A-body-of-mass-0-1kg-dropped-from-a-height-of-8m-onto-a-hard-floor-bounces-back-to-a-height-of-2m-Calculate-the-change-of-momentum-If-the-body-is-in-contact-with-the-floor-for-0-1s-then-what-is-the-

Question Number 48194 by Necxx last updated on 20/Nov/18

$${A}\:{body}\:{of}\:{mass}\:\mathrm{0}.\mathrm{1}{kg}\:{dropped}\:{from} \\ $$$${a}\:{height}\:{of}\:\mathrm{8}{m}\:{onto}\:{a}\:{hard}\:{floor} \\ $$$${bounces}\:{back}\:{to}\:{a}\:{height}\:{of}\:\mathrm{2}{m}. \\ $$$${Calculate}\:{the}\:{change}\:{of}\:{momentum}. \\ $$$${If}\:{the}\:{body}\:{is}\:{in}\:{contact}\:{with}\:{the} \\ $$$${floor}\:{for}\:\mathrm{0}.\mathrm{1}{s}\:{then}\:{what}\:{is}\:{the} \\ $$$${force}\:{exerted}\:{on}\:{the}\:{body}? \\ $$$$\left[{g}=\mathrm{10}{ms}^{−\mathrm{2}} \right] \\ $$

Commented by Necxx last updated on 20/Nov/18

$${please}\:{help} \\ $$

Answered by mr W last updated on 21/Nov/18

v_1 =−(√(2gh_1 )) v_2 =(√(2gh_2 )) Δp=m(v_2 −v_1 )=m(√(2g))((√h_2 )+(√h_1 )) =0.1×(√(20))((√2)+(√8))=1.897 Ns F=Δp/Δt=1.897/0.1=18.97 N

$${v}_{\mathrm{1}} =−\sqrt{\mathrm{2}{gh}_{\mathrm{1}} } \\ $$$${v}_{\mathrm{2}} =\sqrt{\mathrm{2}{gh}_{\mathrm{2}} } \\ $$$$\Delta{p}={m}\left({v}_{\mathrm{2}} −{v}_{\mathrm{1}} \right)={m}\sqrt{\mathrm{2}{g}}\left(\sqrt{{h}_{\mathrm{2}} }+\sqrt{{h}_{\mathrm{1}} }\right) \\ $$$$=\mathrm{0}.\mathrm{1}×\sqrt{\mathrm{20}}\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{8}}\right)=\mathrm{1}.\mathrm{897}\:{Ns} \\ $$$${F}=\Delta{p}/\Delta{t}=\mathrm{1}.\mathrm{897}/\mathrm{0}.\mathrm{1}=\mathrm{18}.\mathrm{97}\:{N} \\ $$

Commented by Necxx last updated on 20/Nov/18

$${oh}…..\:{I}\:{thought}\:{its}\:{meant}\:{to}\:{be} \\ $$$${the}\:{difference}\:{between}\:{the}\:{final} \\ $$$${and}\:{initial}\:{momentum}\:{not}\:{the} \\ $$$${reverse}.{Why}\:{did}\:{you}\:{take}\:{this} \\ $$$${step}? \\ $$$$ \\ $$$${Thanks} \\ $$

Commented by Necxx last updated on 21/Nov/18