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Experiment 4
Inelastic Collisions
4.1 Objectives
- Measure the momentum and kinetic energy of two objects before and after a perfectly inelastic one-dimensional collision.
- Observe that the concept of conservation of momentum is in- dependent of conservation of kinetic energy, that is, the total momentum remains constant in an inelastic collisions while the kinetic energy changes.
- Calculate the percentage of KE which will be lost (converted to other forms of energy) in a perfectly inelastic collision between an initially stationary mass and an initially moving mass.
4.2 Introduction
One of the most important concepts in the world of physics is the concept of conservation. We are able to predict the behavior of a system through the conservation of energy (energy is neither created nor destroyed). An interesting fact is that while total energy is always conserved, kinetic energy is not. However, momentum is always conserved in both elastic and inelastic collisions. In this experiment and the following experiment, we will see how momentum always remains a conserved quantity while kinetic energy does not.
- Inelastic Collisions
4.3 Key Concepts
As always, you can find a summary on-line at Hyperphysics.^1 Look for keywords: elastic collision, and inelastic collision.
4.4 Theory
The following two experiments deal with two different types of one-dimensional collisions. Below is a discussion of the principles and equations that will be used in analyzing these collisions. For a single particle, momentum is defined as the product of the mass and the velocity of the particle:
p = mv (4.1) Momentum is a vector quantity, making its direction a necessary part of the data. For the one-dimensional case, the momentum would have a direction in either the +x direction or the −x direction. For a system of more than one particle, the total momentum is the vector sum of the individual momenta:
p = p 1 + p 1 + ... = mv 1 + mv 2 + ... (4.2) So you just add the momentum of each particle together. One of the most fundamental laws of physics is that the total momentum of any system of particles is conserved, or constant, as long as the net external force on the system is zero. Assume we have two particles with masses m 1 and m 2 and velocities v 1 and v 2 which collide with each other without any external force acting. Suppose the resulting velocities are v 1 f and v 2 f after the collision. Conservation of momentum then states that the total momentum before the collision (pinitial = pi) is equal to the total momentum after the collision (pf inal = pf ):
pi = m 1 v (^1) i + m 2 v (^2) i pf = m 1 v (^1) f + m 2 v (^2) f pi = pf (4.3)
In a given system, the total energy is generally the sum of several different forms of energy. Kinetic energy is the form associated with motion, and for a single particle (^1) http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html
- Inelastic Collisions
KEi =
m 1 v^21 i 2
m 2 v^22 i 2
m 1 v (^12) i 2
because Cart 2 is initially at rest (v (^2) i = 0). The final kinetic energy is defined as
KEf =
(m 1 + m 2 )v f^2 2
because the two carts have stuck together after the collision (vf = v 1 f = v (^2) f is the common velocity of the two carts). Using the conservation of momentum, we can calculate the final momen- tum as
m 1 v (^1) i + m 2 v (^2) i = m 1 v (^1) i = (m 1 + m 2 )vf (4.9) Using Eqs. 4.7, 4.8, and 4.9, we arrive at the equation for KEf in terms of KEi.
KEf =
m 1 (m 1 + m 2 )
KEi (4.10)
This is the prediction for the final kinetic energy of a perfectly inelastic collision.
4.5 In today’s lab
Today you will get to see how inelastic collisions work while you vary the masses on two colliding carts. You will then see how there is a significant energy loss in these types of collisions and will try to figure out where this energy goes.
4.6 Equipment
4.7. Procedure
Figure 4.1: Equipment used in lab fully set up.
- Two carts one with needle and one with clay (carts are sometimes called gliders)
- Photogate Circuit
- 4 - 50g masses
4.7 Procedure
Do not move the carts on the air track when the air is not turned on. It will scratch the track and ruin the “frictionless” environ- ment we need to get accurate data.
- Start by making sure that the air track is level. Your instructor will demonstrate how at the beginning of class.
- Set up the photogates such that there is sufficient room for the collision to happen in the middle and enough room on the remainder of the track for the carts to move freely.
- Set the photogates to GATE mode.
- We will define Cart 1 as the cart with the fin and Cart 2 as the cart without. We will always push Cart 1 for each trial and will always start with Cart 2 stationary (v (^2) i = 0 cm/s ) in the middle. Before placing the carts on the track, measure the mass of them without the extra masses. Record the empty cart masses data on the given results sheet.
4.8. Uncertainties
- Repeat this trial one more time and record the results.
- Repeat steps 6–15 for the cases when you have:
- 2 mass disks on Cart 1 and 2 mass disks on Cart 2
- 2 mass disks on Cart 1 and no mass disks on Cart 2
- Be sure to include hand calculations for the light blue boxes in excel.
4.8 Uncertainties
In today’s experiment we have already input all of the equations into excel for you out of the interest of brevity, but it is important to understand the uncertainties for the values you used in this experiment. The uncertainty for velocity is:
δv = v
δL L
δt t
The uncertainty for momentum is:
δP = P
δv v And the uncertainty for kinetic energy is:
δKE = 2KE
δv v The uncertainties for the differences for the momenta and kinetic energies are then:
δPdiff = δPf + δPi and δKEdiff = δKEf + δKEi
4.9 Checklist
- Excel sheets
- Questions
- Hand Calculations
- Inelastic Collisions
- Compare one of your measured KEf trials with the KEf calc prediction of equation 4.10 for a perfectly inelastic collision. Use your measured masses and KEi value. Are they compatible? The uncertainty of KEf calc is: ( δKEf calc = KEf calc
δKEi KEi
- Combine equations 4.7, 4.8 and 4.9 to obtain the expression in equation 4.10. Hint: solve equation 4.9 for vf , then substitute this into equation 4.8.
