Question

At given instant of time two particles are having the position vectors $4\hat{i}-4\hat{j}+7\hat{k}$ and $2\hat{i}+2\hat{j}+5\hat{k}$ m respectively. If the velocity of the first particle is $0.4\hat{i}$ m/s, the velocity of second particle in m/s if they collide after $10$ sec is

• A. $6(\hat{i}-\hat{j}+\frac{1}{\sqrt{5}}\hat{k})$
• B. $0.6(\hat{i}-\hat{j}+\widehat{\frac{k}{3}})$
• C. $0.6(\hat{i}+\hat{j}+\widehat{\frac{k}{3}})$
• D. $0.6(\hat{i}+\hat{j}-\widehat{\frac{k}{3}})$

Answer 1

The correct option is B $0.6(\hat{i}-\hat{j}+\widehat{\frac{k}{3}})$

After $10s$, co-ordinates of particle A is: $\overrightarrow{r_A}=4\hat{i}-4\hat{j}+7\hat{k}+0.4\hat{i}\times 10$
Co-ordinates of particle B is$\overrightarrow{r_B}=2\hat{i}+2\hat{j}+5\hat{k}+\overrightarrow{v_A}\times 10=4\hat{i}-4\hat{j}+7\hat{k}+0.4\hat{i}\times 10$
$\overrightarrow{r_A}=\overrightarrow{r_B}$
$\Rightarrow 2\hat{i}+2\hat{j}+5\hat{k}+\overrightarrow{v_A}\times 10=8\hat{i}-4\hat{j}+7\hat{k}$

$\overrightarrow{v_A}=\frac{6\hat{i}-6\hat{j}+2\hat{k}}{10}=0.6\hat{i}-0.6\hat{j}+0.2\hat{k}=0.6(\hat{i}-\hat{j}+\frac{\hat{k}}{3})$