Question Number 64697 by Rio Michael last updated on 20/Jul/19
$${i}\:{need}\:{some}\:{help}\:{here}.\: \\ $$$$\:{An}\:{object}\:{of}\:{mass}\:\:\:{m}\:\:\:{falls}\:{from}\:{a}\:{height}\:\:{h}_{\mathrm{1}} \:{and}\:{rebound} \\ $$$${to}\:{a}\:{height}\:{of}\:{h}_{\mathrm{2}} .\:{write}\:{an}\:{expression}\:{for}\:{its}\:{momentum}. \\ $$
Answered by peter frank last updated on 20/Jul/19
$${impulse}\:={change}\:{of}\:{momentum} \\ $$$${F}={m}.\Delta{v} \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}{gh}_{\mathrm{1}} \\ $$$${v}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{2}{gh}_{\mathrm{2}} \\ $$
Commented by Rio Michael last updated on 20/Jul/19
$${thanks},\:{F}\:={m}.\bigtriangleup{v} \\ $$$$\:\:{v}_{\mathrm{1}} =\:\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }\:{and}\:{v}_{\mathrm{2}} =\sqrt{\mathrm{2}{gh}_{\mathrm{2}} }\: \\ $$$${and}\:{F}\:=\:{m}\left({v}_{\mathrm{2}} −{v}_{\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:{F}=\:{m}\left(\sqrt{\mathrm{2}{gh}_{\mathrm{2}} }\:−\:\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }\right)\:?\:{correct}? \\ $$
Commented by peter frank last updated on 20/Jul/19
$${yes} \\ $$