Momentum

Purpose

 To investigate the behavior of objects colliding in elastic and inelastic collisions

 To investigate momentum and energy conservation for a pair of colliding carts

 To investigate the effect of the type of collision on the change in kinetic energy of a system of colliding carts

Equipment

Virtual Momentum Track VPL Grapher PENCIL

See Video Overview at: http://virtuallabs.ket.org/physics/apparatus/07_mom/

Explore the Apparatus

Open the Virtual Momentum Lab on the website. A lot should look familiar after having worked with the Dynamics Track.

Figure 1 – The Momentum Track

Let’s take a trial run with the apparatus.

Toss the carts around a bit to see how they interact. You’ll see that they bounce nicely off of each other and lose a bit of

speed when they collide with the orange bumpers. Turn on the “Velcro®” and try it again. Hopefully they act as you’d

expect.

Turn the “Spring Plunger” back on. Start both carts moving. Now click on either cart. The other one stops too. But when you

release the mouse button the unclicked cart starts moving again. The best way to stop them both is with the stop carts button.

Try adding some mass to one cart. You have to release the mass while it’s over the mast (the rod holding up the flag). Bang

the carts around a bit to see how the mismatched carts behave.

Now try the launchers. Here’s how they’re operated: Drag a cart to the launcher bumper and continue dragging it until it

appears to go behind the bumper. Release the cart. This engages the cart and bumper. Now cock the launcher spring by

pulling back the white circle on the side of the launcher. Click the release button to fire. Try again with the launcher spring

pulled back different distances. Try launching both at once (release both).

This would be a good time to try out the “Motion Sensor.” It’s a bit trickier with this lab since you have two carts to work

with as well as abrupt changes in speed and direction. Try this. Double the mass of the red cart by adding 250 grams. Attach

it to its launcher and pull back the spring about 50%. Put the blue cart at mid-track. Switch to the “Spring Plunger” if you still

have the “Velcro®” on.

You’ll soon need to change the masses on the carts. You remove masses by double-clicking on the stack of masses. In

computer lingo the order is LIFO. “Last in, first out.” The last mass added (the top one) is the first one removed by clicking.

http://virtuallabs.ket.org/physics/apparatus/07_mom/
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 Turn on the “Motion Sensor” by clicking it.

 Release the left launcher.

 Once the carts collide, click the “Motion Sensor” to turn it off.

 You should see a graph similar to the one shown in Figure 2.

Figure 2: Dual x-t Graphs

Let’s identify all the parts of this interaction on our graph.

When you turn on the “Motion Sensor,” each cart is stationary until you

click the launcher release button. During this time each cart has an

initial, constant position (zero velocity) as indicated by a horizontal (t, x)

lines.

When the launcher is fired the red cart is given an initial velocity. (We

ignore acceleration in these graphs).

The red cart collides with the blue cart.

The red cart is pushed to the left, slowing it down.

The blue cart is pushed to the right with an equal, but opposite force,

thus giving it a positive velocity which appears to be somewhat faster

than the red cart’s initial velocity.

Figure 3: Before and After Collision

Let’s analyze our data. You have five velocities, but you’re interested in only four of them. You are only interested in the

velocities just before and just after interactions–collisions. The blue cart is at rest before, and traveling at some positive velocity after the collision. The red cart has a positive velocity before and a smaller positive velocity after the collision. The

red cart’s initial zero velocity before launch is not related to the collision. You’ll always ignore similar segments.

The blue cart’s initial zero velocity is obvious from the graph. Let’s see how to find the other velocities using Grapher.

1. In the momentum apparatus, click Copy Data to Clipboard.

2. Open Grapher. Click in the “paste data” box and use Ctrl+V to paste it into the data table.

3. You should see three columns of data labeled “X”, “Blue”, and “Red.” These hold the data for time, blue cart position,

and red cart position, respectively. You should also see two graphs – one position-time graph for each cart. Use manual

scaling to make sure that each graph starts at (0, 0) and has the same maximum x and t values. A range of 0.0 m to 2.0 m

is good. Label the axes of the top graph “Blue Position (m)” and “Time (s).” Label the bottom graph similarly. Give each

graph an appropriate name.

5. Turn on “Linear Fit.” Determine the constant velocities of the carts prior to the collision by finding the four slopes of

their (t, x) graphs. You should know how to use the “Linear Fit” tool from your kinematics labs or by studying the Lab

Guide. You’ll note that when you drag across a certain portion of the graph a pink line of best fit is drawn for that range

of points. The data in the linear fit data box also corresponds to that data range. Thus you can determine before and after

collision velocities for each cart. The longer the line segment you drag across, the better your results will be.

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You should now have something like Figure 4. (Actually Figure 4 has been cobbled together from both graphs to make it

easier to compare the graphs). The actual velocity values depend on your initial plunger setting.

The slopes produced by Grapher are as follows.

v(Blue)o = 0 m/s

v(Red)o = .434 m/s

v(Blue)f = .577 m/s

v(Red)f = .148 m/s

These velocities and the masses are the only values

needed to analyze the momenta and energies of our

carts.

Figure 4: Finding Velocities with Grapher

Theory

In class and in your reading you’ve seen the following development of the concepts of momentum, momentum conservation,

and their connection to energy conservation. (Vectors are shown in bold type).

The momentum, p of a body of mass, m, and velocity, v is written as

= (1)

Case 1: A net external force acts on a body

When a (net) force acts on a body its velocity and hence its momentum will change according to equation 2.

∆ = ∆ (2)

where ∆ is referred to as the impulse, J.

Note that the impulse is a vector and is equal to the difference in two momentum vectors pf and po.

In words, Equation 2 says

impulse = change in the momentum (3)

Case 2: An internal force acts between two bodies. (Or more than two, but that’s out of our league).

As you know, forces come in equal and opposite pairs. So you know that in case 1 you were just focusing on one of the

bodies involved. That’s great for determining things like the effect of the force on a golf ball hit by a golf club.

There are also many situations where it’s useful to look at both bodies involved. If you want to slow down or change the

course of an incoming asteroid you need to know how big an object you need to throw at it, how fast you need to throw it and

in what direction. In these cases the impulse on the system is zero since all forces are internal to the system.

When there is a no net, external force,

F Δt = Δp = 0 (4)

That is, there is no impulse so there will be no change in momentum. Thus the momentum of the system is conserved.

p(sys)o = p(sys)f (5)

So, if there is no net, external force on the system, the momentum of the system will remain unchanged even if the

parts of the system exert forces on one another and the individual parts change their momenta.

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An excellent example of this behavior is the system of two carts that you collided together earlier. Their total momentum

before the collision, po, was equal to their total momentum after the collision, pf. Try using the data in Figure 4 to verify the

following.

For a system of two carts you can expand p(sys)o = p(sys)f to

p1o + p2o = p1f + p2f (6)

After rearranging,

Δp1 = – Δp2 (7)

That is, the loss in momentum by cart one equals the negative of the gain in momentum of cart 2.

Kinetic Energy

You’ve also become familiar with the kinetic energy, KE of an object. The equation for kinetic energy involves the same

terms, m and v.

KE = 1

2 2 (9)

So, shouldn’t the kinetic energy of the system always be conserved too? That is, shouldn’t it be the same before and after an

interaction. No. There’s a subtle difference between the equation mv for momentum and 1

2 2 for kinetic energy. The in

kinetic energy is just the speed, a scalar. So kinetic energy is a positive scalar. So we could never add up two kinetic energies

for a result of zero. Kinetic energy is not necessarily conserved. Momentum is always conserved.

In this lab we’ll observe the momentum of a system of two carts in several situations. We’ll also investigate their kinetic

energies.

Procedure

I. Conservation of Momentum in a collision

NOTE: In most of this lab you’ll use your data to answer questions. Even non-numerical questions.

Equations 6 and 7 say the same thing but in two different ways. Let’s explore both.

 Load the blue cart with 500 g, giving it a total mass of 750 g.

 Position the blue cart right between the release both and stop carts buttons.

 Engage the red cart with its launcher. Cock the launch spring about half-way

back.

 Turn on the “Motion Sensor.”

 Release the left cart.

 Somewhat before the blue cart hits the right bumper, turn off the “Motion

Sensor.” Your graph should look something like Figure5.

There may be additional lines due to collisions with the plungers. Ignore them.

Dragging the big white ball below the graph will adjust the time display scale.

p1o + p 2o = p 1f + p 2f (6)

Δp1 = – Δp2 (7)

Figure 5: A Collision

Before finding actual values for the velocities, let’s think about what the graph tells us about the signs of the momenta.

1. What is the sign of the red cart’s initial momentum, mR vRo, its momentum just before the collision? (+/-)

2. What is the sign of the red cart’s final momentum, mR vRf, its momentum just after the collision? (+/-)

3. What is the sign of the change in the red cart’s momentum, Δ(mRvR) = mR(ΔvR)? (+/-)

4. The blue cart’s initial momentum, mB vBo doesn’t have a sign. Its momentum before the collision is?

5. What is the sign of the blue cart’s final momentum, mB vBf, its momentum just after the collision? (+/-)

6. What is the sign of the change in the blue cart’s momentum, Δ(mBvB) = mB(ΔvB)? (+/-)

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So the red cart had a negative change in momentum while the blue cart had a positive change in momentum. If the total

momentum must be constant, then the sum of these changes, the total change, must add to zero. Let’s look at the numbers.

7. Copy/Paste your data from the apparatus into Grapher as before. Using the “Linear Fit” tool determine the carts’

velocities before (vo) and after (vf) the collision. Record these in the first two empty columns in Table 1.

8. Using the known masses calculate and record the momentum of each cart before and after the collision. (4 values)

9. Add the initial momenta of the carts to determine the initial momentum (p(sys)o) of the system. Similarly determine the

final momentum (p(sys)f) of the system. Record these two values in Table 1.

10. Finally, calculate and record the change in momentum of the system (Δpsys) from (p(sys)f) – (p(sys)o).

Table 1 Momentum before and after a collision (Note: careful with mass units!)

Mass of Red Cart kg

Mass of Blue Cart kg

Cart vo

(m/s)

vf

(m/s)

po

(kg • m/s)

pf

(kg • m/s)

Δp

(kg • m/s)

Red

Blue

p(sys)o = p(sys)f = Δpsys =

11. Equation 6 says that p(Red)o + p(Blue)o = p(Red)f + p(Blue)f. How did that work out? Quote your numbers.

12. Equation 7 says that p(Red)f – p(Red)o = -(p(Blue)f – p(Blue)o). How did that work out? Quote your numbers.

p(Red)f – p(Red)o-(p(Blue)f – p(Blue)f)

II. Elastic Collisions

With the “Spring Plunger” the carts seemed to bounce nicely off of each other. Could this have been an elastic collision, that

is, one where the kinetic energy of the system was unchanged during the collision? Let’s do the math and see.

1. Table 2 works just like Table 1 and uses the same data. By filling in Table 2, you’ll calculate the total KE of the system

of two carts before (KE(sys)o) and after (KE(sys)f) the collision. You’ll also calculate the change in kinetic energy of the

system, ΔKEsystem which is KE(sys)f – KE(sys)o. Note that in the equation for KE, only the speed is considered. That is, KE

has no direction associated with it. This actually doesn’t matter in your calculations since all the velocities are squared.

Table 2 Kinetic Energy before and after a collision

Mass of Red Cart kg

Mass of Blue Cart kg

Cart vo

(m/s)

vf

(m/s)

KEo

(J)

KEf

(J)

ΔKEsys (J)

Red

Blue

KE(sys)o = KE(sys)f = ΔKEsys =

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Show calculations of total KEo, total KEf, ΔKE here.

2. Does the kinetic energy of the system appear to be conserved in this collision? How do you know? Quote your numbers.

III. Totally Inelastic Collisions

In the previous section we observed carts making elastic collisions. We found that both momentum and kinetic energy were

conserved. Our carts can have another type of collision. The springy bumpers between them can be replaced by sticky

“Velcro®” bumpers. In this case we would have a totally inelastic collision, one where the carts stick together after the

collision and share a common speed.

1. Let’s start with the simplest case – a collision between carts of equal mass and equal, but opposite, velocities. Attach

both carts to their launchers. Remove all extra masses. Fully cock both launchers (to provide equal initial speeds). Using

the usual procedure, produce a graph of their head-on collision. You’ll need to use the release both button.

a. Copy and paste your data into Grapher. Manually adjust the y-axis scale on both graphs to range from 0.0 m to

2.0 m. Add axis labels and an appropriate title to each graph.

b. Take a single Screenshot of the pair of graphs from Grapher and upload it as “Mom_III1.png”.

c. Draw the position-time graph produced by the apparatus in Graph III1. You could also print out your Screenshot and

paste it over the blank graph provided.

d. Explain how you know from the graph(s) that momentum was conserved, but kinetic energy was not. You should

be able to tell just by looking at the graphs but you can use Grapher to assist you.

III1. Equal Head-on equal speed collision

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2. Now try a similar collision, but this time with the blue cart initially sitting still at the center of the track.

a. Copy and paste your data into Grapher. Again manually adjust the y-axis scale on both graphs to range from 0.0 m

to 2.0 m. Add axis labels and an appropriate title to each graph.

b. Take a single Screenshot of the pair of graphs from Grapher and upload it as “Mom_III2.png”.

c. Draw the position-time graph produced by the apparatus in Graph III2. You could also print out your Screenshot and

paste it over the blank graph provided.

d. This one’s not so obvious. You’ll need some data and calculations to work with. Use Grapher to get the data you

need. Use the data table below to record your data and calculated values.

Table 3 Momentum and Kinetic Energy before and after a collision

Mass of Red Cart kg

Mass of Blue Cart kg

Cart vo

(m/s)

vf

(m/s)

Po

(kg m/s)

Pf

(kg m/s)

KEo

(kg m2/s2)

KEf

(kg m2/s2)

Red

Blue

P(sys)o = P(sys)f = KE(sys)o = KE(sys)f =

Show the calculations here for P(sys)o, P(sys)f, KE(sys)o, and KE(sys)f.

P(sys)f =

KE(sys)o =

KE(sys)f =

3. Using your data, explain how you know that momentum was conserved, but kinetic energy was not conserved.

III2. Collision with stationary cart

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4. You should have found that the pair of carts moving as one with a mass of 2× mcart had half the speed of the single cart

with a mass of 1× mcart. Why then is the kinetic energy less after the collision? That is, why doesn’t one cart with a

speed of have the same kinetic energy as two carts each with a speed of

?

Here’s a head start to your answer.

Before the collision the object carrying all the momentum and KE had a mass of mcart and a speed of vo.

5. For the situation in Graph III1, what fraction of the kinetic energy of the carts remains after the collision?

That is, what is (KEfinal/KEinitial)? You don’t have any numbers to work with, but you don’t need any.

6. For the situation in Graph III2, what fraction of the kinetic energy of the carts remains after the collision?

That is, compute (KEfinal/KEinitial). You’ll need to use your numbers this time.

Show calculations of total KEfinal/KEinitial here.