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 2.00 m/s. The marble hits the glass plate 6.5 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?

Answer 1

Initial velocity , u = 2m/s

distance = S = 6.5

Final velocity, v = sqrt( u^2 + 2 g S) = sqrt(2^2 + 2*9.81*6.5) = 11.468 m/s

Since collision is elastic , it will bounce back with same velocity

H =v^2/2g = 11.468^2 / 2*9.81 = 6.704 m

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Height the marble will rise after elastic collision = 6.704 m

Relative height = 6.704 - 6.5 = 0.204 m