-field wall is 371 feet (113m) from the plate and is 10 feet (3m) high. You decide to demonstrate your superfast math and physics skills by predicting whether Sammy get a home run before the play is decided.

14. Problem:
We need to find how high the ball is at a distance of 113m away from where it starts.  v h D yo

15. Problem: This is a problem in projectile motion. Choose y axis up.
Choose x axis along the ground in the direction of the hit.
Choose the origin (0,0) to be at the plate.
Say that the ball is hit at t = 0, x = xo = 0, y = yo = 1m  v h D y x

16. Problem... Variables vo = 36.5 m/s yo = 1 m h = 3 m qo = 30º D = 113 m
a = (0,ay)  ay = -g
t = unknown,
Yf – height of ball when x=113m, unknown,
our target

17. Problem... For projectile motion, Equations of motion are:
vx = v0x vy = v0y - g t
x = vx t y = y0 + v0y t - 1/ 2 g t2 y x g  v
v0x
v0y y0
And, use geometry to find vox and voy
Find v0x = |v| cos .
and v0y = |v| sin .

18. Problem... Solve the problem,
The time to reach the wall is: t = D / vx (easy!)
Height at any time: y(t) = y0 + v0y t - g t2/ 2
Combining the two gives, y(t) = y0 + v0y D/vox - g D2/ (2vox2)
And substitute for vox and voy: y(t) = y0 + D tanq - g D2/ 2(vocosq)2
All are known quantities. Solved.
Numbers:
y(t) = (1.0 m) + (113 m)(tan 30) (0.5)(9.8 m/s2)(113 m)2/(36.5 m/s cos 30)2
= ( ) m = 3.6 m

19. Problem...
Think about the answer,
The units work out correctly for a height (m)
It seems reasonable for the ball to be a little over 3m high when it gets to the fence.
But, we haven’t yet answered the question
Since the wall is 3m high, and the ball is 3.26m high when it gets there, Sammy gets a homer.

20. Lecture 6, ACT 3 Motion in 2-D
Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it ? (a) D2 = 2D1 (b) D2 = 4D (c) D2 = 8D1

21. Shooting the Monkey (tranquilizer gun)
Where does the zookeeper aim if he wants to hit the monkey?
( He knows the monkey will let go as soon as he shoots ! )

22. Shooting the Monkey... at the monkey r =v0t r = r0
If there were no gravity, simply aim
at the monkey
r =v0t

23. Shooting the Monkey... monkey! r = r0 - 1/2 g t2
With gravity, still aim at the monkey!
Dart hits the
monkey!
r = v0 t - 1/2 g t2

24. Recap: Shooting the monkey...
x = v0 t
y = -1/2 g t2
This may be easier to think about ,
It’s exactly the same idea!!
x = x0
y = -1/2 g t2

25. Typical questions : (projectile motion; for given v0 and q)
What is the maximum height the ball reaches (h) ?  h L y x v0 P
y :
h = (v0 sin q) t - 1/2 g t2
v = (v0 sin q) - g t
= 0 at P
t = (v0 sin q) / g
t = (v0 sin q) / g !
How long does it take to reach maximum height ?
Would the answers above be any different if the projectile was moving only along y-axis (1-D motion) with the initial velocity: v0 sin (q) ?
( A ) YES ( B ) NO ( C ) CAN’T TELL h y x
v0 sin(q) P

26. Typical questions : (projectile motion; for given v0 and q) h L y x v0 P
What is the range of the ball (L) ?
x :
L = vx0 t = (v0 cos q) t
How long does it take for ball to reach final point (P) ?
y :
y = (v0 sin q) t - 1/2 g t2
= 0 ! when at P
[ (v0 sin q) - 1/2 g t] t = 0
t = 0 ; t = 2 (v0 sin q) / g

27. ( A ) MISS ( B ) HIT ( C ) CAN’T TELL
Problem 2
Suppose a projectile is aimed at a target at rest placed at the same height. At the time that the projectile leaves the cannon the target is released from rest and starts falling toward ground. Would the projectile miss or hit the target ?
( A ) MISS ( B ) HIT ( C ) CAN’T TELL y x v0
t = 0
TARGET
PROJECTILE
t = t1

28. ( A ) MISS ( B ) HIT ( C ) CAN’T TELL
Problem 3
Suppose a projectile is aimed at a target at rest somewhere above the ground as shown in Fig. below. At the same time that the projectile leaves the cannon the target falls toward ground. Would the projectile miss or hit the target ?
( A ) MISS ( B ) HIT ( C ) CAN’T TELL  y x v0
t = 0
TARGET
PROJECTILE
t = t1

29. Inertial Reference Frames:
A Reference Frame is the place you measure from.
It’s where you nail down your (x,y,z) axes!
An Inertial Reference Frame (IRF) is one that is not accelerating.
We will consider only IRF’s in this course.
Valid IRF’s can have fixed velocities with respect to each other.
More about this later when we discuss forces.
For now, just remember that we can make measurements from different vantage points.

30. Lecture 6, ACT 4 Relative Motion
Consider an airplane flying on a windy day.
A pilot wants to fly from New Haven to Bradley airport. Having asked a friendly physics student, she knows that Bradley is 120 miles due north of New Haven and there is a wind blowing due east at 30 mph. She takes off from New Haven Airport at noon. Her plane has a compass and an air-speed indicator to help her navigate. She uses her compass at the start to aim her plane north, and her air speed indicator tells her she is traveling at 120 mph with respect to the air.
After one hour,
She is at Bradley
She is due east of Bradley
She is southeast of Bradley

31. Lecture 6, ACT 5 Relative Motion
You are swimming across a 50m wide river in which the current moves at 1 m/s with respect to the shore. Your swimming speed is 2 m/s with respect to the water. You swim across in such a way that your path is a straight perpendicular line across the river.
How many seconds does it take you to get across? a) b) c) d)
2m/s
1m/s
50m

32. Recap of today’s lecture
Homework 2: Due Friday Sept. 16 @ 6:00PM
Ch.3: # 2, 11, 18, 20, 25, 32, 36, 46, 50, and 56.
Chapter 4: Motion in 2-D
Review of vectors
Projectile motion
Relative velocity 27