Chapter 10

Collisions

27/19/04

Review

Momentum:

vmp

If Fext = 0, then momentum does not change

fi pp

f

iexif M

M+v=vv ln

For continuous momentum transfer (Rockets):

37/19/04

Rockets: Continuous Momentum Transfer

47/19/04

Momentum in a Collision

Total momentum is conserved

In a collision, objects only exert forces on each other, so Fext=0.

57/19/04

Impulse

During a collision, the momentum on an object changes

This change in momentum is called “Impulse”

When objects A and B collide

if pppJ

BA JJ

67/19/04

Impulse

Recall:

t

pF

tFpJ

In the limit of small t:

f

i

f

i

t

t

p

pdtFpdJ

(constant force)

(changing force)

77/19/04

Impulse in a Collision

Different collisions with the same total impulse:

Blue

Red

Large F: p changes rapidly

Small F: p changes slowly

p/t Large

p/t Small Mo

men

tum

t

pF

87/19/04

Example: The Impulsive Spiderman

Spiderman, who has a mass of 70 kg, jumps from a train 5 meters high moving at 20 m/s (about 40 mph).

He lands standing up, taking t = 0.1 s to stop himself after making contact with the ground. How much force did his knees feel?

97/19/04

ExampleTreat as collision between Spiderman and the ground

Initial: p = mvtotal

Final: p = 0

Get force from the impulse:

t

mv

t

pF total

tFpJ

107/19/04

Example

t

mv

t

pF total

ghv

mghmv

KU

y

y

finalinitial

22

2

21

Need to find vy:

sm

msmsm

vvv yxtotal

/3.22

)5)(/8.9(2)/20( 22

22

Ns

smkgF 000,16

1.0

)/3.22)(70(

If he wasn’t a superhero, he’d break his legs!

117/19/04

What if he rolls on landing for t = 2 sec?

Much easier on the knees!

Example

Ns

smkgF 780

2

)/3.22)(70(

127/19/04

Cannon RecoilCannon: mc=1134 kg

Ball: mb=13.6 kg

Ball shot at ~ speed of sound vb = 340 m/s

The cannon and ball are initially at rest:

ballcannon

cannonball

fi

pp

pp

pp

0

0

pball = mballvball = (13.6kg)(340 m/s) = 4620 kg m/s

So, pcannon= -4620 kg m/s

137/19/04

Cannon Recoil

Cannon recoil stopped in ~2 s by ropes. What is the tension in the ropes?

A rope can easily handle this much force withoutbreaking

TT

pc

t

pTFnet

2

lbsNs

smkg

t

pT 260~1160

2

/4620

2

1

2

1

147/19/04

Momentum Conservation in Different Frames

m mv -v

Simple 1D problem

PTOT = mv - mv =0

2m

Sticktogether v=0

157/19/04

Momentum Conservation in Different Frames

m m2v

Same 1D problem viewed from right hand block, or with right hand block at rest

PTOT = 2mv + 0 = 2mv

2 m v

167/19/04

Changes in Momentum Independent of Frame

Left

Right

Case 1 Case 2

i f i f

mv

-mv

0

0

2mv

0

mv

mv

PTf – PTi = 0 – 0 = 0 PTf – PTi = 2mv – 2mv = 0

177/19/04

Center of Momentum Frame

There is always a frame of reference where PTOT=0.

‘Center of mass’ frame

187/19/04

A Limitation of Momentum

V=30 MPH

Before

V=0

m/s kg,

m/s. kg,

MPH lbsp

= m/s kg

MPH) lbsp

truck,i

car,i

29221

)413)(5891(

)30)(3500(

0)0)(681(

)0)(1500(

m/s kg,

ppp truck,icar,itotal,i

29221

truck,fcar,ftotal,f p pp

How do we determine

the velocities?

After

vT vc

BOOM!

197/19/04

A Limitation of MomentumConstant truck,fcar,ftotal,f ppp

truck,ftotal,f car,f ppp

There are many possibilities

Conservation of Momentum can’t tell them apart

ptruck

p ca

r

207/19/04

Elastic Collisions

Momentum and kinetic energy are conserved

Two equations:

Good approximation for a lot of collisions, and exact for some

Examples: Billiard Balls, superball on floor…

2,222

12,112

12,222

12,112

1

,22,11,22,11

ffii

ffii

vmvmvmvm

vmvmvmvm

217/19/04

Elastic Collisions in One Dimension

Two conservation laws

Momentum

Energy

(Elastic only - Mechanical Energy is conserved)

m1 m2

Before

V1,i V2,i

m1 m2

After

V1,f V2,f

(Always)

2,222

12,112

12,222

12,112

1ffii vmvmvmvm

ffii vmvmvmvm ,22,11,22,11

227/19/04

A Unique Solution

,i,i,f

,i,i,f

vmm

mmv

mm

mv

vmm

mv

mm

mmv

221

121

21

12

221

21

21

211

2

2

We now have two equations and two unknowns:

Lots of Algebra

2,222

12,112

12,222

12,112

1ffii vmvmvmvm

ffii vmvmvmvm ,22,11,22,11

237/19/04

Limiting Cases

vv

vv

,i,f

,i,f

12

21

m1 = m2

How do we understand what types of motion

these predict?

Consider limiting case:

The two objects simply trade values of velocity!

,i,i,f

,i,i,f

vmm

m mv

mm

mv

vmm

mv

mm

mmv

221

121

21

12

221

21

21

211

2

2

247/19/04

Limiting Cases

What if m1 >> m2?

,i,i,f

,i,f

vvv

vv

212

11

2

Semi truck hits a parked VW bug: Truck keeps going

Bug bounces off with twice truck’s speed!

,i,i,f

,i,i,f

vmm

m mv

mm

mv

vmm

mv

mm

mmv

221

121

21

12

221

21

21

211

2

2

7/19/04

Demonstration

m1>>m2

0

0

vv

vv

tennis,i

basket,i

A Question:

WhatHappens?Before:

After:

0

0

3vv

vv

tennis,f

basket,f

,i,i,f

,i,f

vvv

vv

212

11

2

26

The Slingshot Effect

km/s.

km/s km/s.

vvv probe,ijupiterprobe,f

229

)10()69(2

2

-10 km/s

9.6 km/s

277/19/04

Car-Truck Crash

A 2000 kg car has a head-on collision with a 10,000 kg truck. They each are travelling at 10 m/s and they collide elastically (solid bumpers!).

m1 m2v1i v2i

What are their final velocities?

Choose positive x direction +x

287/19/04

Car-Truck Crash (continued)

m1 m2v1i v2i

v1i = 10 m/s

v2i = -10 m/s

m1 = 2,000 kg

m2 = 10,000 kg

v1f = -23.3 m/sv2f = -3.33 m/s

Truck slows down Car goes flying backwards!

,i,i,f

,i,i,f

vmm

mmv

mm

mv

vmm

mv

mm

mmv

221

121

21

12

221

21

21

211

2

2

297/19/04

Car-Truck Crash (continued)

If the two vehicles are being driven by 60 kg PSU students, what are the impulses they feel?

In truck: J = p = mv = m(v2f - v2i)

= 60(-3.33 – (-10))

= 400 kg m/s

= m(v1f – v1i)

= 60(-23.3 – (10))

= -2000 kg m/s

In car: J = p = mv

307/19/04

Car-Truck (question)

Which would you rather be driving?

Say collision lasts Δt = 0.2 seconds

Force on student is given by F = Δp/Δt

Student in truck feels 2,000 N (survivable)

Student in car feels 10,000 N (not good)

What if instead of a 2000 kg car, she was on a 500 kg motorcycle!

317/19/04

Example: 2-D Elastic Collision

Two billiard balls collide elastically on a table. The initial velocity of the first ball is v1,i=(1 m/s)i+(2 m/s)j. The second ball is initially at rest. Both balls have the same mass. Determine the final velocity of both after the collision.

v1,i=(1 m/s)i+(2 m/s)j

327/19/04

Inelastic Collisions

Momentum is conserved (NOT Kinetic Energy)

Examples: Spit wads, football player being tackled,…

fii vmmvmvm

)( 21,22,11

Two objects stick together

Completely Inelastic:

337/19/04

Inelastic Collisions…

http://www.baylortv.com/streaming/000026/300kbps_ref.mov

347/19/04

Car Crash

v1=20 m/s v2=30 m/s

m1=750 kg m2=1000 kg

Two cars collide and stick together after the collision. What is the final velocity of the system?

357/19/04

Car Crash

v1=20 m/s v2=30 m/s

m1=750 kg m2=1000 kg

Using conservation of momentum:

smv

kgkg

smkgsmkg

mm

vmvmv

vmmvmvm

pp

f

iif

fii

fi

/6.8

)1000()750(

)/30)(1000()/20)(750(

)(

21

,22,11

21,22,11

367/19/04

Basketball Cannon

A ball projected from a cannon hits the trash can such that:

1) It sticks into the trash can.2) It hits the trash can and bounces back.

Will the velocity of the trash can be bigger for case 1, case 2, or exactly the same?

377/19/04

Basketball Cannon

v

m

vtrash=0

M

Consider an elastic collision:

vmM

mv

mM

mMv trash,itrash,f

2

mM

mvvtrash,f

2

387/19/04

Basketball Cannon

v

m

vtrash=0

M

Consider a perfectly inelastic collision:

fvmMmv )(

mM

mvv f

397/19/04

Basketball Cannon

Elastic:

mM

mvv f

2

Inelastic:

mM

mvv f

Elastic collision results in twice the velocity!