Snow Day Blog! Conservation of Momentum

To perfectly demonstrate conservation of momentum, my group decided to utilize exercise balls we had at our house. The medicine ball had a mass of 4 kg and the volleyball had a mass of .281 kg. In our experiment, we had the volleyball rest on the patio 1 meter in front of the medicine ball. While the volleyball was at rest, my sister gave the medicine ball a healthy push and let physics do its thing. Using Vernier Logger Pro, I recorded the video, analyzing it to show the conservation of momentum between this collision

Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum lost by one object equals the momentum gained by the other object.

Above is the video showing Bella pushing the 4 kg medicine ball. As you can see in the video, once the medicine ball collides with the volleyball, they both appear to move with the same velocity in the same direction. To see if this is true, we began to plug data points and analyse the video using Logger

We were able to find the velocity of the medicine ball before the collision by finding the line of best fit once all the points were plotted. The velocity was 0.94 m/s

Medicine Ball before the Collision:

V: 0.94 m/s

M: 4 kg

P = (M)(V)

p= (4)(0.94)

momentum = 3.76 kg m/s

Volleyball before the Collision:

V: 0 m/s

M: .281 kg

momentum = 0 kg m/s

Medicine Ball/Volleyball System after collision

V: 0.73 m/s

M: (m1+m2) = 4.281 kg

momentum= 3.125 kg m/s

As important point to remember is that momentum is always conserved while energy is lost through various means.

Calculating the Energy:

Initial KE of Medicine ball = (.5)mv^2 = (.5)(4)(0.94)^2 = 1.767 J

The volleyball had no KE because it was not moving before the collision

Final KE of the system after the collision = (.5)(m1+m2)vf^2 = (.5)(4.281)(0.73)^2 = 1.141 J

Upon final analysis, can evaluated that this collision was in fact an inelastic collision; energy is lost in the collision yet momentum is conserved

As you can see, a small amount of energy is lost during this collision. There was a loss of 0.626 J

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