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Question Number 16294 by Tinkutara last updated on 20/Jun/17

Two particles, 1 and 2, move with constant velocities v_1 ^(→) and v_2 ^(→) . At the initial moment, their position vectors are equal to r_1 ^(→) and r_2 ^(→) . How must these four vectors be interrelated for the particle to collide?

$$\mathrm{Two}\:\mathrm{particles},\:\mathrm{1}\:\mathrm{and}\:\mathrm{2},\:\mathrm{move}\:\mathrm{with} \\ $$$$\mathrm{constant}\:\mathrm{velocities}\:\overset{\rightarrow} {{v}_{\mathrm{1}} }\:\mathrm{and}\:\overset{\rightarrow} {{v}_{\mathrm{2}} }.\:\mathrm{At}\:\mathrm{the} \\ $$$$\mathrm{initial}\:\mathrm{moment},\:\mathrm{their}\:\mathrm{position}\:\mathrm{vectors} \\ $$$$\mathrm{are}\:\mathrm{equal}\:\mathrm{to}\:\overset{\rightarrow} {{r}_{\mathrm{1}} }\:\mathrm{and}\:\overset{\rightarrow} {{r}_{\mathrm{2}} }.\:\mathrm{How}\:\mathrm{must}\:\mathrm{these} \\ $$$$\mathrm{four}\:\mathrm{vectors}\:\mathrm{be}\:\mathrm{interrelated}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{to}\:\mathrm{collide}? \\ $$

Commented by prakash jain last updated on 21/Jun/17

For the particle to collide they should meet at some time t. r_1 +v_1 t=r_2 +v_2 t ⇒(r_1 −r_2 )=(v_2 −v_1 )t t is a scaler so it implies that r_1 −r_2 is parallel to v_2 −v_1 .

$$\mathrm{For}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{to}\:\mathrm{collide}\:\mathrm{they} \\ $$$$\mathrm{should}\:\mathrm{meet}\:\mathrm{at}\:\mathrm{some}\:\mathrm{time}\:{t}. \\ $$$$\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{v}}_{\mathrm{1}} {t}=\boldsymbol{\mathrm{r}}_{\mathrm{2}} +\boldsymbol{\mathrm{v}}_{\mathrm{2}} {t} \\ $$$$\Rightarrow\left(\boldsymbol{\mathrm{r}}_{\mathrm{1}} −\boldsymbol{\mathrm{r}}_{\mathrm{2}} \right)=\left(\boldsymbol{\mathrm{v}}_{\mathrm{2}} −\boldsymbol{\mathrm{v}}_{\mathrm{1}} \right){t} \\ $$$${t}\:\mathrm{is}\:\mathrm{a}\:\mathrm{scaler}\:\mathrm{so}\:\mathrm{it}\:\mathrm{implies}\:\mathrm{that} \\ $$$$\boldsymbol{\mathrm{r}}_{\mathrm{1}} −\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:\boldsymbol{\mathrm{v}}_{\mathrm{2}} −\boldsymbol{\mathrm{v}}_{\mathrm{1}} . \\ $$

Answered by ajfour last updated on 20/Jun/17

(r_2 ^→ −r_1 ^→ )×(v_1 ^→ −v_2 ^→ )=0 , i think . that is their relative velocity must not be along any other line but that of (r_2 ^→ −r_1 ^→ ).

$$\:\left(\overset{\rightarrow} {{r}}_{\mathrm{2}} −\overset{\rightarrow} {{r}}_{\mathrm{1}} \right)×\left(\overset{\rightarrow} {{v}}_{\mathrm{1}} −\overset{\rightarrow} {{v}}_{\mathrm{2}} \right)=\mathrm{0}\:,\:{i}\:{think}\:. \\ $$$${that}\:{is}\:{their}\:{relative}\:{velocity} \\ $$$${must}\:{not}\:{be}\:{along}\:{any}\:{other}\:{line} \\ $$$${but}\:{that}\:{of}\:\:\left(\overset{\rightarrow} {{r}}_{\mathrm{2}} −\overset{\rightarrow} {{r}}_{\mathrm{1}} \right). \\ $$

Commented by Tinkutara last updated on 20/Jun/17

But answer is ((v_1 ^→ − v_2 ^(→) )/(∣v_1 ^(→) − v_2 ^(→) ∣)) = ((r_2 ^→ − r_1 ^(→) )/(∣r_2 ^(→) − r_1 ^(→) ∣))

$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\frac{\overset{\rightarrow} {{v}}_{\mathrm{1}} \:−\:\overset{\rightarrow} {{v}_{\mathrm{2}} }}{\mid\overset{\rightarrow} {{v}_{\mathrm{1}} }\:−\:\overset{\rightarrow} {{v}_{\mathrm{2}} }\mid}\:=\:\frac{\overset{\rightarrow} {{r}}_{\mathrm{2}} \:−\:\overset{\rightarrow} {{r}_{\mathrm{1}} }}{\mid\overset{\rightarrow} {{r}_{\mathrm{2}} }\:−\:\overset{\rightarrow} {{r}_{\mathrm{1}} }\mid} \\ $$

Commented by Tinkutara last updated on 20/Jun/17

OK. I will recheck.

$$\mathrm{OK}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{recheck}. \\ $$

Commented by Tinkutara last updated on 22/Jun/17

Thanks Sir!

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

Commented by prakash jain last updated on 21/Jun/17

Hi ajfour, i think book′s answer is correct. Cross product could be 0 even if t<0 (see comment in question). book′s answer implies t>0.

$$\mathrm{Hi}\:\mathrm{ajfour}, \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{book}'\mathrm{s}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}. \\ $$$$\mathrm{Cross}\:\mathrm{product}\:\mathrm{could}\:\mathrm{be}\:\mathrm{0}\:\mathrm{even} \\ $$$$\mathrm{if}\:{t}<\mathrm{0}\:\left(\mathrm{see}\:\mathrm{comment}\:\mathrm{in}\:\mathrm{question}\right). \\ $$$$\mathrm{book}'\mathrm{s}\:\mathrm{answer}\:\mathrm{implies}\:{t}>\mathrm{0}. \\ $$

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