10/01/2013PHY 113 C Fall 2013 -- Lecture 101 PHY 113 C General Physics I 11 AM – 12:15 PM MWF Olin...
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Transcript of 10/01/2013PHY 113 C Fall 2013 -- Lecture 101 PHY 113 C General Physics I 11 AM – 12:15 PM MWF Olin...
PHY 113 C Fall 2013 -- Lecture 10 110/01/2013
PHY 113 C General Physics I11 AM – 12:15 PM MWF Olin 101
Plan for Lecture 10
Chapter 9 -- Linear momentum
1. Impulse and momentum
2. Conservation of linear momentum
3. Examples – collision analysis
4. Notion of center of mass
PHY 113 C Fall 2013 -- Lecture 10 210/01/2013
PHY 113 C Fall 2013 -- Lecture 10 310/01/2013
Summary of physics “laws”
fiveconservatinoniiff
fiveconservatinon
fiveconservati
fitotal
iffi
veconservati
if
f
i
fitotal
WUmvUmv
WWW
UUW
mvmvdW
dt
dmm
rr
rr
rF
vaF
22
22
2
1
2
1
:forces veconservatiFor
2
1
2
1
:oremenergy the kinetic-Work
:law second sNewton'
PHY 113 C Fall 2013 -- Lecture 10 410/01/2013
Another way to look at Newton’s second law:
sNm/skg :momentumlinear of Units
:momentum"linear " Define
pv
vvaF
mdt
md
dt
dmm
iclicker question:Why would you want to define linear momentum?
A. To impress your friends.B. To exercise your brain.C. It might be helpful.D. To distinguish it from angular momentum.
PHY 113 C Fall 2013 -- Lecture 10 510/01/2013
if
f
i
f
i
ddt
ddt
dt
d
dt
md
dt
dmm
pppF
pF
pvvaF
:calculus little a Using
Relationship between Newton’s second law and linear momentum:
if
f
i
f
i
dt
dt
ppFI
FI
:impulse"" Define
(if m is constant)
PHY 113 C Fall 2013 -- Lecture 10 610/01/2013
smsmm
I
dtI
m
dt
f
if
f
i
f
i
/20/50
1000
velocity?final its isWhat
rest.at initially kg, 50 mass of
object an toimparted is which Ns 1000
of value thehas figure in thisshown
impulse theSuppose :Example
v
ppF
FI
PHY 113 C Fall 2013 -- Lecture 10 710/01/2013
Suppose that a tennis ball with mass m=0.057 kg approaches a tennis racket at a speed of 45m/s. What is the impulse the racket must exert on the ball to return the ball in the opposite direction at the same speed. Assume that the motion is completely horizontal.
Ns
Ns
t
I
NsmvI if
1.173.0
13.5F
s? 0.3for lastedn interactio theif
racket by the exerted force average theisWhat
13.5)45)(057.0(22
pp
PHY 113 C Fall 2013 -- Lecture 10 810/01/2013
Example: A 1500 kg car collides with a wall, with vi= -15m/s and vf=2.6m/s. What is the impulse exerted on the car?
Ns
Ns
t
I
sm
vvmI ifif
51076.115.0
26400F
car? on the exerted force average
theis what s, 0.15 is timecollision theIf
Ns 26400
/156.21500kg
pp
JvvmmvmvΔE ifif52222 106368.1
2
1
2
1
2
1
collision todue lossEnergy
PHY 113 C Fall 2013 -- Lecture 10 910/01/2013
Example of graphical representation of F(t)
NssNsFdttFI 5.130015.0180002
10015.0
2
1)(
:impulse of Estimate
max
0015.0
001.0
PHY 113 C Fall 2013 -- Lecture 10 1010/01/2013
Physics of composite systems
ii
i
ii
i
ii
iii
ii dt
d
dt
md
dt
dmm p
vvaF
:law second sNewton'
ii
ii
ii
ii
ii
dt
d
final initial
(constant)
0
: then,0 if that Note
pp
p
p
F
PHY 113 C Fall 2013 -- Lecture 10 1110/01/2013
Example – completely inelastic collision; balls moving on a frictionless surface
i
iiv
iviv
vvv
vvv
pp
ˆ 0.125
/5.03.0
ˆ15.0ˆ23.0
ˆ/1 ,ˆ/2
5.0 ,3.0For
21
21
21
2211
212211
final initial
m/s
sm
smsm
kgmkgm
mm
mm
mmmm
f
ii
iif
fii
ii
ii
PHY 113 C Fall 2013 -- Lecture 10 1210/01/2013
Energy loss in this example:
840
15.02
123.0
2
1125.05.03.0
2
1
ˆ 0.125
ˆ/1 ,ˆ/2
5.0 ,3.0For
2
1
2
1
2
1
222
21
21
2
22
2
11
2
21
J .
E
m/s
smsm
kgmkgm
mmmmE
f
ii
iif
iv
iviv
vvv
PHY 113 C Fall 2013 -- Lecture 10 1310/01/2013
Example – completely elastic collision; balls moving on a frictionless surface
iif
iif
ffii
ffii
iii
iii
ii
ii
vmm
mmv
mm
mv
vmm
mv
mm
mmv
vmvmvmvm
mmmm
vmvm
221
121
21
12
221
21
21
211
222
211
222
211
22112211
2final
2initial
final initial
2
2
:algebra someafter motion, d-1For 2
1
2
1
2
1
2
1
2
1
2
1
vvvv
pp
PHY 113 C Fall 2013 -- Lecture 10 1410/01/2013
Completely elastic collision; numerical example:
/25.1/15.03.0
3.05.0
/25.03.0
3.02
/75.1/15.03.0
5.02
/25.03.0
5.03.0
ˆ/1 ,ˆ/2
5.0 ,3.0For
2
2
1
1
21
21
221
121
21
12
221
21
21
211
smsm
smv
smsm
smv
smsm
kgmkgm
vmm
mmv
mm
mv
vmm
mv
mm
mmv
f
f
ii
iif
iif
iviv
PHY 113 C Fall 2013 -- Lecture 10 1510/01/2013
iclicker exercise:We have assumed that there is no net force acting on the system. What happens if there are interaction forces between the particles?
A. Analysis still appliesB. Analysis must be modified
PHY 113 C Fall 2013 -- Lecture 10 1610/01/2013
Example from homework:
?
final initial
Tf
TfTCfCTiTCiC
ii
ii
mmmm
v
vvvv
pp
Ti
T
CfCiCTf m
mv
vvv
PHY 113 C Fall 2013 -- Lecture 10 1710/01/2013
Example from homework: -- continued
2222
2
1
2
1
2
1
2
1
:energy mechanicalin Change
TiTCiCTfTCfCif vmvmvmvmKK
iclicker questionDo you expect Kf-Ki to be
A. >0B. <0
PHY 113 C Fall 2013 -- Lecture 10 1810/01/2013
Another example:
i
ii
i final initial pp
before
vf
after
PHY 113 C Fall 2013 -- Lecture 10 1910/01/2013
Examples of two-dimensional collision; balls moving on a frictionless surface
-- equations more 2 Need
,,, :Unknowns
,, :Knowns
sinsin0
coscos
21
121
2211
221111
final initial
ff
i
ff
ffi
ii
ii
vv
vmm
vmvm
vmvmvm
pp
PHY 113 C Fall 2013 -- Lecture 10 2010/01/2013
Examples of two-dimensional collision; balls moving on a frictionless surface
smsmsm
v
smsmsm
vvv
smsm
vv
vmvm
vmvmvm
smv
smvkgmm
oof
o
fif
o
ff
ff
ffi
f
i
/11.188.17cos
/060.1
88.17sin
/342.0
88.71 060.1
342.0tan
/060.120cos/1/2
coscos
/342.020sin/1
sinsin
sinsin0
coscos
20 ,/1
,/2 ,06.0 :Suppose
1
o
211
21
2211
221111
o2
121
PHY 113 C Fall 2013 -- Lecture 10 2110/01/2013
Example: two-dimensional totally inelastic collision
ji
jiv
vvv
vvv
ˆ/5.12ˆ/375.9
ˆ/204000
2500ˆ/254000
1500
221
21
21
1
212211
smsm
smsm
mm
m
mm
m
mmmm
f
iif
fii
m1=1500kg
m2=2500kg
JΔE
mmmmΔE iif
5
22
22
2
22
2
11
2
21
108.4
2025002
1251500
2
1
5.12375.940002
1
2
1
2
1
2
1
:case in thisenergy kinetic of Loss
vvv
PHY 113 C Fall 2013 -- Lecture 10 2210/01/2013
iclicker exercise:Can this analysis be used to analyze a real collision?
A. Of course! The laws of physics must be obeyed.B. Of course NOT! In physics class we only deal
with idealized situations which never happen.
PHY 113 C Fall 2013 -- Lecture 10 2310/01/2013
Another example of 2-dimensional elastic collision:
m
vi
vf
m0
2
1
2
1
ˆsin2
-- elastic iscollision thecase, In this
22
if
fi
mmE
mv
v
vv
iI
vv
PHY 113 C Fall 2013 -- Lecture 10 2410/01/2013
Energy analysis of a simple nuclear reaction :
HeRnRa 42
22286
22688
Q=4.87 MeV
PHY 113 C Fall 2013 -- Lecture 10 2510/01/2013
Energy analysis of a simple reaction : HeRnRa 42
22286
22688
Q=4.87 MeV
MeV8.498.04/1222/1
4/1
2
4
1
222
1
2222
222
QQm
pE
m
p
m
p
m
pQ
He
HeHe
uHe
He
Rn
Rn
HeRn ppp
PHY 113 C Fall 2013 -- Lecture 10 2610/01/2013
Elastic collision in two dimensions
,,, :Unknowns
,, :Knowns2
1
2
1
2
1
2
1
sinsin0
coscos
21
121
222
211
222
211
2211
221111
final initial
final initial
ff
i
ffii
ff
ffi
ii
ii
ii
ii
vv
vmm
vmvmvmvm
vmvm
vmvmvm
KK
pp
PHY 113 C Fall 2013 -- Lecture 10 2710/01/2013
Elastic collision in two dimensions -- example of elastic proton-proton scattering
,, :Unknowns
37 ;
;/105.3 :Suppose
2
1
2
1
2
1
2
1
sinsin0
coscos
21
21
51
222
211
222
211
2211
221111
final initial
final initial
ff
o
i
ffii
ff
ffi
ii
ii
ii
ii
vv
mmm
smv
vmvmvmvm
vmvm
vmvmvm
KK
pp
PHY 113 C Fall 2013 -- Lecture 10 2810/01/2013
Elastic collision in two dimensions -- example of elastic proton-proton scattering
o
f
f
oi
ffii
ff
ffi
smv
smv
smv
vvvv
vv
vvv
53
/1011.2
/1080.2
:algebra someAfter
37 ;/105.3 :Given
sinsin0
coscos
52
51
51
22
21
22
21
21
211
PHY 113 C Fall 2013 -- Lecture 10 2910/01/2013
The notion of the center of mass and the physics of composite systems
2
2
2
2
2
2
:Define
or
:law second sNewton'
dt
dM
mMM
m
dt
md
dt
dmm
dt
d
dt
md
dt
dmm
CMtotal
ii
ii
iii
CM
i
ii
i
ii
iii
ii
ii
i
ii
i
ii
iii
ii
rFF
rr
rraF
pvv
aF
PHY 113 C Fall 2013 -- Lecture 10 3010/01/2013dt
dM
dt
d
CM
ii
ii
ii
ii
ii
rpp
p
p
FF
final initial
total
(constant)
0
: then,0 If
2
2
:Define
:mass ofcenter for relation General
dt
dM
mMM
m
CMtotal
ii
ii
iii
CM
rFF
rr
PHY 113 C Fall 2013 -- Lecture 10 3110/01/2013
Finding the center of mass
ii
iii
CM mMM
m
rr
ji
jiir
jiir
ˆ00.1ˆ750
4
ˆ22ˆ21ˆ)1)(1(
ˆˆˆ
2 ;1 :example In this
321
332211
321
mm.
mmm
mmm
ymxmxm
kgmkgmm
CM
CM
PHY 113 C Fall 2013 -- Lecture 10 3210/01/2013
Example from webassign :
x
ym1
m2
m3
m4
PHY 113 C Fall 2013 -- Lecture 10 3310/01/2013
Finding the center of massFor a solid object composed of constant density material, the center of mass is located at the center of the object.