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Question Number 109207 by aurpeyz last updated on 21/Aug/20

  A 50g golf ball is struck with a club moving  wit a velocity of 22m/s after it moves 4cm anda  ball accelerates with a velocity of 44m/s   estimate the average force exerted by the clubn  on the ball

$$ \\ $$$$\mathrm{A}\:\mathrm{50g}\:\mathrm{golf}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{struck}\:\mathrm{with}\:\mathrm{a}\:\mathrm{club}\:\mathrm{moving} \\ $$$$\mathrm{wit}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{22m}/\mathrm{s}\:\mathrm{after}\:\mathrm{it}\:\mathrm{moves}\:\mathrm{4cm}\:\mathrm{anda} \\ $$$$\mathrm{b}{a}\mathrm{ll}\:\mathrm{accelerates}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{44m}/\mathrm{s}\: \\ $$$$\mathrm{estimate}\:\mathrm{the}\:\mathrm{average}\:\mathrm{force}\:\mathrm{exerted}\:\mathrm{by}\:\mathrm{the}\:\mathrm{clubn} \\ $$$$\mathrm{o}{n}\:\mathrm{the}\:\mathrm{ball} \\ $$

Answered by aurpeyz last updated on 21/Aug/20

pls help

$${pls}\:{help} \\ $$

Answered by mr W last updated on 22/Aug/20

W=∫_0 ^s Fds=Fs=E_2 −E_1 =((m(v_2 ^2 −v_1 ^2 ))/2)  F=((m(v_2 ^2 −v_1 ^2 ))/(2s))=((0.05×(44^2 −22^2 ))/(2×0.04))=907.5N

$${W}=\int_{\mathrm{0}} ^{{s}} {Fds}={Fs}={E}_{\mathrm{2}} −{E}_{\mathrm{1}} =\frac{{m}\left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$${F}=\frac{{m}\left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}{s}}=\frac{\mathrm{0}.\mathrm{05}×\left(\mathrm{44}^{\mathrm{2}} −\mathrm{22}^{\mathrm{2}} \right)}{\mathrm{2}×\mathrm{0}.\mathrm{04}}=\mathrm{907}.\mathrm{5}{N} \\ $$

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