#### 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 *

*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*

*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*

*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