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UNIT 1: CONSERVATION LAWS

Lesson 8:

Conservation of Momentum (1-D)

CENTRE HIGH: PHYSICS 30

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Recommended Reading:

Heath pp. 244 - 248

Ladner pp. 43 - 44

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Momentum is conserved for collisions and explosions

Collisions:

STICK!

Explosions:

BOOM!

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F. Law of Conservation of Momentum

1 2

When two objects collide:

3rd Law: They exert equal and opposite forces

for the same period of time

So,

Fon 1 = - Fon 2

Fon 1 t = - Fon 2 t

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Fon 1 = - Fon 2

Fon 1 t = - Fon 2 t

Based on the equation F t = m v, it follows

m v1 = - m v2

p1

= - p2

or

p1 + p2 = 0

No overall change in momentum

(conserved)

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F1. Conservation of Momentum (1-D)

For collisions and explosions, there are two equations:

1. Total Momentum is conserved (constant)

pT = pT '

- the total momentum before the collision

is the same as the total momentum after the collision

- this is also true for explosions

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F1. Conservation of Momentum (1-D)

For collisions and explosions, there are two equations:

1. Total Momentum is conserved (constant)

pT = pT '

These are vector sums

Need a reference direction

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2. No overall change in momentum

p1 + p2 = 0

or

p1 = - p2

Equal but opposite impulses

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Conditions for momentum to be conserved :

1. A closed system

- no objects lost / gained

2. Fnet = 0 on the whole system

- the objects must move at a constant velocity

before and after the collision

e.g.

No friction

Level surface

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Consider a collision:

10 kgm/s 6 kgm/s 3 kgm/s p?

Boom!

What is the total momentum before the collision?

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Ref: Right is positive

10 kgm/s -6 kgm/s -3 kgm/s p?

Boom!

Remember, for 1-D vectors,

you must determine a positive direction (i.e. a ref system)

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Boom!

pT = 10 - 6 = 4 kgm/s

What is the total momentum after the collision?

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Boom!

pT = 10 - 6 = 4 kgm/s pT ' = 4 kgm/s

pT = pT ' = 4 kg m/s

Total momentum remains constant throughout the collision!

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10 kgm/s -6 kgm/s -3 kgm/s

Boom!

pT = 10 - 6 = 4 kgm/s pT ' = 4 kgm/s

So, what is the momentum of the blue object after the collision?

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10 kgm/s -6 kgm/s -3 kgm/s 7 kgm/s

Boom!

pT = 10 - 6 = 4 kgm/s pT ' = 4 kgm/s

So, p = +7 kgm/s

The momentum of the two objects must add up to 4 kgm/s

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Boom!

Calculate the change in momentum for object 1 and 2

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Boom!

Also,

p1 = p1 ' - p1 = (-3 kgm/s) - (10 kgm/s)

= -13 kgm/s

p2 = p2 ' - p2 = (7 kgm/s) - (-6 kgm/s)

= +13 kgm/s

Equal but opposite impulses (no overall change)

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Animation:

Air Track collisionshttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassM

echanics/AirTrack/AirTrack.html

Attempt to predict the answers before you run the animation- use the equations to calculate

Be aware that an inelastic collision is when they stick

together
http://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.htmlhttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.htmlhttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.htmlhttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.html

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Ex. 1 A 12.5 kg object, moving East at 4.0 m/s, collides with

a 13.0 kg object moving West at 7.0 m/s. If they stick

together (an inelastic collision), then the magnitude oftheir velocity is _________ m/s.

Your 2-digit answer is

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Solution:

4.0 m/s 7.0 m/s v ?

Stick!12.5 kg 13.0 kg 25.5 kg

Draw a picture first.

Show clearly before and after the collision.

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Solution:

+4.0 m/s -7.0 m/s v ?

Stick!12.5 kg 13.0 kg 25.5 kg

Ref: East is +

pT = pT'

p1 + p2 = pT'

m1 v1 + m2 v2 = mT v

The two masses combine

Treat as one (total) mass

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m1 v1 + m2 v2 = mT v

(12.5 (4.0) + (13.0) (-7.0) = 25.5 v

-41 = 25.5 v

v = -1.6 m/s

So, v = 1.6 m/s West

Numerical Response:

1 . 6

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Ex. 2 A 750 kg car is moving East and collides with a 580 kg car

moving West at 20.00 m/s. After the collision, the 580 kg car is

moving East at 5.00 m/s.

The magnitude of the impulse on Car 1 is b x 10w N s.

Your 3-digit value for b is

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Solution:

Recall, for momentum to be conserved,the impulses have to be equal and opposite

i.e.

Both cars experience the same magnitude of impulse,but in opposite directions

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Ref: East is positive

750 kg 580 kg 1 2

-20.00 m/s Boom! +5.00 m/s

Find impulse on Car 2:

m2 v2 = m2 (v2' - v2)

= 580 kg [ (+5.00 m/s) - (-20.00 m/s) ]

= +1.45 x 104 kg m/s (East)

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The cars will experience equal but opposite impulses

(Cons of momentum)

Thus, Impulse on Car 1 = -1.45 x 104 kg m/s

= 1.45 x 104 Ns West

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Numerical Response format (b 10w)

1.45 x 10

4

w

b

So, the 3-digit value for b is

1 . 4 5

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Practice Problems:

TryLadner p. 46 #1 - 12

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SUMMARY

1. Conservation of momentum is based on which law?

2. What are the two equations for Cons of p?

3. What conditions are required for Cons of p?

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SUMMARY

1. Conservation of momentum is based on which law?

Newton's 3rd Law- equal and opposite forces

- so, equal and opposite impulses

2. What are the two equations for Cons of p?Total p is constant

pT = pT'

No overall change in momentum

p1 + p2 = 0

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3. What conditions are required for Cons of p?

Closed system - no objects lost or gained

Fnet = 0 on system- objects have constant velocity before and after

the collision / explosion

- no friction / level surface

4. Review Animations:

Air Track collisions

2 Block collisions
http://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.htmlhttp://www.geocities.com/thesciencefiles/collisions/collisions.htmlhttp://www.geocities.com/thesciencefiles/collisions/collisions.htmlhttp://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.html

8. cons of p (1-d)

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