A glass marble is dropped down an elevator shaft and hits a thick glass plate on

ID: 1381029 • Letter: A

Question

A glass marble is dropped down an elevator shaft and hits a thick glass plate on top of an elevator that is descending at a speed of 2.60 m/s. The marble hits the glass plate 6.0 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?

A glass marble is dropped down an elevator shaft and hits a thick glass plate on top of an elevator that is descending at a speed of 2.60 m/s. The marble hits the glass plate 6.0 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?

Explanation / Answer

speed of marble when it hits the glass plate, v = sqrt(2*g*h)

= sqrt(2*9.8*6)

= 10.84 m/s

before the collision, relative velosty of marble with resect to elevator = 10.84 - 2.6

= 8.24 m/s

This is an elestic collsion,

so velosity of marble after the collsion = -8.24 + 2.6

= -5.64 m/s

height raised after the collsion, h = 5.64^2/(2*9.8)

= 1.62 m <<<<<<<<-----------Answer

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A glass marble is dropped down an elevator shaft and hits a thick glass plate on

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