First slide

Collisions

Question

Two particles having position vectors $\vec{r_{1}} = (3 \hat{i} + 5 \hat{j})$ meters and $\vec{r_{2}} = (- 5 \hat{i} - 3 \hat{j})$ meters are moving with velocities ${\vec{v}}_{1} = (4 \hat{i} + 3 \hat{j}) m / s$ and ${\vec{v}}_{2} = (α \hat{i} + 7 \hat{j}) m / s$ . If they collide after 2 seconds, the value of $' α'$ is

Moderate

A

2
B

4
C

6
D

8

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 $α = 8$

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