Question

Two particles $1$ and $2$ are moving with constant velocities ${\overrightarrow{v}}_1=4\hat{i}-3\hat{j}\text{m/s}$ and ${\overrightarrow{v}}_2=b\hat{i}-\hat{j}\text{m/s}$ respectively. The position vectors of the particles at time $t=0$ are ${\overrightarrow{r}}_1=5\hat{i}+2\hat{j}\text{m}$ and ${\overrightarrow{r}}_2=-4\hat{i}-4\hat{j}\text{m}$. If they collide, which of the following is correct. ($b$ is a constant)

• A. They collide at $t=3\text{sec}$
• B. Their time of collision cannot be found
• C. $b=7$
• D. $b=-1$

Answer 1

Answer: (A), (C)

$b=7$

At $t=0$
${\overrightarrow{v}}_1=4\hat{i}-3\hat{j}\text{m/s},{\overrightarrow{r}}_1=5\hat{i}+2\hat{j}\text{m},{\overrightarrow{v}}_2=b\hat{i}-\hat{j}\text{m/s}$
${\overrightarrow{r}}_2=-4\hat{i}-4\hat{j}\text{m}$

If they collide, then ${\overrightarrow{v}}_1-{\overrightarrow{v}}_2$ must be along ${\overrightarrow{r}}_1-{\overrightarrow{r}}_2\dots \dots \left(1\right)$
${\overrightarrow{r}}_1-{\overrightarrow{r}}_2=\left(9\right)\hat{i}+\left(6\right)\hat{j}\dots \dots \left(2\right)$

By equation (1) and (2)
$\frac{4-b}{9}=\frac{-2}{6}\Rightarrow 4-b=-3\Rightarrow b=7$
After time ${}^{\prime }{t}^{\prime }$
${\overrightarrow{r}}_1=(5+4t)\hat{i}+(2-3t)\hat{j}$
${\overrightarrow{r}}_2=(-4+7t)\hat{i}+(-4-t)\hat{j}$

Also the final coordinates are to be same for the collision to happen
$5+4t=-4+7t\Rightarrow 3t=9$
$2-3t=-4-t\Rightarrow 2t=6$
$t=3\text{sec}$