Lab 15 - Collisions in Two dimensions

 The above picture is the glass table setup that we used to generate collisions. The long metal stand is where we attached our phones to record the collisions in 240frames per second.

 Purpose: The purpose of this lab is to show that in collisions, momentum is conserved even though it can be distributed into two dimensions.

Apparatus/Procedure: We used two glass marbles and a heavier metal ball. One ball was positioned stationary on top of a level glass table, while the other one was thrown at it to create a collision. There was a ring stand positioned directly above the glass table with a clip in order to attach a cell phone. The cell phone was used to record the collision at 240 frames per second. This would allow us to get a clearer picture when we uploaded the collisions to logger pro. We ran one trial with two glass marbles, and another trial with the heavier one colliding into one glass one. We then used logger pro to track the paths of the marbles both before, and after they collided. We set up a coordinate system, and logger pro was able to separate both the paths into x and y components. We then took these values for velocities and checked to see if momentum was conserved. 

Theory: The theory behind this is that momentum is conserved in collisions, regardless of the amounts of dimensions. 

The momentum equations the two collisions: 

m= mass of glass balls : 0.0193kg

3m=mass of metal ball : 0.0667kg

Collision with same mass balls:

Po= pf

m (1.282m/s)+m (0)=m (0.4098m/s)+m (0.6795m/s)

0.0247426kgm/s=0.02102kgm/s

Momentum not conserved along y-axis

Ke before=ke after

Ke before=.5mv^2
Ke=.5*0.0193*1.74^2=0.0292j
Ke after=.5*0.0193*0.42^2 +.5*0.0193*1.05^2
      =0.0123j
0.0292j=/=0.0123j
Ke not conserved
The system lost kinetic energy after the first collision with two balls of equal masses.

m (1.173m/s)+m (0)=m (0.1061m/s)+m (0.7994m/s)

0.0226389kgm/s=0.017476kgm/s

Momentum not conserved along x-axis

Collision with different- mass balls: 
Po=pf

3m (1.332m/s)+m (0)=3m (0.5054m/s)+m (1.490m/s)

0.08884kgm/s=0.062467kgm/s

Momentum not conserved along y-axis

The system lost momentum along the y-axis.

3m (1.871m/s)+m (0)=3m (1.204m/s)+m (0.4905m/s)

0.124796kgm/s=0.08977kgm/s

Momentum not conserved along x-axis

Ke before=0.1764j

Ke after=0.0786

Ke not conserved. 

The system lost about half the ke to heat after the second collision with two ballstat of different mass.

Below are all of the graphs on logger pro with all of the paths of all the balls both before and after collision.

The above screen shots are of the of the  collisions. The top one is the collision with same masses, and the one below is the screen shot of the collision with different masses. 

Conclusion: As it turns out, not even in this lab is momentum perfectly conserved. This is because of the fact that after one ball impacted the other, they would slide, then catch friction and start rolling. As the balls catch friction, they also lose some energy to friction and thus, lose momentum. When the two balls collide, there is also kinetic energy that is converted into thermal energy from the impact. That is another reason why momentum s not conserved.  In reality, there is no collision in which momentum will be perfectly conserved.

My partner for this lab is Elliot Sandoval.