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i-need-some-help-here-An-object-of-mass-m-falls-from-a-height-h-1-and-rebound-to-a-height-of-h-2-write-an-expression-for-its-momentum-




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_1  and rebound  to a height of h_2 . write an expression for its momentum.
$${i}\:{need}\:{some}\:{help}\:{here}.\: \\ $$$$\:{An}\:{object}\:{of}\:{mass}\:\:\:{m}\:\:\:{falls}\:{from}\:{a}\:{height}\:\:{h}_{\mathrm{1}} \:{and}\:{rebound} \\ $$$${to}\:{a}\:{height}\:{of}\:{h}_{\mathrm{2}} .\:{write}\:{an}\:{expression}\:{for}\:{its}\:{momentum}. \\ $$
Answered by peter frank last updated on 20/Jul/19
impulse =change of momentum  F=m.Δv  v_1 ^2 =2gh_1   v_2 ^2 =2gh_2
$${impulse}\:={change}\:{of}\:{momentum} \\ $$$${F}={m}.\Delta{v} \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}{gh}_{\mathrm{1}} \\ $$$${v}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{2}{gh}_{\mathrm{2}} \\ $$
Commented by Rio Michael last updated on 20/Jul/19
thanks, F =m.△v    v_1 = (√(2gh_1 )) and v_2 =(√(2gh_2 ))   and F = m(v_2 −v_1 )          F= m((√(2gh_2 )) − (√(2gh_1 ))) ? correct?
$${thanks},\:{F}\:={m}.\bigtriangleup{v} \\ $$$$\:\:{v}_{\mathrm{1}} =\:\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }\:{and}\:{v}_{\mathrm{2}} =\sqrt{\mathrm{2}{gh}_{\mathrm{2}} }\: \\ $$$${and}\:{F}\:=\:{m}\left({v}_{\mathrm{2}} −{v}_{\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:{F}=\:{m}\left(\sqrt{\mathrm{2}{gh}_{\mathrm{2}} }\:−\:\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }\right)\:?\:{correct}? \\ $$
Commented by peter frank last updated on 20/Jul/19
yes
$${yes} \\ $$

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