Question

Two particles 1 and 2 move with constant velocities $v_1$ and $v_2$. At the initial moment their radius vectors are equal to $r_1$ and $r_2$. How these four vectors must be interrelated for the particles to collide?

Answer 1

In this question the initial position and the velocity of the two particles are given and we know for particle to collide their final position must be same so we will find the final position of both the particles and equate both the equation to find the relation between them.

Complete step by step answer: Given,
The velocity of the particle A$= {v_1}$
The velocity of the particle B$= {v_2}$
The radius vector of the particle A$= {r_1}$
The radius vector of the particle B$= {r_2}$
Now let us consider particle A whose velocity$= {\vec v_1}$
The initial position of the particle ${\left( {{{\vec r}_1}} \right)_i} = {\vec r_1}$
Now since the particle is moving with velocity ${\vec v_1}$ from the initial position ${\vec r_1}$, so its final position after time t will become
${\left( {{{\vec r}_1}} \right)_f} = {\vec r_1} + {\vec v_1}t - - (i)$
Now consider particle B whose velocity$= {\vec v_2}$
The initial position of the particle ${\left( {{{\vec r}_2}} \right)_i} = {\vec r_2}$
Now since the particle is moving with velocity ${\vec v_2}$ from the initial position ${\vec r_2}$, so its final position after time t will become
${\left( {{{\vec r}_2}} \right)_f} = {\vec r_2} + {\vec v_2}t - - (ii)$
Now we know that for the collision of the two particles the position of the two particles must be equal, hence we say the final position of particle A is equal to the final position of particle A, we can write
${\left( {{{\vec r}_1}} \right)_f} = {\left( {{{\vec r}_2}} \right)_f}$
Now substitute the values from equation (i) and (ii)

$${{\vec r}_1} + {{\vec v}_1}t = {{\vec r}_2} + {{\vec v}_2}t \\\ \Rightarrow {{\vec r}_1} - {{\vec r}_2} = {{\vec v}_2}t - {{\vec v}_1}t \\\ \Rightarrow {{\vec r}_1} - {{\vec r}_2} = t\left( {{{\vec v}_2} - {{\vec v}_1}} \right) - - (iii) \\\$$

We can further write equation (iii) as
$t = \dfrac{{{{\vec r}_1} - {{\vec r}_2}}}{{\left( {{{\vec v}_2} - {{\vec v}_1}} \right)}} - - (iv)$
From equation (iv) we can see $t$ is a scalar quantity hence we can write equation (iv) as
$t = \dfrac{{\left| {{{\vec r}_1} - {{\vec r}_2}} \right|}}{{\left| {{{\vec v}_2} - {{\vec v}_1}} \right|}} - - (v)$
Now substitute the value of $t$ from equation (v) in the equation (iii), we get
${\vec r_1} - {\vec r_2} = \left( {{{\vec v}_2} - {{\vec v}_1}} \right)\left( {\dfrac{{\left| {{{\vec r}_1} - {{\vec r}_2}} \right|}}{{\left| {{{\vec v}_2} - {{\vec v}_1}} \right|}}} \right)$
By further solving this equation we get
$\Rightarrow {\dfrac{{{{\vec r}_1} - \vec r}}{{\left| {{{\vec r}_1} - {{\vec r}_2}} \right|}}_2} = \dfrac{{\left( {{{\vec v}_2} - {{\vec v}_1}} \right)}}{{\left| {{{\vec v}_2} - {{\vec v}_1}} \right|}}$

So the four vectors must be related as-
${\dfrac{{{{\vec r}_1} - \vec r}}{{\left| {{{\vec r}_1} - {{\vec r}_2}} \right|}}_2} = \dfrac{{\left( {{{\vec v}_2} - {{\vec v}_1}} \right)}}{{\left| {{{\vec v}_2} - {{\vec v}_1}} \right|}}$

Note: Students must be careful while substituting the values of the radius vector and velocity vectors as when they collide the energy will be shared between them according to the velocity and size.