Kinetic Energy & Momentum of a Ball released from Spring

[Physics Person]

Homework Statement

"A spring with a spring constant "k" is compressed 10 cm from equilibrium. A ball of mass 100 g is at rest next to it. The spring then decompresses quickly back to its equilibrium position causing the ball to shoot forward. If the spring constant is 500 N/m, what is the ball's kinetic energy? And what is the ball's momentum after it is struck by the spring?"

Homework Equations

KE = 1/2mv^2
PEspring = 1/2kx^2

The Attempt at a Solution

I have attempted to solve this by using the conservation of mechanical energy to surmise that the kinetic energy of the ball after it is released would be equivalent to the potential energy of the spring before the ball is released. However, what is confusing me is that, after the ball is released from the spring, wouldn't its velocity be higher at first, and then gradually peter off? Thus, since kinetic energy equals 1/2mv^2, and momentum equals mv, wouldn't the kinetic energy and the momentum of the ball taper off over time, as well? So I'm not sure which moment of time they want me to find the kinetic energy and momentum for. Should I just assume it's the maximum amount of kinetic energy and momentum, right at the moment when the ball is released from the spring and starts moving?

 

[PeroK]

Unless you have some information about the subsequent resisting forces, it's not possible to calculate the speed of the ball at any later time.

 

[RPinPA]

However, what is confusing me is that, after the ball is released from the spring, wouldn't its velocity be higher at first, and then gradually peter off?

Only if there's a force acting on it. A diagram is needed. If the ball is moving on a horizontal surface under the usual frictionless assumption, then there's no force acting to change its velocity.

 

[Physics Person]

Alright, I see now. But would my assumption that the kinetic energy of the ball would be equivalent to the potential energy of the spring, given by the equation PEspring = 1/2kx^2, due to conservation of mechanical energy, still be correct?

 

[RPinPA]

Depends on where the ball is. Imagine the ball is at a position where the spring never touches it. Then obviously the kinetic energy obtained by the ball is 0.

The kinetic energy of the ball comes from work that the spring does on the ball. The spring can only do work on the ball when it is touching the ball.

Again, a diagram is needed.

 

[haruspex]

Depends on where the ball is. Imagine the ball is at a position where the spring never touches it. Then obviously the kinetic energy obtained by the ball is 0.

The kinetic energy of the ball comes from work that the spring does on the ball. The spring can only do work on the ball when it is touching the ball.

Again, a diagram is needed.

It says

ball of mass 100 g is at rest next to it.

I agree it is confusing then to refer to the spring "striking" the ball.

 

[haruspex]

Alright, I see now. But would my assumption that the kinetic energy of the ball would be equivalent to the potential energy of the spring, given by the equation PEspring = 1/2kx^2, due to conservation of mechanical energy, still be correct?

Yes.