chapter 8: momentum conservation
DESCRIPTION
Chapter 8: Momentum Conservation. Impulse. Work. Distance, l. K = (1/2) m v 2 Work-Energy Theorem Energy Conservation. p = m v Impulse-Momentum Theorem Momentum Conservation. 1D Collision. M. m. M. m. Elastic Collision. Energy Conservation. Loss of energy as thermal and - PowerPoint PPT PresentationTRANSCRIPT
Momentum Conservation
Chapter 8: Momentum Conservation
K = (1/2) m v2
Work-Energy TheoremEnergy Conservation
p = m vImpulse-Momentum TheoremMomentum Conservation
WorkImpulse
Distance, l
Momentum Conservation
Momentum Conservation
1D Collision
M
m
m M
Momentum Conservation
Elastic Collision
2 '22
2 '11
222
211 2
1
2
1
2
1
2
1 :Energy Kinetic vmvmvmvm
'22
'112211 :Momentum vmvmvmvm
Momentum Conservation
Energy Conservation
f2,f1,2,i1,i K KKK
QK KKK f2,f1,2,i1,i
Loss of energy as thermal andother forms of energy
Momentum Conservation
Example 2
m v1 + m v2 = m v1’ + m v2’
Before collision After collision
v1’ = v2’
(totally inelastic collision)
Momentum Conservation
Momentum Conservation
Impulsive Force
Impulsive Force
Ver
y la
rge
mag
nit
ud
e
Very short time
[Example] an impulsive force ona baseball that is struck with a bathas:
<F> ~ 5000 N & t ~ 0.01 s
[Note] The “impulse’’ conceptis most useful for impulsiveforces.
Momentum Conservation
Impulse-Momentum Theorem
f
i
f
i
p
p
t
t
if p -p ptFJ
p t F
t
p
t
m
t m a m F
d (1)d
dd
d
d
d
)d(
d
d vv
ifif
if
if
p -pt -t F
t -t
p -p
t
p F
)(
|J |
F
)(tF
1D2D
Momentum Conservation
Momentum Conservation
Momentum Conservation)(tF2 x
)(1 tF x
2x,if2x,2x p -pJ
1x,if1x,1x p -pJ
x
y
)()( tt 2x1x F - F 2x1x J-J
f2x,f1x,2x,i1x,i p ppp
Momentum Conservation
Example 3(A
) M
omen
tum
Con
serv
atio
n
(B) Energy Conservation
(A) mv = (m+M) v’(B) K1+Ug1 = K2+Ug2
Express v and v’ in terms ofm, M, g, and h.
1
2
Momentum Conservation
Momentum Conservation
Momentum Conservation
Momentum Conservation
Momentum Conservation
Momentum Conservation
Momentum Conservation
Momentum Conservation
Example 1
vi = 28 m/s
vf = 28 m/s
What is the impulse given thewall ? Note: m = 0.060 kg.px, and py for the ballJ(on the wall) = - J(on the ball)
px,i
px,f
py,f
x
y
(1) Coordinates(2) J(on the ball)
px = px,f - px,i = - 2 x px,i py = py,f - py,i = 0
where: px,i = m vi sin 1.2 N*s
py,i
Momentum Conservation
1D/2D “Explosion’’
1 2 (or more)
Momentum Conservation
Center of mass
Center of Mass (c.m. or CM)
The overall motion of a mechanical system can be described in terms of a special point called “center of mass” of the system:
system. on the exerted forces
theall of sum vector theiswhere F
a M F
system
cmsystemsystem
Momentum Conservation
Momentum Conservation
Momentum Conservation
CM Position (2D)
m1 m2 + m3
m1 + m2
m3
X
Xycm = 0.50 m
xcm = 1.33 m
Momentum Conservation
CM Position and Velocity
m v1 + m v2 = (m + m) v’(totally inelastic collision)
m m
m m
m m
m m
21
2211cm
21
2211cm
vv
vrr
r
;
m m
m m
m m
m m
21
21cm
m/s 12
m 24
vvv
xxx
xcm,
48.0 m
x
m m
m m
m m
m m
21
21
21cm
m/s 12
m 0
vvv
xxx
xcm,
t = -2 s
t = 0 s
Momentum Conservation
2D Collision