# Conservation of momentum/energy in a two particle system

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1. Sep 28, 2014

### ghostfolk

1. The problem statement, all variables and given/known data
A billiard ball of mass M is initially at rest on a horizontal frictionless table. Another ball of mass m < M and velocity $\vec{v}$ the positive x-direction hits the first ball in a perfectly elastic collision. After the collision, the balls move with (unknown) velocities $\vec{U}$ and $\vec{u}$ respectively (not necessarily in the x-direction).

b) Find the maximum amount of kinetic energy ∆T that the second ball can impart on
the first ball.
c) Show that the angle between $\vec{u}$ and $\vec{U}$ will always be greater than $\frac{\pi}{2}$.
d) Find the maximum component of the velocity $U_y$ that the larger ball can have in the
y-direction (normal to the x-direction on the table).
I'm really focused on parts b and d.

2. Relevant equations
$\rho_i=\rho_f$, $K_i=K_f$

3. The attempt at a solution

I'm not too confident on how to solve any of them but I did write down somethings.
Since it is a perfectly elastic collision on a frictionless table, then
$\rho_i=\rho_f \Rightarrow m\vec{v}=m\vec{u}+M\vec{U}$

$K_i=K_f\Rightarrow\frac{1}{2}m\vec{v}^2=\frac{1}{2}(m\vec{u}^2+M\vec{U})^2$

So if I square both sides of the first equation $(m\vec{v})^2=(m\vec{u}+M\vec{U})^2=(m^2\vec{u}^2+2Mm(\vec{u}\vec{U})+M^2\vec{U}^2)$.

Therefore,$\frac{1}{2}m\vec{v}^2=\frac{1}{2}(m\vec{u}^2+2M(\vec{u}\vec{U})+\frac{M^2}{m}\vec{U}^2)=\frac{1}{2}(m\vec{u}^2+M\vec{U}^2)$.

Then $\vec{u}=(\frac{M}{2m}-\frac{M}{2})\vec{U}$

2. Sep 29, 2014

### tms

Is there a part a) that matters?

First, somewhat trivially, you should use $p$ instead of $\rho$ for momentum, to avoid any possible confusion.

What does it mean to square a vector?

Again, what do you mean by multiplying vectors like this?

It would be better to work with the $x$ and $y$ components of momentum separately. For energy, you need to use the magnitudes of the velocity vectors.

3. Sep 29, 2014