Spring compression -- Ball colliding with a spring-mounted platform

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SUMMARY

The discussion focuses on the physics of a ball colliding elastically with a spring-mounted platform, specifically analyzing the energy equations involved. The equation presented, -(1/2)kx^2 + mpgx + (1/2)mpvp^2 = 0, attempts to equate the spring's potential energy with gravitational and kinetic energy. Participants emphasize the need for clarity in the derivation of the velocity value, v_p, and the implications of elastic collisions in this context. The conversation highlights the complexities of solving the resulting quadratic equation.

PREREQUISITES
  • Understanding of elastic collisions in physics
  • Familiarity with potential and kinetic energy equations
  • Basic knowledge of quadratic equations and their solutions
  • Concept of spring constant (k) and its role in energy storage
NEXT STEPS
  • Study the principles of elastic collisions and their mathematical representations
  • Learn about the conservation of energy in mechanical systems
  • Explore the derivation and solutions of quadratic equations in physics problems
  • Investigate the role of spring constants in oscillatory motion and energy transfer
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Physics students, educators, and anyone interested in understanding the dynamics of elastic collisions and energy conservation in mechanical systems.

dge
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Homework Statement
What is the lowest position attained by the platform after it collides elastically with a ball.
Relevant Equations
1/2mv^2=KE
1/2kx^2=Us
mgx=Ug
1639430782496.png

I've tried establishing a systems of equations with -(1/2)kx^2+mpgx+(1/2)mpvp^2=0, but this leads to an annoying quadratic that when solved does not give one of the listed answers. My thought was that the spring's potential energy is going to be equal to the gravitational potential energy and kinetic energy.
 
Last edited:
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dge said:
I've tried establishing a systems of equations
Don't let us guess. Show your work !
Also: what does it mean that the ball collides elastically with the platform ?

##\ ##
 
dge said:
-(1/2)kx^2+mpgx+(1/2)mpvp^2=0
How did you get the value of ##v_p##?
 

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