How many significant figures are there in 0.0143 10-4 cm ?

• 9

• 5

• 4

• 3

What does the prefix “M” in front of a unit mean?

• Multiply by 10-6

• Multiply by 10-3

• Multiply by 103

• Multiply by 106

How many professors are there at Otterbein?

• 10

• 100

• 1000

• 10000

Somebody claims the correct equation to calculate the position of an object from its (constant) velocity v and the elapsed time t is x(t) = x(t=0) + v t2.

Can this be correct?• No

• Yes

• Not enough information

A red and an orange car make a race. When do they have the same

velocity?

At point ABetween point A and BAt point BNever

t

x

A B

A red and an orange car make a race. Who wins?

Red carYellow carInsufficient information

t

x

Motion at constant acceleration

• If (and only if) acceleration is constant, we can calculate x,v,a in any instant by using

a(t) = a(t=0)

v(t) = v(t=0) + a(t=0) t

x(t) = x(t=0) + v(t=0) t + ½ a(t=0) t2 Green represent independent variable, darker hues initial values.

Additionally, we may use

v2(t) = v2(t=0) + 2 a(t=0) [x(t) - x(t=0)]

In the equation below, the following is NOT true…

v2(t) = v2(t=0) + 2 a(t=0) [x(t) -x(t=0)]

• We need to know the position of the object in the moment at which we want to calculate its velocity

• We need to know the initial velocity of the object to calculate its velocity at time t

• This equation is only true if the acceleration of the object is constant.

• The equation is only true for positive velocities, since we have to take the square root eventually.

Example for derivation

• Constant acceleration means linear rising (or falling) velocity

velocity at later time is velocity at earlier time plus slope times time elapsed.

• Slope of velocity curve is acceleration

v(t2) = v(t1) + a (t2-t1)

t1 t2 t

v(t)v(t2)

v(t1) ∆v

∆t

A car accelerates at constant (non-zero) rate. Which of the following motion diagrams is NOT correct?

v

v

t

a

tx

t t

An dropped object accelerates downward at 9.8 m/s2. If instead you

throw it downward, its downward acceleration after release is …

• less than 9.8 m/s2

• exactly 9.8 m/s2

• more than 9.8 m/s2

• Insufficient information

Solving Kinematic Problems

1. Read problem carefully2. Draw diagram3. List knowns and unknowns4. What physics principles do apply?5. Find equations that do apply6. Solve algebraically (with variables, not values!)7. Calculate numerically8. Check results: Numbers reasonable? Units

correct?

Vectors

• “Directions with magnitudes” that can be shifted around

Addition: Tail-to-Tip or Parallelogram

Subtraction by adding negative vector

a-b = a + (-b)

3D Vectors

• 3D vector as a sum of multiples of the three unit vectors i,j,k

• Example: a=1.5 i + 2.5j +3 k

To which quadrant does the following vector belong?

A = - 4 i + 6.5 j

• 1st Quadrant• 2nd Quadrant• 3rd Quadrant• 4th Quadrant

To which quadrant does the following vector belong?

|B| = 4.5, φB = -45º

• 1st Quadrant• 2nd Quadrant• 3rd Quadrant• 4th Quadrant

To which quadrant does the following vector belong?

|c| = 4.5m/s, φc = 3.10

• 1st Quadrant• 2nd Quadrant• 3rd Quadrant• None of the above

To which quadrant does the following vector belong?

• 1st Quadrant• 2nd Quadrant• 3rd Quadrant• None of the above

m

mD

7.6

9.0

To which quadrant does the following vector belong?

• 1st Quadrant• 2nd Quadrant• 3rd Quadrant• Not enough

information

22

9.44.3

42

ts

mm

ts

mm

x

What is the length of the following vector?

• -1 m/s• 1 m/s• 5 m/s• 25 m2/s2

sm

smv

/4

/3

What is the length of this

vector?

• Upper formula• Middle formula• Bottom formula• None of the above

22

9.44.3

42

ts

mm

ts

mm

x

tts

m

s

mmx

29.444.5||

22

2

2

9.44.342||

t

s

mmt

s

mmx

2

29.44.54|| ts

mmt

s

mx

Which is a correct statement concerning this vector?

|B| = 4.5, φB = -45º

• Its magnitude is negative• Its x component is negative• Its y component is positive• Its x and y component have the same absolute

value

Projectile Motion

By aiming a gun higher (increasing the angle of the barrel w.r.t. the x-

axis), I achieve the following:

• Higher initial position

• Larger initial y-component of velocity

• Higher initial velocity

• None of the above

A gun is accurately aimed at a dangerous terrorist hanging from

the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed v0, the terrorist lets go and drops to the

ground. What happens? • The bullet hits the terrorist regardless of the value of v0

• The bullet hits the terrorist only if v0 is large enough

• The bullet misses the terrorist

• Not enough information

v0

gun

gutter

A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic

paths shown, which ship gets hit first?

• A

• B

• Both at the same time

• Need more information

A B

Circular Motion

An object at the end of an r=1m string in circular motion completes 10 revolutions in one second. How

long does each revolution take? • (2 π r)(1s)

• 1s/(2π r)

• 0.1 s

• Need more information

An object at the end of an r=1m string in circular motion completes 10 revolutions in one second. What

is its speed? • 6.3 m/s

• 3.1 m/s

• 63 m/s

• 0.31 m/s

An object at the end of an r=1m string in circular motion completes 10 revolutions in one second. What

is its frequency? • 10 Hz

• 1 Hz

• 1/10 Hz

• 3.14 Hz

Newton’s first Law

• In the absence of a net external force, a body either is at rest or moves with constant velocity.

– Motion at constant velocity (may be zero) is thus the natural state of objects, not being at rest. Change of velocity needs to be explained; why a body is moving steadily does not.

Mass & Weight

• Mass is the property of an object

• Weight is a force, e.g. the force an object of certain mass may exert on a scale

Mass: On the surface of the Moon a standard block of lead …

• … has a different mass

• … has a different weight

• Its weight is unchanged, but g has a different value

• Need more information

Newton’s first law states that objects remain at rest only when they no net force acts on them. A book on a table

is subject to the force of gravity pulling it down. Why doesn’t it move?

• Newton’s first law does not apply (obstacle!)

• There must be another force opposing gravity

• Table shelters book from force of gravity

• Not enough information

Newton’s second Law

• The net external force on a body is equal to the mass of that body times its acceleration

F = ma. • Or: the mass of that body times its acceleration is

equal to the net force exerted on it

ma = F• Or: a=F/m • Or: m=F/a

Newton II: calculate Force from motion

• The typical situation is the one where a pattern of Nature, say the motion of a projectile or a planet is observed: – x(t), or v(t), or a(t) of object are known, likely

only x(t)

• From this we deduce the force that has to act on the object to reproduce the motion observed

Calculate Force from motion: example • We observe a ball of mass m=1/4kg falls to the

ground, and the position changes proportional to time squared.

• Careful measurement yields: xball(t)=[4.9m/s2] t2

• We conclude v=dx/dt=2[4.9m/s2]t a=dv/dt=2[4.9m/s2]=9.8m/s2

• Hence the force exerted on the ball must be • F = 9.8/4 kg m/s2 = 2.45 N

– Note that the force does not change, since the acceleration does not change: a constant force acts on the ball and accelerates it steadily.

Newton II: calculate motion from force

• If we know which force is acting on an object of known mass we can calculate (predict) its motion

• Qualitatively: – objects subject to a constant force will speed up (slow

down) in that direction

– Objects subject to a force perpendicular to their motion (velocity!) will not speed up, but change the direction of their motion [circular motion]

• Quantitatively: do the (vector) algebra!

Newton II: calculate motion from force

• Say a downward force of 4.9N acts on a block of 1 kg (we call the coordinate y and start at y=0)

• We conclude that the block will be accelerated:

ay = F/m= -4.9N / 1kg = -4.9 m/s2

• We use this constant acceleration to calculate

y = -1.225m/s2 t2 + b t + c , where b &c are constants, c=0

Q: What physical situation is this?

Newton II: calculate mass from force and acceleration

• Pretty simple, if you know force and acceleration we have

m=F/a