Spring shooting ball bearing

[schrock]

I am convinced I have not received enough info to solve this prob. Please help me get started. A small steel ball bearing with a mass of 20.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.35 m. Calculated the horizontal distance the ball would travel if the same spring were aimed 33.0 degrees from the horizontal.

 

[rcgldr]

My guess is that you have to assume the spring has no internal inertial so that all of the spring force goes into accelerating the ball. Since it's not included, the problem statement implies that the spring constant doesn't affect the outcome.

 

[baba89]

Hello,

Thank you for sharing your problem with me. I am happy to help you get started on solving this problem.

Firstly, let's review the information we have been given. We know that there is a small steel ball bearing with a mass of 20.0 g. We also know that the ball bearing is on a short compressed spring and when it is released, it is sent to a height of 1.35 m. This information suggests that the spring is acting as a projectile launcher, sending the ball bearing into the air.

To solve this problem, we need to use the principles of projectile motion. This involves breaking down the motion of the ball bearing into its horizontal and vertical components. We can then use equations of motion to calculate the horizontal distance the ball bearing would travel if the spring were aimed at 33.0 degrees from the horizontal.

To get started, we can use the following equation to calculate the maximum height reached by the ball bearing:

Hmax = (v^2 * sin^2θ) / 2g

Where Hmax is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

From the given information, we know that the maximum height reached by the ball bearing is 1.35 m. We also know that the initial velocity is zero, as the spring is compressed and not moving before the release of the ball bearing. We can rearrange the equation to solve for θ:

θ = arcsin(sqrt(2Hmax * g) / v)

Substituting the values we have, we get:

θ = arcsin(sqrt(2 * 1.35 * 9.8) / 0)

θ = arcsin(3.87)

θ = 75.4 degrees

This means that the angle of projection for the spring aimed vertically is 75.4 degrees.

Now, to calculate the horizontal distance, we can use the following equation:

Range = (v^2 * sin2θ) / g

Where Range is the horizontal distance.

Substituting the values we have, we get:

Range = (0^2 * sin(2 * 75.4)) / 9.8

Range = 0 m

This result makes sense, as when the spring is aimed vertically, the ball bearing will only travel vertically and not horizontally.

To calculate the horizontal distance when the