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

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Question Number 22193 by NECx last updated on 13/Oct/17

$$\mathrm{A}\mathrm{body}\mathrm{of}\mathrm{mass}0.1\mathrm{kg}\mathrm{dropped}$$$$\mathrm{from}\mathrm{a}\mathrm{height}\mathrm{of}8\mathrm{m}\mathrm{onto}\mathrm{a}\mathrm{hard}$$$$\mathrm{floor}\mathrm{and}\mathrm{bounces}\mathrm{back}\mathrm{to}\mathrm{a}\mathrm{height}$$$$\mathrm{of}2\mathrm{m}.\mathrm{Calculate}\mathrm{the}\mathrm{chaange}\mathrm{in}$$$$\mathrm{momentum}.\mathrm{If}\mathrm{the}\mathrm{body}\mathrm{is}\mathrm{in}$$$$\mathrm{contact}\mathrm{with}\mathrm{the}\mathrm{floor}\mathrm{for}0.1\mathrm{s},$$$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{force}\mathrm{exerted}\mathrm{on}\mathrm{the}$$$$\mathrm{body}?(\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2)$$$$$$$$$$$$$$

Commented by NECx last updated on 13/Oct/17

$$\mathrm{please}\mathrm{help}\mathrm{with}\mathrm{explanatory}$$$$\mathrm{solution}\left(\mathrm{s}\right)\dots .\mathrm{Thanks}$$

Answered by ajfour last updated on 13/Oct/17

$$v_{approch}=-\sqrt{2{gh}_1}=-\sqrt{2\times 10\times 8}$$$$=-4\sqrt{10}m/s$$$$v_{separation}=\sqrt{2{gh}_2}=\sqrt{2\times 10\times 2}=2\sqrt{10}m/s$$$$F=\frac{△p}{△t}=\frac{m(v_{sep}-v_{app})}{△t}=\frac{0.1\times 6\sqrt{10}}{0.1}N$$$$=6\sqrt{10}N\approx 6\times 3.16\approx 19N.$$

Commented by NECx last updated on 14/Oct/17

$$\mathrm{whats}\mathrm{the}\mathrm{meaning}\mathrm{of}{\mathrm{v}}_{\mathrm{approach}}\mathrm{and}$$$${\mathrm{v}}_{\mathrm{seperation}}?$$

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