Conservation of momentum/energy in a two particle system

  • #1
59
1

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}## 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.

Homework Equations


##\rho_i=\rho_f##, ##K_i=K_f##

The Attempt at a Solution


[/B]
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}##
 

Answers and Replies

  • #2
tms
644
17
b) Find the maximum amount of kinetic energy ∆T that the second ball can impart on
the first ball.
Is there a part a) that matters?

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}##
First, somewhat trivially, you should use [itex]p[/itex] instead of [itex]\rho[/itex] for momentum, to avoid any possible confusion.

##K_i=K_f\Rightarrow\frac{1}{2}m\vec{v}^2=\frac{1}{2}(m\vec{u}^2+M\vec{U})^2##
What does it mean to square a vector?

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)##.
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
59
1
Is there a part a) that matters?
There isn't.

What does it mean to square a vector?
It's the dot product with itself

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.
 

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