Question

Two particles having position vectors $\overrightarrow{r_1}=(3\hat{i}+5\hat{j})\text{m}$ and $\overrightarrow{r_2}=(-5\hat{i}+3\hat{j})\text{m}$ are moving with velocities $\overrightarrow{v_1}=(4\hat{i}-4\hat{j}){\text{ms}}^{-1}$ and $\overrightarrow{v_2}=(a\hat{i}-3\hat{j}){\text{ms}}^{-1}$. If they collide after $2\text{s}$, the value of the constant '$a$' is

• A. $8$
• B. $6$
• C. $4$
• D. $2$

Answer 1

The correct option is A $8$

The position vector $\overrightarrow{r^{\prime }}$ of a particle initially at position $\overrightarrow{r}$ moving with a velocity $\overrightarrow{v}$ after a time $t$ is given by

$\overrightarrow{r^{\prime }}=\overrightarrow{r}+\overrightarrow{v}t$

If the two given particles are colliding at $t=2\text{s}$ their position vector at that point will be the same,

$\Rightarrow \overrightarrow{r}=\overrightarrow{v_1}t+\overrightarrow{r_1}=\overrightarrow{v_2}t+\overrightarrow{r_2}$

Given: $\overrightarrow{r_1}=(3\hat{i}+5\hat{j})\text{m}$

$\overrightarrow{r_2}=(-5\hat{i}+3\hat{j})\text{m}$

$\overrightarrow{v_1}=(4\hat{i}-4\hat{j}){\text{ms}}^{-1}$

$\overrightarrow{v_2}=(a\hat{i}-3\hat{j}){\text{ms}}^{-1}$

$\Rightarrow (4\hat{i}-4\hat{j})2+(3\hat{i}+5\hat{j})=(a\hat{i}-3\hat{j})2+(-5\hat{i}+3\hat{j})$

$11\hat{i}-3\hat{j}=(2a-5)\hat{i}-3\hat{j}$

Comparing the coefficients of $\hat{i}$ on both sides we get,

$11=2a-5$

$2a=16$

$a=8$