Midterm 1 Review

Midterm 1• Sept 25 (Wed)• 7:30‐8:45 pm• Sec 510, 511, 512 – HELD 109• Sec 513, 514, 561 – HELD 111• Q1: 4 short multiple choice questions (20% total)• Q2‐5: 4 work out problems (80% total)• Show your work for possible partial credits• Review my slides for quick recap!

Vector in Unit Vector Notation

jΘ |V| iΘ |V| V

j V i V V

V V V

Θ|V|||VΘ|V|||V

YX

YX

Y

X

ˆsinˆcos

ˆˆ

sincos

General Addition Example

•Add two vectors using the i‐hats, j‐hats and k‐hats

k km 0 j km 5 i km 10 D

k km 0 j km 5 i km 0 D

k km 0 j km 0 i km 10 D

DD D

R

2

1

21R

Subtraction• Subtraction is same as addition, except the second vector is first made negative and then the two are added together

)V(- V V V 1212

V2

V1

V2 + V1

V2 - V1V1

V2

How do we Multiply Vectors?• First way: Scalar Product or Dot Product

– Why Scalar Product? • Because the result is a scalar (just a number)

– Why a Dot Product? 

• Because we use the notation A.B

• A.B = |A||B|Cos

Vector Cross Product

• This is the last way of multiplying vectors we will see

• Direction from the “right‐hand rule”

• Swing from A into B!

SinB ACB A C

Vector Cross Product Cont…•Multiply out, but use the Sin to give the magnitude, and RHR to give the direction

)1(sin ˆˆˆ)1(sin ˆˆˆ)0(sin 0ˆˆ

jki

kji

ii

jki

+ikj

_

SinB ACB A C

x

z

yji

k

Motion in One Dimension Cont…

• Velocity: “Change in position during a certain time period”

• Calculate from the Slope: The “Change in position as a function of time”

– Change in Vertical  – Change in Horizontal

• Change: D• Velocity  DX/Dt

Moving CarA harder example:

X = ct2

What’s the velocity at t=1 sec?

Want to calculate the “Slope” here.

Instantaneous Velocity

What would my speedometer read?

v(t) limt 0x(t t) x(t)

t

dxdt

Acceleration

• Acceleration is the “Rate of change of velocity”

• Said differently: “How fast is the Velocity changing?” “What is the change in velocity as a function of time?”

a vt

V2 V1t2 t1

d vdt

Important Equations of Motion

•If the acceleration is constant

•Position, velocity and Acceleration are vectors. More on this in Chap 3

221

00

0

tatv x x

tav v

Equations of motion for constant acceleration

x(t) x0 v0t 12 at 2

v(t) v0 at a(t) = a

Just says that acceleration is constant in time; it is not useful in terms of algebra.

From now on writing x or v implies x(t) or v(t) so as not to write (t) all the time.

It follows from the one above if you take the derivative wrt time

Algebra reminder: if you have a set of equations and a set of unknown variables their number have to be the same to be able to solve it.

x, x0,v,v0,t,aWe have TWO independent equations and 6 variables

Truly we really have only 5 because x and x0 always come together as x-x0

One objectYou are going along Texas Av. at 35 mph (texting on the phone). Suddenly you see an old lady crossing the street about 100 meters away and hit the breaks. Your car decelerates uniformly at 1.25 m/s2. Did you run the poor lady over?

x x0 = v v0 a t =

1st step x x0 = ? v 0

v0 35mph 15.64ms

a 1.25 ms2

t =

2nd stepBecause it is assume you are stopped at the end

Unit conversion

We have THREE unknowns so we can now find the other two.

3rd step

x x0 = v0t 12 at 2

v v0 at v2 v0

2 2a(x x0)

Which equation contains the three unknowns and the one we want?

(x x0) v2 v0

2

2a

02 (15.64m /s)2

2(1.25m /s2) 97.8m

So the old lady survives the car

Speeder•A speeder passes you (a police officer) sitting by the side of the road and maintains their constant velocity V. You immediately start to move after the speeder from rest with constant acceleration a.•How much time does it take to ram the speeder?•How far do you have to travel to catch the speeder? •What is your final speed?

XPolice Officer Speeder

Throw a Ball up• You throw a ball upward into the air with initial 

velocity V0. Calculate:• c) Time it takes for the ball to come back to your 

hand

y y0 = 0 v v0 v0

a g t = ?

1st and 2nd step

Key part

y y0 = v0t 12 at 2

v v0 at v2 v0

2 2a(y y0)

3rd stepWhich equation contains the three unknowns and the one we want?

0 = v0t 12 (-g)t 2 t(v0

12 gt)

t = 0 and t =2v0

g

y y0 = v0t 12 at 2

v v0 at v2 v0

2 2a(y y0)

3rd stepWhich equation contains the three unknowns and the one we want?

Free fall exampleQ: A ball is thrown upwards with an initial velocity v0. What is the speed of the ball when it hits the ground?v0

h

y y0 = -h v ? v0 v0

a g t =

1st and 2nd step

v v02 2(-g)(h) v0

2 2hgNotice that solution does NOT depend on the sign of v0

Q: A ball is thrown upwards with an initial velocity v0. What is maximum height it gets from the top of the building?

This is a new problem. Start from the beginning b/c the end point is different!

y y0 = ? v 0 v0 v0

a g t =

1st and 2nd step

Key is to realize that v=0 at the top

y y0 = v0t 12 at 2

v v0 at v2 v0

2 2a(y y0)

3rd stepWhich equation contains the three unknowns and the one we want?

(y y0) 02 v0

2

2(g)

v02

2g

Graphs

• Describe motion in each point:– Direction– Velocity– Acceleration 

C

Bx

(cm

)

t (s)

D E F

Position in 3 dimensions

kZ(t) jY(t) iX(t) (t)R

kZ jY iX R

dimensions 3in origin the torelative

position,our as R Write

Velocity in 3 dimensions

ka ja ia

kdtdV jdt

dV idt

dV

dt)kV jV id(V

dtVd a

zyx

zyx

zx

y

dtkd dt

jd 0 dtid used have We :Note

kzV jyV ixV

kdtdZ jdt

dY idtdX

dt)kZ jY id(X

dtRd V

then position, theis R If

x

z

yji

k

Projectile Motion

The horizontal and vertical equations of the motion behave 

independentlyProblem solving:

The trick for all these problems is to break them up into the X and Y directions.

ay=‐g    and ax=0

Ball Dropping

• ay = g (downwards)• ax = 0

Constant for both cases!!!

vx = 0 vx>0

Equations of motion under constant acceleration

Projectile motion sets xo = 0 and yo = 0 then obtains the specific results shown at right.

x = (vocosαo)t

y = (vosinαo)t  1/2gt2

vx = vocosαo

vy = vosinαo gt

Rock and Cliff Problem

A ball is thrown horizontally out off a cliff of a height h above the ground. The ball hits the ground a distance D from the base of the cliff. Assuming the ball was moving horizontally at the top and ignoring air friction, what was its initial velocity?

h

Dy=0

As a ship is approaching the dock at 45.0 cm/s, an important piece of landing equipmentneeds to be thrown to it before it can dock. This equipment is thrown at 15.0 m/s at 60.0oabove the horizontal from the top of a tower at the edge of the water, 8.75 m above theship's deck. For this equipment to land at the front of the ship, at what distance from thedock should the ship be when the equipment is thrown? Air resistance can be neglected.

R

h

y‐component

y y0 hv

0 y v0 sin

vy

ay gt

x‐component

x x0 Rvx v0 cost

y y0 v0yt 12

ayt2

h v0 sint 12

gt 2

t 2 2v0 sin

gt

2hg 0

t

2v0 sing

4v 2

0 sin 2

g2 8hg

2 T

T

R v0 cosTD R vBT

Uniform Circular Motion

• Fancy words for moving in a circle with constant speed

• We see this around us all the time– Moon around the earth– Earth around the sun– Merry‐go‐rounds

Uniform Circular Motion ‐ Velocity

• Velocity vector = |V| tangent to the circle

• Is this ball accelerating?–Why?

Centripetal Acceleration

• Vector difference V2 ‐ V1 gives the direction of acceleration a

dtvvdtvda /)(/ 12

a

R

Centripetal Acceleration

• “Center Seeking”• Accel vector points towards the center

• Acceleration is perpendicular to velocity

r a v 2

R(ˆ r )

R

Circular Motion: Get the speed!

Speed = distance/timeDistance in 1 revolution divided by the 

time it takes to go around once

Speed = 2r/TNote: The time to go around once is known as the Period, or T

Relative velocity in one dimension

– If point P is moving relative to reference frame A, we denote the velocity of P relative to frame A as vP/A.  

– If P is moving relative to frame B and frame B is moving relative to frame A, then the x‐velocity of P relative to frame A is vP/A‐x = vP/B‐x + vB/A‐x.

Relative velocity in two or three dimensions• We extend relative velocity to two or three dimensions by using

vector addition to combine velocities.• In Figure 3.34, a passenger’s motion is viewed in the frame of

the train and the cyclist.

Flying in a crosswind• A crosswind affects the motion of an airplane.• An airplane’s compass indicates it is header north, and its

airspeed is 240 km/hr. If there is a 100 km/hr wind from west to east, what is the velocity of the airplane relative to earth?

In which direction should the pilot head to travel due north and what will be her velocity relative to earth?

Translate: Newton’s Second Law• The acceleration is in the SAME 

direction as the NET FORCE This is a VECTOR equation

It applies for EACH direction If I have a NET force, what is 

my acceleration?  More force →more 

acceleration More mass → less acceleration

gmWWeight

ma F ,ma F

am F

:EquationVector

yyxx

Notice that the 2nd law also implies the 1st law, if a=0 then the sum of all forces on an object is zero

Force to stop a car: combining kinematics and Newtwon’s laws

•You are a car designer. You must develop a new braking system that provides a constant deceleration. What constant net force is required to bring a car of mass m to rest from an speed of V0 within a distance of D?

X0 = 0 XF = D

V0 V = 0

Strategy: first find the acceleration from kinematics (Ch 2-3) and then connect it to the force via Newton’s 2nd law

x x0 Dv0 v0

v 0t a

v 2 v02 2a(x x0)

0 v02 2aD

a v0

2

2D

1st step in kinematic problems 2nd step in kinematic

problems

Connect this result to Newton’s 2nd law

Fnet ma

Fnet mv0

2

2D

2 boxes connected with a string• Two boxes with masses m1 and m2 are placed on a frictionless horizontal surface and pulled with a Force FP. Assume the string between doesn’t stretch and is massless.

a)What is the acceleration of the boxes? b)What is the tension of the strings between the boxes?

M2 M1

Figure 5.3

A crate on an inclined plane

An object on an inclined plane will have components of force in x and y space. 

Person in an elevator

Moving in an arbitrary direction

Connected ‐ Two bodies with same acceleration magnitude 

Draw the two-free body diagrams. The tension from the mass hanging is the same force that draws the cart up the ramp.

Box on an inclined plane• A box with mass m is placed on a frictionless incline with angle 

and is allowed to slide down. a) What is the normal force?b) What is the acceleration of the box?c) How long does it take to fall if the ramp has a length L? 

FN

wB=mgmg cos

mg sinx : mgsin max

y : FN mgcos 0ax gsin