Master Elastic Collision Problems: Basic Help | Homework Solution

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SUMMARY

The discussion focuses on solving elastic collision problems in physics, specifically addressing the conservation of linear momentum and kinetic energy. The user presents a scenario involving two particles, m1 and m2, where m1 is moving at 4 m/s and m2 is initially at rest. The user seeks confirmation on using the equations of momentum and kinetic energy to find the final velocities (v1' and v2') after the collision. Key formulas discussed include momentum conservation (m1v1 = m1v1' + m2v2') and kinetic energy conservation (0.5m1v1^2 = 0.5m1v1'^2 + 0.5m2v2'^2).

PREREQUISITES
  • Understanding of elastic collisions in physics
  • Knowledge of linear momentum conservation
  • Familiarity with kinetic energy equations
  • Basic algebra for solving equations with multiple variables
NEXT STEPS
  • Study the derivation of elastic collision formulas, specifically v1' and v2'
  • Practice solving various elastic collision problems using different mass ratios
  • Explore online resources such as HyperPhysics for visual aids and examples
  • Review conservation laws in physics to strengthen foundational knowledge
USEFUL FOR

Students studying physics, particularly those struggling with elastic collision problems, as well as educators looking for effective teaching strategies in momentum and energy conservation.

fluidistic
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Homework Statement


Some days ago and specially yesterday I realized I have a strong weakness when it comes to elastic collision problems. Unfortunately I disregarded it and I couldn't finish an exercise in a test today.


Homework Equations


Linear momentum is conserved ([tex]\vec{P_{i}}=\vec{P_f}}[/tex]) and so is the kinetic energy of the system. (Two bodies)


The Attempt at a Solution



Well in fact say you have 2 particles, [tex]m_1[/tex] and [tex]m_2[/tex]. [tex]m_2[/tex] is initially at rest while [tex]m_1[/tex] has a speed of [tex]\frac{4m}{s}[/tex].
If [tex]m_1[/tex] hits [tex]m_2[/tex], what would be the speed of [tex]m_2[/tex]?

I get that [tex]m_1v_1=m_1v_1'+m_2v_2'[/tex] [1]. One equation, two unknowns: [tex]v_1'[/tex] and [tex]v_2'[/tex].
Have I to work it out using the equation of the kinetic energy? [tex]\frac{m_1v_1^2}{2}=\frac{m_1v_1^2'}{2}+\frac{m_2v_2^2'}{2}[/tex]. I solve for [tex]v_1'[/tex] and back substitute it in [1]. Is it the way to find out [tex]v_1'[/tex] and [tex]v_2'[/tex]?
If yes then I thought about it during the test but I was too lazy to do it... (it doesn't count for my average but still, I should have done that).
Thanks for your confirmation or infirmation.
 
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Thank you very much iseidthat!
That doesn't answer my question I think, but at least I've learned that [tex]v_1'=\left( \frac{m_1-m_2}{m_1+m_2}\right) v_1[/tex] and that [tex]v_2'[/tex] can be determinate by a given formula.
Ok, I'll try to answer my question by my own.
 

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