Momentum and Energy gives different answer

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SUMMARY

The discussion centers on the discrepancy between momentum and kinetic energy calculations in a collision scenario involving two balls with masses of 1g and 100g. The momentum equation, m1v1 = m2v2, yields a final velocity of 0.1 m/s for the second ball, while the kinetic energy equation suggests a final velocity of 1 m/s. The key conclusion is that the collision is inelastic, meaning kinetic energy is not conserved, which explains the differing results from the two calculations.

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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:

[tex]m_1 v_{1,\text{init}} = m_1 v_{1,\text{final}} + m_2v_2[/tex]

and

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

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