Download Collisions

Conservation of Momentum
Purpose: Gain an understanding of the physics of collisions
Materials: PHET software: http://phet.colorado.edu/en/simulation/collision-lab
Discussion: Newton’s third law states that every action is accompanied by an equal and
opposite reaction. When you push on an object, it pushes back. If you stand on a
skateboard and throw a heavy ball forward, you are pushed backward with and equal
amount of force. When two objects collide, equal and opposite forces push on the two
objects for equal times. Their motion is changed inversely to their mass. If their masses
are equal, their motion is changed by the same amount, but in opposite directions.
Collisions are categorized by their elasticity. A collision is elastic if no kinetic energy is
lost. Atoms and molecules collide in this way. The collision is considered inelastic if
kinetic energy is converted to heat. Billiard balls and air hockey pucks collide with only a
slight loss of kinetic energy, while cars collide with a great loss of kinetic energy. In a
perfectly inelastic collision, the objects stick together when they collide.
Momentum, p, is the product of mass times velocity: p=mv. Momentum is a vector
quantity that points in the same direction as velocity. The change in momentum of each
object resulting from a collision is equal to the impulse, force times time, on that object.
As per Newton’s third law, these must be equal and opposite. This leads to the law of
conservation of momentum. If the impulse is equal and opposite, so is the change in
momentum on each object, so the net change in momentum must be zero regardless of
any conversion of mechanical energy to heat.
Σp = Σp’ (primes are used for after the collision.)
For elastic and inelastic collisions,
m1v1 +m2v2 = m1v’1 + m2v’2
For perfectly inelastic collisions,
m1v1 +m2v2 = (m1+ m2) v
For elastic collisions, conservation of energy may also be used:
KEbefore=KEafter
m1v2 + m2v2 =m1 v’2 + m2v’2
Collisions in two dimensions: If the collision happens in a two dimensional frame, the
collision may be broken into its x and y components. In this way, the momentum can be
considered along each axis. Total momentum is conserved: Σp = Σp’. Component
momentum is also conserved: Σpx = Σp’x and Σpy=Σp’y. Momentum is a vector quantity
and must be added like vectors, not like scalars.
Part 1: Collisions in 1-D
Procedure: Use the PHET software http://phet.colorado.edu/en/simulation/collision-lab
Set up the collisions listed in the table below. Set the collision type to 1 dimension and
select the “more data” button. You should now see a one dimensional set up with an
enlarged data table to include the velocities of the objects. All units are MKS and assume
at least 2 significant figures. Pause immediately after the collision and record the
velocities of objects 1 and 2 in the table below. The elasticity relates to the mechanical
kinetic energy converted to heat in the collision. 100% elasticity is an elastic collision.
All others are inelastic. 0% elasticity is a perfectly inelastic collision.
The following tables serve as guides only. Build your own tables with correct units in the
headers.
trial
1
2
3
4
mass1
1.0
1.0
3.0
3.0
mass2
2.0
2.0
2.0
2.0
vx1
1.0
1.0
2.0
2.0
vx2
-1.0
-1.0
-1.0
-1.0
v’x1
v’x2
elasticity
100%
50%
50%
0%
Using the appropriate collision type, calculate total momentum before, Σp, and after, Σp’.
Also calculate the change in momentum, Δp=final - initial. Using the kinetic energy
equation: KE = .5mv2, calculate total kinetic energy before and after each collision; as
well as, the change in kinetic energy, ΔKE=final-initial .
trial
1
2
3
4
Σp
Σp’
Δp
ΣKEbefore
ΣKEafter
ΔKE
type of collision
Results for Part 1: Fill in the above table and display in the results section and discuss
change in momentum and energy in context of the type of collision.
Part 2: Collisions in 2-D
Choose 2 Dimensions and set the following initial conditions: elasticity at 50%
mass
2.0
3.0
ball 1
ball 2
x
1.0
2.0
y
0
.50
vx
2.0
0
vy
0
-1.0
vx’
vy’
Results for part 2: Complete the following table and discuss your findings with regard to
change in momentum and energy.
px
ball 1
ball 2
Sum
py
p
px’
py’
p’
∆p
KEbefore
KEafter
∆KE
Supplemental Questions: write the questions, circle your answers, and show all work.
1. In some of the above collisions, KE is less after the collision than before. Since energy
cannot be lost, what happened to it? What could you do to show this is true?
2. A 2.00g bullet strikes a 2.00kg block hanging from a long cord. The bullet imbeds
itself in the block causing the block with bullet embedded to swing upwards to a height of
50.0 cm above its original position.
A. What type of collision best describes the bullet striking the block?
B. Do a walk-through of the energy of the system starting just before the bullet strikes the
block and continue until the block and bullet reach max height.
C. Find the velocity of the bullet before it collides with the block.
3. A 450g rubber ball is dropped from 2.0m above the floor. After the bounce, the ball
only rises to a height of 1.5m above the floor.
A. Determine the change in momentum of the ball when it hits the floor and bounces up.
(hint: either use motion equations or conservation of energy to determine the speed of the
ball when it reaches the floor.)
B. Calculate the amount of heat generated when the ball hits the floor.
C. Explain how momentum is conserved when the ball collides with the floor.