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Question: Two particles having position vectors \(\vec{r_1} = (3\hat{i} +5\hat{j})\)m and \(\vec{r_2} = (-5\ha...

Two particles having position vectors r1⃗=(3i^+5j^)\vec{r_1} = (3\hat{i} +5\hat{j})m and r2⃗=(−5i^+3j^)\vec{r_2} = (-5\hat{i}+3\hat{j})m are moving with velocities v1⃗=(4i^−4j^)  ms−1\vec{v_1} = (4\hat{i}-4\hat{j})\;ms^{-1} and v2⃗=(ai^−3j^)  ms−1\vec{v_2} = (a\hat{i}-3\hat{j})\;ms^{-1}. If they collide after 2 seconds, the value of a is:
A.-2
B.-4
C.-6
D.-8

Explanation

Solution

For the two particles to collide, they must be incident on each other. This means that they travel equivalent distances from their respective positions to collide with each other at 2 seconds. Determine the distance that each particle travels by taking the product of time and their velocity. The equivalent distance that they travel will be the additive distance they cover from their original position. To this end, equate the two equivalent distances and arithmetically arrive at the appropriate value for a.
Formula Used:
Equivalent Distance d⃗=r⃗+v⃗t\vec{d} = \vec{r} + \vec{v}t

Complete answer:
Let us begin by listing out what is given to us.
We have two particles whose locations are defined by the position vectors:
r1⃗=(3i^+5j^)\vec{r_1} = (3\hat{i} +5\hat{j})m and r2⃗=(−5i^+3j^)\vec{r_2} = (-5\hat{i}+3\hat{j})m respectively.
They move from these positions at velocities:
v1⃗=(4i^−4j^)  ms−1\vec{v_1} = (4\hat{i}-4\hat{j})\;ms^{-1} and v2⃗=(ai^−3j^)  ms−1\vec{v_2} = (a\hat{i}-3\hat{j})\;ms^{-1} respectively.
We are given that the two particles collide after time t=2  st=2\;s. This means that they travel equivalent distances from their initial positions in order from them to coincidentally collide with each other, i.e.,
d1⃗=d2⃗\vec{d_1} = \vec{d_2}
⇒r1⃗+v1⃗t=r2⃗+v2⃗t\Rightarrow \vec{r_1}+\vec{v_1}t = \vec{r_2}+\vec{v_2}t
Plugging in values for the above vectorial quantities and time, we get:
(3i^+5j^)+(4i^−4j^)2=(−5i^+3j^)+(ai^−3j^)2(3\hat{i} +5\hat{j}) +(4\hat{i}-4\hat{j})2 = (-5\hat{i}+3\hat{j}) + (a\hat{i}-3\hat{j})2
⇒(3i^+5j^)+(8i^−8j^)=(−5i^+3j^)+(2ai^−6j^)\Rightarrow (3\hat{i} +5\hat{j}) +(8\hat{i}-8\hat{j}) = (-5\hat{i}+3\hat{j}) + (2a\hat{i}-6\hat{j})
⇒(3i^+8i^+5i^−2ai^)=(3j^−6j^−5j^+8j^)\Rightarrow (3\hat{i} + 8\hat{i} + 5\hat{i}-2a\hat{i}) = (3\hat{j}-6\hat{j}-5\hat{j}+8\hat{j})
⇒16i^−2ai^=0j^\Rightarrow 16\hat{i}-2a\hat{i} = 0\hat{j}
⇒2ai^=−16i^\Rightarrow 2a\hat{i} = -16\hat{i}
⇒a=162=−8  ms−1\Rightarrow a = \dfrac{16}{2} = -8\;ms^{-1}

Therefore, the correct choice would be D. -8

Note:
It is important to understand that the distance that the two particles might travel may be different, but the equivalent distance that they travel while accounting for their respective positions as well as the rate at which they cover a certain distance will be identical, which is why we were able to equate the equivalent distances of the two particles when they collide after a certain time. We are essentially assuming the same final position for both the particles, which would mean that we only need to concern ourselves with the distance they travel from their initial positions.