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Question Number 223724 by fantastic last updated on 03/Aug/25
A rubber ball of mass m radius R and densityρ is released from a depth h under a fluid of density σ(σ>ρ) i) How high will the ball bounce on the the fluid ?Ignore any obstacles ii)time taken by the ball to reach the surface iii)time in air
$$\: \\ $$$$\mathrm{A}\:\mathrm{rubber}\:\mathrm{ball}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\: \\ $$$$\mathrm{radius}\:\mathrm{R}\:\mathrm{and}\:\mathrm{density}\rho\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{a} \\ $$$$\mathrm{depth}\:\mathrm{h}\:\mathrm{under}\:\mathrm{a}\:\mathrm{fluid}\:\mathrm{of}\:\mathrm{density}\:\sigma\left(\sigma>\rho\right) \\ $$$$\left.\mathrm{i}\right) \\ $$$$\:\mathrm{How}\:\mathrm{high}\:\mathrm{will}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{bounce}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{the}\:\mathrm{fluid}\:?\mathrm{Ignore}\:\mathrm{any} \\ $$$$\mathrm{obstacles} \\ $$$$\left.\mathrm{ii}\right)\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\: \\ $$$$\mathrm{surface} \\ $$$$\left.\mathrm{iii}\right)\mathrm{time}\:\mathrm{in}\:\mathrm{air} \\ $$
Answered by mr W last updated on 03/Aug/25
assume σ>ρ, R≪h ii) a=((σ/ρ)−1)g=acceleration upwards t_1 =(√((2h)/a))=(√((2h)/(((σ/ρ)−1)g))) iii) v^2 =2ah 2gh_2 =v^2 =2ah h_2 =((ah)/g)=((σ/ρ)−1)h t_2 =2(√((2h_1 )/g))=2(√(((σ/ρ)−1)((2h)/g))) i) ((σ/ρ)−1)h
$${assume}\:\sigma>\rho,\:{R}\ll{h} \\ $$$$\left.{ii}\right) \\ $$$${a}=\left(\frac{\sigma}{\rho}−\mathrm{1}\right){g}={acceleration}\:{upwards} \\ $$$${t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}{h}}{{a}}}=\sqrt{\frac{\mathrm{2}{h}}{\left(\frac{\sigma}{\rho}−\mathrm{1}\right){g}}} \\ $$$$\left.{iii}\right) \\ $$$${v}^{\mathrm{2}} =\mathrm{2}{ah} \\ $$$$\mathrm{2}{gh}_{\mathrm{2}} ={v}^{\mathrm{2}} =\mathrm{2}{ah} \\ $$$${h}_{\mathrm{2}} =\frac{{ah}}{{g}}=\left(\frac{\sigma}{\rho}−\mathrm{1}\right){h} \\ $$$${t}_{\mathrm{2}} =\mathrm{2}\sqrt{\frac{\mathrm{2}{h}_{\mathrm{1}} }{{g}}}=\mathrm{2}\sqrt{\left(\frac{\sigma}{\rho}−\mathrm{1}\right)\frac{\mathrm{2}{h}}{{g}}} \\ $$$$\left.{i}\right) \\ $$$$\left(\frac{\sigma}{\rho}−\mathrm{1}\right){h} \\ $$
Commented by mr W last updated on 03/Aug/25
Answered by fantastic last updated on 03/Aug/25
i) volume of the ball=(4/3)πR^3 or (m/ρ) fluid′s volume displaced by the ball=(m/ρ) weight of that fluid=((mσ)/ρ)g weight of ball =mg buoyancy =mg(σ/ρ)−mg=mg((σ/ρ)−1) F=ma a=(F/m)⇒((mg((σ/ρ)−1))/m)=g((σ/ρ)−1) at h depth u=0, a=g((σ/ρ)−1), s=h velocity at surface⇒ v^2 =u^2 +2as v=(√(2×g((σ/ρ)−1)h)) let the heigth it will go upwards be x v=0 ,s=x, u=(√(2gh((σ/ρ)−1))),a=g v^2 =u^2 −2as⇒ 0=2gh((σ/ρ)−1)−2xg or 2gx=2gh((σ/ρ)−1) So x=h((σ/ρ)−1) ii) let the time be t u=0 , v=(√(2gh((σ/ρ)−1))),a=g((σ/ρ)−1) v=u+at ⇒t=(v/a)=(((√(g((σ/ρ)−1)))×(√(2h)))/(g((σ/ρ)−1)))=(√((((2h)/(g((σ/ρ)−1)))))) So t=(√((((2h)/(g((σ/ρ)−1)))))) iii) let the total time in air be 2t_(air) when going upwards or downwards the ball will take t_(air) time u=(√(2gh((σ/ρ)−1))) ,v=0 , a=g v=u−at ⇒t_(air) =(((√(2gh((σ/ρ)−1)))/g))=((√((2h((σ/ρ)−1))/g))) So total time =2t_(air) =2((√((2h((σ/ρ)−1))/g)))
$$\left.\mathrm{i}\right) \\ $$$${volume}\:{of}\:{the}\:{ball}=\frac{\mathrm{4}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \:{or}\:\frac{{m}}{\rho} \\ $$$${fluid}'{s}\:{volume}\:{displaced}\:{by}\:{the}\:{ball}=\frac{{m}}{\rho} \\ $$$${weight}\:{of}\:{that}\:{fluid}=\frac{{m}\sigma}{\rho}{g} \\ $$$${weight}\:{of}\:{ball}\:={mg} \\ $$$${buoyancy}\:={mg}\frac{\sigma}{\rho}−{mg}={mg}\left(\frac{\sigma}{\rho}−\mathrm{1}\right) \\ $$$${F}={ma} \\ $$$${a}=\frac{{F}}{{m}}\Rightarrow\frac{\cancel{{m}g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}{\cancel{{m}}}={g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right) \\ $$$${at}\:{h}\:{depth}\:{u}=\mathrm{0},\:{a}={g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right),\:{s}={h}\: \\ $$$${velocity}\:{at}\:{surface}\Rightarrow \\ $$$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{as} \\ $$$${v}=\sqrt{\mathrm{2}×{g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right){h}} \\ $$$${let}\:{the}\:{heigth}\:{it}\:\:{will}\:{go}\:{upwards}\:{be}\:{x} \\ $$$${v}=\mathrm{0}\:,{s}={x},\:{u}=\sqrt{\mathrm{2}{gh}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)},{a}={g} \\ $$$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}{as}\Rightarrow\:\mathrm{0}=\mathrm{2}{gh}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)−\mathrm{2}{xg} \\ $$$${or}\:\cancel{\mathrm{2}{g}x}=\cancel{\mathrm{2}{g}h}\left(\frac{\sigma}{\rho}−\mathrm{1}\right) \\ $$$${So}\:{x}={h}\left(\frac{\sigma}{\rho}−\mathrm{1}\right) \\ $$$$\left.\mathrm{ii}\right) \\ $$$$\:\:\:{let}\:{the}\:{time}\:{be}\:{t} \\ $$$$\: \\ $$$${u}=\mathrm{0}\:,\:{v}=\sqrt{\mathrm{2}{gh}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)},{a}={g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right) \\ $$$${v}={u}+{at} \\ $$$$\Rightarrow{t}=\frac{{v}}{{a}}=\frac{\sqrt{{g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}×\sqrt{\mathrm{2}{h}}}{{g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}=\sqrt{\left(\frac{\mathrm{2}{h}}{{g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}\right)} \\ $$$${So}\:{t}=\sqrt{\left(\frac{\mathrm{2}{h}}{{g}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}\right)} \\ $$$$\left.\mathrm{iii}\right) \\ $$$$\:\:\:\:\:{let}\:{the}\:{total}\:{time}\:{in}\:{air}\:{be}\:\mathrm{2}{t}_{{air}} \\ $$$$\:\:{when}\:{going}\:{upwards}\:{or}\:{downwards}\:{the}\:{ball}\:{will}\:{take}\:{t}_{{air}} \:{time} \\ $$$${u}=\sqrt{\mathrm{2}{gh}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}\:,{v}=\mathrm{0}\:,\:{a}={g} \\ $$$${v}={u}−{at} \\ $$$$\Rightarrow{t}_{{air}} =\left(\frac{\sqrt{\mathrm{2}{gh}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}}{{g}}\right)=\left(\sqrt{\frac{\mathrm{2}{h}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}{{g}}}\right) \\ $$$${So}\:{total}\:{time}\:=\mathrm{2}{t}_{{air}} =\mathrm{2}\left(\sqrt{\frac{\mathrm{2}{h}\left(\frac{\sigma}{\rho}−\mathrm{1}\right)}{{g}}}\right) \\ $$

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