We can define momentum as the mass times the velocity

$\vec{p}=m\vec{v}$

Momentum has units $\mathrm{kg\,ms^{-1}}$

Unlike kinetic energy momentum is a vector quantity.

The most fundamental aspect of momentum is it's connection to force. We can express Newton's second law as

$\Sigma \vec{F} = \frac{d\vec{p}}{dt}=m\frac{d\vec{v}}{dt}=m\vec{a}$

Over a change in momentum $\Delta \vec{p}$ from $m\vec{v}_{1}$ to $m\vec{v}_{2}$ in a time interval $\Delta t$ during which the instanteous force may vary the average force can also be a useful quantity to consider

$\vec{F}_{ave}=\frac{\Delta \vec{p}}{\Delta t}=\frac{m\vec{v}_{2}-m\vec{v}_{1}}{\Delta t}$

In a collision in which no external force acts the total momentum of the system is a conserved quantity. This can be seen to be a consequence of Newton's third law.

Consider a collision between two objects

$m_{A}\vec{v}_{A}+m_{B}\vec{v}_{B}=m_{A}\vec{v}'_{A}+m_{B}\vec{v}'_{B}$

The force exerted by object A on object B must be

$\vec{F}_{AB}=\frac{d\vec{p}_{B}}{dt}$

And the force exerted by object B on object A must be

$\vec{F}_{BA}=\frac{d\vec{p}_{A}}{dt}$

But we know from Newton's Third Law that

$\vec{F}_{AB}=-\vec{F}_{BA}$

so that

$\frac{d\vec{p}_{A}}{dt}+\frac{d\vec{p}_{B}}{dt}=0$

More generally, the total momentum of a system

$\vec{P}=\Sigma_{i} p_{i}=\Sigma_{i} m_{i}v_{i}$

The change of this quantity with time

$\frac{d\vec{P}}{dt}=\Sigma_{i}\frac{d\vec{p_{i}}}{dt}=\Sigma_{i}F_{i}$

$F_{i}$ represents the net force on the $i$th body. All internal force have equal and opposite reaction force on some other body within the system (Newton's 3rd Law) and cancel out in the sum.

So

$\frac{d\vec{P}}{dt}=\Sigma_{i}F_{ext}$

**In the absence of an external force the total momentum of a system is constant.**

During a collision we can say that the net force on an object $\vec{F}$

$\vec{F}=\frac{d\vec{p}}{dt}$

During an infinitesimal time interval $dt$ the change in momentum $d\vec{p}$ is

$d\vec{p}=\vec{F}dt$

Integrating this over the change from the initial state $i$ to the final state $f$

$\int_{i}^{f}d\vec{p}=\vec{p_{f}}+-\vec{p_{i}}=\int_{t_{i}}^{t_{f}}\vec{F}dt$

We define the integral over the force as the impulse $\vec{J}$

$\vec{J}=\int_{t_{i}}^{t_{f}}\vec{F}dt$

and we can see that the change of momentum of an object is equal to the impulse acting on it

$\Delta\vec{p}=\vec{J}=\int_{t_{i}}^{t_{f}}\vec{F}dt$

Another way of expressing this would be in terms of the average force $F_{Ave}$ during the collision. If the total time for the collision is $t$, then

$\Delta\vec{p}=\vec{F}_{Ave}t$

We throw a tennis ball (mass 57g) at a wall so that it is incident and comes off at a 45^{o} angle. If it’s speed both before and after is 10ms^{-1} what is the magnitude of the change in

- The x component of the momentum.
- The y component of the momentum.

Impulse

$\vec{F}_{Ave}=\frac{\Delta\vec{p}}{\Delta t}$

The general law of conservation of energy tells us that the total energy (including all forms of energy) in a collision is conserved.

However we saw before that this does necessarily mean that mechanical energy is conserved. Further, a collision usually takes place at a constant position with respect to potential energy. Therefore the most important question that we need to ask about a collision is: Is **kinetic** energy conserved?

We call collisions that conserve kinetic energy **elastic**, and for these we can write that.

$ \Sigma_{i}\frac{1}{2}m_{i}v^{2}_{i}=\Sigma_{i}\frac{1}{2}m_{i}v'^{2}_{i}$

In collisions where some of the energy is converted from kinetic energy to some other kind of energy, which we call **inelastic** collisions, we can only say that

$ \Sigma_{i}\frac{1}{2}m_{i}v^{2}_{i}=\Sigma_{i}\frac{1}{2}m_{i}v'^{2}_{i}+\textrm{other forms of energy}$

Which, unless we can quantify the other forms of energy, is a less useful equation.

For a one dimensional elastic collision

$m_{A}v_{A}+m_{B}v_{B}=m_{A}v'_{A}+m_{B}v'_{B}$

**and**

$\frac{1}{2}m_{A}v^{2}_{A}+\frac{1}{2}m_{B}v^{2}_{B}=\frac{1}{2}m_{A}v'^{2}_{A}+\frac{1}{2}m_{B}v'^{2}_{B}$

Rewrite

$m_{A}v_{A}+m_{B}v_{B}=m_{A}v'_{A}+m_{B}v'_{B}$ as $m_{A}(v_{A}-v'_{A})=m_{B}(v'_{B}-v_{B})$

and

$\frac{1}{2}m_{A}v^{2}_{A}+\frac{1}{2}m_{B}v^{2}_{B}=\frac{1}{2}m_{A}v'^{2}_{A}+\frac{1}{2}m_{B}v'^{2}_{B}$

as

$m_{A}(v^{2}_{A}-v'^{2}_{A})=m_{B}(v'^{2}_{B}-v^{2}_{B})$

$m_{A}(v_{A}-v'_{A})(v_{A}+v'_{A})=m_{B}(v'_{B}-v_{B})(v'_{B}+v_{B})$

Divide the kinetic energy by the momentum equation

$(v_{A}+v'_{A})=(v'_{B}+v_{B})$ or $(v_{A}-v_{B})=(v'_{B}-v'_{A})$ or $(v_{A}-v_{B})=-(v'_{A}-v'_{B})$

The relative speed of the objects before and after the collision have the same magnitude but different signs.

Conservation of Momentum

$m_{A}v_{A}+m_{B}v_{B}=m_{A}v'_{A}+m_{B}v'_{B}$

Conservation of Energy

$\frac{1}{2}m_{A}v^{2}_{A}+\frac{1}{2}m_{B}v^{2}_{B}=\frac{1}{2}m_{A}v'^{2}_{A}+\frac{1}{2}m_{B}v'^{2}_{B}$

or

$(v_{A}-v_{B})=-(v'_{A}-v'_{B})$

Consider collisions where $v_{B}=0$

$m_{A}v_{A}=m_{A}v'_{A}+m_{B}v'_{B}$

$(v_{A})=-(v'_{A}-v'_{B})$

We want to know $v'_{A}$ and $v'_{B}$ given the masses of the objects and the initial velocity $v_{A}$

$v'_{A}=-v_{A}+v'_{B}$

$m_{A}v_{A}=-m_{A}v_{A}+m_{A}v'_{B}+m_{B}v'_{B}$

$m_{A}v_{A}+m_{A}v_{A}=m_{A}v'_{B}+m_{B}v'_{B}$

$v'_{B}=\frac{2m_{A}}{m_{A}+m_{B}}v_{A}$

$v'_{B}=v_{A}+v'_{A}$

$m_{A}v_{A}= m_{A}v'_{A}+m_{B}v'_{A}+m_{B}v_{A}$

$m_{A}v_{A}-m_{B}v_{A}=m_{A}v'_{A}+m_{B}v'_{A}$

$v'_{A}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}v_{A}$

A bullet hitting an object is a good example of an **inelastic** collision.

We can however use conservation of momentum for the collision and conservation of energy for the **subsequent motion** in a ballistic pendulum to find the velocity of a bullet fired in to a block from the height that the bullet and block rises to.

$mv=(M+m)v'$

$\frac{1}{2}mv^2\neq\frac{1}{2}(M+m)v'^{2}$

but

$\frac{1}{2}(M+m)v'^{2}=(M+m)gh$ → $\frac{1}{2}v'^{2}=gh$

so

$v=\frac{M+m}{m}\sqrt{2gh}$