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?

18

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

26

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

32

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

33

Example

Elastic Collisions

34

Elastic Collisions Where One Object is at Rest Before Collision.

if vmm

mmv 1

21

211

if vmm

mv 1

21

12

2

35

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

36

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