Procedure:
1) Set up the
apparatus as shown in fig 1. Use the bubble level to verify that track is level.
Measure the mass of each object:
m1 =
0.5024 kg
m2 =
0.5134 kg
Connect the
motion detector to the computer and open logger pro.
2) Check to make sure the motion detector is working properly by pressing collect then moving the cart nearest the detector back and forth. Does it provide a reasonable graph of position vs time?
Yes, as the cart is moved away from
the detector, the functional value increases steadily, and as the cart is moved
towards the detector, the functional value decreases.
Position the carts so that their
Velcro pads are facing each other.
3) With the second cart
(m2) at rest, give the first cart
(m1) a gentle push away from the
detector. Observe the position vs. time graph before and after the collision.
Pictured in fig 2 is an educated guess as to what the position vs. time graph
should look like.
 |
| Estimation of the position versus time graph |
What should these graphs look
like?
The slope of the graph after the
collision should be less than the slope of the graph after the collision.
Pictured in fig 2 is my prediction of what the graph should look
like.
The slope of the position vs. time
graphs will give us our velocities directly before and after the collision. To
avoid dealing with friction, we will find velocities at the instant before and
after the collision.
Is this a good approximation, why
or why not?
Yes, because it will give us the
instantaneous velocity directly before and after impact.
Select a very small range of data
directly before the collision and apply a linear fit to the range. Record the
slope (velocity) of this line. Repeat for a small range of data points directly
after the collision.
v1 =
0.4171 m/s
V = 0.2305
m/s
Pictured in fig
3 is the position vs. time graph of one trial run we completed.
 |
| Actual position versus time using the computer software |
4) Repeat for
two more collisions. Calculate the momentum of the system the instant before and
after the collision for each trial and find percent difference.
The following
calculations are for the first trial in this part of the
experiment.
Pi =
Pf
m1v1
+ m2v2 = (m1 +
m2)V
where:
m1 =
0.5024 kg
m2 =
0.5134 kg
v1 =
0.4171 m/s
V = 0.2305
m/s
Pi =
(0.5024 kg)( 0.4171 m/s) + (0.5134 kg)(0)
Pi =
0.21 kg m/s
Pf =
(0.5024 kg + 0.5134 kg)(0.2305 m/s)
Pf =
0.23 kg m/s
Percent
difference:
[(0.23 kg m/s -
0.21 kg m/s)/ 0.23 kg m/s]*100 = 10.5%
Pictured in fig
4 is data table of each trial and the momentum calculations
Data table.
5) Add 500 g of
mass to cart number 2 and repeat steps 3 and 4.
m2 =
1.01 kg
Pictured in fig
5 is the data table for the additional mass on cart two.
data table where weight is in cart 2
What do the
graphs of velocity vs. time, and acceleration vs. time look like?
Pictured in fig
6 are my predictions of what velocity vs. time and acceleration vs. time should
look like.
predicted graphs
6) remove the
mass from cart two, and now add the mass to cart one.
m1 =
1.00 kg
Pictured in fig
7 is the data collected for the trials with mass added to cart one.
data table where weight is in cart one
The average of
all the percent differences that we had found was 10 % difference from the law
of conservation of momentum.
How well is the
law obeyed based on the results of your experiment?
The law of
conservation of momentum was reasonably obeyed. Although we did encounter some
larger percent differences, it may have been due to the range of data that was
selected.
8) For each of
the above nine trials, calculate the kinetic energy of the system before and
after the collision. Find the percent kinetic energy lost during each collision.
Show sample calculations here:
∆K/Ki
* 100 = percent difference
[(Kf
- Ki) /Ki ]*100
[(1/2*(m1
+ m2)*V2 – ½* m1*v12)/ ½*
m1*v12]*100
[((m1
+ m2)*V2)/ m1*v12) –
1]*100
[((0.5024 kg +
0.5134 kg)* 0.23052 m/s)/ 0.5024 kg * 0.41712 m/s) –
1]*100 = 38 % difference
Pictured in fig
8 is the data showing the kinetic energy before and after collision, as well as
the percent kinetic energy lost.
data for kinetic energy
9) Do
theoretical calculations for ∆K/Ki
* in a perfectly inelastic collision
1- a mass (m) colliding with an identical mass (m) initially at
rest
2- a mass (2m) colliding with a mass (m) initially at rest
3- a mass (m) colliding with a mass (2m) initially at rest
Conclusion:
During this lab, we were able to analyze the motion of two low friction carts during an inelastic collision and verify that momentum is conserved. We learned that if we can take a system where there are no outside forces acting within the system, the momentum will be conserved. Possible sources of error in this lab would include the following:
1) Selecting a bad range of data from the position vs. time function, resulting in an inaccurate velocity value.
2) The carts may have taken a few milliseconds to actually stick together.
3) Friction was present in the system, resulting in a slight loss of momentum.
Further investigations that could be followed after this lab would be to determine the effect of the impulse on the momentum, or maybe even taking friction into account.
During this lab, we were able to analyze the motion of two low friction carts during an inelastic collision and verify that momentum is conserved. We learned that if we can take a system where there are no outside forces acting within the system, the momentum will be conserved. Possible sources of error in this lab would include the following:
1) Selecting a bad range of data from the position vs. time function, resulting in an inaccurate velocity value.
2) The carts may have taken a few milliseconds to actually stick together.
3) Friction was present in the system, resulting in a slight loss of momentum.
Further investigations that could be followed after this lab would be to determine the effect of the impulse on the momentum, or maybe even taking friction into account.
