Bed or Tiles — Same Fall, Different Stop

Mecânica Iniciante Grátis VR · AR
Bed or Tiles — Same Fall, Different Stop – Mecânica
Bed or Tiles — Same Fall, Different Stop – Mecânica

Why is a phone less likely to break when it lands on a bed than on a tiled floor, dropped from the same height? Two identical phones hang side by side, each the same height above its own surface — one over a bed, one over floor tiles. Press play to let them go together: the drop plays once, and the timeline can be dragged back to any moment. Sliders set the drop height, the phone's mass and how far the mattress gives. The fall is shown in slow motion, slowing further at the moment of impact, and two force–time graphs on one time axis record how each surface brings its phone to rest.

Como usar esta simulação

O que observar

A física por trás

Each phone reaches its surface at v = √(2gh) and stops without bouncing, so both lose the same momentum, Δp = mv: with the default 0.18 kg phone dropped 1.00 m, v = 4.43 m/s and Δp = 0.80 N s. What differs is the stopping time. In the model each surface pushes harder the faster the phone is still moving into it. The mattress lets the phone sink the chosen distance (8 cm by default) and stops it in about 0.050 s; the phone's case and the tiles give only about 1.3 mm between them and stop it in about 0.0008 s. Newton's second law in momentum form gives the average resultant force, F = Δp/Δt: about 16 N on the bed against roughly 1000 N on the tiles, some 60 times more. The graphs plot the resultant force — the surface's push minus the phone's weight — so the area under each pulse is exactly Δp: a low, wide pulse for the bed and a tall, narrow one for the tiles. A higher drop or a heavier phone raises Δp and both forces; a mattress that gives more lengthens the bed's stop and lowers its force. In the model the glass cracks when the push on it passes 700 N.

impulsemomentumF = Δp/Δtstopping timeforce–time graphphone drop3D