Falling onto a mattress

Problem

Explain why it is more dangerous to fall from a certain height onto a rigid floor than to fall from the same height onto a mattress, even though the impact velocity is the same. Use the impulse–momentum theorem to justify your answer.

Solution

The impulse–momentum theorem relates the average force acting during the collision to the change in momentum: $\vv{F}_{\text{average}}\,\Delta t = \Delta \vv{p}$ In both cases the body arrives with the same impact velocity $v$ and comes to rest ($v_f = 0$). The change in momentum is therefore identical in the two cases: $\Delta p = m\,(0 - v) = -m\,v \qquad (\text{same value for floor and mattress})$ For the same $\Delta p$, the average force depends only on the stopping time $\Delta t$: $F_{\text{average}} = \frac{\abs{\Delta p}}{\Delta t} = \frac{m\,v}{\Delta t}$ The rigid floor stops the body almost instantaneously: $\Delta t$ is tiny, so $F_{\text{average}}$ is huge. The mattress, instead, deforms and lengthens the stopping time: $\Delta t$ large, so $F_{\text{average}}$ small. It is the force (not the change in momentum) that causes injury, so the mattress is safer. $\ev{\text{For the same }\Delta p,\ \text{the mattress lengthens }\Delta t\ \text{and reduces }F = \Delta p/\Delta t}$

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Topics: Quantità di moto e urti Concepts: Impulso · Quantità di moto