Conservation of momentum

Conservation of momentum was demonstrated through the collision of 2 tennis balls. The tennis balls were set on a track as shown. One ball at rest and one ball given force to start the motion.

It is seen that this is not a perfect example of conservation of momentum. Lots of momentum and energy was lost in the initial collision when the tennis ball bounced up. The energy expended in the y direction wasn't reflected in the x direction. This energy was expended over the course of a few small bounces that were difficult to pinpoint to measure. Although the ball moved vertically, the vertical displacement was equal to zero at the end of the system.

Regarding the conservation of momentum:

v1 (initial): 0.9128 m/s

v1 (final): 0.2285 m/s

v2 (initial): 0 m/s

v2 (final): 0.3519 m/s

mass of tennis ball: 0.0585kg

(equation for elastic collision because trying to find conserved momentum)

m1*v1 (initial) + m2*v2 (initial) = m1*v1 (final) + m2*v2 (final)

Masses are all cancelled out because both tennis balls have equal weight

v1 (initial) + v2 (initial) = v1 (final) + v2 (final)

0.9128 + 0 = 0.2285 + 0.3519

0.9128 = 0.5804

0.3324 unaccounted for (velocity lost)

KE = 1/2(mv^2)

m = mass of tennis ball (0.0585kg)

v = velocity lost (0.3324)

KE = (1/2)(0.0585)(0.3324)^2

KE = 0.003232 J

0.003232 J of Kinetic Energy was lost in this system

The lost kinetic energy could be due to air resistance and friction. Other potential sources of error were that the tracks for the tennis ball weren't perfectly parallel, altering the final or initial speeds of the tennis balls, and also the accuracy of calculating data through logger pro was likely not exact.

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LSA E Mech 2017-18

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