15. April. 2015: Impulse Momentum activity

Purpose: We want to prove that impulse is equal to the change in momentum.

Equations to know:

Impulse(J) = Force*Time
Momentum (p) = mass*velocity
Impulse(J) = Change in Momentum (p) = m(v(f)-v(o))
Also know that since Impulse is force * time it can also be represented by the area under force vs. time graph.

Set up:

track
clamps
cart
motion sensor
force sensor
cart with spring
balance
logger pro


Set it up so that when the moving cart runs into the clamped cart the force sensor is the one pushing into the springy part.


Procedure:
First I calibrated the force sensor and then took the weight of the entire cart ( with force sensor ). The mass is something we are going to need to plug into our equations. We are going to be looking at velocity v time and the force v time graph. I used the Impulse and Momentum file in logger pro which has the force and motion data a 50 data points per second and it also already set the positive direction on the force probe and motion detector as well. After that give the cart a little push and make sure you record little before and after collision.

In this picture we took the integral of the collision from the Force v time and as we recall that gives us the impulse. Now we want to look at the minimum and maximum of the velocity v time graph which is data to find the change in momentum. SO let's look at the equation:

Impulse = change in momentum = m ( vf-vo )  = .76 (.5125 - (- .5545) =  .818 and our impulse was .814

I think the difference was the from what integral to what integral. I could not align the times to be just right so it was kind of off from each other. But overall, if I were to find the exact time of both graphs it would match.

We ran this experiment again but we replaced the car(with spring) with a block with clay. So that when the cart runs into the clay it will just stay there.

Predictions I made before experiment: I predicted that the impulse was going to be larger than the experiment we did before because in this one velocity final is equal to zero. I predict that even under this circumstance that the impulse and  change in momentum will be equal to each other. 

So we used the same logger pro file and we ran the cart into the clay. It just stayed after that and this was our graph: 

Lets test out to see if impulse is still equal to change in momentum: 

Impulse = change in momentum = m ( vf-vo )  = .76 (.01087-(- .4058) =  .316 and our impulse was .309

So this confirms that the impulse momentum theorem is true. I was wrong the impulse is actually less than if there were to have a v(final) not equal to 0 at the end. BECAUSE when v is zero it means that the cart has fully compressed into the "spring" and the force would be at its greatest.

Conclusion: 

In this lab, there was errors probably in mass. The directions said to add weight to the cart (I'm guessing to have a better reading on the force sensors) however we didn't add any extra weight and it would have messed up our consistency if we did the clay one with weights and the other one without. Questions to think about was what was the net force exerted on the cart just before it starts to collide? I think the net force was around zero because if you look at the graph before the collision it went up down up down but all near zero so adding up the areas of those would have been around zero. The magnitude of force exerted on the cart was when it was fully compressed and velocity was zero. I think the net force after the equation was also zero. And that the time that the collision takes is almost half a second from our graphs.