Answer to Question #304257 in Mechanics | Relativity for Pretty

Question

1.   One morning, while little Lenlen’s mother was busy in the kitchen, Little Lenlen rolls off the bed that is 0.50 m above the floor. If the floor is hardwood, little Lenlen’s head is brought to rest in approx. 2.0 mm. If the floor is carpeted, this stopping distance is increased to about 1.0 cm. Calculate the magnitude and duration of the acceleration in both cases, to determine the risk of injury. Assume that little Lenlen remains horizontal during the fall to the floor. (Note: A more complicated fall could result in a head velocity greater or less than the speed you calculate. Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s² lasting for any length of time will not cause injury, whereas an acceleration greater than 1000m/s² lasting for at least 1 ms will cause injury.)

Answer 1

Explanations & Calculations

• Apply $\small v^2=u^2+2as$ vertically downwards for her fall.

$\qquad\qquad \begin{aligned} \small v^2&=\small 0+2g(0.50)\\ \end{aligned}$

• Then apply the same equation for the stopping duration after getting in contact with the floor.

$\qquad\qquad \begin{aligned} \small 0&=\small v^2+2(a)(h)\\ \small a&=\small \frac{-v^2}{2h}\\ &=\small \frac{-g(0.50)}{h} \end{aligned}$

• Apply $\small s =\large\frac{(v+u)t}{2}$ for the period of stopping.

$\qquad\qquad \begin{aligned} \small h&=\small \frac{(0+\sqrt{2g(0.50)})t}{2}\\ \small t&=\small 0.639h \end{aligned}$

• For the hardwood floor, it is

$\qquad\qquad \begin{aligned} \small a_h&=\small \frac{-9.8\times0.50}{2\times10^{-3}m}\\ &=\small 2450\,ms^{-2}\\ \\ \small t_h&=\small 0.639\times(2\times10^{-3})\\ &=\small 1.3\times10^{-3}\,s\\ &=\small 1.3\,ms \end{aligned}$

• For the carpeted floor,

$\qquad\qquad \begin{aligned} \small a_c&=\small 490\,ms^{-2}\\ \\ \small t_c&=\small 6.4\,ms \end{aligned}$

• The carpet floor is safer.