lecture 12

Physics 207: Lecture 12, Pg 1

Lecture 12

Goals:Goals:

Assignment: Assignment: HW6 due Wednesday 3/3HW6 due Wednesday 3/3 For Tuesday: Read all of chapter 10For Tuesday: Read all of chapter 10

• Chapter 9: Momentum & ImpulseChapter 9: Momentum & Impulse Solve problems with 1D and 2D Collisions Solve problems having an impulse (Force vs. time)

• Chapter 10Chapter 10 Understand the relationship between motion and energy Define Potential & Kinetic Energy Develop and exploit conservation of energy principle


Momentum Conservation

Momentum conservation (recasts Newton’s 2nd Law when net external F = 0) is an important principle (usually when forces act over a short time)

It is a vector expression so must consider Px, Py and Pz if Fx (external) = 0 then Px is constant if Fy (external) = 0 then Py is constant if Fz (external) = 0 then Pz is constant

constant that implies 0 PP

dtd

dt

d

dt

md

dt

dmamEXT

P)vvF

( 0 if and EXTF

PPP


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 final speed V.

In terms of m, M, and V :

What is the momentum of the bullet with speed v ?

vV

before after

x


Inelastic collision in 1-D: Example

What is the momentum of the bullet with speed v ?

Key question: Is x-momentum conserved ?

vV

before after

x

aaaa

vm

V)( 0 M v Mmm

P BeforeP Before P AfterP After

V)/1( v mM


Exercise Momentum is a Vector (!) quantity

A. Yes

B. No

C. Yes & No

D. Too little information given

A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction

In regards to the block landing in the cart is momentum conserved?



Let a 2 kg block start at rest on a 30° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart

What is the final velocity of the cart?

x-direction: No net force so Px is conserved. y-direction: Net force, interaction with the ground so

depending on the system (i.e., do you include the Earth?)

Py is not conserved (system is block and cart only)

5.0 m30°

7.5 m

10 kg

2 kg



Initial Final

Px: MVx + mvx = (M+m) V’x M 0 + mvx = (M+m) V’x

V’x = m vx / (M + m) = 2 (8.7)/ 12 m/s

V’x = 1.4 m/s

x-direction: No net force so Px is conserved y-direction: vy of the cart + block will be zero and

we can ignore vy of the block when it lands in the cart.

5.0 m

30°

7.5 m

N

mg

1) ai = g sin 30°

= 5 m/s2

2) d = 5 m / sin 30°

= ½ ai t2

10 m = 2.5 m/s2 t2

2s = t

v = ai t = 10 m/s

vx= v cos 30°

= 8.7 m/s

i

j

x

y

30°


A perfectly inelastic collision in 2-D

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

vv1

vv2

VV

before after

m1

m2

m1 + m2

If no external force momentum is conserved. Momentum is a vector so px, py and pz


A perfectly inelastic collision in 2-D

vv1

vv2

VV

before after

m1

m2

m1 + m2

x-dir px : m1 v1 = (m1 + m2 ) V cos y-dir py : m2 v2 = (m1 + m2 ) V sin

If no external force momentum is conserved. Momentum is a vector so px, py and pz are conseved


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


Billiards

Consider the case where one ball is initially at rest.

ppa

ppb

FF PPa

beforeafter

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

vvcm


Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11)

Conservation of Momentum x-dir Px : m vbefore = m vafter cos + m Vafter cos y-dir Py : 0 = m vafter sin + m Vafter sin

ppafter

ppb

FF PPafter

before after


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 (-kx) towards its equilibrium position (Chapter 10)

Collisions often involve a varying force

F(t): 0 maximum 0 We can plot force vs time for a typical collision. The

impulse, JJ, of the force is a vector defined as the integral of the force during the time of the collision.


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

F

ptt pddtdtpddtFJ )/(

J a vector that reflects momentum transfer

t

ti tf

t

Impulse JJ = area under this curve !

(Transfer of momentum !)

Impulse has units of Newton-seconds


Force and Impulse Two different collisions can have the same impulse since

JJ depends only on the momentum transfer, NOT the nature of the collision.

t

F

t

F

tt

same area

t big, FF smallt small, FF big


Average Force and Impulse

t

F

t

F

tt

t big, FFavav smallt small, FFavav big

FFav av

FFav av


Exercise 2Force & Impulse

A. heavier

B. lighter

C. same

D. can’t tell

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 F light heavy


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


Back of the envelope calculation

(1) marm~ 7 kg (2) varm~7 m/s (3) Impact time t ~ 0.01 s

Question: Are these reasonable? Impulse J = p ~ marm varm ~ 49 kg m/s Favg ~ J/t ~ 4900 N (1) mhead ~ 6 kg ahead = F / mhead ~ 800 m/s2 ~ 80 g ! Enough to cause unconsciousness ~ 40% of a fatal

blow Only a rough estimate!

tFdtFJ t avg


Chapter 10: Energy (Forces over distance)

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 acts over a distance:

Can we identify another useful quantity?


Lecture 12

Assignment: Assignment: HW6 due Wednesday 3/3HW6 due Wednesday 3/3 For Tuesday: Read all of chapter 10For Tuesday: Read all of chapter 10

lecture 12

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