Impulse-Momentum activity

Objective:

• To prove the impulse-momentum theorem that the amount of momentum change for an object is equal the amount of the net impulse acting on the object.

The lab is divide into three parts:

1. Part 1: Observing collision forces that change with time
2. Part 2: Observing collision forces when the momentum becomes larger
3. Part 3: Impulse-Momentum theorem in an inelastic collision

Part 1: Observing collision forces that change with time

Base Theory:

• Impulse is the product of the applied force and the time interval over which the force acts. Thus, an equation to calculate impulse is J = F∆t.
• If given a graph of force vs. time, we can calculate impulse by calculating the area created by the force and time (usually it's a rectangular)
• In case, the applied force is not constant, we still can find the impulse by dividing the region into familiar shape that we can use formula to calculate like rectangular, rectangular, or trapezoid. At the end, add them up will give the value of impulse.
• Even if the applied force varies continuously, we still find the impulse by integrating the area under force vs. time graph. J = ∫ F∆t

Set-up:

• Let a horizontal track lie on the table bench.
• Use a C-clamp to secure a vertical rod to the edge of the table.
• Use another clamp to secure the dynamic cart to a rod. Extend the spring plunger on the dynamics cart.

Figure 1: Extend the plunger of the stationary cart and make sure the other cart will hit it as it move toward the stationary cart.

• Mount a force sensor on another dynamics cart with a rubber stopper replacing the hook mounted on the protruding part of the force sensor.
• Make sure that as the cart move toward the stationary dynamics cart, its rubber stopper will hit the spring plunger of that cart. 
• Collide the cart with the plunger several times and observe what happens to the spring plunger and the moving cart (whether it bounces back or not)

Below is the apparatus of the experiment

Figure 2: The apparatus of the experiment.

What needed to be measured:

• To calculate the impulse, we will need the force and time, but these two can be measured by Logger Pro, we don't actually need to measure them.
• To calculate the momentum, we will need the mass of the cart and velocity. The velocity will be recorded by Logger Pro, so we just need to measure the mass of the cart.

Performing the experiment:

• Open the experiment file called Impulse and Momentum (L08A2-2).cmbl. This experiment has been set up to record force and motion data at 50 data points per second. The software also has been set up to record a push on the force probe as a positive force and velocity toward the motion detector as positive because the positive direction is toward the right.
• Calibrate the force probe with the hook, call the setting zero, then hanging a 1kg mass to get the second calibration point.
• Zero the motion detector again after it's horizontal and connected to the cart.
• When everything is ready, hit Collect button on Logger Pro. As soon as we hear the clicking of the motion detector, give the cart a gentle push to let it move toward the stationary dynamics cart.
• Do the calculations necessary afterward to verify the impulse-momentum theorem

Data Analysis

As the cart moves toward the stationary dynamics cart, the motion detector helps us record the value of time, position, and velocity. While the force probe helps us record the value of force. All values are shown in Logger Pro.

Figure 3: The data of time, force, position, and velocity

We then set up a New Calculated Column in Logger Pro to calculate momentum.

Figure 4: Setting up in Logger Pro to calculate momentum.

The result we got after calculating momentum.

Figure 5: The result of momentum besides all previous data.

Since we have the value of momentum, all we need is the value of impulse which we can calculate by integrating the area under the force vs. time graph.

Figure 6: Integrating the area under force vs. time graph to find the value of impulse.

J = 0.5290 N*s

The last step is find the change in momentum and compare it to the value of impulse.

Calculating initial momentum:

Figure 7: Calculating initial momentum 

mv_i = -0.273 kg * m/s

Calculating final momentum:

Figure 8: Calculating final momentum.

mv_f = 0.232 kg * m/s

Calculating the change in momentum:

∆mv = mv_f - mv_i = 0.232 kg * m/s + 0.273 kg * m/s = 0.505 kg * m/s 

 J = 0.5290 N*s

Since momentum ∆mv = 0.505 kg *m/s and impulse J = 0.5290 N*s, we can conclude that the change in momentum of the cart is approximately equal the impulse acting on the cart

∆mv ≈ J

Part 2: A larger momentum change due to the change in mass 

Set-up:

• We will use the same set-up as part 1, however, we will add some mass on the cart in this part and measure the new total mass of the cart.
  
  Figure 9: Measure the mass of the cart after adding some weight.
• We also follow the same procedure as part 1, and record the appropriate data and graphs.

Data Analysis:

• Setting up an equation to calculate momentum same as part 1, and here is the result we got.

Figure 10: Result table after calculating momentum.

• Integrating the area under force vs. time graph to get the value of impulse

Figure 11: Integrating the area to calculate impulse

J = 0.5950 N*s

• Calculating the change in momentum by getting the final momentum minus the initial momentum
  
  
  Figure 12: Calculating the initial momentum
  mv (initial) = -0.330 kg * m/s
  
  
  
  Figure 13: Calculating the final momentum 
  mv (final) = 0.266 kg * m/s
  
  
  Calculating the change in momentum:

  ∆mv = mv (final) - mv(initial) = 0.266 - (-0.330) = 0.596 kg *m/s

Since momentum ∆mv = 0.596kg * m/s and impulse J = 0.5950 N*s, we can conclude that the change in momentum of the cart is approximately equal the impulse acting on the cart. 

∆mv ≈ J

Part 3: Impulse-Momentum theorem in an inelastic collision
Base theory:
        In part 1 and 2, we performed the experiment in a nearly elastic collision, thus the amount of momentum change for the moving cart is approximately equal to the amount of the net impulse acting on the cart. In other words, the cart would recoil with almost exactly the same magnitude of momentum that it had before the collision. But what if we replace the spring plunger with a clay that makes the cart stop after it collides. Will the impulse-momentum theorem be still applied?
Set-up:

• Replace the dynamics cart in part 1 with some clay attached to a vertical piece of wood clamped to the lab table.
• Replace the rubber stopper with a nail.
• Leave some mass on the cart so that its mass is the same as the one in part 2.

Below is the apparatus of part 3:

Figure 14: The apparatus of part 3

Performing the experiment:

• Following the same procedure in part 1, but in this part we don't expect the cart to bounce back after it collides. Instead, the cart will stick to the wall and come to rest after the collision.
• Using the same experimental file, Impulse and Momentum (L08A2-2) and record the appropriate data any graphs.

Data Analysis:

• We will do exactly what we did in part 1 and part 2. First, calculating the initial and final momentum, then calculate the change in momentum. The next step is to calculate the impulse by integrating the area. Lastly, compare the value of the change in momentum and impulse acting on the object.
• Calculating initial momentum

Figure 15: mv (initial) = -0.356 kg * m/s

• Calculating final momentum

Figure 16: mv (final) = 0 kg * m/s

• Calculating the change in momentum
• ∆mv = mv (final) - mv (initial) = 0 - (-0.356) = 0.356 kg * m/s
• Calculating impulse acting on the object

Figure 17: Impulse = 0.3692 N*s

• Compare the change in momentum and impulse.
• Since ∆mv = 0.356 kg * m/s and Impulse = 0.3692 N*s, we could conclude that the change in momentum is approximately equal the impulse acting on the object considering some errors may happen during the experiment.
• ∆mv ≈ J

Error & Explanation:

• In part 1 and part 2, to prove the impulse-momentum theorem, the value of impulse and momentum should be equal, though they are actually not. Thus, there have to be some errors that make the result not come out the way it supposed to be.
• First of all, when we zero the motion detector and force probe, their values are actually not exactly zero. Thus, they affect the way motion detector and force probe record the value of velocity and force.
• When we hit Collect button in Logger Pro, we waited a little bit longer to give the cart a gentle push rather than as soon as we heard the clicking of the motion detector.
• The cart's spring bumper is not perfect, we can only produce a nearly elastic collision, not perfectly elastic collision.
• In part 3, the difference between the change in momentum and impulse is bigger than that of part 1 and part 2. Thus, we come up with reasonable error may happen during the experiment
• The integration of the force graph is done from point to point, while the force sensor records the force by frame per frame, and each frame has milliseconds gap between them, thus it may lead to some errors in calculation.

Conclusion

• In this lab, we perform experiments to verify the impulse-momentum theorem. We did it in two condition: one is in nearly elastic collision, and the other is inelastic collision.
• In a nearly elastic collision, we divide the lab into two part, the only difference between two parts is the mass of the cart.
• Part 1: The result we got for the change in momentum is 0.505 kg * m/s and the impulse is J = 0.5290 N*s
• Part 2: The result we got for the change in momentum is 0.596 kg * m/s and the impulse is J = 0.5950 N*s
• Overall, the result of both parts show that the change in momentum is approximately equal the impulse acting on the cart.
• In inelastic collision, the result we got for the change in momentum is 0.356 kg *m/s and the impulse is J = 0.3692 N*s. Both results are roughly equal each other considering some reasonable errors may happen during the experiment.
• Overall, the impulse-momentum theory is universal, it is both applied to elastic and inelastic collision. Moreover, the heavier mass and faster speed will result in larger impulse. Last but not least, the inelastic collision produces smaller impulse than the elastic collision.