Glass marble is dropped down an elevator shaft

[tandoorichicken]

A glass marble is dropped down an elevator shaft and hits a thick glass plate on top of an elevatior that is descending at a speed of 2.0 m/s. The marble hits the glass plate 3.0 m below the point from which it was dropped. If the collision is elastic how high will the marble rise, relative to the point from which it was dropped?

So far, I let the point of impact have potential energy = 0, which would make the top of the fall have a PE of 29.4m J. That would mean that the kinetic energy just before impact would be 29.4m = 0.5mv^2. Solving for v I got 7.7 m/s. All I need help with is figuring out the velocity just after impact, and I think I can take it from there. I thought it was just (7.7 - 2)m/s, but I'm not sure.

 

[jamesrc]

Originally posted by tandoorichicken
All I need help with is figuring out the velocity just after impact, and I think I can take it from there. I thought it was just (7.7 - 2)m/s, but I'm not sure.

That's the velocity of the marble relative to the elevator after impact. Now find the velocity with repect to ground before finding the final height.

 

[M98Ranger]

Great job so far! You are correct in thinking that the velocity just after impact would be (7.7 - 2)m/s, as the marble would have lost some of its initial velocity due to the collision with the glass plate.

To continue with the problem, we can use the law of conservation of energy to find the marble's maximum height after the collision. Since the collision is elastic, we know that the total energy (kinetic + potential) before and after the collision will be equal.

Before the collision, the marble had only kinetic energy, which we calculated to be 29.4m J. After the collision, the marble will have both kinetic and potential energy, since it will have some height above the point of impact.

We can set up the following equation to represent the conservation of energy:

29.4m J = 0.5mv^2 + mgh

Where m is the mass of the marble, v is the velocity just after impact, g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height reached by the marble.

Solving for h, we get:

h = (29.4m J - 0.5mv^2) / mg

Plugging in the values we know, we get:

h = (29.4m J - 0.5m(7.7-2)^2) / (m)(9.8 m/s^2)

Simplifying, we get h = 6.12 m. Therefore, the marble will rise 6.12 m above the point from which it was dropped.

I hope this helps! Keep up the good work.