Calculating the height after an elastic collision

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SUMMARY

The discussion centers on calculating the height a smaller ball will reach after an elastic collision with a larger ball. The larger ball, with a mass of 100 g, is projected upwards at 5 m/s, while the smaller ball, with a mass of 50 g, is suspended 1.00 m above it. The key equation used is the conservation of kinetic energy: 1/2 M1V1i^2 = 1/2 M1V1f^2 + 1/2 M2V2f^2. The challenge lies in determining the final velocities (V1f and V2f) post-collision to subsequently calculate the height (Δy) the smaller ball will achieve.

PREREQUISITES
  • Understanding of elastic collisions and conservation of momentum
  • Familiarity with kinetic energy equations
  • Basic knowledge of projectile motion
  • Ability to solve algebraic equations with multiple variables
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  • Study the principles of conservation of momentum in elastic collisions
  • Learn how to derive final velocities from collision equations
  • Explore the relationship between kinetic energy and height in projectile motion
  • Practice problems involving multiple objects in elastic collisions
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Homework Statement


A ball of mass 100 g is projected straight up with a speed of 5 m/s from the floor. Another ball of mass 50 g is hung from the ceiling by a light string at a height of 1.00 m directly above the first ball, so that the projected ball collides elastically with it. Calculate the height above the floor to which the smaller ball will rise.


Homework Equations


1/2 M1V1i^2 = 1/2 M1V1f^2 + 1/2 M2V2f^2


The Attempt at a Solution


When I use the equation above it gives me two unknown variables V1f and V2f (after the collision). but I don't know what equation to use to figure out Δy.
 

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You're using conservation of energy. Energy is not the only thing conserved in elastic collisions. In fact, the thing I'm thinking of is conserved even in inelastic collisions.
 

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