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Question Number 198832 by ajfour last updated on 24/Oct/23

Commented by ajfour last updated on 24/Oct/23

The radii are a(green), b(brown), c(blue). Find v and V when green ball comes sliding at speed u its centre is at (a,a,z) z>a. This pushes other two balls along ground and their respective closer vertical wall. (Assume the motion like wise, meaning balls dont detach and move grazing along two ⊥ walls.)

$${The}\:{radii}\:{are}\:{a}\left({green}\right),\:{b}\left({brown}\right), \\ $$$$\:{c}\left({blue}\right). \\ $$$${Find}\:{v}\:{and}\:{V}\:\:{when}\:{green}\:{ball}\:{comes} \\ $$$${sliding}\:{at}\:{speed}\:{u}\:{its}\:{centre}\:{is}\:{at} \\ $$$$\left({a},{a},{z}\right)\:\:\:\:{z}>{a}.\:{This}\:{pushes}\:{other} \\ $$$${two}\:{balls}\:{along}\:{ground}\:{and}\:{their} \\ $$$${respective}\:{closer}\:{vertical}\:{wall}. \\ $$$$\left({Assume}\:{the}\:{motion}\:{like}\:{wise},\right. \\ $$$${meaning}\:{balls}\:{dont}\:{detach}\:{and} \\ $$$$\left.{move}\:{grazing}\:{along}\:{two}\:\bot\:{walls}.\right) \\ $$

Commented by a.lgnaoui last updated on 25/Oct/23

Answered by mr W last updated on 26/Oct/23

Commented by mr W last updated on 26/Oct/23

A(a,a,p) (p=z as given) B(b,q,b) (b−a)^2 +(q−a)^2 +(b−p)^2 =(a+b)^2 ⇒q=(√(4ab+2bp−b^2 −p^2 ))−a AB^(→) =(b−a, q−a, b−p) u_1 ^→ =(0,0,−1) V_1 ^→ =(0,1,0) cos θ=((p−b)/( a+b)) cos ϕ=((q−a)/(a+b)) u cos θ=V cos ϕ ⇒V=((u(p−b))/(a+b))×((a+b)/(q−a))=((u(p−b))/(q−a)) =((u(p−b))/( (√(4ab+2bp−b^2 −p^2 ))−2a)) similarly v=((u(p−c))/( (√(4ac+2cp−c^2 −p^2 ))−2a))

$${A}\left({a},{a},{p}\right)\:\:\:\:\left({p}={z}\:{as}\:{given}\right) \\ $$$${B}\left({b},{q},{b}\right) \\ $$$$\left({b}−{a}\right)^{\mathrm{2}} +\left({q}−{a}\right)^{\mathrm{2}} +\left({b}−{p}\right)^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{q}=\sqrt{\mathrm{4}{ab}+\mathrm{2}{bp}−{b}^{\mathrm{2}} −{p}^{\mathrm{2}} }−{a} \\ $$$$\overset{\rightarrow} {{AB}}=\left({b}−{a},\:{q}−{a},\:{b}−{p}\right) \\ $$$$\overset{\rightarrow} {{u}}_{\mathrm{1}} =\left(\mathrm{0},\mathrm{0},−\mathrm{1}\right) \\ $$$$\overset{\rightarrow} {{V}}_{\mathrm{1}} =\left(\mathrm{0},\mathrm{1},\mathrm{0}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{{p}−{b}}{\:{a}+{b}} \\ $$$$\mathrm{cos}\:\varphi=\frac{{q}−{a}}{{a}+{b}} \\ $$$${u}\:\mathrm{cos}\:\theta={V}\:\mathrm{cos}\:\varphi \\ $$$$\Rightarrow{V}=\frac{{u}\left({p}−{b}\right)}{{a}+{b}}×\frac{{a}+{b}}{{q}−{a}}=\frac{{u}\left({p}−{b}\right)}{{q}−{a}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{{u}\left({p}−{b}\right)}{\:\sqrt{\mathrm{4}{ab}+\mathrm{2}{bp}−{b}^{\mathrm{2}} −{p}^{\mathrm{2}} }−\mathrm{2}{a}} \\ $$$${similarly} \\ $$$${v}=\frac{{u}\left({p}−{c}\right)}{\:\sqrt{\mathrm{4}{ac}+\mathrm{2}{cp}−{c}^{\mathrm{2}} −{p}^{\mathrm{2}} }−\mathrm{2}{a}} \\ $$

Commented by mr W last updated on 26/Oct/23

is this what you have meant sir?

$${is}\:{this}\:{what}\:{you}\:{have}\:{meant}\:{sir}? \\ $$

Commented by ajfour last updated on 26/Oct/23

Thank you sir, all nice & precise!

$${Thank}\:{you}\:{sir},\:\:{all}\:{nice}\:\&\:{precise}! \\ $$

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