Question

51.
1
Two point masses 1 and 2 move with uniform velocities
v, and v2, respectively. Their initial position vectors are
Fi and 7z , respectively. Which of the following should be
satisfied for the collision of the point masses?​

Answer 1

Correct Question:-

Two point masses 1 and 2 move with uniform velocities  $\bf{v_1}$ and $\bf{v_2}$ respectively. Their initial position vectors are  $\bf{r_1}$ and $\bf{r_2}$ respectively. What is the condition for the two point masses to collide with each other?

Solution:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\vector(2,3){20}}\put(20,30){\vector(3,2){30}}\put(0,0){\vector(4,1){40}}\put(40,10){\vector(1,4){10}}\put(0,0){\vector(1,1){50}}\footnotesize\put(5,18){ \bf{r_1} }\put(22,1){ \bf{r_2} }\put(27,43){ \mathbf{v_1}\mathsf{\Delta t} }\put(47,27){ \mathbf{v_2}\mathsf{\Delta t} }\put(27,22){ \bf{r_3} }\end{picture}$

Let the point masses collide each other after $\sf{\Delta t}$ seconds of the beginning of their journey (assume they start to move simultaneously).

So the displacements of the two point masses are $\bf{v_1}\sf{\Delta t}$ and $\bf{v_2}\sf{\Delta t}$ respectively.

As in the fig., the particles attain same position vector $\bf{r_3}$ during collision, which is given by,

$\longrightarrow\bf{r_3=r_1+v_1}\sf{\Delta t\quad\quad\dots(1)}$

Also,

$\longrightarrow\bf{r_3=r_2+v_2}\sf{\Delta t\quad\quad\dots(2)}$

Thus, from (1) and (2),

$\longrightarrow\bf{r_1+v_1}\sf{\Delta t}=\bf{r_2+v_2}\sf{\Delta t}$

$\longrightarrow\bf{v_1}\sf{\Delta t}-\bf{v_2}\sf{\Delta t}=\bf{r_2-r_1}$

$\longrightarrow\bf{(v_1-v_2)}\sf{\Delta t}=\bf{r_2-r_1}$

$\longrightarrow\sf{\Delta t}=\bf{\dfrac{r_2-r_1}{v_1-v_2}}\sf{\quad\quad\dots(3)}$

Since $\sf{\Delta t}$ is scalar, we can take magnitude only in RHS of (3) as,

$\longrightarrow\sf{\Delta t}=\bf{\dfrac{|r_2-r_1|}{|v_1-v_2|}}\sf{\quad\quad\dots(4)}$

From (3) and (4),

$\longrightarrow\bf{\dfrac{r_2-r_1}{v_1-v_2}=\dfrac{|r_2-r_1|}{|v_1-v_2|}}$

By rule of alternendo,

$\longrightarrow\bf{\dfrac{r_2-r_1}{|r_2-r_1|}=\dfrac{v_1-v_2}{|v_1-v_2|}}$

$\longrightarrow\underline{\underline{\bf{\dfrac{r_2-r_1}{|r_2-r_1|}=-\dfrac{v_2-v_1}{|v_2-v_1|}}}}$

This means, the unit vector along the relative position vector of the two point masses should be equal to the negative unit vector along the relative velocity vector.