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




Question Number 22193 by NECx last updated on 13/Oct/17
A body of mass 0.1kg dropped   from a height of 8m onto a hard  floor and bounces back to a height  of 2m. Calculate the chaange in  momentum.If the body is in  contact with the floor for 0.1s,  what is the force exerted on the  body?(g=10m/s^2 )
$$\mathrm{A}\:\mathrm{body}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{0}.\mathrm{1kg}\:\mathrm{dropped}\: \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\mathrm{8m}\:\mathrm{onto}\:\mathrm{a}\:\mathrm{hard} \\ $$$$\mathrm{floor}\:\mathrm{and}\:\mathrm{bounces}\:\mathrm{back}\:\mathrm{to}\:\mathrm{a}\:\mathrm{height} \\ $$$$\mathrm{of}\:\mathrm{2m}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{chaange}\:\mathrm{in} \\ $$$$\mathrm{momentum}.\mathrm{If}\:\mathrm{the}\:\mathrm{body}\:\mathrm{is}\:\mathrm{in} \\ $$$$\mathrm{contact}\:\mathrm{with}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{for}\:\mathrm{0}.\mathrm{1s}, \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{force}\:\mathrm{exerted}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{body}?\left(\mathrm{g}=\mathrm{10m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by NECx last updated on 13/Oct/17
please help with explanatory  solution(s).... Thanks
$$\mathrm{please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{explanatory} \\ $$$$\mathrm{solution}\left(\mathrm{s}\right)….\:\mathrm{Thanks} \\ $$
Answered by ajfour last updated on 13/Oct/17
v_(approch) =−(√(2gh_1 )) =−(√(2×10×8))                =−4(√(10))m/s  v_(separation) =(√(2gh_2 )) =(√(2×10×2))=2(√(10))m/s  F=((△p)/(△t)) =((m(v_(sep) −v_(app) ))/(△t)) =((0.1×6(√(10)))/(0.1)) N    =6(√(10)) N  ≈ 6×3.16 ≈19N .
$${v}_{{approch}} =−\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }\:=−\sqrt{\mathrm{2}×\mathrm{10}×\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{4}\sqrt{\mathrm{10}}{m}/{s} \\ $$$${v}_{{separation}} =\sqrt{\mathrm{2}{gh}_{\mathrm{2}} }\:=\sqrt{\mathrm{2}×\mathrm{10}×\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{10}}{m}/{s} \\ $$$${F}=\frac{\bigtriangleup{p}}{\bigtriangleup{t}}\:=\frac{{m}\left({v}_{{sep}} −{v}_{{app}} \right)}{\bigtriangleup{t}}\:=\frac{\mathrm{0}.\mathrm{1}×\mathrm{6}\sqrt{\mathrm{10}}}{\mathrm{0}.\mathrm{1}}\:{N} \\ $$$$\:\:=\mathrm{6}\sqrt{\mathrm{10}}\:{N}\:\:\approx\:\mathrm{6}×\mathrm{3}.\mathrm{16}\:\approx\mathrm{19}{N}\:. \\ $$
Commented by NECx last updated on 14/Oct/17
whats the meaning of v_(approach)  and  v_(seperation) ?
$$\mathrm{whats}\:\mathrm{the}\:\mathrm{meaning}\:\mathrm{of}\:\mathrm{v}_{\mathrm{approach}} \:\mathrm{and} \\ $$$$\mathrm{v}_{\mathrm{seperation}} ? \\ $$

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