scilearner Posted May 26, 2010 Share Posted May 26, 2010 Hello everyone, Let's say there was gun and a bullet with a mass of 1kg at rest. The bullet and guns speeds were 2 m/s in opposite directions after firing if I right the equations. Kinetic energy 0.5m1v12 + 0.5m2v12 = 0.5m1v22 + 0.5m2v22 Why doesn't this equation work when objects are at rest. Mathematically I can understand but why physcially. Can anyone explain what is happening to kinetic energy in this scenario. Why does conservation of momentum equation hold in in inelastic collisions but not this one. I understand some energy is lost but why is it not affecting momentum. Link to comment Share on other sites More sharing options...
timo Posted May 26, 2010 Share Posted May 26, 2010 The bullet and guns speeds were 2 m/s in opposite directions after firing if I right the equations. Kinetic energy 0.5m1v12 + 0.5m2v12 = 0.5m1v22 + 0.5m2v22 You should elaborate more on this part: There is only one equation, the 2 m/s does not seem to appear anywhere, your symbols are not explained properly, and you did not say what you did here. TeX-Version of your equation would probably be [math] \frac 12 m_1 v_1^2 + \frac 12 m_2 v_1^2 = \frac 12 m_1v_2^2 + \frac 12 m_2v_2^2[/math], btw. Link to comment Share on other sites More sharing options...
swansont Posted May 26, 2010 Share Posted May 26, 2010 Conservation of momentum does hold. It is zero before and after the gun is fired. Link to comment Share on other sites More sharing options...
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