Download - 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