Momentum and Energy gives different answer

AI Thread Summary
The discussion highlights a scenario involving two balls of different masses colliding, leading to differing results when calculating final velocities using momentum and kinetic energy equations. The momentum calculation suggests a final velocity of 0.1 m/s for the second ball, while the kinetic energy calculation indicates a final velocity of 1 m/s. The discrepancy arises because the collision is inelastic, meaning kinetic energy is not conserved, whereas momentum is conserved. In elastic collisions, both momentum and kinetic energy are conserved, resulting in consistent outcomes. Understanding the nature of the collision is crucial for correctly applying these physical principles.
halloweenjack
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So this may be an easy question, I'm not sure, but I could not find an answer online. Basically here is the scenario.

There are two balls, one has a mass of 1g and the other has a mass of 100g. The 1g ball is traveling at 10m/s, and strikes the other ball. According to momentum, m1v1=m2v2. Using this equation, we have 1g(10m/s)=100g(v2). v2 then equals 0.1 m/s.

The problem I have is that when using kinetic energy, I get a different velocity of the second ball. The first ball has an energy of .5(1g)(10m/s)^2, which is 50 joules. If the first ball then transfers these 50 joules to the second ball, using the same equation I get the second ball having a final velocity of 50=.5(100g)(v^2), or 1 m/s.

Why am I getting different answers? Shouldn't both equations give the same answer?
 
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The answer is simple: The collision as you have established it is not elastic. In other words, energy isn't conserved. In an elastic, head-on collision the small ball will bounce back with nearly the same speed it had before the collision and the big ball will move forward with a tiny velocity:

m_1 v_{1,\text{init}} = m_1 v_{1,\text{final}} + m_2v_2

and

m_1 v_{1,\text{init}}^2 = m_1 v_{1,\text{final}}^2 + m_2v_2^2
 
Thanks for you help. This helps me in some practice I am getting into be better in mathematics.
 
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