02 momentum, impulse and collision

8/16/2019 02 Momentum, Impulse and Collision

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Impulse (J) :

Refers to the product of the force andthe time interval it acts on the body.

J = F(t2 – t1) = F(Δt)where :F – applied constant force ( in

Newtons) Δt – time interval (in seconds) J – Impulse ( in Ns)

F

m m

v

1

v

2

t1 t

2


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Mometum (p) :

Dened as the product of the mass andthe velocity of an object.

p = mvwhere :m – mass (kg)v – velocity (ms) p – momentum (kg!

ms)

F

m m

v

1

v

2

t1 t

2


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Both Momentum & Impulse are vector quantities, thus they have

both horizontal & vertical components. And follow standard signconventions

X - component

px = mvx Jx = Fx(t2 – t1) =

Fx(Δt) Y - component

py = mvy Jy = Fy(t2 – t1) =

Fy(Δt)

p = px2 + py

2 J = Jx2 + Jy

2


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Fm m

v

1

v

2

t1 t2

!el"t#os$#p %etee Mometum & Impulse

J = ΔpFΔt = mv2 – mv1

“The change in momentum at any time intervalequals the impulse of the force applied during thattime interval.”

F'Δt = mv2' – mv1'

FΔt = mv2 – mv1


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2. rob.3)12 4 A bat strikes a 0.15$ kg baseball. ust before impact& the ball istraveling hori6ontally to the right at $0 m%s& and it leaves the bat traveling tothe left at an angle of #07 above the hori6ontal 'ith a speed of 8$m%s. "f theball and bat are in contact for 1.9$ msec& nd the hori6ontal / verticalcomponents of the average force on the ball.

SAMPLE PR!LEMS

v1 = 50 m/s

Fx

Fy

F

30°

v 2 = 6 5 m

/ s

"!component + vectors to the right :!

F'Δt = m(v2' – v1')F' = m(v2' – v1') / Δt

( - 0.15$kg!8(09.m/s)(os;) –

(-.m/s)< % 0.0019$ s!

F' = 0 6 N y!component + vectors going up :!FΔt = m(v2 – v1)

F = m(v2 – v1) / Δt

y - 0.15$kg!8(-9.m/s)(s#;) – < %

0.0019$ s!

F = -26952*5 N


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#. rob.3)1 4a! ;hat is the magnitude of the momentum of a 10&000 kg truck 'hose speed

is 12 m%s<b! ;hat speed 'ould a 2&000 kg =>? have to attain in order to have

i. @he same momentum as the truck!ii. @he same kinetic energy as the truck!

SAMPLE PR!LEMS


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>ONSE!?@TION OF MOMENTUM A When two bodies interact only with each other, their

total momentum is constant. > In an isolated system one where e!ternal forces are

absent" the total momentum will be constant.

Tot"l p%eBo7e = Tot"l p"Bte7m1v1 - m2v2 = m1u1 -

m2u2'here +v initial velocitiesu nal velocitiesm1 mass of object 1

m2 mass of object 2


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1. An open-topped freight car with a mass of 10,000 g is coasting witho!t friction a"ong a "eve" trac. #t

is raining ver$ hard, and the rain is fa""ing vertica""$ downward. %he car is origina""$ empt$ andmoving with a speed of 3 m/s. &hat is the speed of the car after it has trave"ed "ong eno!gh to

co""ect 1,000 g of rainwater'

m1v1 + m2v2 = m1"1+m2"2

Let# m1 = m$%% o& &'e*t c$' = 1,

SAMPLE PR!LEMS

.ven # Re/"'e0 #"c$'

m1 - 10&000kg v1 =

m/s

m2- 1&000

kg u1= 3

m2 = m$%% o& '$n $te' = 1,

m1v1 + m2v2 = m1"1+m2"2

(1, )(m3% )+()( m3%) = (1, )"1+(1, )"2, -m3% = (1, )"1+(1, )"2

By logic u 1 = u 2 , since rain water moving along with the car

, -m3% = (1, )"1+(1, )"1, -m3% = "1 (1,+1, )

, -m3% = "1 (11,)

(, -m3%)311, = "1

24525 m3% = "1

1 2


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mAvA + m!v! = mA"A+m!"!

SAMPLE PR!LEMS

24 .ven # Re/"'e0 #"A

(1)( m3%)+(2)( m3%) = (1)"A+(2)(46 m3%)

"A = 147 m3% to t*e 8e&t

12

A

m A= 1g m( = 2 g

! A

! A = ' !( = 0.)0 m/s

!

= (1)"A+147-m3%

- 147 -m3% = (1)"A(- 147 -m3%)31 = "A

- 147 m3% = "A


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*+A%# #*+A%# ++##


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*+A%# #*+A%# ++##

BC"="CE=

! #e$ned as a kind of interaction %etween two%odies wherein there is a strong interactionthat last for a relatively short time&

! 'sually treated as an isolated system %ecausethe interactive forces are greater than thee"ternal forces (i&e& friction)&


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*+A%# #*+A%# ++##

@FG= C BC"="CE=

1. erfectly Glastic Bollision

A(

A(

v A

!(

! A

v(

Hefore and DuringBollision

After Bollision


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*+A%# #*+A%# ++##

@FG= C BC"="CE=

1. erfectly Glastic Bollision

fter collision the colliding %odies move inseparate direction with respective velocities&

*lastic collisions conserve kinetic energy as wellas total momentum %efore and after collision&

m v + m,v , - m u + m,u,

.elocity relationship in a straight line elasticcollision:

u, – u - v – v ,


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*+A%# #*+A%# ++##

@FG= C BC"="CE=

2. erfectly "nelasticBollision

A(

A(

v A

!

v(

Hefore and DuringBollision

After Bollision


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*+A%# #*+A%# ++##

@FG= C BC"="CE=

2. erfectly "nelasticBollision fter collision the colliding %odies merge and

move in one direction at with a common velocity&

Inelastic collisions don/t conserve kinetic energy%ut total momentum %efore and after collision isconserved& 0inetic energy after collision is lessthan that %efore collision&

m v + m,v , - u (m + m, )


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*amp"es

1 12 g mar%le rolls to the left with a velocity of magnitude 2&3 mson a smooth level surface and makes a head on collision with alarger 42 g mar%le rolling to the right with a velocity of magnitude of2&1 ms& If the collision is perfectly elastic $nd the velocity of eachmar%le after the collision& (5ince the collision is head on all themotion is along a line)&

6iven : 7e8uired :u 9 u,

30g

30g

v A = 0. m/s

v( = 0.1 m/s

!( = '

! A = '

(efore

After

10g

10g


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*amp"es

4& 222 kg automo%ile going eastward on ?rtigas ve at ;2 kmhrcollides with a 3222 kg truck which is going northward across?rtigas ve at 2 kmhr& If they %ecome coupled on collision what isthe magnitude and direction of their velocity immediately aftercollision> (Friction forces %etween the cars 9 the road can %eneglected during the collision)&

6iven : 7e8uired :u

vcar = 50 m/hr

vtr = 20 m/hr

! = '

(efore After


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*+A%# #*+A%# ++##

BoeIcient of RestitutionJ! fractional value representing the ratio of velocities

%efore and after an impact& @his determines the %ounce8uality of an o%Aect practical e"amples are %alls used ingolf 9 tennis&

from *lastic Bollision :u, – u - v – v ,

C - (u, – u ) (v – v , )

C - 1 (for Derfectly *lastic Bollision)C - 2 (for Derfectly Inelastic Bollision)


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*+A%# #*+A%# ++##

BoeIcient of RestitutionJ!If one side is stationary, say the floor or wall.

C - (Eu) (v)

v !

h1

h2

h3

h

"sing bounce heights

C - h h1

C - h4

h

C - h3

h4

02 momentum, impulse and collision

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