Momentum and Energy in Two Dimensional Elastic Collisions
A collision in two dimensions occurs when two objects traveling in the same straight line
collide and after the collision their velocities are above or below the horizontal. This
situation can be somewhat approximated in the case of billiard balls. If the rolling friction
is minimal, the collision will approximate an elastic collision.
Linear momentum is defined as m*v and has units of kg*m/s. Mathematically it is
expressed as, p = mv, and is a vector quantity. The bold font is used to call your
attention to the fact that momentum and velocity are vector quantities. If the net force
acting on an object is zero, its momentum is conserved in a collision and can be
expressed as pi = pf, where i and f stand for initial and final. This conservation law
implies that the total momentum of the universe is constant.
When discussing the conservation of momentum, one must distinguish between internal
and external forces. Internal forces consist of an action-reaction pair of forces according
to Newton's third law. External forces are forces that exist due to outside influences, i.e.
gravity or friction. A conservative force such as gravity does not "drain" the system of
any energy. However, a force such as friction or air resistance is a non-conservative
force, which converts some of the mechanical energy into heat. The ideal situation
exists when the net force sums to zero and there is no net external force converting
some of the mechanical energy into heat.
Keeping simple what is simple; we will only consider a collision of two objects. This
leads to:
Δp = 0
pi = p f
m1v1 + m2v2 = m1v1' + m2v2'
Because there are only two objects involved in this collision, there are only two terms
representing the initial momentum and two terms representing the final momentum.
A second important consideration involving collisions is that of energy. For simplicity, we
will consider two objects traveling on a horizontal surface without friction. There are
three categories of collisions: completely elastic, completely inelastic, or somewhere in
the middle. Given that the ΣFnet = 0, momentum will be conserved but this is not
necessarily true about kinetic energy, KE. Remembering that energy can take many
forms i.e. heat, light, and sound, a perfectly elastic collision only exists when ΔKE = 0. If
the kinetic energy of the system equals zero after a collision the collision is said to be
completely inelastic. Through an unfortunate choice of words, it sometimes sounds like
energy is "lost". Energy is not lost but rather converted into undesirable forms which
decrease the kinetic energy of a system. Keeping simple what is simple; we will only
consider a collision of two objects. For an inelastic collision, the following is true:
ΔKE ≠ 0
KEi ≠ KEf
½m1v12 + ½m2v22 ≠ ½m1v1'2 + ½m2v2'2
and for an elastic collision, the following is true
ΔKE = 0
KEi = KEf
½m1v12 + ½m2v22 = ½m1v1'2 + ½m2v2'2
Another must to remember is that energy is a scalar quantity. Unlike momentum which
is a vector that you resolve vector components, energy is a scalar and a negative sign
simply represents a decrease in energy.
In a problem, you are often given directions such as north (N), east (E), south (S), and
west (W). Therefore, N would be 000°, E would be 090°, etc. A bearing of 135° would
be drawn such that the vector would be 45° below the horizontal (090°).
270°
090°
The diagram below is probably familiar to you if you have used the sine (sin),
cosine (cos) and tangent (tan) functions in math. Do not forget these! These will
be used in combination with the one above but not interchangeably. You will
need to use the one below on the left for sign (+ or -) conventions.
Momentum and Energy (Elastic Collisions 2D) Problems
1) A ball with a mass of 5.0 kg moves to the right with a speed of 4.0 m/s and collides
with a stationary ball whose mass is 5.0 kg. The collision is not head-on and after the
collision the first ball moves in a direction of 45° to the left of its original path with a
velocity of 2.8 m/s. The second ball moves in a direction of 45° below the horizontal.
(a) Determine the velocity of each ball after the collision.
m1 = 5.0 kg
v1 = 4.0 m/s
θ1 = 45°
v1' = 2.8 m/s
m2 = 5.0 kg
v2 = 0
θ2 = 45°
The conservation of momentum still applies as indicated by equations (1) and (2).
However, the collision takes place in two dimensions (x and y) so vector addition
must be used.
Δp = 0
pi = p f
The initial momentum of the system is defined by the only ball that is moving.
pi = m1v1 = 5.0 kg*4.0 m/s = 20. kg-m/s
(b) Determine the momentum of each ball after the collision.
After the collision, you must consider the horizontal and vertical components of
momentum of both balls.
The vertical momentum of the red ball is given by:
pv1' = m1v1sinθ1 = 5.0 kg*4.0 m/s*0.707 = 14 kg-m/s
Similarly, the vertical momentum of the blue ball after the collision is -14 kg-m/s.
Keep in mind that momentum is a vector and if up is positive then down must be
negative. This is good news because if the initial vertical momentum of the system
is zero, so must the final vertical momentum be equal to zero.
The horizontal momentum of the red ball after the collision is given by:
ph1' = m1v1'cosθ1 = 5.0 kg*4.0 m/s*0.707 = 14 kg-m/s
Because of symmetry considerations, masses and angles, the momentum of the
blue ball is also 14 kg-m/s.
The velocity of each ball is given by:
pv1' = m1v1'
v1' = pv1' / m1 = 14 kg-m/s/5.0 kg = 2.8 m/s
Because of symmetry considerations, the velocity of the blue ball is also 2.8 m/s.
Don't you wish that everything was so symmetrical?
2) The stationary 8 ball with a mass of 0.16 kg is struck by the cue ball moving at
5.0 m/s. After the collision, the cue ball moves in a direction 60° to the left of its
original path while the 8 ball moves in a direction of 30° below the horizontal. The
mass of the cue ball has the same mass as the 8 ball. What is the velocity of each
ball after the collision?
m1 = 6.0 kg
v1 = 3.0 m/s
v1' = ?
θ1 = 50°
m2 = 6.0 kg
v2 = 0
v2' = ?
θ2 = 40°
Δp = 0
pi = pf
First find the vertical momentum, pv:
m1v1 = m1v1'sin50° + m2v2'sin40°
0 = 6.0 kg*v1'*0.766 + 6.0 kg*v2'*0.643
0 = 4.60v1' + 3.86v2'
(1)
Second find the horizontal momentum, ph:
m1v1 = m1v1'cos50° + m2v2'cos40°
6.0 kg*3.0 m/s = 6.0 kg*v1'*0.643 + 6.0 kg*v2'*0.766
18 m/s = 3.86v1' + 4.60v2'
(2)
Not bad, two equations with two unknowns, so we will solve equation (1) above in
terms of v2':
v1' = -0.839*v2'
Now, all you have to do substitute v1' into equation (2) and solve for v2':
18 m/s = -3.24v2' + 4.60v2'
m2v2' = 18 kg*m/s/1.36 = 13 kg*m/s
v2' = 13 kg*m/s/6.0 kg = 2.2 m/s
Substitute v2' into equation (2) and solve for v1':
18 m/s = 3.86v1' + 4.60*2.2 m/s
v1' = 2.0 m/s
m1v1' = 6.0 kg*2.0 m/s = 12 kg*m/s
Problem 2 was solved somewhat different than problem 1. This was to show that
sometimes there is more than one way to solve a problem. An important consideration
in this problem is that an elastic collision was not assumed. Remember a couple of
ideas:

An isolated system is one in which there is no net force acting on it and the
conservation of momentum always applies for such a system.

The kinetic energy of an isolated system need not be conserved.