Conservation of Momentum

In application of momentum, the most important law to know is the Conservation of Momentum. Momentum is said to be conserved when there is zero net external force acting on the object.

To explain this mathematically:
• Definition of momentum: p = mv
• Newton's 1st law: F = ma => m dv/dt = dp/dt
• Net external force is zero: F = 0 => dv/dt = 0
• dv/dt = 0 => Dp = 0
Conservation of Momentum is mathematically expressed as:
mvi = mvf

When expanded to n-particle system the n-conservative momentum of each particle add:

Now let's take a look at some things to verify what we covered in previous sections.

In Aiming, a statement was made that the angle formed when the cue collides with the object ball which is initially at rest is always 90 degrees. This is so if assuming there is no friction and no rotation on the balls so that no kinetic energy is dissipated. This is the reason why in real life (with friction and rotation) it is never perfectly right angular after collision but very similar. Back to verifying why it is theoretically 90 degrees, we must consider first the conservation of momentum and the conservation of kinetic energy:

All balls are equal in mass and the object ball is initially at rest, v2i = 0

Conservation of momentum gives us:
mv1i = mv1f + mv2f
v1i = v1f + v2f
This is a vector equation and geometrically means that three vectors form a triangle.

(1/2)mv1i2 = (1/2)mv1f2 + (1/2)mv2f2 v1i2 = v1f2 + v2f2
As you can see, the obtained velocities are in a form of Pythagorean theorem and the vector velocity v1i is the hypotenuse of two leg vector velocities, v1f and v2f.
Hence, right angle between the two final vectors.

In Elasticity, we learned that the coefficient of restitution, e = -(v1i - v2i)/(v1f - v2f).
To verify this, we begin again by writing the conservation of kinetic energy:
(1/2)mv1i2 + (1/2)Mv2i2 = (1/2)mv1f2 + (1/2)Mv2f2
After rearrangement of terms:
mv1i2 - mv1f2 = Mv2f2 - Mv2i2
Eq1: m(v1i - v1f)(v1i + v1f) = M(v2f - v2i)(v2f + v2i)

To simplify the equation any further, we must look at two-object conservation of momentum:
mv1i + Mv2i = mv1f + Mv2f
Eq2: m(v1i - v1f) = M(v2f - v2i)

By inspection, we notice that a part of Eq1 may be simplified using Eq2 and obtain:
v1i + v1f = v2i + v2f OR
v1i - v2i = -(v1f - v2f)
Eq3: -(v1i - v2i)/(v1f - v2f) = 1

We notice from Eq3 which is the coefficient of restitution for a perfectly elastic collision, e = 1. This is true because in elastic collisions, the kinetic energy is conserved and hence the linear momentum is also conserved.

Sample Problem
A cue ball collides elastically with a red ball and a blue ball which are both initially at rest at 10 m/s. If the cue ball comes to rest and the blue ball begins to move at 2m/s after collision, what is the speed of the blue ball?