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(D) For two particles to collide,their positions at time $t$ must be equal. Let $\overrightarrow{r_1}(t)$ and $\overrightarrow{r_2}(t)$ be the positio

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) For two particles to collide,their positions at time $t$ must be equal. Let $\overrightarrow{r_1}(t)$ and $\overrightarrow{r_2}(t)$ be the positions of the particles at time $t$.$\overrightarrow{r_1}(t) = \overrightarrow{r_1} + \overrightarrow{v_1}t = (3\hat{i} + 5\hat{j}) + (4\hat{i} + 3\hat{j})t = (3+4t)\hat{i} + (5+3t)\hat{j}$$\overrightarrow{r_2}(t) = \overrightarrow{r_2} + \overrightarrow{v_2}t = (-5\hat{i} - 3\hat{j}) + (\alpha\hat{i} + 7\hat{j})t = (-5+\alpha t)\hat{i} + (-3+7t)\hat{j}$At collision,$\overrightarrow{r_1}(t) = \overrightarrow{r_2}(t)$ at $t = 2 	ext{ s}$.Equating the $\hat{i}$ components:$3 + 4(2) = -5 + \alpha(2)$$3 + 8 = -5 + 2\alpha$$11 = -5 + 2\alpha$$16 = 2\alpha$$\alpha = 8$Equating the $\hat{j}$ components to verify:$5 + 3(2) = -3 + 7(2)$$5 + 6 = -3 + 14$$11 = 11$. This confirms the result.

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