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

$\Delta\vec{p}=\vec{F_{Ave}}t$

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

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

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 $v_{A}$

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

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

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

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

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

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

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

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

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

$m_{A}=m_{B}$

$m_{A}>m_{B}$

$m_{B}>m_{A}$

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

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

The equation of conservation of momentum is a vector equation.

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

Therefore if we have a collision involving more than one dimension we need to consider conservation of each component of momentum, for example in 2 dimensions

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

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

If a collision is perfectly elastic we can add a 3rd equation

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

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

When we hit a billiard ball straight on all the momentum of the incident ball should be transferred to the 2nd ball.

However we know already from experience that if we hit a billiard ball at an angle this does not happen. Can we use our conservation of momentum principle to predict the trajectories of billiard balls after a collision?

If one ball hits another head on, it should come to rest. But if it comes in at a small angle where should it go?

Conservation of momentum $v_{A}=v'_{Ax}+v'_{Bx}$ $0=v'_{Ay}+v'_{By}$ | Conservation of energy $v_{A}^2=v'^{2}_{A}+v'^{2}_{B}$ |

$v'_{Bx}=v'_{B}\cos\theta$ $v'_{By}=v'_{B}\sin\theta$ | $v'_{Ax}=v'_{A}\sin\theta$ $v'_{Ay}=v'_{A}\cos\theta$ |

Square the conservation of momentum equations

$v'^{2}_{Ay}=v'^{2}_{By}$

$v_{A}^2-2v_{A}v'_{Ax}+v'^{2}_{Ax}=v'^{2}_{Bx}$

Add these together

$v_{A}^2+v'^{2}_{A}-2v_{A}v'_{Ax}=v'^{2}_{B}$

Use kinetic energy equation to eliminate $v'_{B}$

$v_{A}^2+v'^{2}_{A}-2v_{A}v'_{Ax}=v_{A}^{2}-v'^{2}_{A}$ → $v'^{2}_{A}=v_{A}v'_{Ax}$

$v'^{2}_{A}=v_{A}v'_{A}\sin\theta$ → $v'_{A}=v_{A}\sin\theta$

and conservation of energy equation gives us $v'_{B}=v_{A}\cos\theta$

In a perfectly inelastic collision the two object stick together after the collision so conservation of momentum now is written

$m_{A}\vec{v_{A}}+m_{B}\vec{v_{B}}=(m_{A}+m_{B})\vec{v}'$

Clearly there is more to this game than we discussed above! To understand better the way we can manipulate the path of a billiard ball we will need to study rotational motion..