LAB PHYSICS LAB 7: CONSERVATION OF MOMENTUM
PHYSICS 182A/195L LAB REPORT – LAB 7: CONSERVATION OF MOMENTUM
Lab 7: Conservation of Momentum San Diego State University Department of Physics Physics 182A/195L
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Lab partner 2: |
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Data has been entered in blue.
Theory
What is momentum? Momentum is the product of a body’s mass and it’s velocity :
Of the fundamental kinematic quantities, mass, position, velocity, acceleration, why does the product of mass and velocity deserve its own name? It turns out that mass times velocity, momentum, is what’s known as a conserved quantity.
Consider Newton’s third law for the forces experienced by two interacting masses and . By replacing each force by , we can show the following:
This equation says that the quantity in the parenthesis does not change with time. Another way to say this is that the term in parenthesis is a constant:
This fact is so important that we give its own name and symbol, :
This important result shows us that the total momentum of a system is constant. We say:
Momentum is always conserved. |
Collisions
While a collision can be extremely complex and involve many forces and bodies (imagine a car crash), conservation of momentum tells us that the total momentum before and after that collision is the same:
The (i) subscript labels the initial momentum (before the collision) and the (f) subscript labels the final momentum (after the collision).
Elastic Versus Inelastic Collisions
Yet another quantity of interest is kinetic energy. Kinetic energy is defined as:
Kinetic energy is only sometimes conserved during a collision. If kinetic energy is conserved during a collision, then we call it an Elastic collision. This only happens for frictionless collisions. Otherwise, if the kinetic energy is not conserved, then we call it an Inelastic collision and kinetic energy is lost due to internal friction.
The Experiment!
In this lab, we will demonstrate that conservation of momentum works as we expect. Imagine we have two carts with masses and . Cart 1 is initially moving with some known velocity , and Cart 2 is at rest . We know both cart’s initial speeds and we want to determine the final speed of each cart, after the collision.
Elastic Collision
In an elastic collision, we can use the fact that both momentum is conserved and kinetic energy is conserved. We therefore have two equations:
and
In our experiment, we assume that we know and we know . So our equations simplify:
and
After a few lines of algebra, we can solve these two equations for both and , which are the final speeds of Cart 1 and Cart 2 after the collision, respectively. The details for solving these equations are shown in the Appendix. The final results are:
Let’s see what these equations predict for three different values of and :
Inelastic Collision
While an elastic collision maintains both the conservation of momentum and kinetic energy, an inelastic collision only conserves momentum. This creates a problem for our equations because we no longer have two sets of equations to work with.
There is one special case where we can still find and , and that’s when . What would this mean? It implies that the two carts stick together after the collision. This results in a perfectly inelastic collision.
If is known and , conservation of momentum tells us
like before. If we plug-in , we get
Now we can easily solve for the unknown :
Let’s see what these equations predict:
Inelastic predictions | Description | ||
Equal masses
[plug in m1=m2] |
Both carts move to the right with half of Cart 1’s initial speed. |
Procedure
Setup
1. Make sure that both carts have magnetic bumpers on them.
2. Make sure the track is level. You can adjust screws on the track feet to change the incline. When you place a cart at rest on the track, give it a little push in each direction. It should not accelerate in either direction.
3. Use a scale to find the mass of each cart. If the carts do not have the same mass, add weights to one of them until they are the same mass.
4. Record the mass of each cart in their respective columns in Table 1.1 on the Data page.
Section 1: (Perfectly) Inelastic Collision
1. Place the red and blue carts at rest with the Velcro® bumpers facing each other. The blue cart should be in the center of the track and the red cart should be on the left end.
2. Start recording and give the red cart a push toward the blue cart. Stop recording before either cart reaches the end of the track.
3. On the velocity vs. time graph, find the velocity of the red cart just before and just after the collision. You can accomplish this using the coordinate tool. The time just before the collision is most easily identified by finding the time when the blue cart first begins to move. Record these velocities in Table 1.2.
4. The initial velocity of the blue cart is zero and its final velocity is the same as the red cart because they stick together. Record the blue cart’s final velocity in Table 1.2.
5. Add together the sum of the initial velocities, as well as the sum of the final velocities, and record these values in Table 1.2.
6. Using the masses in Table 1.1, multiply your carts’ respective masses with their initial and final velocities to find corresponding momentums. Record the values in Table 1.3.
Section 2: Elastic Collision
Part A: m1 = m2
1. Record the masses of each cart in Table 2.A.1.
2. Place the red and blue carts at rest on the track, with the magnetic bumpers facing each other. The blue cart should be in the center of the track and the red cart should be on the left end.
3. Start recording and give the red cart a push toward the blue cart. Stop recording before either cart reaches the end of the track.
4. On the velocity vs. time graph, find the velocity of the red cart just before and just after the collision. The time just before the collision is most easily identified by finding the time when the blue cart first begins to move. Record these values in Table 2.A.2.
5. The initial velocity of the blue cart is zero. Find the final velocity blue cart just after the collision, then record this value in Table 2.A.2.
6. Add together the sum of the initial velocities, as well as the sum of the final velocities, and record these values in Table 2.A.2.
7. Using the masses in Table 2.A.1, multiply your carts’ respective masses with their initial and final velocities to find corresponding momentums. Record the values in Table 2.A.3.
Part B: 2m1 = m2
1. Add mass to Cart 2 (blue cart) until it weighs twice as much as Cart 1 (red cart). To accomplish this you can use the mass bar.
2. Record these new mass values in Table 2.B.1.
3. Repeat steps 2-7 from Part A, except now use the tables for Part B.
Part C: m1=2m2
1. Remove the extra mass on Cart 2 (blue cart) that you added in Part B.
2. Add mass to Cart 1 (red cart) until it weighs twice as much as Cart 2 (blue cart). To accomplish this you can use the mass bar.
3. Record these new mass values in Table 2.C.1.
4. Repeat steps 2-7 from Part A, except now the tables for Part C.
Data
Section 1: (Perfectly) Inelastic Collision
Table 1.1: Cart masses
m1 (red cart) mass (kg) | 0.2732 |
m2 (blue cart) mass (kg) | 0.2712 |
Table 1.2: Velocities
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
vi (m/s) | 0.258 | 0 | |
vf (m/s) | 0.129 | 0.129 |
Table 1.3: Momentums (p=mv)
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
pi (kg m/s) | 0 | ||
pf (kg m/s) |
Table 1.4: Kinetic Energies (KE=0.5mv^2=0.5p^2/m)
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
KEi (Joules) | 0 | ||
KEf (Joules) |
Section 2: Elastic Collisions
Part A: m1 = m2
Table 2.A.1: Cart masses
m1 (red cart) mass (kg) | 0.2732 |
m2 (blue cart) mass (kg) | 0.2712 |
Table 2.A.2: Velocities
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
vi (m/s) | 0.121 | 0 | |
vf (m/s) | -0.002 | 0.121 |
Table 2.A.3: Momentums (p=mv)
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
pi (kg m/s) | 0 | ||
pf (kg m/s) |
Table 2.A.4: Kinetic Energies (KE=0.5mv^2=0.5p^2/m)
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
KEi (Joules) | 0 | ||
KEf (Joules) |
Part B: 2m1 = m2
Table 2.B.1: Cart masses
m1 (red cart) mass (kg) | 0.2732 |
m2 (blue cart) mass (kg) | 0.5423 |
Table 2.B.2: Velocities
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
vi (m/s) | 0.397 | 0 | |
vf (m/s) | -0.124 | 0.257 |
Table 2.B.3: Momentums
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
pi (kg m/s) | 0 | ||
pf (kg m/s) |
Part C: m1=2m2
Table 2.C.1: Cart masses
m1 (red cart) mass (kg) | 0.5422 |
m2 (blue cart) mass (kg) | 0.2712 |
Table 2.C.2: Velocities
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
vi (m/s) | 0.280 | 0 | |
vf (m/s) | 0.082 | 0.365 |
Table 2.C.3: Momentums
Cart 1 (red) | Cart 2 (blue) | Sum (1+2) | |
pi (kg m/s) | 0 | ||
pf (kg m/s) |
Analysis
Section 1: (Perfectly) Inelastic Collision
We will use the tables from the data section to answer questions about which sum of variables is conserved and which is not conserved. Fill out the following table “Is it Conserved”, by deciding whether or not each variable sum is conserved or not. To decide, examine the Sum(1+2) column of each variable table above.
Table A.1: Is it Conserved?
Variable/Quantity | Is it conserved? (is xi=xf?) [Y/N] |
v: velocity | |
p: momentum | |
Section 2: Elastic Collisions
We will repeat the analysis from Section 1 on Section 2 data. Use the tables from Part A to complete the following table “Is it Conserved?”
Table A.2: Is it Conserved?
Variable/Quantity | Is it conserved? (is xi=xf?) [Y/N] |
v: velocity | |
p: momentum | |
Do you think this table would be different for Parts B and C? Explain why or why not:
Questions
1. In Section 2A, , i.e. Cart 1 should come to a rest. Did your cart do this? If not, what is a reason why it may not have been perfectly at rest?
2. Why is it important to make certain we are using a level, frictionless surface?
3. In Section 1, some of the kinetic energy is lost after the collision. Where did the energy go?
Appendix (optional reading)
Full derivation of final velocities for elastic collisions
In an elastic collision, we can use the fact that both momentum is conserved and kinetic energy is conserved. We therefore have two equations:
and
In our experiment, we assume that we know and we know . So our equations simplify:
and
To solve these equations, we first isolate in the first equation by dividing through by :
Now we can substitute this into the conservation of energy equation:
The fraction can be distributed throughout our parentheticals, and the term on the left side can be cancelled out:
Next, we move the remaining terms to opposite sides of the equality and divide by a factor of :
Isolating and cancelling the factor of , we find:
which gives the final result for :
With an expression for found, we substitute this back into the equation at the top of the appendix to solve for :
This leads to the final derivation listed in the theory section for ,
1 Department of Physics