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Question Number 163178 by Tawa11 last updated on 04/Jan/22

Answered by mr W last updated on 05/Jan/22

$$massoftheballs=m$$$$forceonball1atbeginning=f$$$$distancebetweentheballsatbeginning=s=1$$$$v_{1,i,n}=velocityofball1before{n}^{th}collision$$$$v_{1,f,n}=velocityofball1after{n}^{th}collision$$$$v_{2,i,n}=velocityofball2before{n}^{th}collision$$$$v_{2,f,n}=velocityofball2after{n}^{th}collision$$$$a_n=accelerationofball1after{n}^{th}collision$$$$s_n=distancebetween{n}^{th}and{(n+1)}^{th}collision$$$$$$$$asweknow,afteranelasticcollision$$$$fromtwoballswiththesamemass$$$$theyjustexchangetheirvelocities,i.e.$$$$v_{1,f,n}=v_{2,i,n}$$$$v_{2,f,n}=v_{1,i,n}$$$$$$$$\underset{―}{beforethe{1}^{st}collision:}$$$$v_{2,i,1}=0$$$$a_0=\frac{f}{m}$$$$v_{1,i,1}=\sqrt{2{sa}_0}=\sqrt{\frac{2sf}{m}}=usay$$$$\Rightarrow \frac{f}{m}=\frac{u^2}{2s}$$$$$$$$\underset{―}{afterthe{1}^{st}collision:}$$$$v_{2,f,1}=v_{1,i,1}=u$$$$v_{1,f,1}=v_{2,i,1}=0$$$$afterthecollisiontheball1getsa$$$$forcewhichis2timesoftheforce$$$$beforethecollision:$$$$f_1=2f_0=2f$$$$a_1=\frac{f_1}{m}=\frac{2f}{m}=\frac{u^2}{s}$$$$sayafteradistand{s}_{1}bothballs$$$$collidewitheachotheragainafter$$$$atimet.$$$$s_1=v_{2,f,1}t=ut$$$$s_1=v_{1,f,1}t+\frac{1}{2}{a}_1{t}^2$$$$v_{2,f,1}t=v_{1,f,1}t+\frac{1}{2}{a}_1{t}^2$$$$t=\frac{2(v_{2,f,1}-v_{1,f,1})}{a_1}=\frac{2s(u-0)}{u^2}=\frac{2s}{u}$$$$s_1=\frac{2s(v_{2,f,1}-v_{1,f,1})}{u}=\frac{2s(u-0)}{u}=2s$$$$v_{2,i,2}=v_{2,f,1}=u$$$$v_{1,i,2}=v_{1,f,1}+a_{1}t=0+\frac{u^2}{s}\times \frac{2s}{u}=2u$$$$$$$$\underset{―}{afterthe{n}^{th}collision:}$$$$v_{1,f,n}=v_{2,i,n}$$$$v_{2,f,n}=v_{1,i,n}$$$$f_n=2^{n}f$$$$a_n=\frac{f_n}{m}=\frac{2^{n}f}{m}=2^n\times \frac{u^2}{2s}=\frac{2^{n-1}{u}^2}{s}$$$$sayafteradistance{s}_{n}thebothballs$$$$collidewitheachotheragainafter$$$$timet.$$$$s_n=v_{2,f,n}t$$$$s_n=v_{1,f,n}t+\frac{1}{2}{a}_n{t}^2$$$$v_{2,f,n}t=v_{1,f,n}t+\frac{1}{2}{a}_n{t}^2$$$$t=\frac{s(v_{2,f,n}-v_{1,f,n})}{2^{n-2}{u}^2}$$$$s_n=\frac{s(v_{2,f,n}-v_{1,f,n})v_{2,f,n}}{2^{n-2}{u}^2}$$$$v_{2,i,n+1}=v_{2,f,n}$$$$v_{1,i,n+1}=v_{1,f,n}+a_{n}t=v_{1,f,n}+\frac{2^{n-1}{u}^2}{s}\times \frac{s(v_{2,f,n}-v_{1,f,n})}{2^{n-2}{u}^2}$$$$v_{1,i,n+1}=2v_{2,f,n}-v_{1,f,n}$$$$v_{1,i,n+1}=2v_{1,i,n}-v_{1,f,n}$$$$v_{1,i,n+1}=2v_{1,i,n}-v_{2,i,n}$$$$v_{1,i,n+1}=2v_{1,i,n}-v_{2,f,n-1}$$$$v_{1,i,n+1}=2v_{1,i,n}-v_{1,i,n-1}$$$$v_{1,i,n+1}-v_{1,i,n}=v_{1,i,n}-v_{1,i,n-1}$$$$=v_{1,i,2}-v_{1,i,1}=2u-u=u$$$$\Rightarrow v_{1,i,n}=nu$$$$\Rightarrow v_{2,i,n}=v_{2,f,n-1}=v_{1,i,n-1}=(n-1)u$$$$s_n=\frac{s(v_{2,f,n}-v_{1,f,n})v_{2,f,n}}{2^{n-2}{u}^2}$$$$s_n=\frac{s(v_{1,i,n}-v_{2,i,n})v_{1,i,n}}{2^{n-2}{u}^2}$$$$\Rightarrow s_n=\frac{{snu}^2}{2^{n-2}{u}^2}$$$$\Rightarrow s_n=\frac{ns}{2^{n-2}}$$$$$$$$totaldistanceball1hastraveledtill$$$$the{(n+1)}^{th}collision:$$$$S_n=s_0+\underset{k=1}{\sum ^n}{s}_k=s+\underset{k=1}{\sum ^n}\frac{ks}{2^{k-2}}$$$$withs=1m$$$$S_n=1+4\underset{k=1}{\sum ^n}\frac{k}{2^k}$$$$S_n=1+4(2-\frac{n+2}{2^n})$$$$\underset{n\to \mathrm{\infty }}{\lim }{S}_n=1+4\times 2=9m$$$$thatmeansafteralargenumberof$$$$collisionstheball1hastraveleda$$$$totaldistanceof9m.$$$$thesumofdistancesbetweenthe$$$$collisionsis8m.$$$$◼$$

Commented by mr W last updated on 05/Jan/22

$$pleasecomparewithanswerinbook.$$

Commented by Tawa11 last updated on 06/Jan/22

$$\mathrm{Correct}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{more}.\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

Commented by mr W last updated on 06/Jan/22

$$doyouknowtheanswer?$$

Commented by Tawa11 last updated on 06/Jan/22

$$\mathrm{Yes}\mathrm{sir}.\mathrm{The}\mathrm{answer},\mathrm{they}\mathrm{wrote}8\mathrm{d}\mathrm{at}\mathrm{the}\mathrm{back}\mathrm{of}\mathrm{the}\mathrm{book}\mathrm{sir}.$$

Commented by Tawa11 last updated on 06/Jan/22

$$\mathrm{So},\mathrm{I}\mathrm{think}\mathrm{they}\mathrm{mean}8\mathrm{m}.$$

Commented by mr W last updated on 06/Jan/22

$$thequestionsaysthedistancebetween$$$$theballsatt=0is1m.ifthisdistance$$$$isd\left(orsinmyworkings\right),thenthe$$$$sumofthedistancesbetweenthe$$$$collisionsis8d\left(or8sinmyworkings\right).$$

Commented by Tawa11 last updated on 06/Jan/22

$$\mathrm{Yes}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}.$$

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