Momentum and Energy in Two Dimensional Inelastic Collisions
A collision in two dimensions occurs when two objects traveling in the same straight linecollide 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 = pf
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 anunfortunate 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°).
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 notinterchangeably. You will
need to use the one below on the left for sign (+ or -) conventions.
Momentum and Energy (Inelastic Collisions 2D) Problems
1) Two billiard balls each with a mass of 0.40 kg collide with each other near one of the
side pockets. The first ball istraveling 0.20 m/s at a bearing of 000° and the second
ball is traveling at 0.12 m/s at a bearing of 270°. Because of a mysterious and
invisible gluon, they stick together after the collision. What is the momentum of the
system after the collision?
m1 = 0.40 kg m2 = 0.40 kg
v1 = 0.20 m/s v2 = 0.12 m/s
θ1 = 000° θ2 = 270°
Because the angle between the momentum vectors is 90°, the law of cosines breaks
done to the Pythagorean theorem as the cos 90° = 0. Every little bit helps.
(mRvR)2 = (m1v1)2 + (m2v2)2
mRvR = ((0.40 kg*0.20 m/s)2 + (0.40 kg*0.12 m/s)2)1/2
mRvR = 0.093 kg*m/s
Any one of the trig functions may be used. It may be instructive to solve for the
bearing using one of the other trig functions.
tanθ = m1v1/m2v2 = 0.40 kg*0.20 m/s/0.40 kg*0.12 kg
θ = tan-1(0.20/0.12) = 59°
mRvR = 0.093 kg*m/s, 329°
2) On a punt return, a 92.0 kg return specialist is running toward midfield at 2.85 m/s.
He approaches the 50 yard line at a bearing of 305°. A 97.0 kg defensive back
tackles the returner from behind at a speed of 3.85 m/s as he approaches the
returner at an angle of 20° east of north. If the two players stick together and land as
one massive blob:
(a) What is their velocity immediately following the collision?
m1 = 92.0 kg m2 = 105 kg
v1 = 2.85 m/s v2 = 3.85 m/s
θ1 = 305° θ2 = 20°
You cannot use the Pythagoreantheorem because the included angle equals
105°. One approach is to use the law of cosines to find the magnitude and the
law of sines to find the bearing.
(mRvR)2 = (m1v1)2 + (m2v2)2 – 2*m1v1*m2v2*cosθ
mRvR = ((92.0 kg*2.85 m/s)2 + (105 kg*3.85 m/s)2 – 2*92.0 kg*2.85 m/s*105
kg*3.85 m/s*cos 105)1/2
mRvR = 536 kg•m/s
Because the included angle is not 90°, you must use the law of sines.
sinθ1/m2v2 = sinθ/mRvR where θ1 is the angle that the resultant momentum vector
makes with m1v1.
sinθ1/m2v2 = sinθ/mRvR
sinθ1/105 kg*3.85 m/s = sin 105°/536 kg*m/s
sinθ1 = 47°
mRvR = 536 kg*m/s
vR = 536 kg*m/s/(92.0 kg + 105 kg) = 2.72 m/s, 352°
(b)How much kinetic energy is lost as a result of their collision?
ΔKE ≠ 0
KEi ≠ KEf
1/2m1v12 + 1/2m2v22 ≠ 1/2m1v1'2 + 1/2m2v2'2
½*92.0 kg*(2.85 m/s)2 + ½*105 kg*(3.85 m/s)2 ≠ ½*(92.0 kg + 105 kg)*(2.72 m/s)2
1150 J ≠ 730 J
ΔKE ≠ KEf - KEi = 730 J - 1150 J = -420 J
Remember energy is a scalar quantity so the negative sign indicates that some
mechanical energy has been transformed to sound..
3) A skater with a mass of 75.0 kg is traveling at 3.50 m/s at a bearing of 200°. She
collides with another skater having amass of 55.0 kg. The second skater is traveling
at 8.00 m/s at a bearing of 340°. If the skaters remain stuck together, what is their
final velocity?
m1 = 75.0 kg m2 = 55.0 kg
v1 = 3.50 m/s v2 = 8.00 m/s
θ1 = 200° θ2 = 340°
Problem 3 is similar to Problem 2 but an alternate method of solution will be shown.
Unless told otherwise, either method is correct and depends on your preference. The
good news is that the Pythagorean theorem can be used after the resolution of
forces.
Both momenta have to be resolved into two right angle components. Repetitive
calculations will be shown with fewer steps after the initial calculation.
One momentum vector is resolved into a west and south component.
sin20° = (m1v1)H/mRvR = 75.0 kg*3.50 m/s
(m1v1)H = 75.0 kg*3.50 m/s*0.342 = 89.78 kg*m/s, 270°
cos20° = (m1v1)V/mRvR = 75.0 kg*3.50 m/s
(m1v1)V = 246.67 kg*m/s, 180°
The second momentum vector is resolved into a west and north component.
Calculations similar to those for the first momentum vector, give:
(m2v2)H = 150.49 kg*m/s, 270°
(m2v2)V = 413.46 kg*m/s, 360°
Σp = 0
Σ(m2v2)H = 89.78 kg*m/s + 150.49 kg*m/s = 240.27 kg*m/s, 270°
Σ(m2v2)V = 413.46 kg*m/s - 246.67 kg*m/s = 166.79 kg*m/s, 000°
(mRvR)2 = (m2v2)H2 + (m2v2)V2
mRvR = ((240.27 kg*m/s)2 + (166.79 kg*m/s)2)1/2 = 292.49 kg*m/s
sinθ = (m2v2)V/mRvR = 166.79 kg*m/s/292.49 kg*m/s
θ = 35°
mRvR = 292.49 kg*m/s/(75.0 kg + 55.0 kg) = 2.25 m/s, 305°
Using the law of cosines and the law of sines looks to be shorter but there is a
gotcha! The law of sines gives θ = 75° but you can't add the 75° to the bearing of
m1v1 because your final answer would be 275°.
Without doing the problem graphically or analyzing the initial momenta, you may
overlook thissubtlety. If the problem was done graphically, it would be obvious from
the vector diagram that the angle must be 105°. Remember the sin 75° = sin 105°.
The momentum at a bearing of 340° is440 kg*m/s and thebearing at 200° is
262.5 kg*m/s. This means the bearing ofthe resultant should becloser to 340° and
not 200°.