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

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

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

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

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

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..

All real collisions lie somewhere between the perfectly elastic and perfectly inelastic cases we have looked at, but let's consider now the two extreme cases for a collision of 2 objects of equal mass colliding at right angles with equal velocities.

Momentum must be conserved for both the x and y directions

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

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

Squaring and adding our momentum equations lead to

$2v^{2}=v'^{2}_{A}+v'^{2}_{B}+2v'_{Ax}v'_{Bx}+2v'_{Ay}v'_{By}$

In the case of an elastic collision

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

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

which means that

$2v'_{Ax}v'_{Bx}+2v'_{Ay}v'_{By}=0$

And by inspection we can see that the result is a $90^{o}$ change of direction for each object with no change in the speed of the objects.

For an inelastic collision

$2v^{2}=2v'^{2}+2v'^2_{x}+2v'^{2}_{y}=4v'^{2}$

$v'=\frac{v}{\sqrt{2}}$

and inspection shows that the x and y component of the velocity are equal.

We have looked at two quite different 2 dimensional collisions. However there is one way of looking at them which makes them look more similar. If we consider the motion of the center of mass of the objects, instead of the motion of the individual masses we can see that from this perspective the motion is identical!

We can actually consider the motion of any extended mass to be composed of different kinds of motion with respect to the center of mass.

We will now make explicit our implicit assumption till now that we consider the translation motion of an object's center of mass. We will also see in the coming lectures that we can add to this translational motion other motions of the mass around the center of mass, for example rotational or vibrational motion.

In general the displacement vector for the center of mass of a system of particles $m_{i}$ can be written as

$\vec{r}_{CM}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{\Sigma_{i}m_{i}}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{M}$

Differentiating with respect to time gives

$M\frac{d\vec{r}_{CM}}{dt}=\Sigma_{i}m_{i}\frac{d\vec{r}}{dt}$ or $M\vec{v_{CM}}=\Sigma_{i}m_{i}\vec{v}_{i}$

and doing it once more

$M\vec{a_{CM}}=\Sigma_{i}m_{i}\vec{a}_{i}$

$\Sigma_{i}m_{i}\vec{a}_{i}=\Sigma_{i}F_{i}=\Sigma \vec{F}_{ext}$

so we have obtained a new form of Newton's Second Law that works for a system of particles.

$M\vec{a_{CM}}=\Sigma \vec{F}_{ext}$

the total momentum of a system can also be written in these terms

$\vec{P}=M\vec{v}_{CM}$

We can consider the motion of any system of particles or extended mass to be composed of different kinds of motion with respect to the center of mass.

In general the displacement vector for the center of mass of a system of particles $m_{i}$ can be written as

$\vec{r}_{CM}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{\Sigma_{i}m_{i}}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{M}$

Differentiating with respect to time gives

$M\frac{d\vec{r}_{CM}}{dt}=\Sigma_{i}m_{i}\frac{d\vec{r}}{dt}$ or $M\vec{v_{CM}}=\Sigma_{i}m_{i}\vec{v}_{i}$

and doing it once more

$M\vec{a_{CM}}=\Sigma_{i}m_{i}\vec{a}_{i}$

$\Sigma_{i}m_{i}\vec{a}_{i}=\Sigma_{i}F_{i}=\Sigma \vec{F}_{ext}$

so we have obtained a new form of Newton's Second Law that works for a system of particles.

$M\vec{a_{CM}}=\Sigma \vec{F}_{ext}$

the total momentum of a system can also be written in these terms

$\vec{P}=M\vec{v}_{CM}$

For point like objects at given displacement the center of mass vector is

$\vec{r}_{CM}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{M}$

For solid objects it is more useful to consider the center of mass in integral form.

$\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,dm$

which we should recall is the same as having a set of equations for each of the components, ie.

$x_{CM}=\frac{1}{M}\int x\,dm$ , $y_{CM}=\frac{1}{M}\int y\,dm$ , $z_{CM}=\frac{1}{M}\int z\,dm$

For object's of uniform density we can express the mass element over which we integrate as a spatial element which in 3 dimensions is $dm=\rho\, dV$, in 2 dimensions is $dm=\rho_{A}\, dA$ in one dimension is $dm=\lambda\,dx$, where $\rho$, $\rho_{A}$ and $\lambda$ are the density, areal density or linear density of the object we consider.

In cases where we know the total mass and total size (either volume, area or length) of an object the density can be found by dividing the total mass by the total size.

We can consider a uniform rod to be a one dimensional object.

If we want to find the COM we could place the origin of our coordinate system at the center of the rod and then

$\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,dm=\frac{1}{M}\int_{-l/2}^{l/2}\lambda x\,dx=\frac{1}{M}\frac{\lambda}{2}((\frac{l}{2})^2-(-\frac{l}{2})^2)=0$

On the other hand if we were to place the origin at the left end of the rod then we could show that

$\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,dm=\frac{1}{M}\int_{0}^{l}\lambda x\,dx=\frac{1}{M}\frac{\lambda}{2}l^2$

and as $M=\lambda l$

$\vec{r}_{CM}=\frac{l}{2}\hat{i}$

We now look at a 2 dimensional object, in this case a thin rectangular plate.

The mass interval $dm=\rho_{A}\,dA=\rho_{A}\,dx\,dy$

The COM of the plate in the x direction is

$\vec{x}_{CM}=\frac{1}{M}\int x\,dm=\frac{1}{M}\int_{-l/2}^{l/2}\int_{-w/2}^{w/2}\rho_{A}x\,dy\,dx=0$

$\vec{y}_{CM}=\frac{1}{M}\int y\,dm=\frac{1}{M}\int_{-l/2}^{l/2}\int_{-w/2}^{w/2}\rho_{A}y\,dy\,dx=0$

We can see that symmetry often enables us to identify the center of mass of an object, equally distributed mass on either side of the origin cancels out.

If we instead put the bottom left hand corner of the plate at the origin then

$\vec{x}_{CM}=\frac{1}{M}\int x\,dm=\frac{1}{M}\int_{0}^{l}\int_{0}^{w}\rho_{A}x\,dy\,dx=\frac{1}{M}\frac{1}{2}\rho_{A}l^{2}w=\frac{l}{2}\hat{i}$

$\vec{y}_{CM}=\frac{1}{M}\int y\,dm=\frac{1}{M}\int_{0}^{l}\int_{0}^{w}\rho_{A}y\,dy\,dx=\frac{1}{M}\frac{1}{2}\rho_{A}w^{2}l=\frac{w}{2}\hat{j}$

How symmetric do you think the human body is? Here's an experiment to determine the center of mass of people.

For circular objects and rotational motion we will find polar coordinates to be advantageous.

To transform from polar coordinates in to Cartesian coordinates

$x=r\cos\theta$

$y=r\sin\theta$

and from Cartesian to polar

$r=\sqrt{x^{2}+y^{2}}$

$\large\tan\theta=\frac{y}{x}$

$\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,dm=\frac{1}{M}\int\vec{r}\rho_{A}\,dA$

$dm=\rho_{A}\,dA=\rho_{A}r\,dr\,d\theta$

$x=r\cos\theta$

$y=r\sin\theta$

$x_{CM}=\frac{1}{M}\int x\rho_{A} \, dA=\frac{\rho_{A}}{M}\int_0^{2\pi}\int_{0}^{R}r^2\cos\theta\,dr\,d\theta=\frac{\rho_{A} R^3}{3M}\int_0^{2\pi}\cos\theta\,d\theta$ $=\frac{\rho_{A} R^3}{3M}[\sin(2\pi)-\sin(0)]=0$

$y_{CM}=\frac{1}{M}\int y\rho_{A} \, dA=\frac{\rho_{A}}{M}\int_0^{2\pi}\int_{0}^{R} r^2\sin\theta\,dr\,d\theta=\frac{\rho_{A} R^3}{3M}\int_0^{2\pi}\sin\theta\,d\theta$ $=\frac{\rho_{A} R^3}{3M}[-\cos(2\pi)+\cos(0)]=0$

$\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,dm=\frac{1}{M}\int\vec{r}\rho_{A}\,dA$

$x_{CM}=\frac{1}{M}\int x\rho_{A} \, dA=\frac{\rho_{A}}{M}\int_0^{\pi}\int_{0}^{R}r^2\cos\theta\,dr\,d\theta=\frac{\rho_{A} R^3}{3M}\int_0^{\pi}\cos\theta\,d\theta$ $=\frac{\rho_{A} R^3}{3M}[\sin(\pi)-\sin(0)]=0$

$y_{CM}=\frac{1}{M}\int y\rho_{A} \, dA=\frac{\rho_{A}}{M}\int_0^{\pi}\int_{0}^{R}r^2\sin\theta\,dr\,d\theta=\frac{\rho_{A} R^3}{3M}\int_0^{\pi}\sin\theta\,d\theta$ $=\frac{\rho_{A}R^3}{3M}[-\cos(\pi)+\cos(0)]=\frac{2\rho_{A}R^3}{3M}$

We can convert this in to something more useful by considering that $M=\rho_{A}\frac{1}{2}\pi R^{2}$

which tells us that

$y_{CM}=\frac{4R}{3\pi}$