Elastic and Inelastic Collisions

Unit A: Momentum

Elasticity of Collisions

- In 1666, Newton demonstrated his colliding pendula to the Royal Society of London.

- When one hardwood ball was released from a certain height to strike a second stationary ball, the moving ball stopped dead and the second rose to the height of release! (… and back and forth!)

mA = 1

mB = 1

- Momentum is conserved:

*

'BB

'AABBAA vmvmvmvm

- If and , then and 1vA 0vB

0v'A

1v'B

(1)(1) + (1)(0) = (1)(0) + (1)(1)

1 + 0 = 0 + 1

1 = 1

- But the law is also satisfied if:

a)  and 5.0v'A

5.0v'B

(1)(1) + (1)(0) = (1)(0.5) + (1)(0.5)

1 + 0 = 0.5 + 0.5

1 = 1

b) or if and 2.0v'A

8.0v'B

(1)(1) + (1)(0) = (1)(0.2) + (1)(0.8)

1 + 0 = 0.2 + 0.8

1 = 1

*

- Thus, there are an infinite number of possible outcomes that would satisfy the Law of Conservation of Momentum but only one outcome ever happens!

- How do the balls “know” what their final velocities should be?

- In 1668, Christian Huygens had the answer:

- When hard spheres collide, another quantity, mv2, the “vis viva” (living force) was conserved as well as momentum Thus, there are an infinite number of possible outcomes that would satisfy the Law of Conservation of Momentum but only one outcome ever happens!

- Today we call the “vis viva” kinetic energy:

221

k mvE

*

- Huygens’ two types of collisions:

1. Elastic Collisions

2. Inelastic Collisions

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Collision Spectrum

Perfectly

Inelastic

Somewhat

Inelastic

Somewhat

Elastic

Perfectly

Elastic

No Ek

conserved

egg hits floor

bullet imbeds in tree

Some Ek

conserved

squash ball hit by racquet

“fender bender”

Most Ek

conserved

golf ball hit by club

basketball is dribbled

All Ek

conserved

two electrons collide

N2(g) molecule

hits O2(g)

molecule

*

Elasticity of Collisions

Example 1:

Am = 0.532 kg

Av = 3.41 m/s R

Bm = 0.350 kg

= 0 Bv

= 0.914 m/s R 'Av

= 3.72 m/s R 'Bv

*

Before After

A B BA

Compare Ek to Ek:

2BB2

12AA2

1k vmvmE

2212

sm

21

k 0kg350.041.3kg532.0E

J09.3Ek

2

BB212

AA21'

k vmvmE

2

sm

212

sm

21'

k 72.3kg350.0914.0kg532.0E

J64.2J42.2J222.0E '

k

100E

EEE%

k

kklostk

%10009.3

64.209.3E%

lostk

klostk Eofloss%5.14E%

Collision is somewhat elastic!

So 15% Ek was converted to Ep!

*

Example 2:

Before After

B BA A

= 0.350 kg = 0.532 kg = - 0.472 m/s 2.49 m/s Bm Am 'Bv '

Av

= 3.32 m/s E vA = 0Bv

a) Compare to . totalp

total'p

BBAAtotal vmvmp

sm

total 32.3kg350.00kg532.0p

sm

total kg16.1p

*

BBAAtotal' vmvmp

sm

sm

total' 472.0kg350.049.2kg532.0p

sm

total' kg16.1p

%217.0p% lost

b) Compare Ek to Ek:

2vm2vmkE BB21

AA21

2sm32.3kg350.020kg532.0kE 2

121

J93.1kE

*

2vm212

vm21'

kE BBAA

2sm472.0kg350.02

12

sm49.2kg532.02

1'kE

J69.1'kE

%5.12%10093.1

69.193.1E%

lostk

Collision is somewhat elastic!

*

Example 3:

Before After

B A AB

mB = 0.443 kg mA = 0.662 kg stuck together

VB= 0 VA = 4.11 m/s V A+B = 2.48 m/s 

a) Compare to .totalp

total'p

BBAA vmvmtotalp

0kg443.0sm11.4kg662.0totalp

smkg72.2totalp

*

BBAA vmvmtotal'p

sm48.2kg443.0

sm48.2kg662.0total'p

smkg74.2total'p

%720.0%10072.2

)74.2()72.2(lostp%

experimental error!

b) Compare Ek to Ek:

2vm212vm2

1kE BBAA

20kg443.021

2

sm11.4kg662.02

1kE

J59.5kE

*

2vBmAm21or2vm2

12vm21'

kE BBAA

2sm48.2kg443.02

12

sm48.2kg662.02

1'kE

J40.3'kE

%2.39%10059.5

40.359.5E%

lostk

Collision is not very elastic!

*

Elastic Collision Problems

(assume)

Example 1:Before:

mA = 4.0 kg A = +6.0 m/s

mB = 2.0 kg B = 0

BA

After Elastic Collision:

?vA ?vB

Using Conservation of Momentum:

BBAABBAA vmvmvmvm

BA vkg0.2vkg0.40kg0.2sm0.6kg0.4

BA v2v4024

AB v424v2

AB v212v

( isolate a variable ) Equation #1 *

v

v

Using Conservation of Kinetic Energy:

2BB2

12AA2

12BB2

12AA2

1 vmvmvmvm

Multiply by 2 and substitute:

2vkg0.2

2vkg0.4

20kg0.2

2

sm0.6kg0.4 BA

2v2

2v40144 BA

2v

2v272 BA

*

Equation #2

Substitute for from Conservation of Momentum: Bv

Sub Equation #1 into Equation #2

2A

2A v212v272

AA2

A v212v212v272

2AA

2A v4v48144v272

072v48v6 A2

A

012v8v A2

A

Ax2 + Bx + C = 0

A quadratic, so factor:

6v A 2v A ( ) ( ) = 0

*Or if you’re CHICKEN!

a2ac4bb

x2

12

121488v

2

A

2or624

or2

122

48vA

extraneous root

Since was 6.0 m/s, = +2.0  !Av

Avsm

So = 12 – 2    =  12–2(2)  =  +8.0 Bv Avsm

substitute into equation #1

*