## Conservation of Momentum of Systems

To every action there is a reaction that is equal in magnitude and opposite direction ( Newton's third law - see reference 1). Suppose we have two objects O1 (mass m1) and O2 (mass m2) that are moving towards each other along a line on a smooth frictionless surface, they then collide. If F12 (action) is the force exerted by O1 on O2 and F21 (reaction) is the force exerted by O2 on O1, then according to Newton's third law, we write

F12 = - F21 (action and reaction are equal in magnitude and oppsite direction, F12 and F21 are vectors)

F12 = m1 a1 ; a1 acceleration of object O1 and is given by (Δv / Δt)

F21 = m2 a2 ; a2 acceleration of object O2 and is given by (Δv / Δt)

Substitute to obtain

m1 (Δv / Δt)_{O1} = - m2 (Δv / Δt)_{O2}

m1 (Δv)_{O1} = m2 (Δv)_{O1}

Let v1i (initial) and v1f (final) be the speeds of O1 before and after collision respectively and v2i (initial) and v2f (final) be the speeds of O2 before and after collision respectively.

m1 (v1f - v1i) = - m2 (v2f - v2i)

which may also be written as

** m1 v1i + m2 v2i = m1 v1f + m2 v2f
Momentum BEFORE interaction = momentum AFTER interaction **

The above shows that when no external forces are acting on a system, the momenta before and after interaction are equal and therefore there is conservation of momemntum.

### Example 1

A 35 Kg boy jumps (from rest) into a moving trolley of mass 70 Kg and already moving at a velocity of 5 m/s to the right. What is the speed of the trolley after the boy has jumped in ?

__Solution to Example 1__

We have two items: the boy and the trolley.

interaction is when the boy jumps onto the trolley.

Let v1i and v2i be the velocities of the boy and the trolley before the boy jumps in (initial velocities)

v1i = 0 and v2i = 5 m/s

The moment Pi of the system boy-trolley before the jump

pi = moment of boy + moment of trolley = 35 × 0 + 70 × 5 = 350 Kg m/s

Let v be the velocity of the trolley (with the boy in it) , the momentum of the trolley is

p = (35 + 70) v

Conservation of momentum

350 = (35 + 70) v

v = 350 / 105 = 3.3 m/s to the right.

### Example 2

A 35 Kg boy running at a velocity of 2 m/s to the right, jumps onto a trolley at rest of mass 70 Kg. What is the speed of the trolley and the boy after the boy has jumped in?

__Solution to Example 2__

A system of two items: the boy and the trolley and the interaction is when the boy jumps onto the trolley.

Let v1i and v2i be the velocities of the boy and the trolley before the boy jumps in

v1i = 2 and v2i = 0

The moment Pi of the system boy-trolley before the jump

pi = moment of boy + moment of trolley = 35 × 2 + 70 × 0 = 70 Kg m/s

Let v be the velocity of the trolley (with the boy in it), the momentum of the trolley is

p = (35 + 70) v

Momenta are equal before and after intercation: conservation of momenta

70 = (35 + 70) v

v = 70 / 105 = 0.7 m/s to the right.

### Example 3

A 5 Kg gun fires a bullet of 15 grams at a velocity of 1000 m/s to the right. What is the velocity of recoil of the gun?

__Solution to Example 3__

We assume a motion to the right to be the positive.

A system of two objects: the bullet and the gun

Interaction is when the bullet is fired.

Initial velocities: Let v1i and v2i be the velocities of the gun and the bullet before the gun is fired.

v1i = 0 and v2i = 0

The moment Pi of the system (gun and bullet) before firing the gun is

pi = moment of gun + moment of bullet = 5 × 0 + 0.015 × 0 = 0

Final velocities: Let v1f be the velocity of gun and v2f the velocity of the bullet (final velocities), the momentum after the gun is fired is

p = 5 × v1f + 0.015 × 1000

Momenta are equal before and after interaction: conservation of momenta

0 = 5 × v1f + 15

v1f = - 15 / 5 = - 3 m/s to the left (minus sign).

## More References and Links

1 - Higher Level Physics - IB Diploma - Chris Hamper - Pearson

2 - Physics - Raymond A. Serway and Jerry S. Faughn

Holt, Reinehart and Winston - Harcourt Education Company