Conservation of Momentum

Physics 11

Quick Questions to Discuss with neighbour

If you throw a ball against a wall, which of the three impulses is the greatest: throw? Bounce? Catch?

Why?

How is it possible for an object to obtain a larger impulse from a smaller force than from a larger force?

First a quick reminder…

What is momentum? The product of mass and velocity.

What is the formula for momentum? p = mv

What are the units for momentum? kg m/s

First let’s look back…

What is impulse? Impulse is the product of force and the

time interval during which that force acts.

What is the formula for momentum? j = Ft (don’t use J!!! Joules)

What are the units for momentum? Ns

First let’s look back…

How are impulse and momentum related? Impulse is like momentum except we use

it for sudden changes in momentum, like for collisions, explosions, etc.

Impulse is the change in momentum (not just the momentum)

j = mΔv (change in momentum)

Types of Systems

There are 3 types of systems: Open Closed Isolated

Think back to ecosystems in grade 10 science… what do you remember about these???

What is an open system? Open system = matter and energy can enter

and leave Ex: Earth (meteorites, sun’s energy) Ex: Pot without a lid

What is a closed system? Closed system = matter cannot enter/leave

but energy can We simplify that the Earth is a closed

system in physics 11 as meteorites are very small in relation to the Earth’s mass.

Ex: Pressure cooker or pot with a tight lid Ex: Greenhouse

What is an isolated system? Isolated system = neither matter nor energy

can enter/leave Ex: Pot in an insulator (like a cooler)

Conservation of Momentum

In an isolated system, momentum does not change.

This means that if two objects collide, the total momentum before the collision of the 2 objects is the same as the total momentum after the collision.

Law of Conservation of Momentum *copy*

Within an isolated system, the net change in momentum is zero.

OR The final momentum of the system equals the

initial momentum of the system:

Formula: pA + pB = p’A + p’B

Formula: p = p’

Isolated system: neither matter nor energy can enter/leave the system

For our purposes (copy)

We will consider closed, isolated systems as Closed - no objects enter or leave the

system. Isolated - no net external force is exerted

on it.

Example Two freight cars each have a mass of 3.0 x

105 kg. Car B is moving at +2.2 m/s and car A is at rest.

When the two cars collide, they act as one (“couple”) and move away together. Find the final velocities of A and B.

Assume an isolated system (no external forces).

Answer pA + pB = p’A + p’B

mAvA + mBvB = m’Av’A + m’Bv’B

(300000)(0) + (300000)(2.2) = (300000)(v’A) + (300000)(v’A)

66000 = 600000(v’A) v’A= +1.1 m/s

They both travel at 1.1 m/s in the same direction as car B.

Example 2

You are playing pool with a friend. The 0.17 kg cue ball hits a stationary numbered ball (mass 0.15kg) with a velocity of 3.2 m/s. Assuming the cue ball stops once it hits the numbered ball and the other ball moves in the direction of the hit, what is the velocity of the numbered ball? (Assume isolated system)

Answer pA + pB = p’A + p’B

mAvA + mBvB = m’Av’A + m’Bv’B

(0.17)(3.2) + (0.15)(0) = (0.17)(0) + (0.15)(v’B)

0.544 = 0.15(v’B) v’B= +3.6 m/s

The numbered ball moves at a velocity of 3.6m/s. This makes sense as it has less mass so the same momentum will make it move faster.

Recoil Recoil is the interaction that occurs when

two stationary objects push against each other and then move apart.

This means there is no initial momentum and therefore no final momentum… HOW IS THIS POSSIBLE???

Examples: Guns cause recoil

Recoil (copy)

when two stationary objects push against each other and then move apart

no initial momentum (vi = 0)

Examples: revolver, cannon

Example 3 - Recoil

A Winchester .308 cartridge launches a bullet of mass 64.8 mg. The rifle has a mass of 3.8kg. What is the final velocity of the cartridge assuming the velocity of the gun’s recoil is -2.2m/s?

Answer pA + pB = p’A + p’B

mAvA + mBvB = m’Av’A + m’Bv’B

(0.0000648)(0) + (3.8)(0) = (0.0000648)(v’A) + (3.8)(-2.2)

0 = 0.0000648(v’A) – 8.36 8.36 = 0.0000648(v’A) v’A= +129012 m/s = +1.3 x 105 m/s

The bullet moves at a velocity of +1.3 x 105 m/s. This makes sense as it has less mass so the same momentum will make it move faster.

Practice

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