Impulsion and momentum in collsion

Introduction:
The purpose of this activity is to see the relationship between the applied impulse to an object and the change in momentum of the object. This activity was consisted of three parts: in the first part, we created an elastic collision; in the second part, the condition was all the same except the mass of the moving cart had been increased; in the third part, we attached clay at the collision point to create an inelastic collision.
By theory, the relationship between J, impulse, and ΔP, change in momentum, is the following:
J=integral(Fdt)=ΔP

Data and analysis:
The moving cart we used in each part of the activity was the same, and has a mass of 0.043kg.

Part I

We attached an force sensor on the moving cart, so we could detect the force exert on the carts at every moment of the collision. The motion sensor was installed at the other end of the air track to detect the change in velocity before and after the collision.













The following is the curve the logger pro created with the data collected by the motion sensor and the force sensor:

The area under the force vs. t curve during the collision is the impulse exerted on the carts, and it was  0.5322 N*s
The change in momentum, ΔP =m(Vf-Vi)= 0.043(0.536+0.645)=0.508 kg*m/s
% difference = (0.5322-0.508)/0.508=4.8%
The difference was only 4.8%, and that indicated that the theory was correct that the change in momentum is equal to the impulse exerted on the carts.

Part II
The purpose of the part two of the experiment is to see if the theory proved in part one still works when impulse is exerted on a massive mass.
The set-up is completely the same as the set-up in part one except a hanging mass was added to the moving cart.
We added a 50 g hanging mass on the cart, so the new mass was 0.093kg
And the following is the graph we got from the logger pro:

The area under the force vs. t curve during the collision is the impulse exerted on the carts, and it was  1.024 N*s
The change in momentum, ΔP =m(Vf-Vi)= 0.093(0.524+0.623)=1.07 kg*m/s
% difference = (1.07-1.024)/1.024=4.5%
The difference was only 4.5%, and that indicated that the theory was correct that the change in momentum is equal to the impulse exerted on the carts.

Part III
Impulse and change in momentum in an inelastic collision

We change the collision point with a wood attached with a clay:





The mass of the moving cart is back to be 0.043 kg








And, the following is the new graph:

The area under the force vs. t curve during the collision is the impulse exerted on the carts, and it was  0.2533 N*s
The change in momentum, ΔP =m(Vf-Vi)= 0.043(0.568-0)=0.2442 kg*m/s
% difference = (0.2533-0.2442)/0.2442=3.7%
The change in momentum still equals to the impulse applied on the cart though the collision was an inelastic one. However, the energy is not conserve in the cart-collision point system anymore because some of the energy was used to make the moving object stick with the collision point.