Spring compression of a bowling ball

[reminiscent]

Homework Statement

A 5 kg bowling ball moves at 10 m/s on a flat and frictionless surface. It hits a massless bumper with a spring constant of 100 N/cm. How far (in cm) will the spring compress?

Homework Equations

|F| = |kx|
Fnet = ma

The Attempt at a Solution

This is mostly a plug and chug problem, but the only piece missing is obviously force. Since moving at 10 m/s is a given, and on a flat and frictionless surface, this means that acceleration is 0 because it is going at a constant velocity. I would think that you would need to use Newton's 2nd Law - so drawing out a FBD gives me 2 forces acting on the ball in the y-direction, normal force and force due to gravity, but no forces acting on it in the x-direction. Do I even have to incorporate normal force in this problem? I was thinking that the weight of the ball is acting upon the spring, so the force in the spring equation would be easily (5 kg)(9.81 m/s^2) = 49.05 N. But is it that simple? I need some insight of my thinking.

 

[SteamKing]

Homework Statement

A 5 kg bowling ball moves at 10 m/s on a flat and frictionless surface. It hits a massless bumper with a spring constant of 100 N/cm. How far (in cm) will the spring compress?

Homework Equations

|F| = |kx|
Fnet = ma

The Attempt at a Solution

This is mostly a plug and chug problem, but the only piece missing is obviously force. Since moving at 10 m/s is a given, and on a flat and frictionless surface, this means that acceleration is 0 because it is going at a constant velocity. I would think that you would need to use Newton's 2nd Law - so drawing out a FBD gives me 2 forces acting on the ball in the y-direction, normal force and force due to gravity, but no forces acting on it in the x-direction. Do I even have to incorporate normal force in this problem? I was thinking that the weight of the ball is acting upon the spring, so the force in the spring equation would be easily (5 kg)(9.81 m/s^2) = 49.05 N. But is it that simple? I need some insight of my thinking.

The normal force on the bowling ball has nothing to do with this problem. The weight of the bowling ball has nothing to do with this problem.

In fact, this situation could be done in a zero gravity environment without affecting what happens to the spring when it's hit by the bowling ball.

The ball is moving at a constant velocity, which means it also has this other property in a constant amount.

What other property besides force will the spring possesses when its compressed?

 

[mechpeac]

Think about energy...

 

[reminiscent]

The normal force on the bowling ball has nothing to do with this problem. The weight of the bowling ball has nothing to do with this problem.

In fact, this situation could be done in a zero gravity environment without affecting what happens to the spring when it's hit by the bowling ball.

The ball is moving at a constant velocity, which means it also has this other property in a constant amount.

What other property besides force will the spring possesses when its compressed?

Work total and the change in total kinetic energy is 0. I also know that total work = F*d and the change in kinetic energy = final kinetic energy - initial kinetic energy.
I still don't understand how I can find the force...

 

[mechpeac]

You don't need force. Write a conservation of energy equation.

 

[reminiscent]

You don't need force. Write a conservation of energy equation.

delta K =(1/2)m*vf^2 - (1/2)m*vi^2
I don't understand how that helps me though? Where does the spring constant come into place?
I think I am being clueless here because we haven't discussed about this in too much depth.

 

[mechpeac]

The kinetic energy of the mass before it compresses the spring will equal the potential energy of the compressed spring.