# Spring compression of a bowling ball

• reminiscent
In summary: Using the formula for potential energy of a spring, you can solve for the distance the spring will compress.In summary, the problem involves a 5 kg bowling ball moving at a constant velocity of 10 m/s on a frictionless surface, hitting a massless bumper with a spring constant of 100 N/cm. The goal is to determine how far the spring will compress. The key to solving this problem is to use the conservation of energy equation, where the kinetic energy of the ball before it compresses the spring will equal the potential energy of the compressed spring. Using the formula for potential energy of a spring, the distance the spring will compress can be calculated. The normal force and weight of the ball are not relevant in this problem.
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?

|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.

reminiscent said:

## 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?

|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

SteamKing said:
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...

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

mechpeac said:
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.

## 1. What is spring compression of a bowling ball?

Spring compression refers to the amount of energy stored in a bowling ball when it is compressed by the bowler's hand during release. This energy is released when the ball hits the pins, causing them to scatter.

## 2. How does spring compression affect the performance of a bowling ball?

The amount of spring compression can greatly impact the performance of a bowling ball. A ball with high spring compression will have more energy to transfer to the pins, resulting in a stronger impact and potentially more pin action. On the other hand, a ball with low spring compression will have less energy and may not hit the pins with as much force.

## 3. What factors influence the spring compression of a bowling ball?

The main factors that influence spring compression are the core design and coverstock of the bowling ball. A ball with a large, dynamic core and a reactive coverstock will typically have higher spring compression. The release speed and angle of the bowler also play a role in the amount of spring compression.

## 4. How can I adjust the spring compression of my bowling ball?

The spring compression of a bowling ball cannot be directly adjusted, but it can be influenced by changing the surface texture and oil absorption of the coverstock. Applying a polish or sanding the surface can alter the amount of friction and ultimately affect the spring compression.

## 5. Is spring compression the only factor that determines a bowling ball's performance?

No, spring compression is just one aspect that contributes to a ball's overall performance. Other factors such as weight, core design, coverstock material, and lane conditions also play a significant role in a ball's performance on the lanes.

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