Two particles having position vectors r1→=(3i^+5j^) meters and r2→=(−5i^−3j^) meters are moving with velocities v→1=(4i^+3j^) m/s and v→2=(α i^+7j^)m/s. If they collide after 2 seconds, the value of 'α' is

Solution:

It is clear from figure that the displacement vector $Δ \vec{r}$ between particles p₁ and p₂ is $Δ \vec{r} = \vec{r_{2}} - \vec{r_{1}} = - 8 \hat{i} - 8 \hat{j}$

$| Δ \vec{r} | = \sqrt{{(- 8)}^{2} + {(- 8)}^{2}} = 8 \sqrt{2}$ …..(i)
Now, as the particles are moving in same direction $(∵ \vec{v_{1}} and \vec{v_{2}} are + ve)$ , the relative velocity is given by
${\vec{v}}_{rel} = \vec{v_{2}} - \vec{v_{1}} = (α - 4) \hat{i} + 4 \hat{j}$
${\vec{v}}_{rel} = \sqrt{{(α - 4)}^{2} + 16}$ …..(ii)
Now, we know $| {\vec{v}}_{rel} | = \frac{| Δ \vec{r} |}{t}$
Substituting the values of ${\vec{v}}_{rel}$ and $| Δ \vec{r} |$ from equation (i) and (ii) and $t = 2 s$ , then on solving we get $α = 8$

Solution:

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