Two particle system involving conservation laws

[ghostfolk]

Homework Statement

A billiard ball of mass M is initially at rest on a horizontal frictionless table. Another ball
of mass m < M and velocity $\vec{v}$ in the positive x-direction hits the first ball in a pefectly elastic
collision. After the collision, the balls move with (unknown) velocities $\vec{U}$ and $\vec{u}$ respectively
(not necessarily in the x-direction).
Find the maximum amount of kinetic energy $∆T$ that the second ball can impart on
the first ball.

Homework Equations

$T=\frac{1}{2}mv^2$
$\rho_i=\rho_f$
$T_i=T_f$

The Attempt at a Solution

I'm entirely stuck and I'm not sure how to tackle this problem. Any help is appreciated.

 

[Orodruin]

You have written down some conservation laws, did you try applying them?

 

[ghostfolk]

You have written down some conservation laws, did you try applying them?

The thing is I'm not entirely sure how to. I can derive some basic things such as due to conservation of momentum $m\vec{v}=m\vec{u}+M\vec{U}$ and due to the system be isolated $\frac{1}{2}mv^2=\frac{1}{2}(mv^2+MU^2)$. After this I'm not sure where to go to get the maximum kinetic energy trasnferred.

 

[BvU]

You mean $\frac{1}{2}mv^2=\frac{1}{2}(m{\bf u}^2+MU^2)$, right ?

So there are three equations, but four unknowns. That leaves one degree of freedom.
What is the kinetic energy you 'want' to maximize? Write it as a function of this one free variable left over.
Then try to find a maximum for that function.

 

[ghostfolk]

You mean $\frac{1}{2}mv^2=\frac{1}{2}(m{\bf u}^2+MU^2)$, right ?

So there are three equations, but four unknowns. That leaves one degree of freedom.
What is the kinetic energy you 'want' to maximize? Write it as a function of this one free variable left over.
Then try to find a maximum for that function.

Well since the object in question is the first ball, we have to find the a function of $v$ right? We can't use $u$ because that is the velocity after impact.

 

[BvU]

No such freedom exists: $\vec v = (v_x, v_y)$ is a given ! $v_y=0$ and $v_x$ will appear in the answer, but as a parameter, not as a variable.

 

[ghostfolk]

No such freedom exists: $\vec v = (v_x, v_y)$ is a given ! $v_y=0$ and $v_x$ will appear in the answer, but as a parameter, not as a variable.

Could you explain that in more detail?

 

[ehild]

The kinetic energy before impact is 1/2 mv². After impact, it is 1/2 mu²+1/2 M U², u is the speed of the ball with mass m and U is the speed of the billiard ball of mass M.

Write up the conservation laws also for the momentum components. Choose a coordinate system with x-axis parallel with the initial velocity v of the the small ball and express the components with the angle they enclose with the x axis.

ehild

 

[ghostfolk]

The kinetic energy before impact is 1/2 mv². After impact, it is 1/2 mu²+1/2 M U², u is the speed of the ball with mass m and U is the speed of the billiard ball of mass M.

Write up the conservation laws also for the momentum components. Choose a coordinate system with x-axis parallel with the initial velocity v of the the small ball and express the components with the angle they enclose with the x axis.

ehild

Okay, so when exactly can I find the maximum kinetic energy? I'm thinking if it's when the velocity of m is zero.

 

[ehild]

The masses are given, and so is the velocity of the small ball. But the collision is not necessarily central. The direction of the velocities are the free parameters, but the conservation laws make them related, so only one angle is free.

Find the kinetic energy of the billiard ball after the collision, as function of its angle it encloses with the x axis. You can figure out at what angle is the KE maximum.

ehild

 

[ghostfolk]

[QUOTE="ehild, post: 4868343, member: 481" You can figure out at what angle is the KE maximum.

ehild[/QUOTE]How exactly would I do that?

 

[BvU]

Play with coins n a smooth table to get ideas.
Draw a picture, then follow ehild's advice. Exactly.

 

[ghostfolk]

Play with coins on a smooth table to get ideas.
Draw a picture, then follow ehild's advice. Exactly.

Okay, but how exactly would one determine the angle where there is maximum kinetic energy transfer. That part doesn't make sense to me. How would a drawing of vectors show that?

 

[ehild]

The kinetic energy will depend on some angle. To have maximum, its derivative with respect to the angle should be zero.
Drawing the velocity vectors help you to write up equations for the momentum components.

ehild