Collisions

newtons-cradle

A collision is when two objects impact each other over a short space of time.

The momentum of each object can change, but the total momentum does not. We say the momentum is conserved (the total stays the same).

Momentum is Conserved

Conserved: the total stays the same (within a closed system).

closed system

Closed System: nothing transfers in or out, and no external force acts on it.

In our Universe:

Momentum is conserved (in a closed system)

also

Mass is conserved in everyday situations (in chemistry and daily life mass doesn't just appear or disappear) and
Energy is conserved in everyday situations (it changes forms, but the total amount stays the same)

The Bigger Picture: Albert Einstein taught us that Mass and Energy are actually related via E = mc². In nuclear reactions, a tiny bit of mass can turn into huge energy! So, the ultimate rule is that Total Mass-Energy is always conserved.

Momentum and Kinetic Energy

Momentum (p) is calculated using mass times velocity:

p = m v

With:

m as mass (in kg)
v as velocity (in m/s)

Kinetic Energy (KE) is the energy of motion:

KE = ½ m v²

1kg ball at 20m/s vs 10kg ball at 2m/s

Example: A 1kg ball travels at 20 m/s, and a 10 kg ball travels at 2 m/s.

What is the momentum and KE of each?

Momentum

For the 1 kg ball at 20 m/s:

p = m v

p = 1 kg × 20 m/s

p = 20 kg m/s

For the 10 kg ball at 2 m/s:

p = 10 kg × 2 m/s

p = 20 kg m/s

The momentum is the same for each.

Kinetic Energy (KE)

For the 1 kg ball at 20 m/s:

KE = ½ m v²

KE = ½ × 1 kg × (20 m/s)²

KE = 200 kg m²/s² = 200 J

For the 10 kg ball at 2 m/s:

KE = ½ × 10 kg × (2 m/s)²

KE = 20 kg m²/s² = 20 J

The KE of the small ball is much higher!

Summary

	1 kg at 20 m/s	10 kg at 2 m/s
Momentum:	20 kg m/s	20 kg m/s
KE:	200 J	20 J

So the momentum can be the same while the KE is very different.

Because KE uses velocity squared.

Inelastic vs Elastic Collisions

Inelastic collisions are mushy (like dough balls)
Elastic collisions are bouncy (like rubber balls)

inelastic dough vs elastic bouncy ball

In a perfectly Inelastic collision:

the objects stick together and end up sharing a new velocity
the objects get deformed by the collision, so
Kinetic Energy is lost (it gets converted into heat, light and sound)

In a perfectly Elastic collision the objects:

bounce perfectly off each other
the total Kinetic Energy stays exactly the same

inelastic elastic drop

Collisions are typically in between inelastic and elastic.

tennis ball bounces

Example: drop a tennis ball

It won't bounce back to the same height.

Because some kinetic energy is lost on the bounce (and a little is lost due to air resistance)

So the bounce is slightly inelastic (but mostly elastic).

We can have a scale of 0 (Inelastic) to 1 (Elastic). Try the Momentum Animation to see for yourself.

And in the Gravity Animation, the collisions are either inelastic (two objects smashing into each other) or elastic (the objects swing around each other and head off again). So things don't have to hit each other to make a collision!

Direction

Momentum uses velocity and direction matters!

Things can collide head on, or when going the same direction, or at different angles to each other!

Let us look at a simple 1-dimensional collision (everything moving along the same straight line):

Example: A railcar weighing 25,000 kg is rolling at 3 m/s east and hooks onto the back of a locomotive weighing 190,000 kg rolling 1 m/s east. What is the combined new velocity?

rail car loco

The collision is inelastic, as the coupling locks the railcar and loco together.

Momentum of Railcar

p_car = 25,000 kg × 3 m/s east

p_car = 75,000 kg m/s east

Momentum of Loco

p_loco = 190,000 kg × 1 m/s east

p_loco = 190,000 kg m/s east

Momentum is conserved, so the combined momentum is the same as when separated:

p_tot = m_tot v

Which can be rearranged to:

v = p_totm_tot

The totals are:

p_tot = 75,000 kg m/s + 190,000 kg m/s = 265,000 kg m/s

m_tot = 25,000 kg + 190,000 kg = 215,000 kg

(note that we added the p values as they are both in the same direction)

Solving:

v = 265,000 kg m/s215,000 kg

v = 1.2326... m/s

General Formulas

These formulas are for 1-dimensional collisions, where everything moves along the same straight line (like along a track). But angled (2-dimensional) collisions would need vectors to handle motion in different directions.

new v_a = elast × m_b(v_b − v_a) + m_av_a + m_bv_bm_a + m_b

new v_b = elast × m_a(v_a − v_b) + m_av_a + m_bv_bm_a + m_b

Where:

elast = coefficient of restitution = (speed after collision) ÷ (speed before collision) along the line of impact (0 = perfectly inelastic, 1 = perfectly elastic)
v_a and m_a are the velocity and mass of object a
v_b and m_b are the velocity and mass of object b

Example (continued): Instead of getting hooked, they bounce perfectly off each other, what are the new velocities?

rail car loco

The collision is perfectly elastic. We can use an "elast" value of 1:

new v_a = elast × m_b(v_b − v_a) + m_av_a + m_bv_bm_a + m_b

new v_b = elast × m_a(v_a − v_b) + m_av_a + m_bv_bm_a + m_b

Put in the values we know:

new v_a = 1 × 190 × (1 − 3) + 25×3+ 190×125 + 190

new v_b = 1 × 25 × (3 − 1) + 25×3 + 190×125 + 190

Calculate:

new v_a = −0.5349...

new v_b = 1.4651...

The railcar is now moving backwards at about 0.5 m/s, and the loco moving forwards at about 1.5 m/s

Is momentum conserved? The momentum values are now:

p_car = 25,000 kg × −0.5349... m/s = −13,372 kg m/s

p_loco = 190,000 kg × 1.4651... m/s = 278,372 kg m/s

p_tot = −13,372 kg m/s + 278,372 kg m/s = 265,000 kg m/s

That is the same total value as in the earlier example, so yes, momentum is conserved.

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