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Chapter 11 - Collisions


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


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


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


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


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


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



Conservation of Momentum

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


The force exerted by object A on object B must be


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


But we know from Newton's Third Law that


so that


Conservation of Energy in Collisions

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.

Elastic Collisions in one dimension

For a one dimensional elastic collision




Change in velocity in an elastic collision


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






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.

Solving 1-D elastic collision problems

Conservation of Momentum


Conservation of Energy





Collisions with only one body moving initially

Consider collisions where $v_{B}=0$



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


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

Three cases

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




Inelastic collisions

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

Ballistic Pendulum

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.





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




Multi-dimensional collisions

The equation of conservation of momentum is a vector equation.


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



Perfectly Elastic collisions in 2 dimensions

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




Billiards question

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?

Billiards (or pucks)

Conservation of momentum     
Conservation of energy


Square the conservation of momentum equations



Add these together


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 $v'_{B}=v_{A}\cos\theta$


Perfectly Inelastic collisions in 2 dimensions

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


Snooker video

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..

phy131studiof15/lectures/chapter11.1443988840.txt · Last modified: 2015/10/04 16:00 by mdawber
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