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Ch9 Impulse and Momentum

講者： 許永昌 老師

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ContentsAction: Law of CONSERVATION of massMomentum and ImpulseAction: impulse Example 9.1 Student Workbook 9.3Law of Conservation of Momentum

CollisionsExplosions

後面幾章的重點

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Action: Law of Conservation of mass ( 請預讀 P239)

Purpose: Realize the meaning of “conservation”.Realize the significance of “before” and “after”.

Action:Water + Juice Drink

Before: mwater and mjuice. After: mdrink.

Conservation: mwater + mjuice = mdrink.

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Momentum and Impulse ( 請預讀P240~P247)

Momentum: Its definition will be different when we

consider the magnetic field.Newton’s 2nd Law of Motion:

Impulse:

Newton’s 2nd Law of Motion:

p mv

, .net net

dpF ma F

dt

f

i

t

nettJ F dt

.f i i fJ p p p J p

Jx

t(s)

Fx(N)

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Action: ImpulsePurpose:

Understand the relationship between impulse and the change of momentum.

Introduce the momentum bar charts.Nothing new.

Object:One tennis ball.

Action:Analyze the motion of a bouncing ball:

The impulse of the free fall. The impulse of the bouncing.

Free-body diagram or motion diagram impulse.

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Example 9.1A 150 g baseball is thrown with a speed of 20

m/s. It is hit straight back toward the pitcher at a speed of 40 m/s. The interaction force between the ball and the bat is shown in the graph. What maximum force Fmax does the bat exert on the ball?

t(s)

Fx(N)

6.0 ms

Fmax

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Student Workbook P9-3A carnival game requires you to knock over a

wood post by throwing a ball at it. You’re offered a very bouncy rubber ball and a very sticky clay ball of equal mass. Assume that you can throw them with equal speed and equal accuracy. You only get one throw.

Which ball will you choose? Why?

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HomeworkStudent Workbook:

9.19.29.49.79.99.11

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Summary IMomentum: Impulse:

一物體的動量會有所改變是因為有 impulse 作用在其上。不過，嚴格說來，這條式子只是 Newton’s 2nd Law

的延伸，而不是新的。由於 D 所描述的，是一個前後狀態的變化，所以，

在討論守恒律時，常常會看到課本用 BEFORE and AFTER 來描述。

f

i

t

nettJ F dt p

因 果

p mv

fi

AfterBefore

pp J

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Law of Conservation of Momentum ( 請預讀 P247~P252)

System:Total Momentum:

Net force:

Newton’s 2nd Law of Motion:

For a System:

1 2 3 ...P p p p

net on ext on Newton's

3rd Law.

net k k extk k

F F F F

net on ,kk k k

dpF m a

dt

net on .kext k

k k

dp dPF F

dt dt

這就是為何我們常常簡化問題為 particle 的樣子。

When 0, .ext f iF P P

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Collisions ( 請預讀 P253~P255)

It is related to the impulse approximation (P246). Assume that the complex interaction is

happened during a very short time interval. i.e.

Check your experiment results:Exp. 2, Table 4 and 5.

Is Pxf =Pxi?

0 0lim lim 0

i i

i i

t t t t

net extt tt tJ F dt F dt

P

(1) 適當選定 system(2) 使得 system 所受的總外力對微小時間間隔的積分 0(3) 則在這微小時間間隔內，總動量守恆。

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Explosions ( 請預讀 P255~P257)

System: Fuel+RocketBefore: Pi,y=(M+Dm)vi,y.

Final: Pf,y=Mvf,y+Dm(vf,y+vthrust,y).

We get:

, ,

thrust,0

lim .

net y ext y

y yy y

t

F F

dP P dMM m a v

dt t dt

, thrust, .ext y y y

dMF v Ma

dt

thrust,

0

0

0y

m

dM dm

dt dtv

,

,

0,

If and constant, we get

.

If = =constant, we get .

ext y y

y

thrust y

y t

thrust y

F Mg a

M g adMdt

M v

M g aM M e

v

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Examples:Textbook:

9.45: An object at rest on a flat, horizontal surface

explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 8.2 m before stopping. How far does the lighter fragment slide? Assume that both fragments have the same coefficient of kinetic friction.

A 10 g bullet is fired into a 1.0 kg wood block, where it lodges. Subsequently, he block slides 4.0 m across a floor (mk=0.20 for wood on wood). What was the bullet’s speed?

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HomeworkStudent Workbook:

9.149.169.20

Student Textbook:9.259.309.41

一樣，請製作卡片。