1. Status: Unit 2, Chapter 3 Vectors and Scalars
Addition of Vectors – Graphical Methods
Subtraction of Vectors, and Multiplication by a Scalar
Adding Vectors by Components
Unit Vectors
Vector Kinematics
Projectile Motion
Solving Problems in Projectile Motion
Relative Velocity
11/15/2018
Physics 253

2. Section Two Problem Assignment
Q3.4, P3.6, P3.9, P3.11, P3.14, P3.73
Q3.21, P3.24, P3.32, P3.43, P3.65, P3.88
11/15/2018
Physics 253

3. Kinematic Equations for Projectile Motion (+y up, ax =0, ay=-g= -9
Kinematic Equations for Projectile Motion (+y up, ax =0, ay=-g= -9.8m/s2)
11/15/2018
Physics 253

4. 11/15/2018
Physics 253

5. 11/15/2018
Physics 253

6. What angle gives the most range?
This can be cast as a nifty calculus question! And the results may surprise you.
What we want is an expression for x in terms of all the other variables.
Then we take the derivative with respect to the angle to find the maxima.
So let’s get to it, we really just follow the illustrative problem from last lesson but keep results general.
11/15/2018
Physics 253

7. 11/15/2018
Physics 253

8. From the calculus you will recall that finding a maximum or minimum of a function is done by setting the derivative with respect to the independent variable to zero.
This corresponds to finding those points where the slope of the function is zero – the very definition of an extreme point.
11/15/2018
Physics 253

9. 11/15/2018
Physics 253

10. 11/15/2018
Physics 253

11. 11/15/2018
Physics 253

12. http://www.lon capa.org/~mmp/kap3/cd060.htm
11/15/2018
Physics 253

13. Properties of Trajectories w/ x0=y0=0
Maximum Range
At q = 45o
Goes as velocity squared
Goes inversely with acceleration
For all other ranges
Two initial angles.
11/15/2018
Physics 253

14. Problem Solving
Information can be propagated forward and backwards using the equations of motion as long as a minimum amount of information is available.
To keep yourself organized follow the usual steps:
1) Draw a figure with a thoughtful origin and xy coordinate system
2) Analyze the x and y motion separately, remember they share the same time interval
3) Make the known and unknown table with ax=0, ay=-g, vx is constant, and vy=0 at the apex.
4) Select the equations with some forethought.
11/15/2018
Physics 253

15. Finding Initial Variables Given Final Variables
The examples last lesson used information about initial position & velocity to find final location.
The problem can be reversed. In the next example we use final information to derive initial parameters.
Let’s consider that very familiar situation of a hit softball or baseball
A ball player hits a homer and the ball lands in the seats 7.5 m above the point at which the ball was hit. The ball lands with V=36m/s, 28 degrees below the horizontal. Find the initial velocity of the ball.
11/15/2018
Physics 253

16. Since there is no acceleration in the horizontal direction:
Well at first glance this seems difficult. What are we after? Well we need Vox and Voy - from that we can find Vo and the direction of the initial velocity vector using Vo2 = Vx2 + Vy2 and tanq = Voy/Vox.
Remembering to keep the dimensions independent, let’s work with the horizontal or x dimension first.
Since there is no acceleration in the horizontal direction:
Vox = Vx = (36 m/s) (cos28o) = 32 m/s.
Now we are halfway there and can focus on y-direction.
This looks dire but we actually do have Vy = -(36 m/s) (sin28o) = -17 m/s
Which is more than enough to proceed.
Known
Unknown
yo=0
t=?
y=7.5m
v0y=?
g=9.8m/s2
vy=?
Known
Unknown
yo=0
t=?
y=7.5m
v0y=?
g=9.8m/s2
vy=-17m/s
11/15/2018
Physics 253

17. 11/15/2018
Physics 253

18. A Problem with Initial and final Information: The Air-mail Drop
A plane traveling at an altitude of 200 m and a speed of 69m/s makes an air drop.
At what distance “x” must the drop be made so that the package lands near the recipients?
This is a mixed problem we know the initial speed and the final distance of the drop, from this we can get missing information.
11/15/2018
Physics 253

19. So
And the drop distance is just given by the horizontal distance corresponding to that time.
Let’s use the usual up=+y, right = +x coordinate system with the origin on the initial position of the plane.
What we need is the time of the projectile motion which as usual is obtained from the vertical fall. Then our table takes the form:
And we can derive the missing time using the second projectile motion formula for the y direction
Known
Unknown
yo=0m
t=?
y=-200m
vy=?
v0y=0
g=9.8m/s2
11/15/2018
Physics 253

20. Let’s make this a bit more difficult and for external reasons require that the drop occur 400 m before the recipients.
This means some vertical velocity down (since 400 m is less that 440m) will be required to ensure the time taken is correct. What will that velocity be?
11/15/2018
Physics 253

21. Well the table is clearly different now:
And doesn’t have enough information. However we can get the time t from the horizontal information.
And the table now is
And the initial vertical velocity can be found from the 2nd equation
Known
Unknown
yo=0m
y=-200m
v0y=?
t=5.80s
g=9.8m/s2
vy=?
Known
Unknown
yo=0m
t=?
y=-200m
v0y=?
g=9.8m/s2
vy=?
11/15/2018
Physics 253

22. As we expected the package required a downward push. 11/15/2018
Physics 253

23. Projectile Motion = Parabolic Motion
It turns out in the absence of air resistance all projectile motion is simple parabolic motion.
This can be show with combination of the x and y equations of motion through substitution for time.
Setting x0=y0=0 the equations are considerable simplified
11/15/2018
Physics 253

24. Parabolic Mirrors
11/15/2018
Physics 253

25. Relative Velocity
As you can see from past discussions we need to move easily between reference frames.
The tyke tossing a ball from the wagon was a good example.
From the ground’s frame of reference the ball clearly has two velocity components.
From the girl’s reference frame there is only the vertical component.
11/15/2018
Physics 253

26. Gloucestershire Warwickshire Railway
Relative Velocity
Actually moving between frames gets confusing.
This can be addressed with a rather prescriptive approach.
Let’s set it up with the rather easy example of two trains meeting each other and the extending to the more general situation. +X
VB-E
VA-E
Gloucestershire Warwickshire Railway
Cotswolds, England.
11/15/2018
Physics 253

27. Suppose:
Train A has velocity 2m/s relative to the earth: VA-E=2m/s
Train B has velocity -1m/s relative to the earth: VB-E=-1m/s
Then from the trains’ points of view:
The earth has velocity of -2m/s relative to train A: VE-A=-2m/s
The earth has velocity of 1m/s relative to train B: VE-B=1m/s
And the following equations are true:
VB-A=VB-E+VE-A=-1m/s+-2m/s=-3m/s
VA-B=VA-E+VE-B=2m/s+1m/s=+3m/s
Notice the nifty way the notation behaves, we have a sort of cancellation of inner subscripts
Also VB-A=-VB-A which is true for any vectors. +X
VB-E
VA-E
The axes points in the
same direction for
all three frames.
11/15/2018
Physics 253

28. Lets consider a person walking in a train at a single instant
From vector addition it’s pretty clear that
rPE = rPT+rTE
If we simply take the time derivative of this equation:
vPE = vPT+vTE
In words, this simply says that the velocity of the person with respect to earth equals the sum of the velocity of the person with respect to the train and the velocity of the train with respect to the earth.
Because of commutativity we could have written this as
vPE = vPT+vTE
but the “canceling” of the inner products lends some preference to the former eq.
11/15/2018
Physics 253

29. The water moves with respect to the shore according to VWS
Let’s consider the more complicated case of a boat in a current. Here we’ll see the value of the subscripts.
If we want to go straight across the river the boat will need to push at an angle with respect to water along the vector VBW.
The water moves with respect to the shore according to VWS
As a result the boat with respect to the shore will travel along VBS.
Vectorially then,
VBS = VBW + VWS
Note for this equation
The final vector subscripts correspond to the outer subscripts on the right side
The inner subscripts are the same.
This notational rule demonstrated above, and “proven” in the previous slide helps us move quickly and correctly between reference frames.
11/15/2018
Physics 253

30. Example: Relative Velocity between Two Moving Objects
Both cars shown are traveling at 11m/s. What is the relative velocity of car 1 in the reference frame of car 2?
11/15/2018
Physics 253

31. We can also denote the velocity of car 1 relative to the earth as V1E.
Let the velocity of car 1 in the reference frame of car 2 be denoted V12.
We can also denote the velocity of car 1 relative to the earth as V1E.
Likewise the velocity of the earth relative to care 2 is VE2.
Using our rules we propose
V12 = V1E + VE2
The last vector is not something we have but can get with the identity
VE2 = -V2E Or
V12 = V1E – V2E
Which makes sense as it describes a vector moving at 45 degrees at car 2.
The magnitude is simply given by the Pythagorean Theorem as:
11/15/2018
Physics 253

32. Schedule Wednesday: Friday: Monday Feb 12th:
We’ll review the most important information and discus the test
Start on Chapter 4: Dynamics!
Friday:
No class, use the time to study for the test
Monday Feb 12th:
Quiz #1
Problem sets for Units 1 and 2 due.
11/15/2018
Physics 253