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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

mass of the balls=m force on ball 1 at beginning=f distance between the balls at beginning=s=1 v_(1,i,n) =velocity of ball 1 before n^(th) collision v_(1,f,n) =velocity of ball 1 after n^(th) collision v_(2,i,n) =velocity of ball 2 before n^(th) collision v_(2,f,n) =velocity of ball 2 after n^(th) collision a_n =acceleration of ball 1 after n^(th) collision s_n =distance between n^(th) and (n+1)^(th) collision as we know, after an elastic collision from two balls with the same mass they just exchange their velocities, i.e. v_(1,f,n) =v_(2,i,n) v_(2,f,n) =v_(1,i,n) before the 1^(st) collision: v_(2,i,1) =0 a_0 =(f/m) v_(1,i,1) =(√(2sa_0 ))=(√((2sf)/m))=u say ⇒(f/m)=(u^2 /(2s)) after the 1^(st) collision: v_(2,f,1) =v_(1,i,1) =u v_(1,f,1) =v_(2,i,1) =0 after the collision the ball 1 gets a force which is 2 times of the force before the collision: f_1 =2f_0 =2f a_1 =(f_1 /m)=((2f)/m)=(u^2 /s) say after a distand s_1 both balls collide with each other again after a time t. s_1 =v_(2,f,1) t=ut s_1 =v_(1,f,1) t+(1/2)a_1 t^2 v_(2,f,1) t=v_(1,f,1) t+(1/2)a_1 t^2 t=((2(v_(2,f,1) −v_(1,f,1) ))/a_1 )=((2s(u−0))/u^2 )=((2s)/u) s_1 =((2s(v_(2,f,1) −v_(1,f,1) ))/u)=((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+(u^2 /s)×((2s)/u)=2u after the 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 =(f_n /m)=((2^n f)/m)=2^n ×(u^2 /(2s))=((2^(n−1) u^2 )/s) say after a distance s_n the both balls collide with each other again after time t. s_n =v_(2,f,n) t s_n =v_(1,f,n) t+(1/2)a_n t^2 v_(2,f,n) t=v_(1,f,n) t+(1/2)a_n t^2 t=((s(v_(2,f,n) −v_(1,f,n) ))/(2^(n−2) u^2 )) s_n =((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) +((2^(n−1) u^2 )/s)×((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 ⇒v_(1,i,n) =nu ⇒v_(2,i,n) =v_(2,f,n−1) =v_(1,i,n−1) =(n−1)u s_n =((s(v_(2,f,n) −v_(1,f,n) )v_(2,f,n) )/(2^(n−2) u^2 )) s_n =((s(v_(1,i,n) −v_(2,i,n) )v_(1,i,n) )/(2^(n−2) u^2 )) ⇒s_n =((snu^2 )/(2^(n−2) u^2 )) ⇒s_n =((ns)/2^(n−2) ) total distance ball 1 has traveled till the (n+1)^(th) collision: S_n =s_0 +Σ_(k=1) ^n s_k =s+Σ_(k=1) ^n ((ks)/2^(k−2) ) with s=1 m S_n =1+4Σ_(k=1) ^n (k/2^k ) S_n =1+4(2−((n+2)/2^n )) lim_(n→∞) S_n =1+4×2=9 m that means after a large number of collisions the ball 1 has traveled a total distance of 9 m. the sum of distances between the collisions is 8 m. ■

$${mass}\:{of}\:{the}\:{balls}={m} \\ $$$${force}\:{on}\:{ball}\:\mathrm{1}\:{at}\:{beginning}={f} \\ $$$${distance}\:{between}\:{the}\:{balls}\:{at}\:{beginning}={s}=\mathrm{1} \\ $$$${v}_{\mathrm{1},{i},{n}} ={velocity}\:{of}\:{ball}\:\mathrm{1}\:{before}\:{n}^{{th}} \:{collision} \\ $$$${v}_{\mathrm{1},{f},{n}} ={velocity}\:{of}\:{ball}\:\mathrm{1}\:{after}\:{n}^{{th}} \:{collision} \\ $$$${v}_{\mathrm{2},{i},{n}} ={velocity}\:{of}\:{ball}\:\mathrm{2}\:{before}\:{n}^{{th}} \:{collision} \\ $$$${v}_{\mathrm{2},{f},{n}} ={velocity}\:{of}\:{ball}\:\mathrm{2}\:{after}\:{n}^{{th}} \:{collision} \\ $$$${a}_{{n}} ={acceleration}\:{of}\:{ball}\:\mathrm{1}\:{after}\:{n}^{{th}} \:{collision} \\ $$$${s}_{{n}} ={distance}\:{between}\:{n}^{{th}} \:{and}\:\left({n}+\mathrm{1}\right)^{{th}} \:{collision} \\ $$$$ \\ $$$${as}\:{we}\:{know},\:{after}\:{an}\:{elastic}\:{collision}\: \\ $$$${from}\:{two}\:{balls}\:{with}\:{the}\:{same}\:{mass}\: \\ $$$${they}\:{just}\:{exchange}\:{their}\:{velocities},\:{i}.{e}. \\ $$$${v}_{\mathrm{1},{f},{n}} ={v}_{\mathrm{2},{i},{n}} \\ $$$${v}_{\mathrm{2},{f},{n}} ={v}_{\mathrm{1},{i},{n}} \\ $$$$ \\ $$$$\underline{{before}\:{the}\:\mathrm{1}^{{st}} \:{collision}:} \\ $$$${v}_{\mathrm{2},{i},\mathrm{1}} =\mathrm{0} \\ $$$${a}_{\mathrm{0}} =\frac{{f}}{{m}} \\ $$$${v}_{\mathrm{1},{i},\mathrm{1}} =\sqrt{\mathrm{2}{sa}_{\mathrm{0}} }=\sqrt{\frac{\mathrm{2}{sf}}{{m}}}={u}\:{say} \\ $$$$\:\Rightarrow\frac{{f}}{{m}}=\frac{{u}^{\mathrm{2}} }{\mathrm{2}{s}} \\ $$$$ \\ $$$$\underline{{after}\:{the}\:\mathrm{1}^{{st}} \:{collision}:} \\ $$$${v}_{\mathrm{2},{f},\mathrm{1}} ={v}_{\mathrm{1},{i},\mathrm{1}} ={u} \\ $$$${v}_{\mathrm{1},{f},\mathrm{1}} ={v}_{\mathrm{2},{i},\mathrm{1}} =\mathrm{0} \\ $$$${after}\:{the}\:{collision}\:{the}\:{ball}\:\mathrm{1}\:{gets}\:{a} \\ $$$${force}\:{which}\:{is}\:\mathrm{2}\:{times}\:{of}\:{the}\:{force} \\ $$$${before}\:{the}\:{collision}: \\ $$$${f}_{\mathrm{1}} =\mathrm{2}{f}_{\mathrm{0}} =\mathrm{2}{f} \\ $$$${a}_{\mathrm{1}} =\frac{{f}_{\mathrm{1}} }{{m}}=\frac{\mathrm{2}{f}}{{m}}=\frac{{u}^{\mathrm{2}} }{{s}} \\ $$$${say}\:{after}\:{a}\:{distand}\:{s}_{\mathrm{1}} \:{both}\:{balls} \\ $$$${collide}\:{with}\:{each}\:{other}\:{again}\:{after} \\ $$$${a}\:{time}\:{t}. \\ $$$${s}_{\mathrm{1}} ={v}_{\mathrm{2},{f},\mathrm{1}} {t}={ut} \\ $$$${s}_{\mathrm{1}} ={v}_{\mathrm{1},{f},\mathrm{1}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{\mathrm{1}} {t}^{\mathrm{2}} \\ $$$${v}_{\mathrm{2},{f},\mathrm{1}} {t}={v}_{\mathrm{1},{f},\mathrm{1}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{\mathrm{1}} {t}^{\mathrm{2}} \\ $$$${t}=\frac{\mathrm{2}\left({v}_{\mathrm{2},{f},\mathrm{1}} −{v}_{\mathrm{1},{f},\mathrm{1}} \right)}{{a}_{\mathrm{1}} }=\frac{\mathrm{2}{s}\left({u}−\mathrm{0}\right)}{{u}^{\mathrm{2}} }=\frac{\mathrm{2}{s}}{{u}} \\ $$$${s}_{\mathrm{1}} =\frac{\mathrm{2}{s}\left({v}_{\mathrm{2},{f},\mathrm{1}} −{v}_{\mathrm{1},{f},\mathrm{1}} \right)}{{u}}=\frac{\mathrm{2}{s}\left({u}−\mathrm{0}\right)}{{u}}=\mathrm{2}{s} \\ $$$${v}_{\mathrm{2},{i},\mathrm{2}} ={v}_{\mathrm{2},{f},\mathrm{1}} ={u} \\ $$$${v}_{\mathrm{1},{i},\mathrm{2}} ={v}_{\mathrm{1},{f},\mathrm{1}} +{a}_{\mathrm{1}} {t}=\mathrm{0}+\frac{{u}^{\mathrm{2}} }{{s}}×\frac{\mathrm{2}{s}}{{u}}=\mathrm{2}{u} \\ $$$$ \\ $$$$\underline{{after}\:{the}\:{n}^{{th}} \:{collision}:} \\ $$$${v}_{\mathrm{1},{f},{n}} ={v}_{\mathrm{2},{i},{n}} \\ $$$${v}_{\mathrm{2},{f},{n}} ={v}_{\mathrm{1},{i},{n}} \\ $$$${f}_{{n}} =\mathrm{2}^{{n}} {f} \\ $$$${a}_{{n}} =\frac{{f}_{{n}} }{{m}}=\frac{\mathrm{2}^{{n}} {f}}{{m}}=\mathrm{2}^{{n}} ×\frac{{u}^{\mathrm{2}} }{\mathrm{2}{s}}=\frac{\mathrm{2}^{{n}−\mathrm{1}} {u}^{\mathrm{2}} }{{s}} \\ $$$${say}\:{after}\:{a}\:{distance}\:{s}_{{n}} \:{the}\:{both}\:{balls} \\ $$$${collide}\:{with}\:{each}\:{other}\:{again}\:{after} \\ $$$${time}\:{t}. \\ $$$${s}_{{n}} ={v}_{\mathrm{2},{f},{n}} {t} \\ $$$${s}_{{n}} ={v}_{\mathrm{1},{f},{n}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}} {t}^{\mathrm{2}} \\ $$$${v}_{\mathrm{2},{f},{n}} {t}={v}_{\mathrm{1},{f},{n}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}} {t}^{\mathrm{2}} \\ $$$${t}=\frac{{s}\left({v}_{\mathrm{2},{f},{n}} −{v}_{\mathrm{1},{f},{n}} \right)}{\mathrm{2}^{{n}−\mathrm{2}} {u}^{\mathrm{2}} } \\ $$$${s}_{{n}} =\frac{{s}\left({v}_{\mathrm{2},{f},{n}} −{v}_{\mathrm{1},{f},{n}} \right){v}_{\mathrm{2},{f},{n}} }{\mathrm{2}^{{n}−\mathrm{2}} {u}^{\mathrm{2}} } \\ $$$${v}_{\mathrm{2},{i},{n}+\mathrm{1}} ={v}_{\mathrm{2},{f},{n}} \\ $$$${v}_{\mathrm{1},{i},{n}+\mathrm{1}} ={v}_{\mathrm{1},{f},{n}} +{a}_{{n}} {t}={v}_{\mathrm{1},{f},{n}} +\frac{\mathrm{2}^{{n}−\mathrm{1}} {u}^{\mathrm{2}} }{{s}}×\frac{{s}\left({v}_{\mathrm{2},{f},{n}} −{v}_{\mathrm{1},{f},{n}} \right)}{\mathrm{2}^{{n}−\mathrm{2}} {u}^{\mathrm{2}} } \\ $$$${v}_{\mathrm{1},{i},{n}+\mathrm{1}} =\mathrm{2}{v}_{\mathrm{2},{f},{n}} −{v}_{\mathrm{1},{f},{n}} \\ $$$${v}_{\mathrm{1},{i},{n}+\mathrm{1}} =\mathrm{2}{v}_{\mathrm{1},{i},{n}} −{v}_{\mathrm{1},{f},{n}} \\ $$$${v}_{\mathrm{1},{i},{n}+\mathrm{1}} =\mathrm{2}{v}_{\mathrm{1},{i},{n}} −{v}_{\mathrm{2},{i},{n}} \\ $$$${v}_{\mathrm{1},{i},{n}+\mathrm{1}} =\mathrm{2}{v}_{\mathrm{1},{i},{n}} −{v}_{\mathrm{2},{f},{n}−\mathrm{1}} \\ $$$${v}_{\mathrm{1},{i},{n}+\mathrm{1}} =\mathrm{2}{v}_{\mathrm{1},{i},{n}} −{v}_{\mathrm{1},{i},{n}−\mathrm{1}} \\ $$$${v}_{\mathrm{1},{i},{n}+\mathrm{1}} −{v}_{\mathrm{1},{i},{n}} ={v}_{\mathrm{1},{i},{n}} −{v}_{\mathrm{1},{i},{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={v}_{\mathrm{1},{i},\mathrm{2}} −{v}_{\mathrm{1},{i},\mathrm{1}} =\mathrm{2}{u}−{u}={u} \\ $$$$\Rightarrow{v}_{\mathrm{1},{i},{n}} ={nu} \\ $$$$\Rightarrow{v}_{\mathrm{2},{i},{n}} ={v}_{\mathrm{2},{f},{n}−\mathrm{1}} ={v}_{\mathrm{1},{i},{n}−\mathrm{1}} =\left({n}−\mathrm{1}\right){u} \\ $$$${s}_{{n}} =\frac{{s}\left({v}_{\mathrm{2},{f},{n}} −{v}_{\mathrm{1},{f},{n}} \right){v}_{\mathrm{2},{f},{n}} }{\mathrm{2}^{{n}−\mathrm{2}} {u}^{\mathrm{2}} } \\ $$$${s}_{{n}} =\frac{{s}\left({v}_{\mathrm{1},{i},{n}} −{v}_{\mathrm{2},{i},{n}} \right){v}_{\mathrm{1},{i},{n}} }{\mathrm{2}^{{n}−\mathrm{2}} {u}^{\mathrm{2}} } \\ $$$$\Rightarrow{s}_{{n}} =\frac{{snu}^{\mathrm{2}} }{\mathrm{2}^{{n}−\mathrm{2}} {u}^{\mathrm{2}} } \\ $$$$\Rightarrow{s}_{{n}} =\frac{{ns}}{\mathrm{2}^{{n}−\mathrm{2}} } \\ $$$$ \\ $$$${total}\:{distance}\:{ball}\:\mathrm{1}\:{has}\:{traveled}\:{till} \\ $$$${the}\:\left({n}+\mathrm{1}\right)^{{th}} \:{collision}: \\ $$$${S}_{{n}} ={s}_{\mathrm{0}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{s}_{{k}} ={s}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{ks}}{\mathrm{2}^{{k}−\mathrm{2}} } \\ $$$${with}\:{s}=\mathrm{1}\:{m} \\ $$$${S}_{{n}} =\mathrm{1}+\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}}{\mathrm{2}^{{k}} } \\ $$$${S}_{{n}} =\mathrm{1}+\mathrm{4}\left(\mathrm{2}−\frac{{n}+\mathrm{2}}{\mathrm{2}^{{n}} }\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =\mathrm{1}+\mathrm{4}×\mathrm{2}=\mathrm{9}\:{m}\: \\ $$$${that}\:{means}\:{after}\:{a}\:{large}\:{number}\:{of} \\ $$$${collisions}\:{the}\:{ball}\:\mathrm{1}\:{has}\:{traveled}\:{a}\: \\ $$$${total}\:{distance}\:{of}\:\mathrm{9}\:{m}. \\ $$$${the}\:{sum}\:{of}\:{distances}\:{between}\:{the} \\ $$$${collisions}\:{is}\:\mathrm{8}\:{m}. \\ $$$$\blacksquare \\ $$

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

please compare with answer in book.

$${please}\:{compare}\:{with}\:{answer}\:{in}\:{book}. \\ $$

Commented by Tawa11 last updated on 06/Jan/22

Correct sir. God bless you more. I really appreciate.

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

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

do you know the answer?

$${do}\:{you}\:{know}\:{the}\:{answer}? \\ $$

Commented by Tawa11 last updated on 06/Jan/22

Yes sir. The answer, they wrote 8d at the back of the book sir.

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

Commented by Tawa11 last updated on 06/Jan/22

So, I think they mean 8m.

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

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

the question says the distance between the balls at t=0 is 1m. if this distance is d (or s in my workings), then the sum of the distances between the collisions is 8d (or 8s in my workings).

$${the}\:{question}\:{says}\:{the}\:{distance}\:{between} \\ $$$${the}\:{balls}\:{at}\:{t}=\mathrm{0}\:{is}\:\mathrm{1}{m}.\:{if}\:{this}\:{distance} \\ $$$${is}\:{d}\:\left({or}\:{s}\:{in}\:{my}\:{workings}\right),\:{then}\:{the} \\ $$$${sum}\:{of}\:{the}\:{distances}\:{between}\:{the} \\ $$$${collisions}\:{is}\:\mathrm{8}{d}\:\left({or}\:\mathrm{8}{s}\:{in}\:{my}\:{workings}\right). \\ $$

Commented by Tawa11 last updated on 06/Jan/22

Yes sir. Thanks for your time.

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

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