Physics 207: Lecture 2 Notes

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Transcript Physics 207: Lecture 2 Notes

Lecture 12

Goals: • Chapter 9: Momentum & Impulse  Solve problems with 1D and 2D Collisions  Solve problems having an impulse (Force vs. time) • Chapter 10  Understand the relationship between motion and energy  Define Potential & Kinetic Energy  Develop and exploit conservation of energy principle Assignment:  HW5 due Wednesday  For Wednesday: Read all of chapter 10 Physics 207: Lecture 12, Pg 1

Inelastic collision in 1-D: Example

 A block of mass

M

is initially at rest on a frictionless horizontal surface. A bullet of mass

m

is fired at the block with a muzzle velocity (speed)

v

. The bullet lodges in the block, and the block ends up with a speed

V.

In terms of

m, M

, and

V

: What is the momentum of the bullet with speed

v

?

x v V

before after Physics 207: Lecture 12, Pg 2

Inelastic collision in 1-D: Example

What is the momentum of the bullet with speed

v

?

m

 v  Key question: Is x-momentum conserved ?

Before

m

v  M 0

After

aaaa

(

m

M

) V

v

before

V

after Physics 207: Lecture 12, Pg 3

x

Exercise

Momentum Conservation

 Two balls of equal mass are thrown horizontally with the same initial velocity . They hit identical stationary boxes resting on a frictionless horizontal surface.  The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.

 Which box ends up moving fastest ?

A.

Box 1 B.

Box 2 C.

same 1 2 Physics 207: Lecture 12, Pg 4

Exercise

Momentum Conservation

 The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.

 Which box ends up moving fastest ?

 Notice the implications from the graphical solution: Box 1’s momentum must be bigger because the length of the summed momentum must be the same.

 The longer the green vector the greater the speed Before

After

Before After Ball 1

Ball 1

Ball 2 Ball 2 Box 1

Box 1

Box 2 Box 2 Box 1+Ball 1 Box 1+Ball

1

Box 2+Ball 2 Box 2+Ball 2 1 2 Physics 207: Lecture 12, Pg 5

A perfectly inelastic collision in 2-D

 Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).

v

1

V

q

m 1 + m 2 m 1 m 2

v

2

before after  If no external force momentum is conserved.

 Momentum is a vector so p x , p y and p z Physics 207: Lecture 12, Pg 6

A perfectly inelastic collision in 2-D

 If no external force momentum is conserved.

 Momentum is a vector so p x , p y and p z are conseved

V v

1

q

m 1 + m 2 m 1 m 2

v

2

before  x-dir p x : m 1 v 1  y-dir p y : m 2 v 2 =

(m

1 =

(m

1

+ m

2

+ m

2 after

) V cos

q

) V sin

q Physics 207: Lecture 12, Pg 7

Elastic Collisions

 Elastic means that the objects do not stick.

 There are many more possible outcomes but, if no external force, then momentum will always be conserved  Start with a 1-D problem.

Before After Physics 207: Lecture 12, Pg 8

Billiards

 Consider the case where one ball is initially at rest. after before

p

b

p

a

q

F v

cm

P

a

f The final direction of the red ball will depend on where the balls hit.

Physics 207: Lecture 12, Pg 9

Billiards: All that really matters is conservation momentum (and energy Ch. 10 & 11)

   Conservation of Momentum x-dir P y-dir P x y : : m v before = m v after 0 = m v after cos sin q q + m V after + m V after cos f sin f before

p

b

after

p

after

q

F P

after

f Physics 207: Lecture 12, Pg 10

Force and Impulse (A variable force applied for a given time)

 Gravity: At small displacements a “constant” force t  Springs often provide a linear force (-

k

its equilibrium position (Chapter 10) x) towards  Collisions often involve a varying force F(t): 0  maximum  0  We can plot

force vs time

for a typical collision. The impulse,

J

, of the force is a

vector

defined as the integral of the force during the time of the collision.

Physics 207: Lecture 12, Pg 11

Force and Impulse (A variable force applied for a given time)

J

reflects momentum transfer

J

  

t F

dt

 

t

( 

d p

/

dt

)

dt

 

p

d p F

Impulse

J

= area under this curve !

(Transfer of momentum !) Impulse has units of Newton-seconds

t i

t t f

Physics 207: Lecture 12, Pg 12

t

Force and Impulse

 Two different collisions can have the same impulse since

J

depends only on the

momentum transfer

, NOT the nature of the collision.

F F

same area 

t

t

big,

F

small

t t

t

t

small,

F

big Physics 207: Lecture 12, Pg 13

Average Force and Impulse

F av

F

t

t

big,

F av

small

t

F av

F t

t

t

small,

F av

big Physics 207: Lecture 12, Pg 14

Exercise 2

Force & Impulse

 Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force

F

acts on each one for exactly 1 second .

Which box has the most momentum after the force acts ?

F light F heavy

A.

heavier B.

lighter C.

D.

same can’t tell Physics 207: Lecture 12, Pg 15

Boxing: Use Momentum and Impulse to estimate g “force”

Physics 207: Lecture 12, Pg 16

Back of the envelope calculation

J

  

t F

dt

F

 avg 

t

(1) m arm

~ 7 kg (2)

v arm

~7 m/s (3)

Impact time

t

~ 0.01 s Question: Are these reasonable?

Impulse J =

p

~ m arm v arm ~ 49 kg m/s 

F ~ J/

t ~ 4900 N

(1) m head ~ 6 kg 

a head = F / m head ~ 800 m/s 2 ~ 80 g !

 

Enough to cause unconsciousness ~ 40% of fatal blow Only a rough estimate!

Physics 207: Lecture 12, Pg 17

Woodpeckers

During "collision" with a tree

a

head

~ 600 - 1500 g How do they survive?

• Jaw muscles act as shock absorbers • Straight head trajectory reduces damaging rotations (rotational motion is very problematic) Physics 207: Lecture 12, Pg 18

Home Exercise

 The only force acting on a 2.0 kg axis. Notice that the plot is

force

object moving along the x vs

time

.  If the velocity v x is +2.0 m/s at 0 sec, what is v x at 4.0 s ?   p = m  v = Impulse  m  v = J 0,1 + J 1,2 + J 2,4  m  v = (-8)1 N s + ½ (-8)1 N s + ½ 16(2) N s m  v = 4 N s  v = 2 m/s v x = 2 + 2 m/s = 4 m/s Physics 207: Lecture 12, Pg 19

Chapter 10: Energy

 We need to define an “isolated system” ?

 We need to define “conservative force” ?

 Recall, chapter 9, force acting for a period of time gives an impulse or a change (transfer) of momentum  What if a force acting over a distance:

Can we identify another useful quantity?

Physics 207: Lecture 12, Pg 20

Energy

F y  = m a y y(t) = y 0 and let the force be constant + v y0  t + ½ a y  t 2   y = y(t)-y 0 = v y0  t + ½ a y  t 2  v y (t) = v y0 + a y  t  Eliminate  t and regroup  So  y = v y0 (v y - v y0 ) / a y  t = (v + ½ a y y - v (v y 2 y0 ) / a - 2v y y v y0 +v y0 2 ) / a y 2  y = ( v y0 v y v y0 2 ) / a y + ½  y = ( v y0 2 ) / a y + ½ (v (v y y 2 2 2v y v y0 + v y0 2 + v y0 2 ) / a y ) / a y  y = ½ (v y 2 v y0 2 ) / a y Physics 207: Lecture 12, Pg 21

Energy

And now 

y =

½

(v

y 2

-

v y0 2

) / a

y can be rewritten as:

ma

y 

y = ½ m (v

y 2

- v

y0 2

)

And if the object is falling under the influence of gravity then

a

y

= -g

Physics 207: Lecture 12, Pg 22

Energy

-

mg  y= ½ m (v y 2 - v y0 2 )

-

mg ( y f – y i ) = ½ m ( v yf 2 -v yi 2 ) A relationship between

y-displacement

and change in the y-speed Rearranging to give initial on the left and final on the right ½ m v yi 2

+

mgy i = ½ m v yf 2

+

mgy f We now define mgy as the “gravitational potential energy ” Physics 207: Lecture 12, Pg 23

Energy

 Notice that if we only consider gravity as the external force then  the x and z velocities remain constant To ½ m v yi 2

+

mgy i = ½ m v yf 2

+

mgy f  Add ½ m v xi 2 + ½ m v zi 2 and ½ m v xf 2 + ½ m v zf 2 ½ m v i 2

+

mgy i = ½ m v f 2

+

mgy f  where v i 2 = v xi 2 +v yi 2 + v zi 2 ½ m v 2 terms are defined to be kinetic energies (A scalar quantity of motion) Physics 207: Lecture 12, Pg 24

Energy

 If only “conservative” forces are present, the total energy ( sum of potential, U, and kinetic energies, K ) of a system is conserved

E mech = K + U E mech = K + U

=

constant

K

and

U

may change, but

E mech = K + U

remains a fixed value.

E

mech

is called “mechanical energy” Physics 207: Lecture 12, Pg 25

Example of a conservative system: The simple pendulum.

 Suppose we release a mass

m

its lowest possible point.

from rest a distance

h 1

above  What is the maximum speed of the mass and where does this happen ?

 To what height

h 2

does it rise on the other side ?

h 1 m v h 2

Physics 207: Lecture 12, Pg 26

Example: The simple pendulum.

 What is the maximum speed of the mass and where does this happen ?

E = K + U = constant is a minimum.

and so K is maximum when U

y y= h 1 y= 0

Physics 207: Lecture 12, Pg 27

Example: The simple pendulum.

 What is the maximum speed of the mass and where does this happen ?

E = K + U = constant is a minimum and so K is maximum when U E = mgh 1 at top E = mgh 1 = ½ mv 2 at bottom of the swing

y y= h 1 y=0 h 1 v

Physics 207: Lecture 12, Pg 28

Example: The simple pendulum.

To what height

h2

does it rise on the other side?

E = K + U = constant again (when K = 0) and so when U is maximum it will be at its highest point.

E = mgh 1 = mgh 2 or h 1 = h 2

y y= h 1 =h 2 y=0

Physics 207: Lecture 12, Pg 29

 1 st

Cart Exercise Revisited: How does this help?

Part: Find v at bottom of incline a i = g sin 30 ° = 5 m/s 2 j N E mech

is conserved

K i + U i 0 + mgy i (2gy) ½ = K f = ½ mv 2 + U + 0 = v = (100) ½ f m/s v x = v cos 30 ° = 8.7 m/s

One step, no FBG needed

i 5.0 m y mg 30 ° 7.5 m d = 5 m / sin 30 ° = ½ a i  t 2 10 m = 2.5 m/s 2  t 2 2s =  t v = a i  t = 10 m/s v x = v cos 30 ° = 8.7 m/s x Physics 207: Lecture 12, Pg 30

Home exercise

A block is shot up a frictionless 40 ° slope with initial velocity

v

. It reaches a height

h

before sliding back down. The same block is shot with the same velocity up a frictionless 20 ° slope.

On this slope, the block reaches height A.

2h B.

h C.

D.

E.

h/2 Greater than

h

, but we can’t predict an exact value.

Less than

h

, but we can’t predict an exact value.

Physics 207: Lecture 12, Pg 31

Exercise 3: U, K, E & Path

 1.

2.

3.

A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless Ball is dropped Ball slides down a straight incline Ball slides down a curved incline After traveling a vertical distance h , how do the speeds compare?

1 2 3 h (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell Physics 207: Lecture 12, Pg 32