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Momentum conservation (ballistic pendulum)
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[QUOTE="jakec, post: 5640681, member: 611070"] [h2]Homework Statement [/h2] A .01kg bullet is fired into a 1.2kg block hanging from a 1m wire. The bullet exits the block with a speed of 200m/s and the block swings to a height of .2 meters. What is the original velocity of the bullet? What percentage of the original energy of the bullet is no longer in mechanical forms of energy? [h2]Homework Equations[/h2] I know that initial momentum = final momentum but I can't seem to find the velocity of the block after the collision. [h2]The Attempt at a Solution[/h2] I had this on a test and tried energy conservation. Obviously this isn't correct but I can't find the momentum of the block. I've been working this for about 5 hours now so I'm clearly not getting something so a step by step explanation would be great. [/B] GPE of block = (mgh) = (1.2 x 9.8 x 0.2m) = 2.353 Joules. This has resulted from KE of the bullet having swung the block. The bullet exited the block. Its KE after exit = 1/2 (m*v^2) = 1/2 (0.01*200^2) = 200 Joules. The total energy has come from the bullet, = (200 + 2.353) = 202.353 Joules. Original bullet V = sqrt.(2KE/m) = sqrt.(404.706/0.01) = 201.2 m/sec. [/QUOTE]
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Momentum conservation (ballistic pendulum)
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