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Question Number 15102 by Tinkutara last updated on 07/Jun/17

Small steel balls falls from rest through  the opening at A, at the steady rate of  n balls per second. Find the vertical  separation h of two consecutive balls  when the lower one has dropped d  meters.

$$\mathrm{Small}\:\mathrm{steel}\:\mathrm{balls}\:\mathrm{falls}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{through} \\ $$$$\mathrm{the}\:\mathrm{opening}\:\mathrm{at}\:{A},\:\mathrm{at}\:\mathrm{the}\:\mathrm{steady}\:\mathrm{rate}\:\mathrm{of} \\ $$$${n}\:\mathrm{balls}\:\mathrm{per}\:\mathrm{second}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{separation}\:{h}\:\mathrm{of}\:\mathrm{two}\:\mathrm{consecutive}\:\mathrm{balls} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{one}\:\mathrm{has}\:\mathrm{dropped}\:{d} \\ $$$$\mathrm{meters}. \\ $$

Commented by Tinkutara last updated on 07/Jun/17

Answered by mrW1 last updated on 07/Jun/17

ball 1:  d=(1/2)gt_1 ^2   t_1 =(√((2d)/g))  ball 2:  d_2 =(1/2)g(t_1 −(1/n))^2   h=d−d_2 =(1/2)g[t_1 ^2 −(t_1 −(1/n))^2 ]  =(1/2)g(2t_1 −(1/n))((1/n))  =(g/(2n^2 ))(2n(√((2d)/g))−1)

$$\mathrm{ball}\:\mathrm{1}: \\ $$$$\mathrm{d}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2d}}{\mathrm{g}}} \\ $$$$\mathrm{ball}\:\mathrm{2}: \\ $$$$\mathrm{d}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{2}} \\ $$$$\mathrm{h}=\mathrm{d}−\mathrm{d}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left[\mathrm{t}_{\mathrm{1}} ^{\mathrm{2}} −\left(\mathrm{t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{2}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{2t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{n}}\right)\left(\frac{\mathrm{1}}{\mathrm{n}}\right) \\ $$$$=\frac{\mathrm{g}}{\mathrm{2n}^{\mathrm{2}} }\left(\mathrm{2n}\sqrt{\frac{\mathrm{2d}}{\mathrm{g}}}−\mathrm{1}\right) \\ $$

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by ajfour last updated on 07/Jun/17

Δt=(1/n) ;   let s_i  be the height   through which i^( th)  ball has fallen  at a certain instant of time. Let  also that the duration of its fall  till tbat time instant is t.    s_i =(1/2)gt^2   = d  ⇒ 2t= (√((8d)/g))       ....(i)     then  s_(i+1) =(1/2)g(t−𝚫t)^2      Δh= s_i −s_(i+1)         = (g/2)[t^2 −(t−𝚫t)^2 ]        =(g/2)(𝚫t)(2t−𝚫t)       =(g/2)((1/n))((√((8d)/g))−(1/n))  𝚫h=(g/2)((√((8d)/(n^2 g)))−(1/n^2 )) .

$$\Delta{t}=\frac{\mathrm{1}}{{n}}\:;\:\:\:{let}\:{s}_{{i}} \:{be}\:{the}\:{height}\: \\ $$$${through}\:{which}\:\boldsymbol{{i}}^{\:{th}} \:{ball}\:{has}\:{fallen} \\ $$$${at}\:{a}\:{certain}\:{instant}\:{of}\:{time}.\:{Let} \\ $$$${also}\:{that}\:{the}\:{duration}\:{of}\:{its}\:{fall} \\ $$$${till}\:{tbat}\:{time}\:{instant}\:{is}\:\boldsymbol{{t}}. \\ $$$$\:\:\boldsymbol{{s}}_{\boldsymbol{{i}}} =\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:\:=\:\boldsymbol{{d}} \\ $$$$\Rightarrow\:\mathrm{2}\boldsymbol{{t}}=\:\sqrt{\frac{\mathrm{8}\boldsymbol{{d}}}{\boldsymbol{{g}}}}\:\:\:\:\:\:\:....\left({i}\right) \\ $$$$\:\:\:{then}\:\:\boldsymbol{{s}}_{\boldsymbol{{i}}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{g}}\left(\boldsymbol{{t}}−\boldsymbol{\Delta}{t}\right)^{\mathrm{2}} \: \\ $$$$ \\ $$$$\Delta\boldsymbol{{h}}=\:\boldsymbol{{s}}_{\boldsymbol{{i}}} −\boldsymbol{{s}}_{\boldsymbol{{i}}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:=\:\frac{\boldsymbol{{g}}}{\mathrm{2}}\left[\boldsymbol{{t}}^{\mathrm{2}} −\left(\boldsymbol{{t}}−\boldsymbol{\Delta}{t}\right)^{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:=\frac{\boldsymbol{{g}}}{\mathrm{2}}\left(\boldsymbol{\Delta}{t}\right)\left(\mathrm{2}\boldsymbol{{t}}−\boldsymbol{\Delta}{t}\right) \\ $$$$\:\:\:\:\:=\frac{\boldsymbol{{g}}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\boldsymbol{{n}}}\right)\left(\sqrt{\frac{\mathrm{8}\boldsymbol{{d}}}{\boldsymbol{{g}}}}−\frac{\mathrm{1}}{\boldsymbol{{n}}}\right) \\ $$$$\boldsymbol{\Delta{h}}=\frac{\boldsymbol{{g}}}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{8}\boldsymbol{{d}}}{\boldsymbol{{n}}^{\mathrm{2}} \boldsymbol{{g}}}}−\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} }\right)\:. \\ $$

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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