15 April 2015. Impulse-Momentum Theorem.

Purpose

Part 1
to observe collision forces that change with time and finding out how impulse-momentum theorem works.

Part 2
to observe bigger collision forces that change with time, which also means a larger momentum change.

Part 3
to observe impulse-momentum theorem in an inelastic collision.


Base Theory:

Momentum is defined as:

p = mass * velocity

Then, impulse combines the applied force and the time interval over which that force acts. We call this impulse, which is defined as:

J = Force * time

Therefore, since the momentum of the system cannot change without forces (no external forces: friction and gravity could be disregarded), the magnitude of force before and after a collision must be 0, because there is no change in momentum before and after the collision. Thus, we say that momentum of a system is conserved, regardless of elastic or inelastic collisions (which just vary in force-time). This is shown by the Impulse-Momentum, impulse is equal to the change of momentum.

What we "give" to mass is impulse, and what it "gains" is momentum.

J = Δp = integral of F . dt

Apparatus:

1. Cart and plunger. This set-up is used to model the collision.
2. Clay. Is used in expt. 3 to model inelastic collision.

Procedure and Data analysis

for this lab blog, the procedure and data analysis will be described individually under one part section. Since there are three parts of this experiment.


Part 1 (observing impulse-momentum theorem)

Procedure

1. Clamp the dynamics cart to a rod clamped to a lab table. Extend the spring plunger on the dynamics cart.
2. Mount a force sensor on another dynamics cart, with a rubber stopper replacing the hook mounted on the protruding part of the force sensor.
3. Do not forget to calibrate the force sensor with hanging some mass and see whether the number in the laptop make sense.
4. Set things up such that the stopper of the moving cart hits the plunger of the stationary cart when the moving cart gets close to the end of the track,.
5. Collide the cart with the plunger several times and observe what happens to the spring plunger.

5. Also don't forget to weigh the mass of the cart.

Data Analysis

• Now, what we have is the data of position and velocity of the cart from the motion detector and the force from the force sensor.
• Before we start analyzing the data, we need to understand this statement:
• That there is no external forces acting in the system; track is leveled and friction is ignored. Therefore the force sensor would detect 0 N.
• The maximum force would be when the plunger is compressed at the maximum by the cart.
• After the collision in complete. There is also no forces acting.
• The collision would take very little time, at max 1/2 second.
• Okay. Now, from the data we have, we can make a momentum column, which equals to the mass of the cart times the velocity.

• Then, we take the integral of the force graph, which will give us the impulse. J = 0.5170 N*s.
• Comparing this to the change of the momentum. 

P final - P initial = 0.224 - (-0.264) = 0.488 kg*m/s

Momentum final = 0.224 kg*m/s

Momentum initial = -0.264 kg*m/s

• The percent error is 

• Notice that this is a nearly elastic collision. When it is elastic, then the initial momentum would exactly equal to the final momentum. In this case, we are losing a little bit of momentum in the final as you can see from the graph. Then, where does this momentum goes? Well, actually it goes to the earth movement, because the mass of the earth is really huge, we can't really see its movement.
• But, anyway the impulse equals to the change of momentum with 5.61 % error. This proof that the impulse-momentum theorem is true.

Part 2 (A larger momentum change)

Procedure

1. It is basically the set up from the first expt.
2. But, we add some more mass into the cart, in this case we add 500 g into it.

We add more mass into it!

Data Analysis

• Now, same as before, what we have is the data of position and velocity of the cart from the motion detector and the force from the force sensor.
• Because our mass also changes, our formula for momentum also changes:

• Next, this is the graph for the force and the integral = impulse. I = 0.5820 N*s

• And the final and initial momentum.

Momentum final = 0.260 kg*m/s

Momentum final = - 0.323 kg*m/s

• the change of the momentum. 
P final - P initial = 0.260 - (-0.323) = 0.583 kg*m/s
• With percent error:

• Notice that this is a larger momentum change than before (this makes sense because we add more mass to the cart, automatically the momentum will also increase), and yes the change of momentum and impulse still equal to each other (with some error) even though we add more mass to the cart.
• This just proof that this law is universal and can be applied regardless the change of the mass.

Part 3 (Inelastic collision)

Procedure

1. It is basically the same set up from part 2
2. but we change the plunger into a clay, which will produce the inelastic collision.
3. DO NOT change the mass of the cart, and try to set up the speed of the cart as the same as in part 2.

Data Analysis

• Now, same as before, what we have is the data of position and velocity of the cart from the motion detector and the force from the force sensor.
• Turns out, the cart stick to the clay and generate a velocity = 0 at the end.
• In this experiment, we fail to set up the speed of the cart as the same as in part 2, due to limitation of time. Therefore, we can't really compare the impulse and momentum from part 3 to part 2.
• Based on the graph, the impulse is going to be I = 0.4825 N*s.
• and the final and initial of momentum.

Momentum final = 0 kg*m/s

Momentum initial = - 0.483 kg*m/s

• the change of the momentum. 
P final - P initial = 0 - (-0.483) = 0.483 kg*m/s
• With percent error:

• the change of momentum and impulse still equal to each other (with some error) even though this is an inelastic collision.
• This just proof that this law is universal and can be applied regardless whether its elastic or inelastic.

Extra

• Because we fail to set up the same speed. I find some sources that do the elastic and inelastic collision with same mass and speed from http://www.vernier.com/innovate/impulse-comparison-for-elastic-and-inelastic-collisions/.
• This is the graph.

Trial	Change in Momentum (kg m/s)	Impulse (N s)
Elastic Collision	0.32	0.32
Inelastic Collision	0.17	0.18

• The impulse from the elastic collision was very close to twice the impulse of the inelastic collision. This is the result that we sought. The elastic impulse is a little less than twice the inelastic impulse. That the ratio is just under 2 could be due to the fact that the “elastic” collision is losing some energy. We saw this as a slightly smaller speed after the impulse. Perhaps a magnetic bumper would see a more nearly elastic collision, and yield a ratio closer to 2 (http://www.vernier.com/innovate/impulse-comparison-for-elastic-and-inelastic-collisions/)

Conclusions

• Impulse is always going to be equal to the the change in momentum regardless its an elastic or inelastic collision, faster or slower speed, or lighter or heavier mass.
• Faster and heavier mass will produce larger impulse.
• The inelastic collision will produce half as big impulse as the elastic collision.
• The error for this experiment are from:
• the integral of the force graph is done from a recorded point to point (the force sensor read the input by frame per frame, and each frame has milliseconds gap between them), so it is not purely from what really happening.
• there might be very small friction and air resistance acting on the cart as the external forces, but we ignore that.