chapter 7: linear momentum cq: 2 problems: 1, 7, 22, 41, 45, 47. momentum & impulse conservation...
TRANSCRIPT
Chapter 7: Linear Momentum
• CQ: 2
• Problems: 1, 7, 22, 41, 45, 47.
• Momentum & Impulse
• Conservation of Momentum
• Types of Collisions
1
Type of Collision
Kinetic Energy Conserved
Momentum Conserved
Complete Inelastic
X yes
Inelastic X yes
Elastic yes yes
Conserved Quantities
• Energy – has many forms
• Motion – has only one form
• How does Nature conserve motion?
• Speed?
• Velocity?
• Mass x Velocity?
3
4
stops
Conserved ConservedNot
ConservedNot
ConservedNot
Conserved
smooth and level rough
Motion and Its Conservation
5
Momentum
• Momentum = mv
• Symbol: p , P .
• SI Unit: kg·m/s
• Ex. 1000kg car moves at 8m/s.
• mv = (1000kg)(8m/s) = 8,000 kg·m/s
6
Impulse• impulse = Ft
• force x time = change in momentum
• SI Unit: N·s = kg·m/s
• Ex. A 4000N force acts for 0.010s.
• impulse = Ft = (4000N)(0.010s) = 40 N·s
7
Impulse and Momentum
• Impulse = Ft = change-in-momentum
• consequence of Newton’s 2nd law that Ft = change in momentum
• F = ma
• Ft = mat
• Ft = (m)(at)
• Ft = (m)(v)
• N·s = kg·m/s
8
Impulse Example• A braking force of 4000N acts for 0.75s on
a 1000kg car moving at 5.0m/s.
• impulse = Ft = (-4000N)(0.75s) = -3000 N·s
• mv = Ft = -3000 kg·m/s
• v = (-3000 kg·m/s)/1000kg = -3m/s
• vf = vi + v
• = 5m/s + (-3m/s)
• = 2m/s
Collisions
• “Brief” interaction between objects
• Objects share the collision force (N3L)
• One object gets +Ft, the other gets –Ft.
• As a whole, they receive no impulse, thus no change in total momentum of this system (due to the collision)
9
10
11 m
Fa
22 m
Fa
tav 11
tm
Fv
11
tav 22
tm
Fv
22
21 aa
21 vv
|||||| 2211 vmtFvm
1m 2m12 mm
11
1m 2m12 mm
Conservation of Momentum (i.e. Motion):
Initial Momentum (Motion): 02211 iii vmvmP
Final Momentum (Motion): 02211 fff vmvmP
ff vmvm 2211 (Since )
12
Collisions
• elastic: total KE stays same
• inelastic: total KE decreases
• complete inelastic: objects move at same velocity after collision, total KE decreases
13
Conservation of Linear Momentum
ffii vmvmvmvm 22112211
If net-external force = 0 (e.g. level frictionless surface)
14
Ex: 2 objects, complete inelastic
• (m1v1)initial + (m2v2)initial = (m1v1)final + (m2v2)final
• Each car has mass 1000kg
• (1000)(10) + (1000)(0) = (1000)v + (1000)v
• 10,000 = 2000v v = 5 m/s
15
Conservation of Momentum may occur when KE is lost
• Ex: Mass m, Speed v, hits & sticks to another mass m, speed = 0. Final speed of each object is v/2.
• K-initial = ½(m)(v)2 + 0 = ½mv2.
• K-final = ½(m)(v/2)2 + ½(m)(v/2)2 = ¼ mv2.
• Half the original KE converted to heat, sound, etc.
Elastic Collisions
• Approach velocity = Separation velocity
• Equal mass-collisions: exchange velocity
• Ex: straight pool shot
• Ex: car +5m/s bumps car at +3m/s (bumping car +3, bumped +5)
• Ex: (4kg)10m/s (1kg)5m/s
• Result: (4kg)8m/s (1kg)13m/s
16
17
Ex: collision type
• 2kg @ +6m/s hits 1kg @ +3m/s
• After) 2kg @ +4.8ms, 1kg @ ????
• mv-before = mv-after
• (2)(6)+(1)(3) = (2)(4.8)+(1)(v)
• 15 = 9.6 + v v = 5.4
• Is this collision elastic or inelastic?
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fp1
fp2
iP
2 dimensional p conservation
fP
19
Example: m1 = 0.010kg, m2 = 1.0kg, v1i = 200m/s.
• Calculate vf (mom. cons.), Then calculate h (using energy).
fii vmmvmvm )( 212211
fv)01.1()0)(1()200)(010.0( smv f /98.1
ghmmvmm f )()( 212
2121
ghv f 221
mgg
vh f 40.0
98.1 22
Initial momentum
Collision-impulse
p + impulse
Final momentum
1 2
Initial momentum
Collision-impulse
p + impulse
Final momentum
1 2
Initial momentum
Collision-impulse
p + impulse
Final momentum
1 2
Initial momentum
Collision-impulse
p + impulse
Final momentum
1 2
Summary
• momentum = mass x velocity
• Ft = change-in-momentum
• Momentum is conserved when net external forces are negligible
• Momentum conservation may occur for elastic & inelastic collisions
• Elastic: approach & separation vel. equal
24
25
complete inelastic (e.g. 62)
• 2000kg truck vi = +10m/s hits and locks with 1000kg car vi = -4m/s.
• mv-before = mv-after
• (2000)(10) + (1000)(-4) = (3000)v
• 20,000 – 4,000 = 3000v
• 16,000 = 3000v v = 5.33 m/s
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27
28
dropped ball (e.g. 21)
• 1kg ball hits ground at -5m/s and bounces off with +4m/s. Contact time t = 0.33s.
• (Fnet)t = mvf –mvi = m(vf – vi)
• (Fc – mg)t = m(vf – vi)
• (Fc – 9.8)(0.33) = (1)(4 –(-5)) = 9
• (Fc – 9.8) = 27
• Fc = 37 newtons
29
Two masses move on a frictionless horizontal surface. M1 = 1kg, v1i = 4m/s. M2 = 2kg, v2i = 1m/s.
The masses collide along a straight line. Find v1f, if v2f = 2.3 m/s and no other external forces act.
ffii vMvMvMvM 22112211
)/3.2)(2())(1()/1)(2()/4)(1( 1 smkgvkgsmkgsmkg f
smv
vkgsmkg
smkgvkgsmkg
f
f
f
/4.1
))(1(/4.1
/6.4))(1(/6
1
1
1
30
(cont) Calculate the initial and final kinetic energies.
JvmvmK iisysi 9)1)(2()4)(1( 2
212
212
22212
1121
JvmvmK ffsysf 27.6)3.2)(2()4.1)(1( 2
212
212
22212
1121
It is possible for kinetic energy to decrease due to the production of thermal energy in a collision.
In this case 2.73J of Thermal Energy were created by the collision.
31
08-4. A 0.0149kg bullet moves horizontally at 830 feet per second and strikes a 10lb wood block lying at rest on a horizontal surface. The bullet takes 1.0 millisecond to stop inside the block.
a) Convert the data to SI units.
smft
m
s
ft/253
28.3
1830 kg
lb
kglb 55.4
2.2
110
b) Calculate the speed the block moves just after the bullet stops in the block.
sysff
sysi PvsmkgP )0149.055.4()/253)(0149.0(
System momentum conserved when external impulse is negligible.
smv f /825.0
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Calculate the kinetic energy of the bullet before the collision and of the moving block + bullet after the collision. What percent of the original kinetic energy is converted to other energies? What percent is retained as kinetic?
JsmkgK sysi 477)/253)(0149.0( 2
21
JsmkgK sysf 55.1)/825.0)(0149.055.4( 2
21
JKKKU sysi
sysfother 45.47547755.1
%7.99%100477
47.475% converted
%3.0%100477
55.1% retained
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Example
Elastic Collisions
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Elastic Collisions Where One Object is at Rest Before Collision.
if vmm
mmv 1
21
211
if vmm
mv 1
21
12
2
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Elastic Collisions Where One Object is at Rest Before Collision.
0)0.3(121
211
mm
mmv
mm
mmv if
0.3)0.3(2
2
mm
mv f
21 mm smv i /0.31 02 iv
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Elastic Collisions Where One Object is at Rest Before Collision.
0.1)0.3()0.3(22
3
221
2221
2221
121
211
m
m
mm
mmv
mm
mmv if
221
1 mm smv i /0.31 02 iv
67.0)0.3()0.3(22
223
2
2221
221
121
12
m
m
mm
mv
mm
mv if
1a)
• (2000)(10)+(1000)(0) = (3000)vf
• vf = 6.67m/s
• Kf = ½ (3000)(6.67)(6.67) = 66,700J
1b)
• (2000)(10) = (2000)(5) + (1000)(vf)
• 20,000 = 10,000 + 1,000vf
• 10,000 = 1,000vf vf = 10m/s
• Kf-sys = ½ (2000)(5)(5) + ½ (1000)(10)(10)
• 75,000J (Not elastic, but more elastic than previous)
2
• (2000)(5)+(500)(0) = (2500)vf
• vf = 4m/s
• E2 = ½ (2500)(4)(4) = E3 = (fk)(12)
• 20,000 = fk(12)
• fk = 1,667N
• frict.coeff. = fk/FN = 1,667/(500)(9.8)
• = 0.34