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Relative Velocity in 2D

In the given ...

Question

In the given figure, a particle $A$ moves along the line $y = 30 m$ with a constant velocity of magnitude $3 m/s$ and parallel to the $x -$ axis. At the instant particle $A$ passes the $y -$ axis, the particle $B$ leaves the origin with a zero initial speed and a constant acceleration of magnitude $0.40 {m/s}^{2}$ . What angle $θ$ between ${\to a}_{B}$ and the positive direction of the $y -$ axis would result in a collision?

15^{\circ}

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30^{\circ}

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45^{\circ}

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60^{\circ}

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15^{\circ}

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30^{\circ}

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45^{\circ}

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60^{\circ}

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Solution

The correct option is D $60^{\circ}$
Given that:
Vertical distance between $A$ and $B$ , $(y) = 30 m$
Velocity of $A$ , $(u_{A}) = 3 m/s$
Initial velocity of $B$ , $(u_{B}) = 0 m/s$
Acceleration of $B$ , $(a_{B}) = 0.40 {m/s}^{2}$

Let the particles collide at point $P$ .
In vertical direction for $B$ :
$y = \frac{1}{2} a_{y} t^{2}$
$\Rightarrow 30 = \frac{1}{2} (0.40) cos θ t^{2}$
$\Rightarrow 150 = cos θ t^{2}$ ........... $(i)$
In horizontal direction for $A$ and $B$ :
$u_{A} t = \frac{1}{2} a_{x} t^{2}$
$\Rightarrow t = \frac{2 u_{A}}{a_{x}}$
$\Rightarrow t = \frac{2 \times 3}{(0.40) sin θ}$
$\Rightarrow t = \frac{15}{sin θ}$
Substituting $t$ in equation $(i)$ ,
$150 = cos θ \times {(\frac{15}{sin θ})}^{2}$
$\Rightarrow 150 = cos θ \times \frac{225}{1 - {cos}^{2} θ}$
$\Rightarrow 1 - {cos}^{2} θ = \frac{3 cos θ}{2}$
$\Rightarrow 2 - 2 {cos}^{2} θ = 3 cos θ$
$\Rightarrow 2 {cos}^{2} θ + 3 cos θ - 2 = 0$

Solving this, we get
$cos θ = \frac{1}{2}$ or $cos θ = - 2$ $[$ neglect $]$
$\Rightarrow cos θ = \frac{1}{2}$
$\Rightarrow θ = 60^{\circ}$

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