Work Energy And Power Question 122
Question: Two particles having position vectors $ \overrightarrow{r_1}=(3\hat{i}+5\hat{j}) $ metres and $ \overrightarrow{r_2}=(-5\hat{i}-3\hat{j}) $ metres are moving with velocities $ {{\overrightarrow{v}}_1}=(4\hat{i}+3\hat{j})m/s $ and $ {{\overrightarrow{v}}_2}=(\alpha \hat{i}+7\hat{j}) $
$ m/s. $ If they collide after 2 seconds, the value of $ ‘\alpha ’ $ is [EAMCET 2003]
Options:
A) 2
B) 4
C) 6
D) 8
Show Answer
Answer:
Correct Answer: D
Solution:
It is clear from figure that the displacement vector $ \Delta \overrightarrow{r} $
between particles $ p_1 $ and $ p_2 $ is $ \Delta \overrightarrow{r}=\overrightarrow{r_2}-\overrightarrow{r_1}=-8\hat{i}-8\hat{j} $
$ |\Delta \overrightarrow{r}|=\sqrt{{{(-8)}^{2}}+{{(-8)}^{2}}}=8\sqrt{2} $ …(i)
Now, as the particles are moving in same direction $ (\because \ \overrightarrow{v_1}\text{ and }\overrightarrow{v_2}\text{ are }+ve) $ ,
the relative velocity is given by $ {{\overrightarrow{v}} _{rel}}=\overrightarrow{v_2}-\overrightarrow{v_1}=(\alpha -4)\hat{i}+4\hat{j} $
$ {{\overrightarrow{v}} _{rel}}=\sqrt{{{(\alpha -4)}^{2}}+16} $ …(ii)
Now, we know $ |{{\overrightarrow{v}} _{rel}}|=\frac{|\Delta \overrightarrow{r}|}{t} $
Substituting the values of $ {{\overrightarrow{v}} _{rel}} $ and $ |\Delta \overrightarrow{r}| $ from equation (i) and (ii) and $ t=2s $ , then on solving we get $ \alpha =8 $
BETA
