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
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
$2$
$4$
$6$
$8$
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
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 , 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 $\alpha = 8$