11.Inelastic Collisions

Purpose: To analyze the motion of two low friction carts during an inelastic collision and verify that the law of conservation of linear momentum is obeyed

Equipment: Computer with Logger Pro software, lab pro, motion detector, horizontal track,two carts, 500 g masses(3), triple beam balance, bubble level

Introduction: This experiment uses the carts and track as shown in the figure. If we regard thesystem of the two carts as an isolated system, the momentum of this system will beconserved. If the two carts have a perfectly inelastic collision, that is, stick together after the collision, the law of conservation of momentum says

Pi = Pf

m₁v_{1 +} m₂v₂ = (m₁ + m₂) * V

where v₁ and v₂ are the velocities before the collision and V is the velocity of thecombined mass after the collision.

 

Procedure:

1. Set up the apparatus as shown in Figure 1. Use the bubble level to verify that the track is aslevel as possible. Record the mass of each cart. Connect the lab pro to the computer and themotion detector to the lab pro. On the computer, start the Logger Pro software; open theMechanics folder and the Graphlab file.


2. First, check to see that the motion detector is working properly by clicking the Collect buttonto start collecting data. Move the cart nearest the detector back and forth a few times whileobserving the position vs time graph being drawn by the computer. Does it provide a reasonablegraph of the motion of the cart? Remember to be aware of unwanted reflections caused byobjects in between the motion detector and the cart. Also, position the carts so that their velcropads are facing each other. This will insure that they will stick together after the collision.


3. With the second cart (m2) at rest give the first cart (m1) a moderate push away from themotion detector and towards m2. Observe the position vs time graph before and after thecollision. What should these graphs look like? Draw an example:

m1= 501.5g m2= 516.1g

the graph look like two parts of continuous increasing liner curves. The slope of first part should be lower than the second part.

The slope of the position vs. time graph directly before and directly after the collision gives thevelocity directly before and directly after the collision. To avoid the problem of dealing with friction forces (Remember, we are assuming the system is isolated.), we will find the velocity of the carts at the instant before and after the collision.

Is this a good approximation? Why or why not?

It's a good approximation, because we focus on the change of the velocity after the collision, so we should only compare the instant velocity before and after the collision. Due to the friction and air resistance, the velocity will have big error if we use the velocity in a large range of time.

For the velocity before the collision, select a very small range of data points just before the collision. Avoid the portion of the curve which represents the collision. Choose Analyze/LinearFit. Record the slope (velocity) of this line. Repeat for a very small range of data points just afterthe collision. Record this slope (velocity) as well.

p-t graph when m1= 501.5g and m2= 516.1g

Slope before the collision: 0.444
Slope after the collision: 0.150

4. Repeat for two more collisions. Calculate the momentum of the system the instant before andafter the collision for each trial and find the percent difference. Put your results in an Excel datatable. Show sample calculations here:

m1= .5024kg

m2= .5134kg

vbefore is equal to the slope of a very small range of data points before the collision.

vafter is equal to the slope of a very small range of data points after the collision.

pi = m1× vbefore + m2 × 0

pf = (m1+ m2) × vafter

perror = (pf - pi ) / pf

5. Place an extra 500 g on the second cart and repeat steps 3 and 4. Sketch one representativegraph showing the position vs time for a typical collision. (What do velocity vs. time and acceleration vs. time look like?

m1= 501.5g
m2= 516.1g +495g= 1011.1gThe velocity is almost constant before and after the collision, because according to the a-t graph, the acceleration is almost 0. The velocity after the collision is lower than the velocity before the collision.

pi = m1× vbefore + m2 × 0

pf = (m1+ m2) × vafter

perror = (pf - pi ) / pf

6. Remove the 500g from the second cart and place it on the first cart. Repeat steps 3 and 4.

m1= 501.5g +495g= 996.5g
m2= 516.1g

p-t graph when m1= 996.5g and m2= 516.1g

7. Find the average of all of the percent differences found above. This average represents your verification of the law of conservation of linear momentum. How well is the law obeyed based on the results of your experiment? Explain.

Based on our result, we found our average value of momentum has less than 10 percent difference between pf and pi. We can say in the acceptable error range, the momentum before and after the collision is almost the same.

8. For each of the nine trials above calculate the kinetic energy of the system before and afterthe collision. Find the percent kinetic energy lost during each collision. Put this information in aseparate data table.

9. Do a theoretical calculation for ΔK/K in a perfectly inelastic collision for the three situations:

1. a mass, m, colliding with an identical mass, m, initially at rest.

mvb = 2mva ------ vb = 2va
K = (1/2)×m×(vb)^2
ΔK = (1/2)×2m×(va)^2- (1/2)×m×(vb)^2


ΔK/K= -50%


2. a mass, m, colliding with a mass, 2m, initially at rest.

mvb = 3mva ------ vb = 3va

K = (1/2)×m×(vb)^2

ΔK = (1/2)×3m×(va)^2- (1/2)×m×(vb)^2

ΔK/K= -66.7%

3. a mass, 2m, colliding with a mass, m, initially at rest.

2mvb = 3mva ------ vb = (3/2)va

K = (1/2)×2m×(vb)^2

ΔK = (1/2)×3m×(va)^2- (1/2)×2m×(vb)^2

ΔK/K= -0.5%

Conclusions:

Inelastic collisions is a collision in which kinetic energy is not conserved, even though the momentum of the system can be conserved. When there is no external forces in a system, then the momentum of this system is conserved.
An impulse delivered to a particle changes the particle's momentum. The momentum after the impulse is equal to the momentum before the interaction plus the impulse that arises from the interaction.