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
m1v1 + 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.
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| p-t graph when m1= 501.5g and m2= 516.1g |
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= 501.5g
m2= 516.1g
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
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= 1011.1g
The acceleration is almost zero before and after the collision.
The 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.
pf = (m1+ m2) × vafter
perror = (pf - pi ) / pf
m2= 516.1g +495g= 1011.1g
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| p-t graph when m1= 501.5g and m2= 1011.1g |
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| a-t graph when m1= 501.5g and m2= 1011.1g |
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| v-t graph when m1= 501.5g and m2= 1011.1g |
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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
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.
m2= 516.1g
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| 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. Show sample calculations here:
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.
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. Show sample calculations here:
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| the percent kinetic energy lost for each trial |
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= -33.3%
Conclusions:
• What you learned
I learned that momentum is equal to mass times velocity. If 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.
Inelastic collisions is a collision in which kinetic energy is not conserved, even though the momentum of the system can be conserved.
• Sources of error
1. The table is not exactly horizontal, so the component of the gravity force will influence the result.
2. The friction and the air resistance are external forces of the system that can't be avoided, so our graphs are not perfect.
• Compare your experimental numbers calculated in part 8 above with the results of your theoretical calculations in part 9.
We find in part 8, the calculated values of percent energy lost of trials 4 to 9 are close to the theoretical values in part 9, but the values of trail 1,2 and 3 are little bit biased. The reason is may be that in trail 1,2 and 3, we didn't put the bar on either car. So, the mass of the system is the smallest, which is easy to be influenced by external forces, like friction and air resistance.
• If momentum is conserved, is kinetic energy also conserved?
No. Inelastic collisions is a collision in which kinetic energy is not conserved. But if there is no external forces in the system, the momentum of the system can also be conserved.
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