Kinetic energy of inelastic collision problem

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SUMMARY

The discussion centers on calculating the fraction of kinetic energy lost during an inelastic collision between two particles, where one particle is initially at rest. The conservation of momentum is applied, leading to the equation m1u1 = (m1 + m2)u, allowing for the determination of the final velocity u. The initial kinetic energy (Ti) is expressed as ½ m1u1^2, while the final kinetic energy (Tf) is calculated as ½ (m1 + m2)u^2. The fraction of kinetic energy lost is derived as m2/(m1 + m2), with participants confirming the correctness of their calculations.

PREREQUISITES
  • Understanding of conservation of momentum
  • Knowledge of kinetic energy formulas
  • Familiarity with inelastic collisions
  • Basic algebra for manipulating equations
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  • Study the principles of inelastic collisions in detail
  • Learn about energy conservation in different types of collisions
  • Explore real-world applications of momentum conservation
  • Practice problems involving kinetic energy calculations in collisions
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Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for examples of inelastic collisions and energy loss calculations.

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Homework Statement



A particle of mass m1 and velocity u1 collides with a particle of mass m2 at rest. The two particles stick together. What fraction of the original kinetic energy is lost in the collision?

Homework Equations



Conservation of momentum law

The Attempt at a Solution



I think this one is right but was hoping somebody could check for me... It seemed too easy, so I just wanted to make sure I'm understanding it correctly.

m1u1 = (m1 + m2)u by conservation of momentum, so u = m1u1/(m1+m2)

Ti = ½ m u1^2 and Tf = ½ (m1+m2)u^2 = m1^2 u1^2/2(m1+m2) in terms of u1


Then I found Ti – Tf = (m1m2 u1^2)/2(m1 + m2) and then found ratio of this with Ti


I got m2/(m1 + m2). Is this correct? Thanks! :)
 
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I got \frac{m_1}{m_1+ m_2} and worked it the same basic way you did.
 
At least one of you two is obviously wrong. How much energy is lost when you throw a 1 gram spitball at a 75 kg man? What happens when a 75 kg man runs into a 1 gram spitball suspended from the ceiling?
 
Hi, rock.freak667, isn't your answer the amount of the original remaining (not lost)? I'm not sure, I'll check my work again. Man's kinetic energy isn't effected much but spitball's is - I think that's what makes me uncomfortable about these problem w/o numbers - harder to get an intuitive feel for them but making up concrete examples helps. Thanks to both of you.
 

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