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

[physicsman23]

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 = \sqrt{k(x^2)/m}

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 = \sqrt{2k(x^2)/m}


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.

 

[gneill]

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.

 

[physicsman23]

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?

 

[ideasrule]

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.