Question-97628

Question Number 97628 by peter frank last updated on 08/Jun/20

Answered by Rio Michael last updated on 08/Jun/20

Assume the thickness of each piece = h and the acceleration produced by each piece = a from v_f ^2 −v_0 ^2 = 2ah ⇒ v_0 ^2 = −2ah for reducing speed. lets assume we have n pieces the v_0 ^2 = −2nah ⇒ n = ((−v_0 ^2 )/(2ah)) ......(i) The bullet losses (1/(20)) of its speed as it passes through the piece so its final speed is given by : v_f = v_0 − (v_0 /(20 )) ⇒ v_f = ((19v_0 )/(20)) v_f ^2 −v_0 ^2 = 2ah but v_f = ((19v_0 )/(20)) ⇒ (((19v_0 )/(20)))^2 −v_0 ^2 = 2ah ((361v_0 ^2 )/(400)) − v_0 ^2 = 2ah −((39v_0 ^2 )/(400)) = 2ah but from (i), 2ah = −(v_0 ^2 /n) ⇒ n = ((−v_0 ^2 )/(−((39v_0 ^2 )/(400)))) ⇒ n = 10.62 , n ≈ 11 pieces

$$\mathrm{Assume}\:\mathrm{the}\:\mathrm{thickness}\:\mathrm{of}\:\mathrm{each}\:\mathrm{piece}\:=\:{h}\:\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{produced}\:\mathrm{by}\:\mathrm{each}\:\mathrm{piece}\:=\:{a} \\ $$$$\mathrm{from}\:{v}_{{f}} ^{\mathrm{2}} \:−{v}_{\mathrm{0}} ^{\mathrm{2}} \:=\:\mathrm{2}{ah}\:\Rightarrow\:{v}_{\mathrm{0}} ^{\mathrm{2}} \:=\:−\mathrm{2}{ah}\:\mathrm{for}\:\mathrm{reducing}\:\mathrm{speed}. \\ $$$$\mathrm{lets}\:\mathrm{assume}\:\mathrm{we}\:\mathrm{have}\:{n}\:\mathrm{pieces}\:\mathrm{the}\:\:{v}_{\mathrm{0}} ^{\mathrm{2}} \:=\:−\mathrm{2}{nah}\: \\ $$$$\:\:\Rightarrow\:\:{n}\:=\:\frac{−{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{ah}}\:……\left({i}\right) \\ $$$$\mathrm{The}\:\mathrm{bullet}\:\mathrm{losses}\:\frac{\mathrm{1}}{\mathrm{20}}\:\mathrm{of}\:\mathrm{its}\:\mathrm{speed}\:\mathrm{as}\:\mathrm{it}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{piece} \\ $$$$\mathrm{so}\:\mathrm{its}\:\mathrm{final}\:\mathrm{speed}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\::\:{v}_{{f}} \:=\:{v}_{\mathrm{0}} \:−\:\frac{{v}_{\mathrm{0}} }{\mathrm{20}\:}\:\Rightarrow\:\:{v}_{{f}} \:=\:\frac{\mathrm{19}{v}_{\mathrm{0}} }{\mathrm{20}} \\ $$$$\:{v}_{{f}} ^{\mathrm{2}} \:−{v}_{\mathrm{0}} ^{\mathrm{2}} \:=\:\mathrm{2}{ah}\:\mathrm{but}\:{v}_{{f}} \:=\:\frac{\mathrm{19}{v}_{\mathrm{0}} }{\mathrm{20}} \\ $$$$\Rightarrow\:\left(\frac{\mathrm{19}{v}_{\mathrm{0}} }{\mathrm{20}}\right)^{\mathrm{2}} −{v}_{\mathrm{0}} ^{\mathrm{2}} \:=\:\mathrm{2}{ah}\: \\ $$$$\:\:\:\frac{\mathrm{361}{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{400}}\:−\:{v}_{\mathrm{0}} ^{\mathrm{2}} \:=\:\mathrm{2}{ah}\: \\ $$$$\:\:−\frac{\mathrm{39}{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{400}}\:=\:\mathrm{2}{ah}\:\:\mathrm{but}\:\mathrm{from}\:\left({i}\right),\:\mathrm{2}{ah}\:=\:−\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{n}} \\ $$$$\Rightarrow\:{n}\:=\:\frac{−{v}_{\mathrm{0}} ^{\mathrm{2}} }{−\frac{\mathrm{39}{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{400}}}\:\Rightarrow\:{n}\:=\:\mathrm{10}.\mathrm{62}\:\:,\:{n}\:\approx\:\mathrm{11}\:\mathrm{pieces} \\ $$

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