Solve Hooke's Law Question: Velocity of Ball After Release

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Homework Help Overview

The problem involves a ball being released from a compressed spring, with the goal of determining the horizontal velocity of the ball after release. The context is rooted in Hooke's Law and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to equate kinetic and potential energy to find the velocity of the ball, while also applying Newton's second law to analyze the forces involved. Some participants question the validity of using constant acceleration in this scenario, noting that the spring force varies as it compresses and decompresses.

Discussion Status

Participants are exploring different methods to approach the problem, with some affirming the correctness of the energy conservation method while others suggest that a more complex analysis involving integrals may be necessary due to the non-constant nature of the spring force. There is no explicit consensus on the best approach, but productive dialogue is occurring around the implications of using different methods.

Contextual Notes

There is a suggestion that additional information may be needed to fully resolve the problem using force and acceleration methods, although some participants believe the existing information is sufficient for the energy conservation approach.

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Homework Statement



A ball with mass 'm' is pushed against a spring (with a spring constant 'k') and causes a displacement 'x' on the spring. Once the force is released from, the ball is shot horizontally. What is the velocity 'v' horizontally?

Homework Equations



E(k) = 1/2 mv^2
E(s) = 1/2 kx^2

The Attempt at a Solution



I equated both of the relevant equations and get the result that:

v = [tex]\sqrt{k(x^2)/m}[/tex]

This answer seems right. However, when I use Newton's second law, it doesn't seem to work anymore. I state that:

kx = ma (Spring force = Net force) and therefore: a = kx/m

Substitute for a:

a = v^2/2x

equate a = a and get the result that:

v = [tex]\sqrt{2k(x^2)/m}[/tex]


Can anyone tell me where my logic is going wrong? My teacher said that the first one is correct. I don't understand why the second one is not.
 
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The force that the spring applies to the ball is not constant, so the acceleration is not constant either. It starts out at some maximum value and decreases to zero as the spring relaxes. To solve the problem using forces and acceleration, you'd have to set up an integral to sum up the impulse imparted while the spring is in contact with the ball.
 
gneill said:
The force that the spring applies to the ball is not constant, so the acceleration is not constant either. It starts out at some maximum value and decreases to zero as the spring relaxes. To solve the problem using forces and acceleration, you'd have to set up an integral to sum up the impulse imparted while the spring is in contact with the ball.

Would there need to me more information provided in the question, or would this information suffice to solve using this method?
 
The information you have is enough. Your first solution--the one with conservation of energy--is completely correct, and here's why:

When the ball is at maximum displacement, it isn't moving, so kinetic energy is 0. Potential energy is 1/2kx^2, so total energy is 1/2kx^2. After the ball starts moving, the spring keeps on accelerating the ball until it reaches x=0, at which point the spring force changes direction and stops the spring's motion. The ball is still moving at speed v (the speed you're looking for), so at this point, kinetic energy is 1/2mv^2 while potential energy is 0. Equating the total energy of 1/2mv^2 with the previously-found total energy of 1/2kx^2 gives you the answer.
 

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