Two balls of equal masses move toward each other on the x-axis. When they
collide, each ball ricochets 90 degrees, such that both balls are moving away
from each other on the y-axis. What can be said about the final velocity of each
ball?

Initially, since both balls are moving on the x-axis, the y component of the
momentum is zero. Since momentum is conserved, we can state that the momentum of
each ball must be equal and opposite after the collision, as they move along the
y-axis. Since both masses are equal, the velocity of each ball must be equal
and opposite.

Problem :

Two pool balls traveling in opposite directions collide. One ball travels off at
an angle θ to its original velocity, as shown below. Is there any
possible way for the second ball to be completely stopped by this collision? If
so state the conditions under which this could occur.

No, the second ball must also leave the collision at an angle. The first ball
has a component of linear momentum in the y direction after the collision, given
by v_{1f}sinθ. Since both balls were traveling in the x direction before
the collision, there was no initial momentum in the y direction. Thus, for
momentum to be conserved, the second ball must travel in the negative y
direction, to counteract the momentum of the first ball. If the second ball
remained stationary, momentum would not be conserved.

Problem :

Two objects are traveling perpendicular to each other, one moving at 2 m/s with
a mass of 5 kg, and one moving at 3 m/s with a mass of 10 kg, as shown below.
They collide and stick together. What is the magnitude and direction of the
velocity of both objects?

The collision is completely inelastic, and we have two variables, v_{f} and
θ, and the two equations of conservation of linear momentum. We start by
relating momentum before and after the collision in the x direction:

(5kg)(2m/s) = 15v_{f}cosθ

implying that

v_{f}cosθ =

Now equating the y components,

(10kg)(3m/s) = 15v_{f}sinθ

Implying that

2 = v_{f}sinθ

We have two independent equations for v_{f} and θ If we divide the second
one by the first, v_{f} will cancel, and we will be left with an expression for
θ only:

=

Thus

tanθ = 3

And θ = 71.6^{o}. Substituting this in to find v_{f}, we find that:

v_{f} = = = 2.11

Thus the two stuck together objects have a final velocity of 2.11 m/s directed
71.6^{o} above horizontal.

Problem :

A common pool shot involves hitting a ball into a pocket from an angle. Shown
below, the cue ball hits a stationary ball at an angle of 45^{o}, such
that it goes into the corner pocket with a speed of 2 m/s. Both balls have a
mass of .5 kg, and the cue ball is traveling at 4 m/s before the collision.
Recalling that this collision is elastic, calculate the angle with which the cue
is deflected by the collision.

To solve this problem we begin with our familiar momentum equations for both x
and y components. Since we only have two variables (v_{1} and θ) we need
not generate a third equation from conservation of kinetic energy. Thus we
equate the x and y components of linear momentum before and after the collision:

p_{xo}

=

p_{xf}

.5(4)

=

.5v_{1}cosθ + .5(2)cos 45

4

=

v_{1}cosθ +

p_{yo}

=

p_{yf}

0

=

2 sin 45 - v_{1}sinθ

=

v_{1}sinθ

v_{1}

=

Here we have two equations relating θ and v_{1}. To solve we can simply
substitute our expression for v_{1} in terms of θ into our first
equation:

4

=

()cosθ +

4 -

=

(cotθ)

cotθ

=

1.83

θ

=

28.7^{o}

Thus the pool cue will be deflected about 30 degrees from horizontal.