# Lab Activity: Conservation of Momentum

• dani123
In summary: You are told that cart #2 is stationary. ...initially? Initially, both carts are stationary.Is momentum only conserved if it is in an elastic collision? In an inelastic collision, the objects involved lose some kinetic energy, usually in the form of heat or sound. In an elastic collision, the objects involved keep all their kinetic energy. If you come to a complete stop, you've lost all your kinetic energy. That's an inelastic collision.In part d) do we use the fact that total kinetic energy before and after collision are not equal? Yes. Draw the displacement-time graph as requested in part a), and you should be able to see the loss of kinetic energy that you need to show
dani123

## Homework Statement

Suppose we have done a conservation of momentum lab the smart pulley apparatus shown in the figure provided. With this apparatus we can time the motion of cart #1 very accurately. In the beginning cart #2 was stationary. We set cart #1 in motion by giving it a slight push. After the collision we continued to time their motion.

A portion of raw data is given in the table provided. Note cart #1 has a mass of 500g while cart #2 has a mass of 800 g. Analyze and draw conclusions about the lab as follows.

Time in seconds Distance in centimeters
1.1 4.34
1.2 4.7
1.3 5.05
1.4 5.42
1.5 5.76
1.6 6.2
1.7 6.54
1.8 6.9
1.9 7.26
2 7.46
2.1 7.59
2.2 7.72
2.3 7.85
2.4 7.98
2.5 8.11
2.6 8.24
2.7 8.37
2.8 8.5

a) draw a displacement-time graph of the motion from the data provided.

b) from your sketch determine the approximate time of collision.

c) using the slope of the tangent method determine the initial speed of cart #1 and cart #2

d)show that the collision was not elastic

e)determine if momentum was conserved

## Homework Equations

KE=1/2 m*v2

FΔt=mΔv= change in momentum= impulse

## The Attempt at a Solution

For part b) do I use the motion formula inorder to determine acceleration and then use the impulse equation to determine the time of collision?
For cart #1 do I consider it as stationary as well in the beginning or do I consider as in motion initially?

Is momentum only conserved if it is in an elastic collision?

In part d) do we use the fact that total kinetic energy before and after collision are not equal?

if anyone could give me a hand with this problem, it would be greatly appreciated! Thanks so much in advance.

dani123 said:

## Homework Statement

Suppose we have done a conservation of momentum lab the smart pulley apparatus shown in the figure provided. With this apparatus we can time the motion of cart #1 very accurately. In the beginning cart #2 was stationary. We set cart #1 in motion by giving it a slight push. After the collision we continued to time their motion.

A portion of raw data is given in the table provided. Note cart #1 has a mass of 500g while cart #2 has a mass of 800 g. Analyze and draw conclusions about the lab as follows.

View attachment 45600

Time in seconds Distance in centimeters
1.1 4.34
1.2 4.7
1.3 5.05
1.4 5.42
1.5 5.76
1.6 6.2
1.7 6.54
1.8 6.9
1.9 7.26
2 7.46
2.1 7.59
2.2 7.72
2.3 7.85
2.4 7.98
2.5 8.11
2.6 8.24
2.7 8.37
2.8 8.5a) draw a displacement-time graph of the motion from the data provided.

b) from your sketch determine the approximate time of collision.

c) using the slope of the tangent method determine the initial speed of cart #1 and cart #2

d)show that the collision was not elastic

e)determine if momentum was conserved

## Homework Equations

KE=1/2 m*v2

FΔt=mΔv= change in momentum= impulse

## The Attempt at a Solution

For part b) do I use the motion formula inorder to determine acceleration and then use the impulse equation to determine the time of collision?
I don't think there's any part of the lab exercise where you need to determine acceleration.

Draw the displacement vs. time graph as requested in part a). The time of the collision should be pretty obvious. You can probably get a little more accurate if you use a straight edge. (You'll see what I mean after you plot all the points on the graph.)
For cart #1 do I consider it as stationary as well in the beginning or do I consider as in motion initially?
I think you're supposed to consider it as being in motion initially. Ignore the time involving the slight push you gave the cart before the measurements began.
Is momentum only conserved if it is in an elastic collision?
Momentum is always conserved. Always. It doesn't matter if collisions are elastic or inelastic.

But in order to always see that, you'd need to consider the momentum of everything involved. In this case, there is also the momentum of table+Earth that you would need to consider to be totally accurate. In other words, if you wanted the momentum before and after the collision of the carts to be conserved -- and the carts alone -- you'd better be sure there are no external forces involved, including no frictional forces from the table. But of course that's not 100% practical. Let me put it a different way, and ignore the collision for a moment. Think about a single cart and the friction between it and the table. What I'm saying here is that when friction [The friction from the table that is attached to the Earth -- not the friction between the two carts] slows down the cart, thus changing its momentum, it changes the momentum of the table+Earth by the same amount but in the opposite direction. You don't need to worry about that too much for this exercise, but just I wanted to point out that momentum is always conserved, in a general sense.

In this lab exercise, if you find that the numbers don't work out and momentum doesn't "seem" to be conserved, you should probably try to explain why. Experimental error? And what are the possible sources? Was the total mass of each cart measured properly? If the collision caused a rather large, momentary frictional interaction between the carts and table (ignoring the frictional interaction between the carts themselves), how would that effect the combined-carts' momentum?
In part d) do we use the fact that total kinetic energy before and after collision are not equal?
That's right.

Thank you so much for the quick reply! I just did my graph with excel and the only point where i can see there being a difference in the curve would be at 2.0 seconds, is this when the collision occured?

If I am trying to find the final speed of the cart #1 do I do the following?
vf= 0.085m/2.0s= 0.03m/s
m1vi=m1vf
vi=[m1*vf]/m1

or do I do this...

vi=[(m1+m2)*vf]/m1... if i do it this way, Id get a faster initial speed then my final speed.

OR
do I just look at my table and use the first point as my initial? but then again I don't know what the table is measuring... is it the measurements taken from when the two carts are in contact?

dani123 said:
Thank you so much for the quick reply! I just did my graph with excel and the only point where i can see there being a difference in the curve would be at 2.0 seconds, is this when the collision occured?
When I graphed your data, I found that it is close to 2.0 seconds, but not quite. You should be able to estimate it a little more accurate than that.
dani123 said:
If I am trying to find the final speed of the cart #1 do I do the following?
vf= 0.085m/2.0s= 0.03m/s
m1vi=m1vf
vi=[m1*vf]/m1
Umm, no. 0.85 m is the total distance traveled, before and after the collision, and 2.0 seconds is the time stamp somewhere near the collision, but that doesn't really help.

Calculate the rise over the run (the slope). You are looking for Δxt. The important thing, being the deltas.

Pick a part of the line near the collision, but a little before it. I suggest staying way from the data points 1.9 and 2.0, because the collision occurred somewhere right around there. Maybe pick a segment just before that. Pick a small segment of time around this section of plot, and call that Δt. Then find corresponding Δx. (These deltas are not absolute time stamps and absolute lengths -- you need to calculate the differences between points on the graph.)

or do I do this...

vi=[(m1+m2)*vf]/m1... if i do it this way, Id get a faster initial speed then my final speed.
You can't do that because that formula assumes that momentum is perfectly conserved. But conservation of momentum is what you are ultimately trying to check. So instead, you need to calculate the velocities directly from the data.
OR
do I just look at my table and use the first point as my initial?
If I were you, I would stay away from data points that might correspond to the time at which you briefly pushed the cart. Anything just after that you could use for the cart's initial velocity. I'd also stay away from points that might have involved the collision itself.
but then again I don't know what the table is measuring... is it the measurements taken from when the two carts are in contact?
You were the one that took the data. I should be asking you. But I can say from the data that you posted in the original post, the data measures position vs. time.

For what it's worth, it looks to me as though the data corresponds to times well before and well after the collision (including the moment of collision).

I actually wasn't the one who took this data, it was given to me! and thanks for your reply!:)

## 1. What is the purpose of the conservation of momentum lab activity?

The purpose of this lab activity is to demonstrate the principle of conservation of momentum, which states that in a closed system, the total momentum of all objects before a collision is equal to the total momentum of all objects after the collision.

## 2. How is momentum conserved in a closed system?

In a closed system, momentum is conserved because there are no external forces acting on the system. This means that the total momentum of all objects in the system remains constant, even after a collision or interaction between objects.

## 3. What materials are needed for this lab activity?

The materials needed for this lab activity may vary, but typically include: a track or surface for objects to roll on, objects with different masses (such as marbles or toy cars), a stopwatch or timer, and measuring tools such as a ruler or scale.

## 4. Can you explain the difference between elastic and inelastic collisions?

An elastic collision is one in which kinetic energy is conserved, meaning the objects bounce off each other with no loss of energy. In an inelastic collision, kinetic energy is not conserved and some energy is lost, typically in the form of heat or sound.

## 5. How does the conservation of momentum relate to real-life situations?

The conservation of momentum is a fundamental law of physics and applies to all objects in motion. It can be seen in everyday situations such as a game of billiards, a car crash, or even a rocket launch. Understanding this principle can help us predict and explain the motion of objects in the world around us.

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