04/19/23 PHY 113 -- Lecture 10 1

Announcements

1. Test corrections due today.

2. Today’s topics –

Conservation of momentum

Notion of impulse

Analysis of collisions

Elastic

Inelastic

04/19/23 PHY 113 -- Lecture 10 2

dtdp

F :law second sNewton'

Linear momentum: p = mv

Generalization for a composite system:

dt

dM

dt

d comnet

i

i

ii

vF

pF

)constant(;0;0If i

ii

i

ii dt

dp

pF

Conservation of momentum:

Energy may also be conserved (“elastic” collision)

or not conserve (“inelastic” collision)

04/19/23 PHY 113 -- Lecture 10 3

Snapshot of a collision:

Pi

Pf

Impulse:

dttddtd

t )( )( Fpp

F

2

1

2

1

)(t

t

t

t

dttd JFp

04/19/23 PHY 113 -- Lecture 10 4

Example from HW:

2

1

2

1

)(t

t

if

t

t

dttd JFppp

Plot shows model of force exerted on ball hitting a wall as a function of time. Given information about pi and pf , you are asked to determine Fmax.

04/19/23 PHY 113 -- Lecture 10 5

Notion of impulse:

Ft = pdtdp

F

Example:

vi

vf

mi

f

p = mvf - mvi

p=(-mvf sin f - mvi sin i) i

+(-mvf cos f + mvi cos i) j

pf-pi

04/19/23 PHY 113 -- Lecture 10 6

Notion of impulse:

dtdp

F

Example:

vi

vf

mi

f

p = mvf - mvi

p=(-mvf sin f - mvi sin i) i

+(-mvf cos f + mvi cos i) j

pf-pi

p = pf - pi = F t

cause effect

04/19/23 PHY 113 -- Lecture 10 7

Example: Suppose that you (mg= 500 N) jump down a distance of 1m, landing with stiff legs during a time t=0.1s. What is the average force exerted by the floor on your bones?

tp

F

Ns

sNtp

F

sNsNgh

mgmvp

mghmv

floor

ifloor

22601.09.225

9.2258.912

5002

0

221

04/19/23 PHY 113 -- Lecture 10 8

04/19/23 PHY 113 -- Lecture 10 9

One dimensional case:

Conservation of momentum: m1v1i + m2v2i = m1v1f+m2v2f

If energy (kinetic) is conserved, then:

½ m1v1i2 + ½ m2v2i

2 = ½ m1v1f2+ ½ m2v2f

2

Extra credit: Show that

iif

iif

vmmmm

vmm

mv

vmm

mv

mmmm

v

221

121

21

12

221

21

21

211

2

2

Analysis of before and after a collision:

04/19/23 PHY 113 -- Lecture 10 10

More examples in one dimension:

vibefore

vf

totally inelastic collision

Conservation of momentum: m1vi = (m1+m2) vf

if vmm

mv

21

1

Energy lost:

21

22

121

2

21212

121

mmm

vm

vmmvmE

i

fi

04/19/23 PHY 113 -- Lecture 10 11

Statement of conservation of momentum:

inφinθ0

φcosθcos

2211

221111

svmsvm

vmvmvm

ff

ffi

If mechanical (kinetic) energy is conserved, then:

4 unknowns, 3 equations -- need more information

2222

12112

12112

1ffi vmvmvm

Elastic collision in 2-dimensions

04/19/23 PHY 113 -- Lecture 10 12

Example:

Detector measures v2f ,

φvmθvm

φvmθvmvm

ff

ffi

sinsin0

coscos

2211

221111

Conservation of momentum

cossin1

cos

sintan

coscos

sinsin

2

2211

2

22

1

1

2211

22

221111

2211

φvmvmφvmm

vφvmvm

φvm

φvmvmθvm

φvm θvm

fiff

fi

f

fif

ff

Note: energy is not conserved in this case.

04/19/23 PHY 113 -- Lecture 10 13

Special case: m1 = m2

222111

22211111

sinsin0

coscos

θvmθvm

θvmθvmvm

ff

ffi

2

2

2

1

2

1

2

22212

11212

1121

ffiffi vvvvmvmvm

Elastic collision

v1f v2f

v1i

04/19/23 PHY 113 -- Lecture 10 14

Other examples:

Suppose you are given h1, m1, m2 -- find h2 (assume elastic collision)

h2