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Ballistic Momentum: Dalton Ledford And Tiger F

979 Words4 Pages

Ballistic Pendulum Lab
Dalton Ledford and Tiger Frick

The purpose of the lab was to find the velocity of a brass ball as it left the muzzle of a cannon and nearly immediately collided with a pendulum. This teaches us about the transformation of energies and transfers of momentums in collisions.

Background:

In this lab we used conservation of momentum, conservation of energy, and pendulums to complete the lab. Momentum is always conserved in collisions regardless of whether or not kinetic energy is conserved. The law of conservation of momentum states that momentum is always conserved in a collision (Mr. Patterson’s Class). The law of conservation of energy states that energy can not be created or …show more content…

Patterson’s Class) This is because momentum is defined as mass times velocity (mv) (physicsclassroom.com). Anytime 2 or more objects collide they exert equal but opposite forces on each other as defined in Newton’s third law of physics (physicsclassroom.com). Newton’s second law of physics is defined as F=ma where force is any interaction with another object, or mass multiplied by acceleration (physicsclassroom.com). So this force between the 2 or more objects cannot ever add or take away momentum from them because both objects received an equal but opposite force to the other. We also use conservation of energy in two forms kinetic energy and potential energy. Kinetic energy is the energy of an object in motion with a mass and in an isolated system energy cannot be created or destroyed. Also there is potential energy which is the energy that an object has due to gravitation and its height. Pendulums are a swing device that rock back and forth using gravity to keep them in motion. At the bottom of the pendulums motion it has all kinetic energy and no …show more content…

Using triangle above:
(.315)cos(35)= .285 meters
.315-.285= .0569 meters

Velocity of the Ball & Mass system: Initial velocity of the ball:
KE=PE mv=mv
1/2 mv^2=mgh .0085v=(.0085+.035)√(2gh)
(1/m)1/2 mv^2=mgh(1/m) .0085v=(.0085+.035)√(2(9.8)(.0569))
(2)1/2 v^2=gh(2) .0085v= .0435√1.115 v^2=2gh .0085v= .0459 v=√(2gh) v=5.4 m/s

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